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Matrices in Matlab
• MATLAB
Main Question or Discussion Point
Question 1:
Solve the system of equations
-w + 3x - 2y + z = 1
w + 2x - 4y - 3z = 2
x + 2y + z = 4
2w + 3x - y - z = 7
The attempt at a solution
To do this in matlab, this is what I will did:
>> A=[-1,3,-2,1;1,2,-4,-3;0,1,2,1;2,3,-1,-1]; b = [1;2;4;7]
>>x=inv(A)*b
Then matlab should give an answer with the solutions of w, x, y and z written in a separate line…no?
Question 2:
Find the eigenvalues and eigenvectors of the matrix
3, -1, 0, 0
-1, 5, -2, 0
0, -2, 6, -1
0, 0, -1, 2
The attempt at a solution
In matlab do I do this:
>>A=[3,-1,0,0;-1,5,-2,0;0,-2,6,-1;0,0,-1,2] ;
>>[V,D]=eig(A)
Then I should get an answer for V and D, and v is a matrix whose columns are the eigenvectors and D is the diagonal matrix, so we multiply the first column of V with the first diagonal of D etc… am I wrong?
Question 3:
Let
A=
-3, 4, 2, 0
1, -2, 0, 1
-4, 5, 0, 3
0, 1, -4, 2
Write the Matlab commands to construct the following matrix:
3, 0, 0, 0, -3, 1, -4, 0
0, 4, 0, 0, 4, -2, 5, 1
0, 0, 5, 0, 2, 0, 0, -4
0, 0, 0, 6, 0, 1, 3, 2
-3, 4, 2, 0, 5, 5, 5, 5
1, -2, 0, 1, 5, 5, 5, 5
-4, 5, 0, 3, 5, 5, 5, 5
0, 1, -4, 2, 5, 5, 5, 5
The attempt at a solution
In matlab should i do this:
>>A=[-3,4,2,0;1,-2,0,1;-4,5,0,3;0,1,-4,2]
Then to construct the other big matrix do I do this:
>> B=[diag(3:6),A’;A, fives(4,4)]
You’ve probably all figured that I don’t have access to matlab just now, so I would be grateful if someone who has access to matlab could just copy and paste the commands in to see if it works.
Also between each question do I write clear all to clear everything before I start the question?
Last edited:
Related MATLAB, Maple, Mathematica, LaTeX News on Phys.org
J77
For question 2, you don't have to do anything after eig.
D is a diagonal matrix of eigenvalues - just read them off the diagonal, and V is the corresponding matrix of eigenvectors - just take each column separately.
(You also have the relationship A*V=V*D - this is what you may be thinking when you want to multiply...?)
Qu 3 is right but for, there is no such thing as fives: use 5*ones(4,4).
Qu 1 looks good too.
Last edited:
ok thanx
do you know if i did Question 3 right coz i have a feeling it is wrong?
J77
See edit above for Qu3.
ok thanx i really appreciate your help!
and you didn't have to say 'there is no such thing' just correct me lol
J77
and you didn't have to say 'there is no such thing' just correct me lol
That's my new found Dutch abruptness invading my English politeness :tongue: | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/matrices-in-matlab.153583/", "fetch_time": 1596807350000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439737178.6/warc/CC-MAIN-20200807113613-20200807143613-00402.warc.gz", "warc_record_offset": 812111530, "warc_record_length": 16497, "token_count": 976, "char_count": 2572, "score": 3.90625, "int_score": 4, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.8311163783073425, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00056-of-00064.parquet"} |
# Conservation of energy gone for a six?
1. Mar 17, 2008
### manjuvenamma
Let us charge a capacitor and disconnect it from the battery. Let the capacitance, charge and voltage of the capacitor be C, Q and V respectively. Now do some work and reduce the distance of the plates of the capacitor and make it half of the original distance. What are the energies stored in the capacitor before and after the reduction of distance? The energies can be calculated as square(Q)/2C and square(Q)/4C respectively. Where has the energy gone from the capacitor?
2. Mar 18, 2008
### Staff: Mentor
The energy was dissipated by whatever brought the plates together. Note that the plates are oppositely charged, thus one must do negative work to bring the plates together.
A similar situation can be had with gravity by lowering an object. The energy decreases. Where did it go?
3. Mar 19, 2008
### manjuvenamma
You must be right. But I think I am missing a point.
Whien a stone is raised to a point, it has potential energy. If leave it there, it comes down.
If charge a capacitor, leave the plates as they are, they don't come closer on their own. Does the system do any work to bring the plates closer, or an external agency shoud do work?
Sorry, for the basic nature of my question. But I cant help asking, to convince myself.
4. Mar 19, 2008
### Staff: Mentor
5. Mar 19, 2008
### D H
Staff Emeritus
The plates don't come closer on their own because something (nonconducting spacers or some other physical constraint) is preventing that from happening. The exact same situation applies to the rock. Tie a rock to a rope, suspend the rock via a pulley, and tie the free end of the rope to some anchor. Voila, the rock is suspended above the ground with some energy proportional to the height of the rock. Now untie the rope from the anchor, lower the rock halfway to the ground, and retie the rope to the anchor. Has conservation of energy taken a six here? Of course not. You let the rock do work.
Back to the original problem. Just as the rope is needed to keep the rock from falling to the ground, some agent is needed to keep the plates of the capacitor apart. Suppose that agent is a number of nonconducting Hookean springs. Removing half of the springs will half the distance between the plates. Where did the energy go? You took it away by removing the springs.
6. Mar 20, 2008
### manjuvenamma
Great, thanks, my mind is clear now. I understand it now. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/conservation-of-energy-gone-for-a-six.222679/", "fetch_time": 1527233745000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-22/segments/1526794867046.34/warc/CC-MAIN-20180525063346-20180525083346-00622.warc.gz", "warc_record_offset": 823386753, "warc_record_length": 16085, "token_count": 582, "char_count": 2465, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2018-22", "snapshot_type": "longest", "language": "en", "language_score": 0.9417935609817505, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"} |
# How do you divide ( -3x^3+ 9x^2-12x+4 )/(x + 1 )?
Feb 25, 2016
$- 3 {x}^{2} + 12 x - 24 + \frac{28}{x + 1}$
#### Explanation:
These are difficult to answer because of the way that progressive steps have to be demonstrated and the high level of formatting needed.
The principle is that you progressively divide the highest value in the denominator into each of the 'element' of the numerator. In each instance of division a remainder results. This remainder is then combined with the next 'element' in the numerator and so on.
The process is the same as that adopted for numeric long division.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Starting point}}$ $\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\textcolor{b l u e}{\text{Step 1}}$
Using the $x$ from $x + 1$
$- 3 {x}^{3} \div x = \textcolor{g r e e n}{- 3 {x}^{2}}$
Now write:
$\text{ } \textcolor{g r e e n}{- 3 {x}^{2}}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
But $- 3 {x}^{2} \times \left(x + 1\right) = \textcolor{red}{- 3 {x}^{3} - 3 {x}^{2}}$
Now write:
$\text{ } \textcolor{g r e e n}{- 3 {x}^{2}}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{\textcolor{red}{- 3 {x}^{3} - 3 {x}^{2}}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$
Carry out the subtraction giving $\textcolor{g r e e n}{12 {x}^{2}}$
$\text{ } - 3 {x}^{2}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + \textcolor{g r e e n}{12 {x}^{2}}$
Bring down the $\textcolor{red}{- 12 x}$
$\text{ } - 3 {x}^{2}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} \textcolor{red}{- 12 x} + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} \textcolor{red}{- 12 x}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$
The left most value now is $+ 12 {x}^{2}$
Using the $x$ from $x + 1$
$+ 12 {x}^{2} \div x = \textcolor{g r e e n}{+ 12 x}$
Write:
$\text{ } - 3 {x}^{2} \textcolor{g r e e n}{+ 12 x}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} - 12 x$
$\textcolor{g r e e n}{12 x} \times \left(x + 1\right) = \textcolor{red}{12 {x}^{2} + 12 x}$
Write:
$\text{ } - 3 {x}^{2} \textcolor{g r e e n}{+ 12 x}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} - 12 x$
$\text{ } \underline{\textcolor{red}{12 {x}^{2} + 12 x}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$
Do the subtraction and bring down the $\textcolor{red}{+ 4}$
$\text{ } - 3 {x}^{2} + 12 x$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x \textcolor{red}{+ 4}}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} - 12 x$
$\text{ "underline(12x^2+12x) " }$carry out the subtraction
$\text{ "0" "color(green)(-24x)color(red)(+4)" bring down the } \textcolor{red}{+ 4}$
Using the $x$ from $\left(x + 1\right) \text{ } \textcolor{g r e e n}{- 24 x} \div x = \textcolor{b l u e}{- 24}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 5}}$
$\text{ } - 3 {x}^{2} + 12 x \textcolor{b l u e}{- 24}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} - 12 x$
$\text{ } \underline{12 {x}^{2} + 12 x}$
$\text{ "0" } - 24 x + 4$
$\textcolor{b l u e}{\left(- 24\right)} \times \left(x + 1\right) = \textcolor{red}{- 24 x - 24}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 6}}$
$\text{ } - 3 {x}^{2} + 12 x \textcolor{b l u e}{- 24}$
$\text{ } x + 1 \overline{| - 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}$
$\text{ } \underline{- 3 {x}^{3} - 3 {x}^{2}}$
$\text{ } 0 + 12 {x}^{2} - 12 x$
$\text{ } \underline{12 {x}^{2} + 12 x}$
$\text{ "0" } - 24 x + 4$
$\text{ } \underline{\textcolor{red}{- 24 x - 24}}$
Subtracting gives$\text{ } 0 + 28$
The final remainder is 28 so the answer is:$- 3 {x}^{2} + 12 x - 24 + \frac{28}{x + 1}$
Feb 27, 2016
$= - 3 {x}^{2} + 12 x - 24 + \frac{28}{x + 1}$
#### Explanation:
In the following process a factor x+1 (the divisor) is taken in required steps and adjusted with necessary terms.
to make $- 3 {x}^{3}$, I multiplied x+1 with $- 3 {x}^{2}$ and the extra term$- 3 {x}^{2}$ is adjusted by adding $3 {x}^{2}$ . Thus $12 {x}^{2}$ comes ,and then x+1 factor is taken again as $12 x \left(x + 1\right)$and I proceed as earlier til introduction of x+1 factor is not required.
$\frac{- 3 {x}^{3} + 9 {x}^{2} - 12 x + 4}{x + 1}$
$= \frac{- 3 {x}^{2} \left(x + 1\right) + 3 {x}^{2} + 9 {x}^{2} - 12 x + 4}{x + 1}$
$= \frac{- 3 {x}^{2} \left(x + 1\right) + 12 x \left(x + 1\right) - 12 x - 12 x + 4}{x + 1}$
$= \frac{- 3 {x}^{2} \left(x + 1\right) + 12 x \left(x + 1\right) - 24 \left(x + 1\right) + 24 + 4}{x + 1}$
$= - 3 {x}^{2} + 12 x - 24 + \frac{28}{x + 1}$
Is this process wrong? | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 79, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-divide-3x-3-9x-2-12x-4-x-1", "fetch_time": 1585411154000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00242.warc.gz", "warc_record_offset": 718010226, "warc_record_length": 7342, "token_count": 2274, "char_count": 5089, "score": 4.84375, "int_score": 5, "crawl": "CC-MAIN-2020-16", "snapshot_type": "latest", "language": "en", "language_score": 0.4971320629119873, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"} |
### 1: The student will demonstrate an understanding of numbers, operations, and quantitative reasoning.
1.5.1: The student uses place value to represent whole numbers and decimals.
1.5.1.A: use place value to read, write, compare, and order whole numbers through 999,999,999,999;
1.5.1.B: use place value to read, write, compare, and order decimals through the thousandths place.
1.5.2: The student uses fractions in problem-solving situations.
1.5.2.A: generate a fraction equivalent to a given fraction such as 1/2 and 3/6 or 4/12 and 1/3;
1.5.2.C: compare two fractional quantities in problem-solving situations using a variety of methods, including common denominators;
1.5.2.D: use models to relate decimals to fractions that name tenths, hundredths, and thousandths.
1.5.3: The student adds, subtracts, multiplies, and divides to solve meaningful problems.
1.5.3.A: use addition and subtraction to solve problems involving whole numbers and decimals;
1.5.3.C: use division to solve problems involving whole numbers (no more than two-digit divisors and three-digit dividends without technology), including interpreting the remainder within a given context;
1.5.3.D: identify common factors of a set of whole numbers;
1.5.3.E: model situations using addition and/or subtraction involving fractions with like denominators using [concrete objects,] pictures, words, and numbers.
1.5.4: The student estimates to determine reasonable results.
1.5.4.A: use strategies, including rounding and compatible numbers to estimate solutions to addition, subtraction, multiplication, and division problems.
### 2: The student will demonstrate an understanding of patterns, relationships, and algebraic reasoning.
2.5.5: The student makes generalizations based on observed patterns and relationships.
2.5.5.A: describe the relationship between sets of data in graphic organizers such as lists, tables, charts, and diagrams;
2.5.5.B: identify prime and composite numbers using [concrete objects,] pictorial models, and patterns in factor pairs.
### 3: The student will demonstrate an understanding of geometry and spatial reasoning.
3.5.8: The student models transformations.
3.5.8.B: identify the transformation that generates one figure from the other when given two congruent figures on a Quadrant I coordinate grid.
3.5.9: The student recognizes the connection between ordered pairs of numbers and locations of points on a plane.
3.5.9.A: locate and name points on a coordinate grid using ordered pairs of whole numbers.
### 5: The student will demonstrate an understanding of probability and statistics.
5.5.12: The student describes and predicts the results of a probability experiment.
5.5.12.A: use fractions to describe the results of an experiment;
5.5.13: The student solves problems by collecting, organizing, displaying, and interpreting sets of data.
5.5.13.A: use tables of related number pairs to make line graphs;
5.5.13.B: describe characteristics of data presented in tables and graphs including median, mode, and range;
5.5.13.C: graph a given set of data using an appropriate graphical representation such as a picture or line graph.
Correlation last revised: 3/6/2012
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gizmos.explorelearning.com/index.cfm?method=cResource.dspStandardCorrelation&id=696", "fetch_time": 1632409642000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780057424.99/warc/CC-MAIN-20210923135058-20210923165058-00011.warc.gz", "warc_record_offset": 337825337, "warc_record_length": 13895, "token_count": 752, "char_count": 3345, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "latest", "language": "en", "language_score": 0.8724402189254761, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"} |
# Riemann Sums Approximating Area One of the classical
• Slides: 18
Riemann Sums Approximating Area
One of the classical ways of thinking of an area under a curve is to graph the function and then approximate the area by drawing rectangular or trapezoidal regions under the curve (or nearly so).
Finding the area of each region is easy with rectangles, and these areas can be added to approximate the area of the function. f a b Here we approximate the area under f between a and b. Note that the rectangles do not all have the same width.
f a b This area is an approximation of
This method is one of several that can be used if we are unable to find an antiderivative of a function, and therefore cannot evaluate an integral symbolically, as we have been doing.
This methd can be used when an antiderivative can be found, but this is usually done for instructional purposes.
In order to approximate the area under the curve from 0 to 4, we can use a left sum with 2 subintervals. Note the way to write and calulate L 2. The left sum uses the height at the left side of each subinterval.
Similarly we find a right sum with 2 subintervals. The right sum uses the height at the right side of the subinterval.
A midpoint sum uses the midpoint of each subinterval to define the height.
Finally, a trapezoidal sum uses the area of trapezoids instead of reactangles. These may be calculated from the function, or recognized as the average of L 2 and R 2.
If we increase the number of subintervals to 4 we can repeat the processes. The approximation will be better with smaller subintervals.
Similarly with the right sum.
And for the midpoint sum.
And the trapezoidal sum.
As the number of subintervals increase the approximation get closer to the true area under the curve. Usually, the calculations are not made by hand, but utilize software for this purpose.
You have probably noticed that sometimes you can predict whether a given sum will be an over or underestimate. Talk this over and see if you can come up with circumstances where you can predict over- or underestimates.
The heights are always determined by the value of the function, sometimes given in a table, sometimes found from a formula. This is a very brief overview and should be supplemented by reading Section 5. 6 in your text. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://slidetodoc.com/riemann-sums-approximating-area-one-of-the-classical/", "fetch_time": 1624447977000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00467.warc.gz", "warc_record_offset": 481532051, "warc_record_length": 10249, "token_count": 513, "char_count": 2310, "score": 4.75, "int_score": 5, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.8969535231590271, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00019-of-00064.parquet"} |
# Derivation of ideal gas law
I looked up on the ideal gas law which our high school textbook derives with the empirical Combined Gas Law. However, the textbook did give a good explanation for this equation $$pV = \frac{N}{3}m\bar{v^2}$$ with which I only need to verify that $$K.E. = \frac{3}{2}k_BT$$ is true. I further looked up this link Average Molecular Kinetic Energy which deduces the result from the Boltzmann distribution $$f(E)=Ae^{-\beta \epsilon}$$ but I could not read any literature deriving $$\beta = \frac{1}{kT}$$ I was wondering if I am in a correct direction and how to derive the thermodynamic beta $\beta$.
• its a mere definition. $\beta$ is just a concise way of writing $1/kT$. Commented Oct 31, 2015 at 12:54
• but then why is the exponent $\frac{1}{kT}$? Commented Nov 1, 2015 at 2:13
• At some point in statical physics we must define what we mean by temperature. The modern approach is just to take $f_{MB}(E)\equiv Z^{-1} \mathrm e^{-E/kT}$ (where $Z=1/A$ in your notation). This means: we define temperature as the number $T$ that appears in the Boltzmann distribution (also, note that you have the formula wrong: the exponent is $\beta E$ instead of $\beta T$). The former approach (in the beginings of Thermodynamics) is to define temperature through the Ideal Gas Law, that is, we take $pV=nRT$ as an axiomatic rule that has (1/2) Commented Nov 1, 2015 at 11:03
• (2/2) to be obeyed by all gases (at sufficiently dim pressures). From this definition, it is not difficult to prove that $\beta=1/kT$. In any case, one way or another, we have to explicitly say what $T$ is. All definitions are equivalent, but the easier (and more theoretically meaningful) is to take the exponent in the Boltzmann distribution to be $-E/kT$ by definition. Commented Nov 1, 2015 at 11:08
• @qftishard thanks for reminding me of the wrong formula. So to sum up, in classical thermodynamics temperature is defined through average kinetic energy as $U = \frac{3}{2}RT$ but in statistical approach we define $T$ as a part of the exponent in Boltzmann distribution, and by using this newer definition we can derive back the classical definition? Commented Nov 2, 2015 at 14:35
The ideal gas law is a combination of many other laws about gases. Some assume the pressure to be constant, others assume the quantity stays constant and others. Now those laws have been set up mostly after experiment and it people working on it noticed that the pressure $P$ according to the small laws seemed to be proportional to the quantity $n$ (in moles), to the temperature $T$ (in kelvins) and inversely proportional to the volume. Know when we find such proportionality, we always need to multiply by a proportionality constant (here they called it R), that could be different than one and that would obviously nkt change the proportions. That proportionality constant, they have found its value by experiment and putting all this together gives the familiar $$P=nRT/V$$.
In classical thermodynamics, temperature $T$ is defined through ideal gas equation $$pV = nRT$$ from which we conclude that $$K.E. = \frac{3}{2}k_BT$$ is true for any ideal monatomic gas which cannot exist in real life anyways. Statistical mechanics provides postulates that is broader in context. It redefines the temperature through the second law $$dE = TdS$$ Now from $$p_i = \frac{e^{-\beta \epsilon_i}}{Z}$$ obviously we could obtain $$\beta = \frac{d \ln \Omega}{dE}$$ and by Boltzmann's assumption $$S = k_b \ln \Omega$$ we could have $$\beta = \frac{1}{k_BT}$$ so everything boils down to the definition of absolute temperature. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://physics.stackexchange.com/questions/215722/derivation-of-ideal-gas-law", "fetch_time": 1725766367000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00042.warc.gz", "warc_record_offset": 437978081, "warc_record_length": 42755, "token_count": 931, "char_count": 3607, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.9260371923446655, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
# System (rod+ball) collides elastically with a block
1 vote
155 views
The following link directs to the picture of the problem:
https://imgur.com/gallery/1a8cX?s=wa
Questions:
A) Are momentum, angular momentum and energy conserved? Under which circumstances?
B) Angular velocity of the ball immediately before the collision?
C) M/m rate so as to make the system (rod+ball) stay at rest after collision.
D) Velocity of the block after the collision.
E) Obtain an equation for D taking into account there is friction (Mu coefficient) ground-block along distance d and height h.
A) - Linear momentum is conserved before/after the collision, due to the lack of external forces. However, there is a force exerted by the pivot on the system during the collision. Therefore, linear momentum of the system (ball+rod) IS NOT CONSERVED.
- Angular momentum is conserved due to the fact that torque exerted on the rod by the ball is equal and opposite to the torque exerted on the ball by the rod. (Meaning that AM of the rod is equal and opposite to AM of the ball; therefore AM of the system IS CONSERVED).
- Energy IS CONSERVED IF FRICTION of any kind IS NEGLECTED. If we considered friction ground-block along distance d, energy would not be conserved after the elastic collision.
B) First, I obtain system's center of mass: I consider the rod a puntual mass and remembering both rod and ball masses are equal to m: Xcm= (m(l/2)+ml)/2m= 3/4l
As energy is conserved before the collision and system start and ends at rest, we can assure gravitational potential energy transforms to kinetic energy(note omega equals to angular velocity):
2mg(3/4l)=1/2(2m(3/4l)^2)omega^2 Therefore: omega= sqroot(8g/3l)
C) First we obtain CoM's velocity before the collision:
Now to obtain the velocity of the block after the collision we use momentum conservation principle:
2mVcm=MVblock; therefore: Vblock=2m/Msqroot(3gl/2)
Finally, to relate both M and m we use the principle of conservation of kinetic energy at an elastic collision:
1/22mVcm=1/2MVblock; therefore: M/m=2
D) Vblock= sqroot(3gl/2)
E) Energy is not conserved along d, however we now deltaE=muMgd. Therefore, Vblock= (mu)gd.
We have a projectile trajectory. Block has an initial velocity just at its horizontal component. Therefore:
D= d + Vblock(t) and y= h - 1/2g(t^2) Solving for t at second equation: t=sqroot(2h/g)
Finally, D= d(1 + mu(g)sqroot(2h/g))
Please if you could check my answers I would be greatful. If there's anything wrong or that can be improved let me know.
Thanks
edited Apr 6, 2018
Is there some reason you think your answers might be wrong? Is there something you are unsure about?
Is there some text to go with the image?
It is just I want to be sure I am right, as I am a physics students who is preparing himself. This is an exam-type problem (first year of physics degree)
I tried to post the pic as you said but did not work.
The text says the following: there is a system (ball+rod) which collides elastically against a block. The system starts at rest, besides, after the collision the system stays at rest too. The block is at rest until is hit by the system. The block goes along a distance d before falling from a height h, reaching a distance D. There's friction between block and ground. To sum up, I want to point out that even if I'm right with the answers if you wanted to add something in order to improve them, it would be great. Thanks
The instructions for posting images via PasteBoard do work, if you follow them correctly.
Well, I can try it again. I posted the image via PasteBoard and copied the link as instructed, however I will cheak it
A) Your answer is not clear about whether linear momentum is conserved. As you state, there is an external force acting during the collision between the ball $m$ and the block $M$. This external force is the reaction at the hinge (pivot), which is attached to the outside world. This means that linear momentum might not be conserved. However, angular momentum about the hinge is conserved.
B) The question states that "the system" is at rest after the collision. Doesn't this mean the ball + rod? You have calculated the angular velocity immediately before the collision.
I can see that you have equated PE lost to KE gained, using $\frac12 I \omega^2$ for KE. But how have you calculated moment of inertia $I$?
The moment of inertia of the rod about one end is $\frac13 mL^2$, not $m(\frac12 L)^2=\frac14 mL^2$. The moment of the (point) mass $m$ is $mL^2$. So the total moment of inertia of ball+rod is $\frac43 mL^2$ not $\frac98 mL^2$.
Your calculation method for moment of inertia is wrong. The moment inertia of an extended object is not the moment of inertia of its CM. For example, if the rod were pivoted at its CM, your method gives MI=0 instead of $\frac{1}{12}mL^2$. However, the method does work for calculating the change in PE (provided the gravitational field is uniform).
C) Here you need to apply conservation of angular momentum not linear momentum.
The KE of the ball+rod immediately before the collision equals the KE of the block immediately after the collision :
$\frac12 I\omega^2=\frac12 Mu^2$
Also, the angular momentum of the ball+rod equals that of the block :
$I\omega=MuL$
so
$I^2\omega^2=M^2u^2L^2$
Dividing the AM equation by the KE equation we get
$I=ML^2$
However, $I=\frac43 mL^2$ therefore $M=\frac43 m$.
D) The velocity $u$ of the block immediately after the collision is given by $\frac12 Mu^2=32mgL$ where $M=\frac43m$ therefore $u=\frac32\sqrt{gL}$.
E) Your calculation of the velocity of the block as it leaves the table does not seem to be correct.
The initial KE of the block before it slides across the rough part of the table is $\frac12 Mu^2=\frac12 M \frac94 gL=\frac98 MgL$ from above. The work done on the block by friction is $\mu Mgd$. So the velocity $v$ of the block as it leaves the table is given by
$\frac12 Mv^2=\frac98 MgL-\mu Mgd$
$v=\sqrt{(\frac94L-2\mu d)g}$.
It is not possible to simplify this expression unless you are given a value for $\mu d$ in terms of $L$. EG if $\mu d=\frac58 L$ then $v=\sqrt{gL}$.
Judging by the diagram $D$ is measured from the edge of the table, not from the initial position of the block. Then $D=vt$ where $h=\frac12 gt^2$ which you have got right.
answered Apr 2, 2018 by (28,746 points)
selected Apr 9, 2018
Yes, I calculated the angular velocity immediately before the collision.
We know moment of inertia equals to mr^2. Therefore: I=2m(3/4l)^2. I know r=3/4l Thanks to the fact I calculated the system's Center of Mass
That's right, the system is the ball + the rod.
The moment of inertia of the rod about one end is $\frac13 mL^2$, not $m(\frac12 L)^2=\frac14 mL^2$. The moment of the (point) mass $m$ is $mL^2$. So the total moment of inertia of ball+rod is $\frac43 mL^2$ not $\frac98 mL^2$.
Your calculation method is wrong. The moment inertia of an extended object is not the moment of inertia of its CM. For example, if the rod were pivoted at its CM, your method gives MI=0 instead of $\frac{1}{12}mL^2$.
There's a mistake at B) I meant calculating the angular velocity of the ball before the collision. I'll change it
Okey, I Thought I could use 3/4L as the radius to calculate MoI. Using 4/3mL^2 I obtain as the angular velocity of the ball before the collision 3/2sqroot(g/L). Do you agree with it?
Note I did not took 1/2L as the radius. What I did was first, calculate the Center of mass of the system (rod+ball) which equals to 3/4L. That's the value I considered as my 'height'. Therefore as initial potential energy of the system I do have 2mg(3/4L) . Note I wrote down 2m because the system is composed by ball+rod of equal masses. Until this point do you agree with me?
Yes your calculation of change in PE is correct. You can use the CM for that, but not for calculating moment of inertia.
Then you agree with 3/2sqroot(g/L) as angular velocity of the ball before the collision, don't you?
Yes omega=(3/2)sqrt(g/L). Next you need to conserve angular momentum during the collision.
Okey, now I would like to explain you what I did at C. Firstly I calculated the velocity of the center of mass of the system before its collision. As we know: v=omega*radius. Therefore: Vcm=9/8*L*sqroot(g/L)=9/8*sqroot(gL)
Once I got Vcm, as we know linear momentum is conserved before and after the collision we can assure the following: 2mVcm=MVblock.
Therefore: Vblock=(2*m*9*sqroot(g*L))/(8M)
As we know the collision is elastic, kinetic energy is conserved. Therefore:
1/2*2*m(9/8*sqroot(g*L)^2 = 1/2*M*(2m/M*9/8*sqroot(g*L))^2
Therefore M/m=2
Linear momentum is not conserved during the collision, as you acknowledged in your answer to part A. There is an external force (at the pivot) which acts during the collision. Angular momentum is conserved.
Linear momentum is not conserved during the collision, we both agree with that. But note linear momentum IS conserved before and after the collision, due to the lack of external forces. Therefore what I did should be right, as I considered momentum before the collision (2mVcm) to be equal to momentum after the collision (MVblock). Do you agree? As far as I'm concerned we both are right
No we do not agree. Linear momentum of the ball+rod before the collision is not equal to linear momentum of the block after the collision. An external force acts during the collision, ie between "before" and "after".
If conservation of linear momentum gives a different answer to conservation of angular momentum, then we cannot both be right.
Okey, I'm trying to obtain the rate M/m using conservation of angular momentum and conservation of kinetic energy (elastic collision), but I'm struggling to get it. I'm using the following equations:
1) CONSERVATION OF AM: 2m*(3/4L)*9/8sqroot(gL)=I*omega; I=4/3m(L^2)+M(L^2) and omega=3/2sqroot(g/L)
2) CONSERVATION OF KE: 1/2*(2m)*(9/8sqroot(gL)^2)=1/2MVblock.
The problem is that I'm not getting a clear result for M/m
And Vblock should be 9/8sqroot(gL), as the velocity of the block is the one corresponding to the velocity of the system (we negligited friction until section E)
I have updated part D of my answer.
Thanks I realized what I did wrong! I got the velocity of the block (asked to calculate it at D)) as follows:
As the block is at rest before the collision, Therefore Vblock should be equal to Vcm of the system, which equals to 0 after the collision. Then:
What do you think about it?
$\frac12 Mv^2=\frac32 mgL$ and $M=\frac43 m$ therefore $v=\frac32\sqrt{gL}$ is the velocity of the block immediately after the collision.
I get your point, you applied conservation of energy of the system. But why am I wrong? Shouldn't be Vcm of the system before the collision equal to Vblock? Why not then? You got as a result Vblock equals to angular velocity of the system before the collision, which has a lot of sense.
Then to sum up, I want to expose what I did at E)
I got D = Vblockt therefore D= 3/2sqroot(2hL)
Do you agree with this result?
The impulse at the pivot means that linear momentum is not conserved, but angular momentum is conserved. Therefore you cannot use the motion of the CM at impact. Also, if two objects of equal mass collide elastically then their velocities are switched, but in this case the colliding objects have unequal masses.
To avoid confusion I use $u=\frac32\sqrt{gL}$ as the velocity of the block immediately after the collision. This is different from the velocity $v$ with which it leaves the table, after it has moved the distance $d$ and lost some KE due to friction.
I have updated my answer to provide an expression for $v$, which should include $d, \mu$ unless you have been given values for these. $D=vt$ is correct but you are using the wrong value for $v$.
Okey, as momentum is not conserved we can't use Vcm=Vblock; got it Thanks
Then now I obtained for D:
D= sqroot(2*(9/4*Lh - 2*mu*d*h)) do you agree?
But now, if d went to infinity we would get a complex number as our D. How would you explain that case?
Thanks a lot
The block loses kinetic energy because of friction. If $d$ were infinite the the block would have to start with an infinite amount of kinetic energy in order to get to the edge of the table. If the block starts with a finite amount of kinetic energy, it can only go a finite distance before losing all that energy and stopping.
In this case the block stops (ie $v \to 0$) when $\frac94L = 2\mu d$. This gives you the maximum distance which the block can travel. $d$ cannot increase any further than this, so $v$ does not become a complex number.
Oh, You're totally right! V is not able to become a complex number, I was wrong.
Just to finish; do you agree with D=sqroot(2h(9/4L-2(mu)d))?
Yes that is correct.
Thanks A LOT Sammy Gerbil, this is the kind of site I 've been looking for a long time. As you could see, at the end everything was wrong. Thanks to you I improved.
I'll be back soon | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://qandaexchange.com/?qa=2492/system-rod-ball-collides-elastically-with-a-block&show=2538", "fetch_time": 1566236449000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027314852.37/warc/CC-MAIN-20190819160107-20190819182107-00099.warc.gz", "warc_record_offset": 151650047, "warc_record_length": 14700, "token_count": 3402, "char_count": 12921, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "longest", "language": "en", "language_score": 0.899150013923645, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"} |
# Singular value
In mathematics, in particular functional analysis, the singular values, or s-numbers of a compact operator T: XY acting between Hilbert spaces X and Y, are the square roots of non-negative eigenvalues of the self-adjoint operator T*T (where T* denotes the adjoint of T).
The singular values are non-negative real numbers, usually listed in decreasing order (s1(T), s2(T), …). The largest singular value s1(T) is equal to the operator norm of T (see Min-max theorem).
In the case that T acts on euclidean space Rn, there is a simple geometric interpretation for the singular values: Consider the image by T of the unit sphere; this is an ellipsoid, and the lengths of its semi-axes are the singular values of T (the figure provides an example in R2).
The singular values are the absolute values of the eigenvalues of a normal matrix A, because the spectral theorem can be applied to obtain unitary diagonalization of A as A = UΛU*. Therefore, ${\displaystyle {\sqrt {A^{*}A}}={\sqrt {U\Lambda ^{2}U^{*}}}=U\ |\Lambda |\ U^{*}}$ .
Most norms on Hilbert space operators studied are defined using s-numbers. For example, the Ky Fan-k-norm is the sum of first k singular values, the trace norm is the sum of all singular values, and the Schatten norm is the pth root of the sum of the pth powers of the singular values. Note that each norm is defined only on a special class of operators, hence s-numbers are useful in classifying different operators.
In the finite-dimensional case, a matrix can always be decomposed in the form UΣV*, where U and V* are unitary matrices and Σ is a diagonal matrix with the singular values lying on the diagonal. This is the singular value decomposition.
## Basic properties
For ${\displaystyle A\in \mathbb {C} ^{m\times n},}$ and ${\displaystyle i=1,2,\ldots ,\min\{m,n\}}$ .
Min-max theorem for singular values. Here ${\displaystyle U:\dim(U)=i}$ is a subspace of ${\displaystyle \mathbb {C} ^{n}}$ of dimension ${\displaystyle i}$ .
{\displaystyle {\begin{aligned}\sigma _{i}(A)&=\min _{\dim(U)=n-i+1}\max _{\underset {\|x\|_{2}=1}{x\in U}}\left\|Ax\right\|_{2}.\\\sigma _{i}(A)&=\max _{\dim(U)=i}\min _{\underset {\|x\|_{2}=1}{x\in U}}\left\|Ax\right\|_{2}.\end{aligned}}}
Matrix transpose and conjugate do not alter singular values.
${\displaystyle \sigma _{i}(A)=\sigma _{i}\left(A^{\textsf {T}}\right)=\sigma _{i}\left(A^{*}\right)=\sigma _{i}\left({\bar {A}}\right).}$
For any unitary ${\displaystyle U\in \mathbb {C} ^{m\times m},V\in \mathbb {C} ^{n\times n}.}$
${\displaystyle \sigma _{i}(A)=\sigma _{i}(UAV).}$
Relation to eigenvalues:
${\displaystyle \sigma _{i}^{2}(A)=\lambda _{i}\left(AA^{*}\right)=\lambda _{i}\left(A^{*}A\right).}$
### Singular values of sub-matrices
For ${\displaystyle A\in \mathbb {C} ^{m\times n}.}$
1. Let ${\displaystyle B}$ denote ${\displaystyle A}$ with one of its rows or columns deleted. Then
${\displaystyle \sigma _{i+1}(A)\leq \sigma _{i}(B)\leq \sigma _{i}(A)}$
2. Let ${\displaystyle B}$ denote ${\displaystyle A}$ with one of its rows and columns deleted. Then
${\displaystyle \sigma _{i+2}(A)\leq \sigma _{i}(B)\leq \sigma _{i}(A)}$
3. Let ${\displaystyle B}$ denote an ${\displaystyle (m-k)\times (n-l)}$ submatrix of ${\displaystyle A}$ . Then
${\displaystyle \sigma _{i+k+l}(A)\leq \sigma _{i}(B)\leq \sigma _{i}(A)}$
### Singular values of ${\displaystyle A+B}$
For ${\displaystyle A,B\in \mathbb {C} ^{m\times n}}$
1. ${\displaystyle \sum _{i=1}^{k}\sigma _{i}(A+B)\leq \sum _{i=1}^{k}\sigma _{i}(A)+\sigma _{i}(B),\quad k=\min\{m,n\}}$
2. ${\displaystyle \sigma _{i+j-1}(A+B)\leq \sigma _{i}(A)+\sigma _{j}(B).\quad i,j\in \mathbb {N} ,\ i+j-1\leq \min\{m,n\}}$
### Singular values of ${\displaystyle AB}$
For ${\displaystyle A,B\in \mathbb {C} ^{n\times n}}$
1. {\displaystyle {\begin{aligned}\prod _{i=n}^{i=n-k+1}\sigma _{i}(A)\sigma _{i}(B)&\leq \prod _{i=n}^{i=n-k+1}\sigma _{i}(AB)\\\prod _{i=1}^{k}\sigma _{i}(AB)&\leq \prod _{i=1}^{k}\sigma _{i}(A)\sigma _{i}(B),\\\sum _{i=1}^{k}\sigma _{i}^{p}(AB)&\leq \sum _{i=1}^{k}\sigma _{i}^{p}(A)\sigma _{i}^{p}(B),\end{aligned}}}
2. ${\displaystyle \sigma _{n}(A)\sigma _{i}(B)\leq \sigma _{i}(AB)\leq \sigma _{1}(A)\sigma _{i}(B)\quad i=1,2,\ldots ,n.}$
For ${\displaystyle A,B\in \mathbb {C} ^{m\times n}}$ [2]
${\displaystyle 2\sigma _{i}(AB^{*})\leq \sigma _{i}\left(A^{*}A+B^{*}B\right),\quad i=1,2,\ldots ,n.}$
### Singular values and eigenvalues
For ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ .
1. See[3]
${\displaystyle \lambda _{i}\left(A+A^{*}\right)\leq 2\sigma _{i}(A),\quad i=1,2,\ldots ,n.}$
2. Assume ${\displaystyle \left|\lambda _{1}(A)\right|\geq \cdots \geq \left|\lambda _{n}(A)\right|}$ . Then for ${\displaystyle k=1,2,\ldots ,n}$ :
1. Weyl's theorem
${\displaystyle \prod _{i=1}^{k}\left|\lambda _{i}(A)\right|\leq \prod _{i=1}^{k}\sigma _{i}(A).}$
2. For ${\displaystyle p>0}$ .
${\displaystyle \sum _{i=1}^{k}\left|\lambda _{i}^{p}(A)\right|\leq \sum _{i=1}^{k}\sigma _{i}^{p}(A).}$
## History
This concept was introduced by Erhard Schmidt in 1907. Schmidt called singular values "eigenvalues" at that time. The name "singular value" was first quoted by Smithies in 1937. In 1957, Allahverdiev proved the following characterization of the nth s-number :
${\displaystyle s_{n}(T)=\inf {\big \{}\,\|T-L\|:L{\text{ is an operator of finite rank }}
This formulation made it possible to extend the notion of s-numbers to operators in Banach space. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 40, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Singular_value", "fetch_time": 1618474457000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00624.warc.gz", "warc_record_offset": 1203054300, "warc_record_length": 10322, "token_count": 1847, "char_count": 5463, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "longest", "language": "en", "language_score": 0.8116852641105652, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"} |
### The second exam
Problem #1 Problem #2 Problem #3 Problem #4 Problem #5 Problem #6 Problem #7 Problem #8 Problem #9 Total 12 14 15 14 12 12 12 10 12 104 6 5 8 7 0 0 6 1 6 68 10.43 12.71 13.90 10.33 8.57 5.52 9.62 5.33 10.43 86.86 10 13 15 10 10 5 10 5 11 86
21 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:
Letterequivalent Range A B+ B C+ C D F [85,100] [75,84] [65,74] [55,64]
### Discussion of the grading
Generally arithmetic errors will be penalized only minimally. If, however, your error makes the problem much simpler, more credit will be deducted and your answer may not be eligible for all of the credit of the problem.
Problem 1 (20 points)
a) (4 points) Graph of the line (1 point) and the parabola (3 points). b) (9 points) Finding the intersections (3 points); setup of the integral (3 points); computation of the integral (3 points). c) (5 points) 2 points off for drawing too much of the parabola, and 1 point off for drawing the "line" insufficiently linear. d) (2 points) 1 point for the correct answer and 1 point for a relvant supporting reason.
Problem 2 (16 points)
Partial fractions. Full credit for the correct answer.
Problem 5 (16 points)
a) (8 points) Points off for partial or incorrect solutions.
b) (8 points) Points off for partial or incorrect solutions.
Problem 6 (16 points) a) (10 points) A correct curve.
b) (6 points) 2 points for identifying two critical points; 2 points for discerning what type they are; 2 points for finding an algebraic relationship. People should know what the phrase critical point means.
Problem 7 (16 points) Separating the variables: 2 points; antidifferentiation of y: 2 point; antidifferentiation of x: 5 points; getting the constant: 2 points; solving for y as a function of x: 2 points; stating the domain: 3 points. a) (7 points) I deducted 2 points for people who forgot +C at an appropriate place. I deducted 1 point for bad algebra (incorrected exponential manipulation, for example).
b) (8 points) I tried to follow the student's solution in a) here, even if incorrect. 6 points for the correct particular solution, again deducting points for bad algebra. 2 points for the domain. If the student's version of the solution unduly simplified the domain question, then no points could be earned. No points were earned if the student confused the domain and the range. I took off a point if the correct particular solution resulting in a domain answer which had only the restriction that x be non-zero. The solution curve which passes through the point (1,0) only has domain x>0.
Problem 8 (16 points)
8 points each for sections a) and b), which were graded similarly. Luckily no one in the class worked directly with the algebraic fourth derivative in b), although a few people worked with the algebraic second derivative in a). I took off 1 point for reversing inequalities (I graded the sections independently), and 1 point for other minor errors. Most people indeed did get the bounds on the derivatives from the given graphs, and did use the correcte error formulas. Of course, the answers given for the number of partitions were quite different. The largest answer was n=101000, which got full credit since enough supporting evidence was given. I just hope that the person giving that answer never has a computation task which takes 101000 steps!
Problem 9 (20 points)
Yes, this was a proof and verification for one or two values for the sequence of initial steps is not logically definitive.
Parts a), b) c) d), e), and f) each were worth 2 points. I then hoped that students would use this information to conclude that the sequence defined in f) was increasing and bounded above, and therefore must converge. f) was worth 8 points. The graph is of a function one of whose roots is the limit of the sequence defined.
Problem 10 (20 points)
Each part was worth 10 points. Again, a reversed inequality lost a post. More costly were errors in logic which I penalized more strictly. Part a) had been done in front of most students several times. Part b) was a different logical take, but again, things like it had been done in class a number of times. In b) either Taylor's Theorem with n=4 or 5 or condsideration of the alternating series nature of cosine's Taylor series are good approaches.
Problem 3 (15 points)
a) (5 points) I gave 2 points for recognizing the statement: first term=5 and fourth term=3 (1 point each). Completing the problem was worth the balance of credit. I deducted 1 point for faulty algebra.
b) (10 points) The "setup" earned 6 points: this is the computation of several of the (different) square side lengths and assembling them into a sum. 2 more points were earned by the recognition that this was a geometric series and the final 2 points for the sum of the series. Faulty algebra lost 1 point.
Problem 4 (12 points)
a) (8 points) The Ratio Test correctly used earned 6 points. The Root Test could also have served. Then identification of the interior of the interval and the radius of convergence was the other 2 points. I deducted 1 point if there was no clear or correct statement of the radius of convergence. I deducted 1 point for faulty algebra.
b) (4 points) 2 points for each endpoint. 1 point was earned in each case for correct insertion of the value to get a series of constants. 1 point was earned in each case for the correct conclusion with some sufficient reason. Please note that for x=-1/3, the Alternating Series Test only shows that the series converges conditionally, but the correct answer is that it converges absolutely, so additional reasoning must be given.
Problem 5 (12 points)
a) (6 points) The Alternating Series Test appllies, but I insisted that all three criteria be checked (or at least mentioned!). Students who omitted one of the criteria (frequently that the absolute value of the terms decreased) lost 2 of the 6 points. It is also possible in this case to strip away the sign change and conclude (using the Integral Test, say) that the resulting series converges and then, using the implication {absolute convergence} implies {conditional convergence} conclude that the series converges.
b) (6 points) If the series satisfies the conditions of the Alternating Series Test, then the first omitted term supplies an error estimate. Generally, even if a series converges, the first omitted term supplies no information about the accuracy of the partial sum. Students who did the problem but omitted the connection with alternating series were penalized 2 points. Students who did not give a partial sum lost 1 point. It is also possible, as in a), to delete the (-1)n and estimate the infinite tail using an integral. Students who incorrectly estimated powers of log with powers of n and then used an easier integral estimate did not receive credit. They simplified the problem too much.
Problem 6 (12 points)
a) (6 points) The Integral Test easily applies. Both the Ratio Test and the Root Test, when the appropriate limits are correcatly computed, do not give sufficient information for any conclusions (the limits are 1). Students who indicated a desire (?) to use the Integral Test were generally given at least 2 points.
b) (6 points) Again, an estimate using an integral works. I deducted 1 point because of the common error in the upper bound (N instead of N+1). Errors in inequalities were penalized at least 2 points. More serious errors in logic (overestimates instead of underestimates, for example, or {A>B} and {A>B} implies some relationship between B and C) generally received no credit.
Problem 7 (12 points)
a) (6 points) Most people compared the series to a convergent geometric series. Other valid approaches, such as comparison to a p-series with p=2, are also possible.
b) (6 points) Comparing the infinite tail to the infinite tail of a geometric series was done by most people. The geometric series can be computed exactly, and it also can be overestimated by an integral which was done by some. Just asserting that the Nth or (N+1)st term is less than a desired tolerance earns no points. I allowed 2 points if this was connected with an infinite tail.
Problem 8 (12 points)
a) (4 points) 1 point for "Yes" and then 3 points for explaining why. A generally succesful explanation would need to detour via the implication {absolute convergence} implies {conditional convergence}. The Ratio Test is not itself valid for the series, since various terms can equal 0.
b) (4 points) 1 point for "Yes" and then 3 points for an explanation. A likley explanation would begin with the 2Pi periodicity of cosine, and that earned a point. No more credit was earned unless all of the different cosine functions were somehow analyzed.
c) (4 points) 1 point for "Yes" and the other 3 points for some sort of explanation. Please note that the question is an effort to explain the graph, and therefore a verification can't use the graph, which could be incorrect or an artifact:
artifact
1. a product of human art and workmanship.
2. [Archaeol.] a product of prehistoric or aboriginal workmanship as distinguished from a similar object naturally produced.
3. [Biol. etc.] a feature not naturally present, introduced during preparation or investigation (e.g. as in the preparation of a slide). Maple could be showing things that aren't correct. It can draw misleading graphs.
Problem 9 (12 points)
a) (6 points) 2 points for "True" and 4 points for a general explanation. An example or two gains only 1 more point. Asserting that the function goes to 0 so the reciprocal of the function ... without more specification does not earn credit. Discussion of an's denominator or numerator (for an abstract an) similarly earns no credit, since I don't understand what that means.
a) (6 points) 2 points for "False" and 4 points for a vaild example. Several people told me that an=x and thus an=1/x, but the context was sufficiently unclear that I did not understand whether the x was suppose to be n or a constant. I then deducted 1 point, although either assertion would have led to a correct example. Clarity should be rewarded and its lack, penalized. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://shell.math.rutgers.edu/~greenfie/mill_courses/math192/xm2.html", "fetch_time": 1537429545000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-39/segments/1537267156418.7/warc/CC-MAIN-20180920062055-20180920082055-00019.warc.gz", "warc_record_offset": 223675747, "warc_record_length": 4996, "token_count": 2399, "char_count": 10249, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2018-39", "snapshot_type": "latest", "language": "en", "language_score": 0.856282651424408, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00053-of-00064.parquet"} |
# Abc Triangle Right Angled Line Through Mid Point Hypotenuse Parallel Bc Intersects Ac Ad
ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC … AC (iii) CM =…
Given: A △ABC , right – angled at C. A line through the mid – point M of hypotenuse AB parallel to BC intersects AC at D.
To Prove:
(i) D is the mid – point of AC
(ii) MD | AC
(iii) CM = MA = 1 / 2 AB.
Proof : (i) Since M is the mid point of hyp. AB and MD | | BC .
⇒ D is the mid – point of AC .
(ii) Since ∠BCA = 90°
and MD | | BC [given]
⇒ ∠MDA = ∠BCA
= 90° [corresp ∠s]
⇒ MD | AC
(iii) Now, in △ADM and △CDM
MD = MD [common]
∠MDA = ∠MDC [each = 90°]
AD = CD [∵ D the mid – point of AC]
⇒ △ADM ≅ △CDM [by SAS congruence axiom]
⇒ AM = CM
Also, M is the mid – point of AB [given]
⇒ CM = MA = 1 / 2 = AB. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://writersanswers.com/questions/abc-triangle-right-angled-line-through-mid-point/", "fetch_time": 1679443086000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00539.warc.gz", "warc_record_offset": 710745815, "warc_record_length": 20630, "token_count": 328, "char_count": 852, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2023-14", "snapshot_type": "latest", "language": "en", "language_score": 0.7703047394752502, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00012-of-00064.parquet"} |
# AS-Level Maths
:
Mechanics 1
for Edexcel
M1.3 Kinematics
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1 of 37
Contents
Motion graphs
Motion graphs
Formulae for constant acceleration
Examination-style questions
2 of 37
Kinematics
Kinematics involves the study of how things move.
It is only concerned with the motion itself, not the forces that
cause this motion.
The kinematics of an object is described in terms of its
distance,
displacement,
speed,
velocity,
acceleration.
3 of 37
m.Distance and displacement Distance is a scalar quantity. The distance a body has travelled is literally the amount of ‘ground’ it has covered during its motion. Displacement describes how far a body is from its starting point and in what direction. 4 of 37 © Boardworks Ltd 2005 . Distance and displacement are measured in metres. Displacement is a vector quantity.
Speed and velocity are measured in metres per second. velocity and acceleration Speed is a scalar quantity. Acceleration is the rate of change of speed or velocity.Speed. ms –1. It is the rate at which a body changes its position. It is measured in metres per second per second. 5 of 37 © Boardworks Ltd 2005 . Velocity is a vector quantity. The velocity of a body relates to how fast the body is travelling and in what direction. Acceleration can be a scalar or a vector quantity. ms –2. Negative acceleration is often called deceleration or retardation. The speed of a body relates to how fast the body is travelling.
The most common graphs are position-time. The gradient of a distance-time graph gives speed. The area under an acceleration-time graph gives the change in velocity. 6 of 37 © Boardworks Ltd 2005 . velocity-time and acceleration-time graphs. speed-time.Motion graphs The kinematics of a body can be represented graphically. The area under a velocity-time graph gives the change in displacement. The gradient of a velocity-time graph gives acceleration. The area under a speed-time graph gives the distance travelled. The gradient of a displacement-time graph gives velocity.
3 metres per minute Time (mins) 7 of 37 © Boardworks Ltd 2005 . velocity = 30 = –33. The person then returns to their starting position. 1000 velocity = 20 = 500 metres per minute 600 400 200 0 0 20 40 60 80 100 120 For the second part of the journey the velocity is zero. 1000 For the last part. 800 For the first part of the journey. The gradient of this graph 1000 gives velocity.Displacement-time graph Displacement (m) This graph shows a journey of 2000 m. It includes a stop of 1 hour after travelling 1000 m metres.
3 metres per minute Time (mins) 8 of 37 © Boardworks Ltd 2005 . Distance (m) 2000 1600 1200 800 400 0 0 20 40 60 80 100 120 For the first part of the journey. The gradient of this graph gives speed. 1000 speed = 20 = 500 metres per minute For the second part of the journey the speed is zero. 1000 For the last part. for this graph there is no indication of direction.Distance-time graph This graph also shows a journey of 2000 m with a 1 hour stop. speed = 30 = 33. However.
The first part of the graph shows an acceleration of 0. the second part 0 and the last part a deceleration of 1. 9 of 37 © Boardworks Ltd 2005 .Velocity-time graph Velocity (ms–1) The area under a velocity-time graph gives displacement.5 10 7. In this example. the area under the graph is given by a trapezium with height 12. 12.5 = 1375 m The gradient of the graph gives acceleration.5 0 0 20 40 60 80 100 120 140 Time (s) Displacement = 21 (130 + 90)×12.42 ms –2.25 ms –2.5 and parallel sides of length 130 and 90.5 5 2.
10 of 37 © Boardworks Ltd 2005 . The last part shows constant velocity. The second part shows a deceleration of 6 ms–2.Acceleration-time graph This graph shows constant acceleration. Acceleration (ms–2) 6 4 2 0 –2 2 4 6 8 10 12 14 Time (s) –4 –6 The first part shows an acceleration of 4 ms–2.
Acceleration (ms–2) 6 4 2 0 –2 2 4 6 8 10 12 14 Time (s) –4 –6 For the first part. change in velocity = 4 × 8 = 32 ms–1 For the second part. change in velocity = –6 × 3 = –18 ms–1 There is no change in velocity for the last part. 11 of 37 © Boardworks Ltd 2005 .Acceleration-time graph The area under an acceleration-time graph gives change in velocity.
Graphs example 1 A man travels in a lift from the top floor of a hotel to reception on the ground floor. The lift accelerates with a constant acceleration of 1 ms –2 until it reaches a constant velocity of 4 ms–1. It then travels at this constant velocity for t seconds before decelerating with a constant deceleration of 2 ms–2 until it reaches the ground floor. a) sketch the velocity-time graph of the lift and use it to find t b) sketch the acceleration-time graph of the lift. 12 of 37 © Boardworks Ltd 2005 . Given that the man has descended 44 m.
Graphs solution 1 The distance travelled is given by the area under the graph. so 4 ( 21 × 4 × 4) + (4t ) + ( 21 × 4 × 4) = 44 4 t Time (s) The acceleration-time graph for the lift can then be sketched as follows: 13 of 37 8 + 4t + 4 = 44 4t = 32 t = 8 secs 2 Acceleration (ms–2) Velocity (ms–1) The velocity-time graph for the lift can be sketched as follows: 1 0 2 4 6 8 10 12 14 Time (s) –1 –2 © Boardworks Ltd 2005 .
The motorcycle overtakes the car after they have both travelled 3700 m.Graphs example 2 A car and a motorcycle are travelling along a straight road. Draw a speed-time graph and use it to find the time when the motorcycle overtakes the car and how long the car was initially accelerating for. The car accelerates from rest to a constant speed of 28 ms –1. The motorcycle is unaffected by the traffic and maintains his speed. The motorcycle accelerates from rest to a constant speed of 25 ms–1 in 10 seconds. 14 of 37 © Boardworks Ltd 2005 . After travelling for 90 seconds the car hits traffic and decelerates to a constant speed of 22 ms–1 in 5 seconds.
The motorbike overtakes the car after travelling 3700 m so.Velocity (ms–1) Graphs solution 2 t1 28 25 22 0 Motorbike Car 10 Time (s) 90 95 Let t1 be the time for which the motorbike is travelling with a constant speed before it overtakes the car. ( 21 ×10 × 25) + (t1 × 25) = 3700 125 + 25t1 = 3700 25t1 = 3575 t1 = 143 seconds 15 of 37 © Boardworks Ltd 2005 .
Graphs solution 2 Velocity (ms–1) So the motorbike accelerates for 10 seconds and then travels at a constant speed for 143 seconds before overtaking the car. 28 25 22 0 t2 Motorbike Car 10 Time (s) 90 95 153 This area represents 3700 m 16 of 37 © Boardworks Ltd 2005 . The motorbike travels for 153 seconds before overtaking the car. Let t2 be the time for which the car is initially accelerating.
8 s (to 3 s.Graphs solution 2 Since the area under the graph for the car between 0 and 153 seconds is equal to 3700 so we can write. ( 21 × t2 × 28) + ((90 t2 )× 28) + ( 21 (22 + 28)× 5) + (58 × 22) = 3700 14t2 + 2520 – 28t2 + 125 + 1276 = 3700 14t2 = 221 t2 = 15.8 (to 3 sf) Therefore the car was initially accelerating for 15.f.) 17 of 37 © Boardworks Ltd 2005 .
Contents Formulae for constant acceleration Motion graphs Formulae for constant acceleration Examination-style questions 18 of 37 © Boardworks Ltd 2005 .
Formulae for constant acceleration If a particle is moving in a straight line with a constant acceleration then there are five equations of motion that can be used to determine missing quantities. 19 of 37 © Boardworks Ltd 2005 . 9. v = u + at s = 21 (u + v )t s = ut + 21 at 2 s = vt 21 at 2 v 2 = u 2 + 2as Where s = displacement in metres u = initial velocity in ms–1 v = final velocity in ms–1 a = acceleration in ms–2 t = time taken in seconds These are sometimes called the suvat formulae.8ms–2. For vertical motion acceleration due to gravity is g.
acceleration is the rate of change of velocity. v The constant acceleration a is therefore given by the gradient of the graph. Velocity (ms–1) By definition. So u vu t at = v – u a= t 20 of 37 Time (s) v = u + at © Boardworks Ltd 2005 .v = u + at The motion of an object with initial velocity u and final velocity v over time t can be illustrated using a velocity-time graph.
Velocity (ms–1) This distance is given by the area under the graph. v This area is a trapezium with parallel sides of length u and v and width t. So u s = 21 (u + v )t This can also be written as t 21 of 37 Time (s) u + v s = t 2 © Boardworks Ltd 2005 .s = ½(u + v)t We can use the same graph to find the distance s travelled by an object with initial velocity u and final velocity v over time t.
s = ut + ½at2 distance travelled = area of rectangle A + area of triangle B = ut + 21 t (v u ) vu a= so at = v – u t This gives us distance travelled = ut + 21 t ( at ) So 22 of 37 s = ut + 21 at 2 © Boardworks Ltd 2005 .
s = vt – ½at2 distance travelled = area of rectangle C – area of triangle D = vt 21 t (v u ) We have shown that at = v – u This gives us So 23 of 37 distance travelled = vt 21 t ( at ) s = vt 21 at 2 © Boardworks Ltd 2005 .
v2 = u2 + 2as We can show that v2 = u2 + 2as as using 1 v = u + at u + v s= t 2 2 Rearranging equation 1 to make t the subject gives (v u ) t= a Substituting this into equation 2 u + v v u s = a 2 2as = (u + v )(v u ) 2as = v 2 u 2 So 24 of 37 v 2 = u 2 + 2as © Boardworks Ltd 2005 .
Find the maximum height above the ground that the stone reaches and find the time taken for the stone to reach the ground. and for t when s = –15.8) The question is asking for s when v = 0. 25 of 37 © Boardworks Ltd 2005 .Constant acceleration example 1 A stone is thrown vertically upwards with a speed of 10 ms–1 from a point 15 m above the ground. Taking to be positive. the information given in the question is: u = 10 a = –g (Take g to be 9.
6s = 100 s = 5.). 26 of 37 © Boardworks Ltd 2005 .10 (to 3 s.8)(s) 0 = 100 – 19.f.) Therefore.1 m (to 3 s.6s 19. the maximum height above the ground that the stone reaches is 20.f. given u and a requires the use of v2 = u2 + 2as.Constant acceleration solution 1 To calculate s when v = 0. 02 = 102 + 2(–9.
05 s (to 3 s.05 or t = –1. 27 of 37 © Boardworks Ltd 2005 .9t2 Arranging all the terms on the left gives us the following quadratic equation 4.) Therefore the stone reaches the ground after 3.01 (to 3 s.).8)t2 –15 = 10t – 4. –15 = 10t + 21 (–9.Constant acceleration solution 1 To calculate t when s = –15.9t2 –10t – 15 = 0 b b 2 4ac Using gives the solution 2a t = 3.f.f. given u and a requires the use of s = ut + 21 at2.
Calculate the velocity at A and the distances AB and BC. via point B.Constant acceleration example 2 A particle moves in a horizontal line from a point A to a point C. 28 of 37 © Boardworks Ltd 2005 . This requires the use of v = u + at. We can sketch the situation as follows A B t=6 C t = 10 v = 50 Taking to be positive the question firstly asks for u when a = 1. It has a constant acceleration of 1 ms –2 and passes point B after 6 seconds and point C after a further 4 seconds. t = 10 and v = 50. Its velocity at C is 50 ms–1.
a = 1 and t = 6. requiring the use of s = ut + 21 at2.Constant acceleration example 2 v = u + at 50 = u + (1)(10) 50 = u + 10 u = 40 Therefore the particle passes A with a velocity of 40 ms–1. s = ut + 21 at2 s = 40(6) + 21 (1)(6)2 s = 240 + 18 = 258 Therefore AB is 258m. 29 of 37 © Boardworks Ltd 2005 . The next part of the question asks for s when u = 40.
a = 1 and t = 10. 30 of 37 © Boardworks Ltd 2005 .Constant acceleration example 2 The final part of the question asks for s when u = 40. s = ut + 21 at2 s = 40(10) + 21 (1)(10)2 s = 400 + 50 = 450 Therefore AC is 450 m and so BC is 192 m. again requiring the use of s = ut + 21 at2.
s = ut + 21 at2 1 s = (0)(3.2 seconds later.8 t = 3.2)2 s = 50.Constant acceleration example 3 A ball falls off a cliff and lands on the beach 3.2) + 2 (9.2 u=0 To calculate s requires the use of the formula s = ut + 21 at2.2 m (to 3 s.2 (to 3 s.) 31 of 37 © Boardworks Ltd 2005 .f.f ) Therefore the height of the cliff is 50. How high is the cliff? Taking to be positive the information given in the question is a = 9.8)(3.
Contents Examination-style questions Motion graphs Formulae for constant acceleration Examination-style questions 32 of 37 © Boardworks Ltd 2005 .
Examination-style question 1 A ball is thrown vertically upwards with an initial velocity of ms–1 from a point 1.2 m above the ground. 33 of 37 © Boardworks Ltd 2005 u . It reaches a maximum height of 23 m above the ground. Calculate a) the initial velocity b) the velocity with which the ball strikes the ground c) the total time the ball is in the air.
8 v2 = u2 + 2as 0 = u2 + 2(-9. s = 23 v2 = u2 + 2as v2 = 02 + 2(9.2 (to 3 s.8)(23) v2 = 450.Using downward motion only Taking to be positive.8 v = ± 21. u = 0. s = 21. a = 9.f.8.f.) initial velocity is 20.f. v = 0.7ms–1 (to 3 s.8)(21.Solution 1 a) Taking as positive. a = –9.) b) Method 1 .8) 0 = u2 – 427.28 (to 3 s.) 34 of 37 The ball hits ground with a velocity of 21.) © Boardworks Ltd 2005 .28 u = ±427.2 ms–1 (to 3 s.8.f.
u = –427.28 + 2(9.f.8 v = u + at (v u ) t= a ( 450.8.8)(1. s = 1.28 s (to 3 s.) 35 of 37 © Boardworks Ltd 2005 .28 (to 3 s.f.) ball strikes ground with a velocity of 21.) c) Taking as positive.28 ) t= 9.2 ms-1 (to 3 s.2) v2 = 450.28. v = 450.28.8 v = 21.8 427.f. a = 9. u = –427.8 t = 4.) ball is in the air for 4.8.2 v2 = u2 + 2as v2 = 427.f.Solution 1 b) Method 2 – Using whole motion Taking to be positive.2 (to 3 s. a = 9.
Examination-style question 2 A car is travelling with a uniform acceleration along a straight road. Find the velocity with which the car passes the mid-point of AB. It passes a point A with a velocity of 8 ms–1 and 5 seconds later it passes a point B with velocity 25 ms–1. v = 25. 36 of 37 © Boardworks Ltd 2005 . The distance AB needs to be found first: u = 8. t = 5 s = 21 (u + v)t 1 s = 2 (8 + 33)(5) s = 82.5 the distance AB is 82.25 m from A.5 m and so the mid-point of AB is at a distance of 41.
t = 5 (v u ) a = v = u + at t (25 8) a= 5 a = 3.) 37 of 37 © Boardworks Ltd 2005 .5 v = ±344. a = 3.Solution 2 Acceleration now needs to be found: u = 8.4 ms–2.25) = 344.25 v2 = u2 + 2as v2 = 82 + 2(3.6 ms–1 (to 3 s.4)(41. s = 41.f.5 the car passes the mid-point of AB with a velocity of 18. v = 25. The velocity of the car as it passes the mid-point of AB can now be found: u = 8.4 the acceleration of the car is 3.4. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.scribd.com/presentation/335557079/3-kinematics", "fetch_time": 1537326557000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-39/segments/1537267155817.29/warc/CC-MAIN-20180919024323-20180919044323-00192.warc.gz", "warc_record_offset": 855889563, "warc_record_length": 45570, "token_count": 4664, "char_count": 14629, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2018-39", "snapshot_type": "latest", "language": "en", "language_score": 0.9303781986236572, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"} |
# Part Two - Base Line Data on the Soda Cup Lander
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## Objective
SWABT use a scale to find the mass and a ruler to take the linear measurements of their 'soda cup lander'.
#### Big Idea
Tools of the trade: students learn how to use a scale and ruler, so they can take accurate data for their Sea Cup Lander in the next lesson.
## NGSS Connections and Lesson Preparation
30 minutes
Connection to NGSS
Science and Engineering Practices
- SP 5 Using Mathematics and Computational Thinking
In order for my students to take mathematical data on their Sea Cup Lander, they will need to know how to use a scale and ruler. In this lesson students learn how to use a scale, and review how to take linear measurements to the nearest cm.
Next week students apply these skills when they take data on the 'scientific instruments' they will be evaluating for their Soda Cup Lander (SCL).
Materials
12 Soda Cup Landers - for linear and mass measurements
6 gram weight scales
6 rulers
6 washers
Class Preparation
metric ruler demonstration to show how to take accurate cm measurements
Demonstration area to show how to use a scale
Sentence strips with 'how to steps' for linear and mass measurements. Number the steps on the back and cover with a post-it.
2 pocket charts to use with the 'how to steps'
Set up for Each Linear Stations (6 stations; 2 students/station)
1 Sea Cup Lander
1 ruler
Set up for Each Mass Measurement Station (6 stations; 2 students/station)
1 Sea Cup Lander
1 scale
1 washer
gram weights
Groups
Students work in pairs. Before the lesson I have selected the 'scale' and 'linear' expert for each group.
Each student will need their lab book, pages 5 and 6, to record their SCL measurements.
## Question for the Day
5 minutes
Question for the Day: What information can we get about our sea cup lander, (SCL) from a scale and a ruler? Along with the question is a picture of a scale and a ruler, to support students who may not know what either of these tools are.
Science begins with a 'question of the day' for a couple of reasons. It is an established routine for science that helps to bring the class together efficiently. I use a question to hook the students participation with an aspect of today's lesson.
I direct students to take out their white board and marker and write their answer to the question on their white board. Students hold up their white board to show me their answer.
This gives me a quick assessment as to what students may know about taking measurements with these tools.
While students are writing their answers on their whiteboard, I project the 'expert groups' on the smartboard.
After I view student answers, students to sit on the rug with their partner and share answers to the 'question for the day' with their partner.
Partner sharing provides a way for students to check in with their assigned partner before starting the next part of the lesson.
## Linear and Scale Measurement Experts
15 minutes
"Today each of you will become an expert on how to use the scale or ruler and then teach your partner how to use the tool. Afterwards you will practice what you learned by taking different measurements of the Sea Cup Lander."
Rather than students learn 2 measuring methods from me, how to use a ruler and scale. I decided to split the class into 2 groups, one group would learn how to use the scale and the other group would learn how to use the centimeter side of the ruler.
"Each 'expert group' will go to a designated area of the room and watch a short demonstration on how to use the measurement tool. Afterwards you will work with your expert team to order the 'how to steps'."
Each step is printed on a sentence strip and is in random order in a pocket chart. Once the expert groups have arranged the steps, each team will have steps to refer to when taking measurements.
"When everyone agrees with the order, your team may check your sort. On the back of the sentence strip, underneath the post -it is the order number for that step. Only check it when your expert group has agreed with the order."
I chose to have the students arrange the steps in a pocket chart rather than individuals write the steps to save time. Some of my kiddos need a lot of support with transferring information to their paper and some just need lots of time to write. Ordering the sentence strips removes the writing anxiety from some of my students.
The activity promotes collaboration, because the students are working toward a common goal. It also gives them independence with their learning because they decide when their work is ready to be checked and they get to check their work themselves.
The sequenced sentence strips are displayed for students to reference when sharing their information with their partner.
I explain that the expert stations will run for 10 mins. 5 mins. for the 'expert workshop' and 5 mins to organize the 'how to steps'.
While students are ordering the steps, I set up the measuring stations. I have the materials for each station in tubs so that I can easily move the materials to the desks.
## Sharing Scale and Linear Expertise
25 minutes
By now the sequenced 'how to steps' are displayed in the room and students are on the rug sitting with their team partner.
"Congratulations! Each team member has successfully completed their technical training and have earned the title 'Expert'!
Experts, you will teach your team member how to how to take accurate measurements with the scale or ruler, using the 'how to steps' as a guide. Teams will work together to take Sea Cup Lander measurements.
I cannot get to each student while they are taking measurements to check that their measurements are accurate, but my experts can assist by being next to their partner and checking that they are setting the scale or ruler correctly.
6 teams will start with the scales and the other 6 teams will start with the rulers. After 10 mins. I will ask teams to organize their center, and then move to the other measurement station."
"At each station there is a measurement tool, the Soda Cup Lander and parts you will measure. In your lab book, pages .... each of you will write your measurements."
I project the lab pages and point out each part that will be measured.
I set a visible timer so that students can pace their work. I circulate around the room to check that students are taking accurate measurements and to clarify any directions.
When the timer rings, students organize their station and move to the other station.
## Wrap Up
5 minutes
After students have organized the materials at the 2nd measurement station, I ask them to bring their measurement data to the rug.
"Should our measurements be the same? why or why not? Turn to someone other than your team partner and discuss."
"Now turn to a different shoulder partner and share your data."
This is my back up strategy for students to check that they made accurate measurements, and to reflect on why they may or may not have gotten the same measurement as the other team.
I give students a couple of minutes to share and then I ask the class, show me with thumbs up if you got the same measurements. If you did not get the same, why do you think this happened? Teams how could you check your measurements?"
Students noticed that the scale measurements seemed to be more varied than than length measurements. I used this opportunity to discuss how our scales are not built as accurate as a ruler.
After this discussion, I direct students to place their lab book in their science folder. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://betterlesson.com/lesson/resource/3160046/scl2-experts-share", "fetch_time": 1591266267000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00407.warc.gz", "warc_record_offset": 255445010, "warc_record_length": 32312, "token_count": 1579, "char_count": 7614, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.9102008938789368, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"} |
# Cissoid
## History
Cissoid is the generalization of Cissoid of Diocles. Who generalized it? Around what era?
## Description
Cissoid is a method of deriving a new curve based on two (or one) given curves C1, C2, and a fixed point O. A curve derived this way may be called the cissoid of C1 and C2 with the pole O.
Step-by-step description:
1. Given two curves C1 and C2, and given a fixed point O.
2. Let P1 be a point on C1. Draw a line L passing O and P1. Let the intersection of L and C2 be P2.
3. Mark a point Q on line L, such that distance[O,Q]==distance[P1,P2].
4. The locus of Q (as P1 moves on C1) is the cissoid of C1 and C2 with the pole O.
Note: There are two points on line L such that distance[O,Q]==distance[P1,P2]. The two points are symmetric around point O, so either one will generate the same cissoid. Also, if L and C2 have more than one intersections, then we can label additional points P3, P4,… and the cissoid may have loops.)
The cissoid method can also be used on a single curve or multiple curves. Given a curve (or curves) and a point O, let a line L passing O sweeps the plane in a complete revolution. Let P_1, P_2, …P_n be the intersections of L and the given curves. Mark points on L with lengths of Length[vector[P_i,P_j]] for all possible combinations of i ,j ≤ n. (consistently mark them on one side of L) The locus of these points is the cissoid of the given curves with pole O. If the line L never makes more than one intersection with the given curves, then we may say that there is no cissoid of such curves with respect to O. The essence of the cissoid method is that it construct a poler curve that measures the distance of (two) given curves as a line sweeps by. For example, we can have the cissoid of a circle, or a cissoid of a parabola.
## Properties
### Cissoids of two Lines or two Circles
The cissoid of two concentric circles with pole on center is two concentric circles centered on pole. The cissoid of two circles in general is a combinations of various oval, figure-eight, or droplet-shaped curves. analyze the number of loops, branches, nodes, cusps, …etc.
The cissoid of two parallel lines is a line. The cissoid of two non-parallel lines with pole not on the line is a hyperbola-like curve. The curve has two asymptotes parallel to the given lines. If the point is on one line, then the cissoid is a line. prove or disprove that cissoid of two lines is a hyperbola.
### Cissoids of a Line and a Circle
If C1 is a circle, and C2 is a line tangent to C1 at point A, and O is the point on C1 opposite A, then the cissoid of C1, C2 and the pole O is called Cissoid of Diocles. If O is a arbitrary point on the circle, the curve is a oblique cissoid. Oblique Cissoid
If the line passes through the center of the circle, and pole on the circle, then the resulting curve is a strophoid, and if the pole is a point on the circle furthest from the line, it's a right strophoid.
The cissoid of a line and a circle, with pole on the center of circle, is any member of conchoid of Nicomedes.
### Tangent at Pole
Let there be a cissoid based on given curves c1, c2, and point O. If c1, c2 intersect at point P, then line[O,P] is a tangent of the cissoid at point O.
### Algebraic Curve
The cissoid of a algebraic curve and a line is itself algebraic. from to Robert C Yates. Needs proof. Check J Dennis Lawrence's book.
### Curve relations by cissoid
Base Curves Pole Cissoid
two parallel lines point not on line line
a line and a circle center of circle conchoid of Nicomedes
a circle and a tangent on circumsference oblique cissoid
circle and a tangent opposite point of tangency cissoid of Diocles
circle and line through center on circumsference (oblique) strophoid
two concentric circles center two concentric circles
a single circle (as two arcs) point r*Sqrt[2] distant from center one loop of lemniscate of Bernoulli | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://xahlee.info/SpecialPlaneCurves_dir/Cissoid_dir/cissoid.html", "fetch_time": 1675292757000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764499953.47/warc/CC-MAIN-20230201211725-20230202001725-00363.warc.gz", "warc_record_offset": 85758123, "warc_record_length": 4042, "token_count": 1049, "char_count": 3892, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.9064304232597351, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"} |
# The jones family plans a trip to the amusement park
Posted By Admin @ November 26, 2022
#### Question:
The Jones family plans a trip to the amusement park and purchase a family package for $128.00 They also want to buy lunch and souvenir cups. The chicken platter costs$8.50 each and the burger
platter costs $9.50 each. The souvenir cups each cost$10.00, which includes unlimited refiis. Mr.
and Mrs. Jones have a budget and are trying to decide who will order which platter and who will buy a
souvenir cup. Let x represent the number of chicken platters, y represent the number of burger
platters, and z represent the number of souvenir cups. Which of the following algebraic expressions
can be used to describe the amount of money the Jones family will spend, before sales tax, based on
the number of each meal and souvenir cups they purchase?
A-128+38(x+y+z)
B-128(8.50x+9.50y+10z)
C-128+8.50x+9.50y+10z
D-128+x+y+z
The algebraic expressions that can be used to describe the amount of money the Jones family will spend, before sales tax is given by: Option C-128+8.50x+9.50y+10z
### How to form mathematical expression from the given description?
You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.
For this case, we're provided that:
• Jones family purchase a family package for $128.00 • The chicken platter costs$8.50 each and
• The burger platter costs $9.50 each • The souvenir cups each cost$10.00
• x represent the number of chicken platters
• y represent the number of burger platters
• z represent the number of souvenir cups
1 chicken platter = \$8.50
x chicken platter = 8.50 + 8.50 + ... + 8.50 (x times) = dollars
Similarly, we get:
y burger platters = dollars
z souvenir cups = dollars
Total money spend on food before tax = Family package cost + food cost
= (in dollars)
(we don't write symbols like of currency generally, and understand it from context. Also, sign of multiplication is often hidden if there are non numeric symbols and numbers being multiplied are written together)
Thus, the algebraic expressions that can be used to describe the amount of money the Jones family will spend, before sales tax is given by: Option C-128+8.50x+9.50y+10z
brainly.com/question/11938672
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The equivalent expression of 2^4x=8^x-3 is 2^4x=2^(3x-9)How to determine the equivalent equation?The equation is given as: 2^4x=8^x-3Express 8 as 2^3 2^4x=2^3(x-3)Open the bracket 2^4x=2^(3x-9)Hence, the … | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://cupsix.com/30971/the-jones-family-plans-a-trip-to-the-amusement-park/", "fetch_time": 1680129057000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00198.warc.gz", "warc_record_offset": 246508253, "warc_record_length": 10670, "token_count": 1354, "char_count": 5589, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2023-14", "snapshot_type": "latest", "language": "en", "language_score": 0.8828515410423279, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"} |
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1. The cubic equation x^3 + ax^2 + bx -26 = 0 has 3 positive, distinct, integer roots. Find the values a and b.
I don't understand how to work this out.
2. Have you tried getting everything factorised
Is that what you mean?
3. What factor would I use?
4. If the equation has three distinct roots, you will be able to factorise it into something of the form:
Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, -fgh. You know that this must equal -26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full.
5. (Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that
I wish I understood this.
6. (Original post by porkstein)
If the equation has three distinct roots, you will be able to factorise it into something of the form:
Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, -fgh. You know that this must equal -26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full.
Thanks.
7. I got 1,2,13, can't it be -1,-2, -13?
8. (Original post by Math12345)
I got 1,2,13, can't it be -1,-2, -13?
Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*
If -1, -2, and -13 are the roots, then you'll have the cubic equation:
But , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
9. (Original post by Blazy)
Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*
If -1, -2, and -13 are the roots, then you'll have the cubic equation:
But , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
Got it, thanks.
10. (Original post by Math12345)
If are the roots of the polynomial , then . Use the fact that if , and that
I wish I understood this.
TheMagicMan is being almost deliberately confusing, introducing notation that simply isn't necessary for this question. He's essentially just expressing what the question says in words with symbols, presumably because that satisfies him or something. I can't see how he could've thought it would help you in any way.
It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):
Means:
a4 just represents some general constant that is multiplied by the x4 term. You could call it "the x4 coefficient". As such the expression above involving that capital sigma (the big E-like symbol) is just a way of expressing some general polynomial (of degree n).
, which should really be written as
Means:
So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a0, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (-1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on just how many things you multiply together.
if
This is just a fancy way of saying that the roots are all different. The second root of the equation, r2, is not equal to the fifth root of the equation, r5. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.
This is just a fancy way of saying the roots are integers.
11. (Original post by porkstein)
TheMagicMan is being almost deliberately confusing, introducing notation that simply isn't necessary for this question. He's essentially just expressing what the question says in words with symbols, presumably because that satisfies him or something. I can't see how he could've thought it would help you in any way.
It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):
Means:
a4 just represents some general constant that is multiplied by the x4 term. You could call it "the x4 coefficient". As such the expression above involving that capital sigma (the big E-like symbol) is just a way of expressing some general polynomial (of degree n).
, which should really be written as
Means:
So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a0, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (-1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on how many things you multiply together.
if
This is just a fancy way of saying that the roots are all different. The second root of the equation, r2, is not equal to the fifth root of the equation, r5. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.
This is just a fancy way of saying the roots are integers.
There's really only one hint that's helpful in this problem...that the product of the roots is equal to the coefficient of ...as for the notation, it's a lot quicker than words, and most users of this site would understand it.
Also, when there is an 'obvious' range for the summation and product limits, there is really no need to put them on.
Anyhow, like most people here, I'm just trying to help, so...
12. (Original post by TheMagicMan)
...
It's certainly not quicker than words - I highly doubt it took you longer to write that 'helpful hint' than the mess of LaTeX you gave him. Given the OP is asking a question about additional maths, which even if one didn't know was a qualification for 16 year-olds is (from the question) evidently not particularly advanced stuff, I find it highly unlikely that (s)he would be comfortable throwing around summations and products and terms with subscript i in that fashion.
And if you're going to leave the limits off, at least be consistent in doing so.
edit: nevermind
14. I am sorry but why is everone confusing this kid on such a basic question.
If it is a cubic and the constant is -26, then the factors have to make -26.
Now let's take all the factors of -26; they are -1,-2,-13,-26.
A factor can not be -26, since then you would have -1 and -1 as the other two factors which aren't distinct. Therefore, the three factors are -1,-2,-13.
Therefore we see the equation can be written in this format:
(x-1)(x-2)(x-13)=0
The roots of the above equation are x=1,2,13: Three positive, distinct, integer roots!
Now simply expand the brackets and you get:
x^3–16x^2+41x–26 = 0.
Therefore, a =-16 and b=41
15. (Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that
Don't try and act really clever with this. You are being annoying by trying to look so great and confusing the kid, when there is a very easy solution which you have failed to find with your "complex" method.
16. (Original post by Blazy)
Notice the condition upon the question - they are three POSITIVE, distinct integer roots.*
If -1, -2, and -13 are the roots, then you'll have the cubic equation:
But , so -1, -2 and -13 can't be the three roots (plus they're not even positive in the first place)
i think he is confused between factors and roots.
17. (Original post by GreenLantern1)
Don't try and act really clever with this. You are being annoying by trying to look so great and confusing the kid, when there is a very easy solution which you have failed to find with your "complex" method.
It's not complex...it's exactly the same as whatever else has been suggested
18. (Original post by TheMagicMan)
It's not complex...it's exactly the same as whatever else has been suggested
No you have suggested a complex notation that the kid is clearly not going to understand when he is only 15/16 years old an is doing maths harder than the GCSE normally taken at his age.
19. (Original post by TheMagicMan)
It's not complex...it's exactly the same as whatever else has been suggested
You know what he means. There was just no need to use higher level notation for this question.
When you help someone with maths (in TSR or anywhere else), you should first consider the level and ability of the person that you are helping. Sometimes, you may think that you're helping out but really, you're just making them more confused.
20. (Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that
Did it make you feel better to give me a negative rating when you clearly were being an arrogant, pompous fool!
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# Unlocking the Mystery of the 135° Interior Angle of a Regular Polygon
## Introduction: Overview of Regular Polygons and an Interior Angle of 135 Degrees
Regular polygons are defined as having all sides of equal length and all interior angles being equal to each other. When talking about the interior angle of 135 degrees, we must first discuss how it relates to an equilateral triangle, which has three sides, three vertices and most importantly, three interior angles measuring 60 degrees each. This is a basic proof demonstrating why the interior angle of 135 degrees will not appear in regular polygons.
To begin with, an equilateral triangle is a special type of polygon because it has three equal sides and therefore three equivalent angles inside the shape. When the sum of the measure for these angles is added together it should be 180° (three times 60°). So if multiple triangles make up a larger polygon, then this would mean that the sum within one full rotation of a regular polygon containing only triangles as its sides would have to be 360° since there would be six individual angles at work here (60°+60°+60°+60°+60°+60°=360). In other words, an equilateral triangle will always maintain its 3×60=180 degree interior angle configuration; regardless if it’s part of a group or separated alone. Even though some neighboring polygonal shapes such as squares have four 90-degree corners resulting in their collective interior angle addition output being 360 degrees this combination does not apply for any trapezoid which could contain both acute and obtuse angles combined together or parallel lines forming bigger quadrilaterals with non-functional alternating pairs correlating back towards a single linear trajectory when measured from corner to corner .
In conclusion, given that every consecutive edge around their respective perimeters binds together by using two intersecting vertices on each side then based on geometric properties even if you think about adding or subtracting one or two more angular measurements into the equation doesn’t necessarily guarantee another configuration closer towards 135° being reached simply because nothing else meets this
## How does the Interior Angle Measure of a Regular Polygon Influence its Geometry?
The interior angle measure of a regular polygon directly affects the geometry of the shape. A regular polygon is one whose sides are all equal and its angles are all congruent (the same). When the number of sides is increased, the size of its internal angles decreases; when the number of sides is decreased, the size of its internal angles increases. This means that increasing or decreasing the interior angle measure of a regular polygon will have an effect on the overall geometry.
For example, when discussing triangles, we know that a triangle with three interior angles measuring 60° each is an equilateral triangle – meaning it has three equal length sides and three equal length angles. Thus, changing this angle measurement to 30° instead would result in a different shape entirely – namely, a right triangle – which has two equal lengths and two right angles.
This same concept applies to any other shape with more than three sides (such as squares, pentagons and beyond). As such, by altering the size of any given regular polygon’s internal angle measure – via increasing or decreasing it accordingly – it effectively alters its geometrical makeup from one shape to another (or vice versa).
It should be noted that for any arbitrary number of sides on a regular polygon (say five for instance), conventional wisdom states that this would typically correspond with an interior angle measure totaling 360° divided by five; resulting in each individual side being 72° in measurement. Remember this equation well as it’s often useful when trying to accurately define what type and/or size of regular polygon you’re dealing with at any given time!
## What Type of Polygon Has an Interior Angle Measure Equal to 135 Degrees?
A polygon is any two-dimensional figure with three or more straight sides. The interior angles of a polygon measure the angle formed between two of its side lines at a single corner. Every polygon has an exterior and an interior angle measurement, and when these measurements add up accurately, it is possible to determine the type of polygon you are dealing with.
The type of polygon that has an interior angle measure equal to 135 degrees is known as an octagon. An octagon consists of eight total sides and eight interior angles, each measuring 135 degrees for a grand total of 1080 degrees – the sum for all sides in any polygon regardless of how many edges it may have.
When forming an octagon from scratch, each internal angle can be calculated by taking 360° – 1080°/ 8 = 135° . As long as individual interior angles adhere to this formula then you’ve got yourself and correct octagonal shape! Taking into account the sum total of all sides and corollary internal angles is essential when completing your own polygons so that each point measures properly; otherwise, your angled shape will be off balance and hold less visual appeal than it might otherwise have done had the math been executed accurately. This is why it pays off to properly identify which types of polygons will work best within your project based on their individual measurements before moving forward into creating something more permanent!
## Step-by-Step Guide: Understanding the Inner Workings of a Regular Polygon with an Interior Angle of 135 Degrees
This step-by-step guide will serve as an introduction to the concept of polygons, namely regular polygons, and explain how one with an interior angle of 135 degrees is constructed.
Let’s begin by defining a polygon. A polygon is a shape composed of line segments connecting at their endpoints in a closed loop. The most basic example would be a triangle, which has three points connected in a certain way to create a shape with three angles and three sides.
Regular polygons are those that have the same length for each side and the same interior angle across all vertices – or corners – of its perimeter. As it stands, this means that we can calculate the number of sides in a regular polygon given two numbers: the degree measurement of its interior angles and its total number of sides.
Now let’s convert this theoretical understanding into practical application – starting with our focus: drawing the perfect regular polygon with an interior angle of 135 degrees!
The first step is to apply some basic geometry principles by determining how many sides must make up this regular polygon. If we use an equation like S=180(n-2) / n, where ‘S’ represents the degree measurement for all interior angles (in our case, it would be 135), ‘n’ represents all possible numbers for different standard points along the perimeter (two or more), then subtracting 2 from ‘n’ helps us figure out exactly how many sides must compose this particular regular polygon to maintain congruency and balance (the actual answer turns out 8).
Since we now know there should be eight different points around the perimeter forming our particular shape, connecting them so they maintain their exact restriction via equal length measurements requires further math and calculations not required here during a basic introduction; suffice it to say they must remain neat lines drawn on paper or digitally in such manner as if they completed an entire
## FAQs: Common Questions about Understanding the Essential Characteristics of a Regular Polygon with an Interior Angle totaling 135 Degrees
Q: What is a regular polygon?
A: A regular polygon is a 2-dimensional, closed shape composed of straight line segments. The sides of the polygon are all the same length and each corner angle has an equal measure. Common examples of regular polygons include triangles, parallelograms, and hexagons.
Q: What is the interior angle totaling 135 degrees in a regular polygon?
A: When all angles inside a regular polygon add up to 135 degrees, it means that each interior angle has an equal measure of 15 degrees. This type of regular polygon can be broken down into nine equilateral triangles made from joining three corner points at each triangle’s apex.
## Top 5 Facts: Uncovering Truths Behind a Regular Polygon with an Interior Angle Summing up to Be exactly 135 Degrees
1. A regular polygon is an n-sided shape with all sides the same length and all angles between its sides equal. When a regular polygon has interior angles that sum up to exactly 135 degrees, it has five sides, and is therefore called a pentagon.
2. Five-sided polygons are not just limited to those with a 135-degree interior angle sum. Because each angle in the pentagon is usually measured in degrees, any triangular shape can be classified as a number of possible polygons as long as the sum of its interior angles adds up to 360 degrees.
3. The most interesting fact about a regular pentagon whose interior angles sum up to 135 degrees is that it’s actually composed of two half equilateral triangles joined together at their vertices, making it mathematically equivalent to those shapes known more conventionally as a hexagon or an octagon with different interiors sums of 180 or 240 degrees respectively.
4. While it may sound curious that this five sided shape can also be represented by two forms usually associated with six and eight sided figures respectively, there is one requirement for them to remain mathematically consistent; both the hexagonal and octagonal variations must have three 120 degree internal angles – by logically combining these two figures together tessellating around each other without having any uneven portions then results in the formation of our rare135 degree regular pentagons shown initially.
5. It might come completely intuitive then from analysis of this unorthodox method for deducing such shapes that skilled mathematicians could use similar tactics within more complex problems involving geometry; parts of irregular puzzles snatched from their individual elements recombined into a larger structure swiftly reassembled coherently but never neglected within entirety where variables have values appraised close intimately still considered differently somehow wedged notably upon small matters bled together forming new surmises per field widened drastically throughout which quickened response aside held trends perceived enhanced understanding now unlocking further whispers collected
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# 410920 (number)
410,920 (four hundred ten thousand nine hundred twenty) is an even six-digits composite number following 410919 and preceding 410921. In scientific notation, it is written as 4.1092 × 105. The sum of its digits is 16. It has a total of 5 prime factors and 16 positive divisors. There are 164,352 positive integers (up to 410920) that are relatively prime to 410920.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 16
• Digital Root 7
## Name
Short name 410 thousand 920 four hundred ten thousand nine hundred twenty
## Notation
Scientific notation 4.1092 × 105 410.92 × 103
## Prime Factorization of 410920
Prime Factorization 23 × 5 × 10273
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 102730 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 410,920 is 23 × 5 × 10273. Since it has a total of 5 prime factors, 410,920 is a composite number.
## Divisors of 410920
16 divisors
Even divisors 12 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 924660 Sum of all the positive divisors of n s(n) 513740 Sum of the proper positive divisors of n A(n) 57791.2 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 641.03 Returns the nth root of the product of n divisors H(n) 7.11042 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 410,920 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 410,920) is 924,660, the average is 57,791,.25.
## Other Arithmetic Functions (n = 410920)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 164352 Total number of positive integers not greater than n that are coprime to n λ(n) 10272 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 34611 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 164,352 positive integers (less than 410,920) that are coprime with 410,920. And there are approximately 34,611 prime numbers less than or equal to 410,920.
## Divisibility of 410920
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 6 0 7
The number 410,920 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
## Base conversion (410920)
Base System Value
2 Binary 1100100010100101000
3 Ternary 202212200021
4 Quaternary 1210110220
5 Quinary 101122140
6 Senary 12450224
8 Octal 1442450
10 Decimal 410920
12 Duodecimal 179974
20 Vigesimal 2b760
36 Base36 8t2g
## Basic calculations (n = 410920)
### Multiplication
n×y
n×2 821840 1232760 1643680 2054600
### Division
n÷y
n÷2 205460 136973 102730 82184
### Exponentiation
ny
n2 168855246400 69385997850688000 28512094236804712960000 11716189763787792649523200000
### Nth Root
y√n
2√n 641.03 74.3451 25.3186 13.2664
## 410920 as geometric shapes
### Circle
Diameter 821840 2.58189e+06 5.30474e+11
### Sphere
Volume 2.90643e+17 2.1219e+12 2.58189e+06
### Square
Length = n
Perimeter 1.64368e+06 1.68855e+11 581129
### Cube
Length = n
Surface area 1.01313e+12 6.9386e+16 711734
### Equilateral Triangle
Length = n
Perimeter 1.23276e+06 7.31165e+10 355867
### Triangular Pyramid
Length = n
Surface area 2.92466e+11 8.17722e+15 335515
## Cryptographic Hash Functions
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# BTU to Joules conversion
BTU (British Thermal Unit) to joules (J) conversion is the process of converting energy values from BTU to joules. To convert BTU to joules, you need to multiply the value in BTU by the conversion factor 1055.06.
Enter energy in BTU: BTU(IT) Result in Joules: J
Related
## What is BTU to Joules conversion
To convert British thermal units (BTU) to joules (J), you can use the conversion factor 1055.06.
Joules = BTU * 1055.06
For example, let's convert 10 BTU to joules:
Joules = 10 BTU * 1055.06 Joules ≈ 10550.6 J
Therefore, 10 BTU is approximately equal to 10550.6 joules.
## How to convert BTU to Joules
How to convert BTU (IT) to Joules:
1 BTUIT = 1055.05585262 J
So the energy conversion from BTU to joules is given by the formula:
E(J) = 1055.05585262 ⋅ E(BTU)
## BTU to Joules conversion table
Energy (BTUIT) Energy (J)
1 BTU 1055.055853 J
2 BTU 2110.111705 J
3 BTU 3165.167558 J
4 BTU 4220.223410 J
5 BTU 5275.279263 J
6 BTU 6330.335116 J
7 BTU 7385.390968 J
8 BTU 8440.446821 J
9 BTU 9495.502674 J
10 BTU 10550.558526 J
20 BTU 21101.117052 J
30 BTU 31651.675579 J
40 BTU 42202.234105 J
50 BTU 52752.792631 J
60 BTU 63303.351157 J
70 BTU 73853.909683 J
80 BTU 84404.468210 J
90 BTU 94955.026736 J
100 BTU 105505.585262 J
200 BTU 211011.170524 J
300 BTU 316516.755786 J
400 BTU 422022.341048 J
500 BTU 527527.926310 J
600 BTU 633033.511572 J
700 BTU 738539.096834 J
800 BTU 844044.682096 J
900 BTU 949550.267358 J
1000 BTU 1055055.852620 J
10000 BTU 10550558.526200 J | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.onlinecalculatorsfree.com/convert/btu-to-joules.html", "fetch_time": 1695984522000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00112.warc.gz", "warc_record_offset": 1028427968, "warc_record_length": 12499, "token_count": 598, "char_count": 1519, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.5797924995422363, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"} |
# Fractions
In Mathematics, fractions are defined as the parts of a whole. The whole can be an object or a group of objects. In real life, when we cut a piece of cake from the whole of it, then the portion is the fraction of the cake. A fraction is a word that is originated from Latin. In Latin, “Fractus” means “broken”. In ancient times, the fraction was represented using words. Later, it was introduced in numerical form.
The fraction is also termed as a portion or section of any quantity. It is denoted by using ‘/’ symbol, such as a/b. For example, in 2/4 is a fraction where the upper part denotes the numerator and the lower part is the denominator. In this article, we are going to learn the definition of fractions in Maths, types of fractions, conversion from fractions to decimals, and many solved examples with complete explanation.
## What are Fractions?
Definition 1: A fraction represents a numerical value, which defines the parts of a whole.
Definition 2: A fraction is a number that represents a part of a whole.
Generally, the fraction can be a portion of any quantity out of the whole thing and the whole can be any specific things or value.
The basics of fractions explain the top and bottom numbers of a fraction. The top number represents the number of selected or shaded parts of a whole whereas the bottom number represents the total number of parts.
Suppose a number has to be divided into four parts, then it is represented as x/4. So the fraction here, x/4, defines 1/4th of the number x. Hence, 1/4 is the fraction here. It means one in four equal parts. It can be read as one-fourth or 1/4. This is known as fraction.
Fractions play an important part in our daily lives. There are many examples of fractions you will come across in real life. We have to willingly or unwillingly share that yummy pizza amongst our friends and families. Three people, four slices. If you learn and visualize fractions in an easy way, it will be more fun and exciting. For example, slice an apple into two parts, then each part of the sliced apple will represent a fraction (equal to 1/2).
## Parts of Fractions
The fractions include two parts, numerator and denominator.
• Numerator: It is the upper part of the fraction, that represents the sections of the fraction
• Denominator: It is the lower or bottom part that represents the total parts in which the fraction is divided.
Example: If 3/4 is a fraction, then 3 is the numerator and 4 is the denominator.
## Properties of Fractions
Similar to real numbers and whole numbers, a fractional number also holds some of the important properties. They are:
• Commutative and associative properties hold true for fractional addition and multiplication
• The identity element of fractional addition is 0, and fractional multiplication is 1
• The multiplicative inverse of a/b is b/a, where a and b should be non zero elements
• Fractional numbers obey the distributive property of multiplication over addition
## Types of Fractions
Based on the properties of numerator and denominator, fractions are sub-divided into different types. They are:
• Proper fractions
• Improper fractions
• Mixed fractions
• Like fractions
• Unlike fractions
• Equivalent fractions
### Proper Fractions
The proper fractions are those where the numerator is less than the denominator. For example, 8/9 will be a proper fraction since “numerator < denominator”.
### Improper Fractions
The improper fraction is a fraction where the numerator happens to be greater than the denominator. For example, 9/8 will be an improper fraction since “numerator > denominator”.
### Mixed Fractions
A mixed fraction is a combination of the integer part and a proper fraction. These are also called mixed numbers or mixed numerals. For example:
### Like Fractions
Like fractions are those fractions, as the name suggests, that are alike or same.
For example, take ½ and 2/4; they are alike since if you simplify it mathematically, you will get the same fraction.
### Unlike Fractions
Unlike fractions, are those that are dissimilar.
For example, ½ and 1/3 are unlike fractions.
### Equivalent Fractions
Two fractions are equivalent to each other if after simplification either of two fractions is equal to the other one.
For example, ⅔ and 4/6 are equivalent fractions.
Since, 4/6 = (2×2)/(2×3) = 2/3
## Unit Fractions
A fraction is known as a unit fraction when the numerator is equal to 1.
• One half of whole = ½
• One-third of whole = 1/3
• One-fourth of whole = ¼
• One-fifth of whole = ⅕
## Fraction on a Number Line
We have already learned to represent the integers, such as 0, 1, 2, -1, -2, on a number line. In the same way, we can represent fractions on a number line.
For example, if we have to represent 1/5 and 3/5 parts of a whole, then it can be represented as shown in the below figure.
Since the denominator is equal to 5, thus 1 is divided into 5 equal parts, on the number line. Now the first section is 1/5 and the third section is 3/5.
Similarly, you can practice marking more of the fractions on the number line, such as 1/2, 1/4, 2/11, 3/7, etc.
## Rules for Simplification of Fractions
There are some rules we should know before solving the problems based on fractions.
Rule #1: Before adding or subtracting fractions, we should make sure that the denominators are equal. Hence, the addition and subtraction of fractions are possible with a common denominator.
Rule #2: When we multiply two fractions, then the numerators are multiplied as well as the denominators are multiplied. Later simplify the fraction.
Rule #3: When we divide a fraction from another fraction, we have to find the reciprocal of another fraction and then multiply with the first one to get the answer.
The addition of fractions is easy when they have a common denominator.
For example, ⅔ + 8/3 = (2+8)/3 = 10/3
Hence, we need to just add the numerators here.
### Adding Fractions with Different Denominators
If the denominators of the two fractions are different, we have to simplify them by finding the LCM of denominators and then making it common for both fractions.
Example: ⅔ + ¾
The two denominators are 3 and 4
Hence, LCM of 3 and 4 = 12
Therefore, multiplying ⅔ by 4/4 and ¾ by 3/3, we get;
8/12 + 9/12
= (8+9)/12
= 17/12
## Subtracting Fractions
The rule for subtracting two or more fractions is the same as for addition. The denominators should be common to subtract two fractions.
Example: 9/2 – 7/2 = (9-7)/2 = 2/2 = 1
### Subtracting with Different Denominators
If the denominators of the two fractions are different, we have to simplify them by finding the LCM of denominators and then making it common for both fractions.
Example: ⅔ – ¾
The two denominators are 3 and 4
Hence, LCM of 3 and 4 = 12
Therefore, multiplying ⅔ by 4/4 and ¾ by 3/3, we get;
8/12 – 9/12
= (8-9)/12
= -1/12
## Multiplication of Fractions
As per rule number 2, we have discussed in the previous section, when two fractions are multiplied, then the top part (numerators) and the bottom part (denominators) are multiplied together.
If a/b and c/d are two different fractions, then the multiplication of a/b and c/d will be:
(a/b) x (c/d) = (axc)/(bxd) = (ac/bd)
Example: Multiply ⅔ and 3/7.
(⅔) x (3/7) = (2×3)/(3×7) = 2/7
## Division of Fractions
If we have to divide any two fractions, then we will use here rule 3 from the above section, where we need to multiply the first fraction to the reciprocal of the second fraction.
If a/b and c/d are two different fractions, then the division a/b by c/d can be expressed as:
Example: Divide ⅔ by 3/7.
(⅔) ÷ (3/7) = (⅔) x (7/3) = (2×7)/(3×3) = 14/9
## Real-Life Examples of Fractions
Let us visualize some of the fractions examples:
1. Imagine a pie with four slices. Taking two slices of pie for yourself would mean that you have two out of the four. Hence, you represent it as 2/4.
2. Fill half a glass of water. What do you see? 1/2 glass is full. Or 1/2 glass is empty. This 1/2 is fractions where 1 is the numerator that is, the number of parts we have. And 2 is the denominator, the number of parts the whole glass is divided into.
## How to Convert Fractions To Decimals?
As we already learned enough about fractions, which are part of a whole. The decimals are the numbers expressed in a decimal form which represents fractions, after division.
For example, Fraction 1/2 can be written in decimal form as 0.5.
The best part of decimals are they can be easily used for any arithmetic operations such as addition, subtraction, etc. Whereas it is difficult sometimes to perform operations on fractions. Let us take an example to understand;
solution: 1/6 = 0.17 and 1/4 = 0.25
Hence, on adding 0.17 and 0.25, we get;
0.17 + 0.25 = 0.42
### How to Simplify Fractions?
To simplify the fractions easily, first, write the factors of both numerator and denominator. Then find the largest factor that is common for both numerator and denominator. Then divide both the numerator and the denominator by the greatest common factor (GCF) to get the reduced fraction, which is the simplest form of the given fraction. Now, let us consider an example to find the simplest fraction for the given fraction.
For example, take the fraction, 16/48
So, the factors of 16 are 1, 2, 4, 8, 16.
Similarly, the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
Thus, the greatest common factor for 16 and 48 is 16.
i.e. GCF (16, 48) = 16.
Now, divide both the numerator and denominator of the given fraction by 16, we get
16/48 = (16/16) / (48/16) = 1/3.
Hence, the simplest form of the fraction 16/48 is 1/3.
## Solved Examples on Fractions
Example 1:
Is 12/6 a fraction?
Solution:
Yes, it is. It is called an improper fraction.
Example 2:
Convert 130.1200 into a fraction.
Solution:
Here will use the concept of how to convert decimals into fractions
130.1200 = 130.1200/10000
= 13012/100
Example 3:
Solution:
3 /5 + 10/15
LCM of 5 and 15 is 15
= (9 + 10)/15
= 19/15
Example 4:
Which of the following fraction is the largest?
(a) 29/23
(b) 29/27
(c) 29/25
(d) 29/30
Solution:
To find whether the largest fractions among the given options, first convert the fractional value to the decimal value.
(a) 29/23 = 1.261
(b) 29/27 = 1.074
(c) 29/25 = 1.16
(d) 29/30 = 0.967
Thus, 29/23 is the largest fraction among the given options.
Hence, option (a) 29/23 is the correct answer.
Example 5:
Reduce the fraction 15/65 to the simplest form.
Solution:
Given fraction: 15/65.
Factors of 15: 1, 3, 5 and 15
Factors of 65: 1, 5, 13, and 65
Hence, the greatest common factor of 15 and 65 is 5.
i.e. GCF (15, 65) = 5.
Now, divide both the numerator and the denominator of the given fraction (16/65) by 5, we get
15/65 = (15/5) / (65/5) = 3/13.
Hence, the simplest form of the fraction 15/65 is 3/13.
## Practice Questions on Fractions
Solve the following:
1. 3/7+9/2-8.
2. 22/7+8/11.
3. 32/9 x 81/4.
4. 44/9 ÷ 36/4.
5. Reduce 35/84 to the simplest form.
6. Convert the fraction 81/63 to the reduced form.
## Frequently Asked Questions on Fractions
### What are fractions in Maths?
Fractions are the numerical values that are a part of the whole. A whole can be an object or a group of objects. If a number or a thing is divided into equal parts, then each part will be a fraction of the whole. A fraction is denoted as a/b, where a is the numerator and b is the denominator.
### How to solve fractions?
To add or subtract fractions, we have to check if the denominators are the same or different. For the same denominators, we can directly add or subtract the numerators, keeping the denominator common. But if the denominators are different, then we need to simplify them by finding the LCM.
### What are the 3 types of fractions in Maths?
The 3 types of fractions in Maths are Proper fractions, Improper fractions, and Mixed fractions.
### Give real-life examples of fractions.
If a watermelon is divided into four equal parts, then each part is a fraction of ¼.
Similarly, if a pizza is divided into three equal parts, then each part shows 1/3rd of pizza.
### What is a unit fraction?
A fraction with numerator 1 is called a unit fraction. Examples are ½, ⅓, ¼, ⅕, 1/7, 1/10, etc. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathlake.com/Fractions", "fetch_time": 1721314394000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00576.warc.gz", "warc_record_offset": 333198607, "warc_record_length": 8244, "token_count": 3268, "char_count": 12269, "score": 4.59375, "int_score": 5, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.9359038472175598, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
# Currying in Lambda Calculus and OCaml
### Currying
Recall that in lambda calculus, a function can have more than one input, each preceded by a λ symbol. Another way of thinking about more than one input is curryingCurrying a function of two inputs turns that function into a function with one input by passing one of the inputs into it. In other words, currying turns f(x,y) to g(y) in which g is f with x passed into it. And g only takes one input, y. For example:
```f(x,y) = x + y ; if x = 3 then
f(3,y) = 3 + y
```
Similarly in lambda calculus:
```λx.λy.(x+y) 3 y
= λy.(3+y) y```
One can curry recursively, and turn a function of any number of input to a function of that number of input minus one. For example:
```λx.λy.λz.(x+y+z) 3 4 5
= λy.λz.(3+y+z) 4 5
= λz.(3+4+z) 5
= (3+4+5)```
Line 1 is a function that takes in 3 inputs (x,y,z). Line 2 is a function that takes in 2 inputs (y,z). Line 3 is a function that takes in 1 input (z).
### Currying in OCaml
Currying is also applied in popular functional languages including Haskell and OCaml. For example:
```let fun1 x y z = x + y + z;;
val fun1 : int -> int -> int -> int = <fun>```
In line 1 I defined a function in OCaml that takes inputs x,y,z. Line 2 is the OCaml output, displaying the type signature of my function named fun1. How do we read it?
The first 3 int are the input types of x,y,z. The last int is the type of the output of fun1. You can see that OCaml treats fun1 as a function that takes and passes in x (an int), and returns a function (with x passed in) that takes and passes in y, which returns a function (with x and y passed in) that takes in z. Lastly, the function (with all x,y,z passed in) returns an int.
Let’s take a look at another example:
```let is_larger_than fun1 y = fun1 y > 100;;
val is_larger_than : ('a -> int) -> 'a -> bool = <fun>```
The first input type is (‘a -> int). No currying is involved here. It is a function that takes one input of a generic type (‘a), and returns an int. OCaml treats is_larger_than as a function that takes in fun1, and returns a function (with fun1 passed in) that takes and passes in (‘a). The output of is_larger_than is a bool. If fun1 were to take more than one input, OCaml curries it:
```let is_larger_than fun1 x y z = fun1 x y z > 100;;
val is_larger_than : ('a -> 'b -> 'c -> int) -> 'a -> 'b -> 'c -> bool = <fun>
```
Again, the first input is a function. fun1 takes in 3 inputs x,y,z. OCaml curries it, similar to above. OCaml treats is_larger_than as a function that takes and passes in fun1, and returns a function (with fun1 passed in) that takes in (‘a), and so on. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://thealmarty.com/2018/08/23/currying-in-lambda-calculus-and-ocaml/", "fetch_time": 1725810629000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00652.warc.gz", "warc_record_offset": 545685981, "warc_record_length": 15380, "token_count": 807, "char_count": 2651, "score": 3.8125, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.7532244920730591, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"} |
# GATE2005-48
1.8k views
Consider the following system of linear equations : $$2x_1 - x_2 + 3x_3 = 1$$ $$3x_1 + 2x_2 + 5x_3 = 2$$ $$-x_1+4x_2+x_3 = 3$$ The system of equations has
1. no solution
2. a unique solution
3. more than one but a finite number of solutions
4. an infinite number of solutions
rank of matrix $=$ rank of augmented matrix $=$ no of unknown $=$ $3$
so unique solution..
Correct Answer: $B$
edited
0
Can case C arise? If Yes, how shall we determine?
0
when rank of matrix = rank of augmented matrixno of unknown
3
then it is infinite solutions. r < n, that is option D. I'm asking about option C
1
i think more than one but a finite number of solutions will never arise
as we have only 3 cases r=n,r<n and r>n
2
yes c option case can never arise
0
@Angkit rank(r)>n this case will never arise
0
How can we find determinant of augmented matrix ?
Determinant of matrix =14 which is non zero
If The determinant of the coefficient matrix is non zero then definitely the system of given equation has a unique solution
so option B
1
in matrix $[A]_{3\times3},$ if $|A|_{3\times3}\neq0$ then rank should be $3$
1
if we get |A|=0 then we have to check for either infinite solⁿ or no solution ...so we have to go with our fundamental method..
Then i think finding determinant is not fruitful
## Related questions
1
7.6k views
Let $c_{1}.....c_{n}$ be scalars, not all zero, such that $\sum_{i=1}^{n}c_{i}a_{i}$ = 0 where $a_{i}$ are column vectors in $R^{n}$. Consider the set of linear equations $Ax = b$ where $A=\left [ a_{1}.....a_{n} \right ]$ ... set of equations has a unique solution at $x=J_{n}$ where $J_{n}$ denotes a $n$-dimensional vector of all 1. no solution infinitely many solutions finitely many solutions
Consider the system, each consisting of $m$ linear equations in $n$ variables. If $m < n$, then all such systems have a solution. If $m > n$, then none of these systems has a solution. If $m = n$, then there exists a system which has a solution. Which one of the following is CORRECT? $I, II$ and $III$ are true. Only $II$ and $III$ are true. Only $III$ is true. None of them is true.
If the following system has non-trivial solution, $px + qy + rz = 0$ $qx + ry + pz = 0$ $rx + py + qz = 0$, then which one of the following options is TRUE? $p - q + r = 0 \text{ or } p = q = -r$ $p + q - r = 0 \text{ or } p = -q = r$ $p + q + r = 0 \text{ or } p = q = r$ $p - q + r = 0 \text{ or } p = -q = -r$
Let $Ax = b$ be a system of linear equations where $A$ is an $m \times n$ matrix and $b$ is a $m \times 1$ column vector and $X$ is an $n \times1$ column vector of unknowns. Which of the following is false? The system has a solution if and only if, both $A$ ... system has a unique solution. The system will have only a trivial solution when $m=n$, $b$ is the zero vector and $\text{rank}(A) =n$. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://gateoverflow.in/1173/gate2005-48", "fetch_time": 1596842639000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00353.warc.gz", "warc_record_offset": 344759426, "warc_record_length": 17899, "token_count": 899, "char_count": 2851, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.8450114130973816, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"} |
# 02 - Finite State Machines #2
1. A finite state machine is a form of abstraction because it models the behaviour of a system by showing states and transitions.
FALSE
TRUE
2. An example of a FSM: Consider a lift (elevator). What are the possible states of the system?
There can only be a single state in a FSM
static on floor 1, moving up, static on floor 2, moving down, etc.
boolean off and boolean true
on and off
3. Representing a system as a finite state machine is very powerful because the model allows us to ________________________________.
demonstrate the behaviour clearly
break down the problem into smaller parts of the same problem
represent every detail of the system, thereby illustrating the complexity
4. Applications of finite state machines are found in many sciences. Mainly engineering, biology and most commonly in linguistics, where they are used to describe languages.
TRUE
FALSE
5. Read the excerpt below and in relation to the diagram, fill in the blanks.
```In this diagram, we can see that it starts in state S1,
an input of 1 will keep it in state S1, and an input of
0 will move it to state S2.
Once in S2 an input of 1
will keep it there, and an input of 0 will
____________________________.
This means that the following
inputs are valid:
110011001
001110011```
switch it back to a 0
switch it to S2 (for good)
switch it back to S1.
switch it off
6. A Mealy machine does not output values where as a finite state automaton does output values.
TRUE
FALSE
7. The following diagram represents a finite state machine which takes as input a binary number from the least significant bit. Which of the following is true?
The FSM decrements the input number
The FSM increments the input number
The FSM computes the 1s complement of the input string
The FSM computes 2s complement of the input string
8. Finite state automatons are unique in that they have stable and definite outputs.
FALSE
TRUE
9. For example, the finite state machine in the diagram has three states. If the machine is in state 1 then an A moves it to state 2 and a ___________________.
B moves it to state 3
B moves it back and forth between state 1,2 and 3
B moves it back to state 1
A moves it also to state 3
10. A _____________________ is a method of graphically representing finite-state machine.
state data flowchart
transition flowchart
state transition diagram
transition table flow chart
11. Many FSMs have a final state known as the ___________________. This is indicated by a double circle.
finishing state
creating state
accepting state
finite state
12. The sequence baaaba would produce 100010 as the output. The sequence ababb would produce _______as the output.
`Some FSMs produce output. These are identified by looking for two values seperated by a / symbol.`
13. Many communications protocols, such as USB can be defined by a finite state machine’s diagram showing what happens as different pieces of information are input.
FALSE
TRUE
14. You can build a finite state machine that can recognize palindromes and as it can count infinitely, it would be able to recognize all palindromes.
TRUE
FALSE
15. A finite state machine can not only count but count for an infinite number (e.g. never stop)
FALSE
TRUE | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.testandtrack.io/index.php/studenttest/givetest/547", "fetch_time": 1603381314000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00585.warc.gz", "warc_record_offset": 925650681, "warc_record_length": 19936, "token_count": 747, "char_count": 3277, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2020-45", "snapshot_type": "latest", "language": "en", "language_score": 0.8757668137550354, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00006-of-00064.parquet"} |
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Homework Statement CoCl4 2-+6H2O --> Co(H2O)62++4Cl-+energy If I add silver nitrate, which direction does it shift? I think it is no effect, because adding solid does not affect the equilibrium?... Am i right? Btw, what happen if i add aqueous to X gas --> Z gas + Y gas, vise verse... or gas...
26. ### Extremely Hard Physic Question (Ray diagram)
Homework Statement http://img163.imageshack.us/img163/7010/78614537.jpg [Broken] o= student The position of the teacher is (d) from the mirror. mirror is 6m a) Calculate the maximum distance(d) of the corridor from the mirror b) Trace the diagram. Draw the ray diagram to support your... | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/search/2055404/", "fetch_time": 1597140697000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439738746.41/warc/CC-MAIN-20200811090050-20200811120050-00522.warc.gz", "warc_record_offset": 781350868, "warc_record_length": 15061, "token_count": 2162, "char_count": 7542, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.7914073467254639, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00059-of-00064.parquet"} |
CCSS.MATH.CONTENT.4.OA.C.5 : Fourth Grade Math Worksheets
Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule "Add 3" and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way.
Here is a collection of our common core aligned worksheets for core standard 4.OA.C.5.
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4.OA.C 4.OA.C.5
Patterns
Predict and extend a growing pattern.
Core Standard: 4.OA.C.5
Repeating Patterns
Find the pattern unit and predict what comes next. Color the picture.
Core Standard: 4.OA.C.5
Patterns
Predict what comes next in the pattern. Cut and glue the pictures in the boxes to complete each pattern.
Core Standard: 4.OA.C.5
Identify Patterns
Find the pattern unit and predict what comes next. Color what is missing.
Core Standard: 4.OA.C.5
Patterns
Identify the pattern in each row, then extend the pattern.
Core Standard: 4.OA.C.5
Patterns
Find the pattern in each row.Then circle the picture to complete the pattern.
Core Standard: 4.OA.C.5
Color the Pattern
Identify the pattern in each row, then color the last shape to extend that pattern.
Core Standard: 4.OA.C.5
Patterns ( Fruits Theme )
Cut and glue the pictures in the boxes to create ABC pattern.
Core Standard: 4.OA.C.5
Halloween Patterns
Find the pattern in each row, then write ABC, ABAB, AAB, or ABB to describe each pattern.
Core Standard: 4.OA.C.5
Create the Patterns ( Animal Theme )
Color in the pictures to create your own patterns.
Core Standard: 4.OA.C.5
Create the Patterns
Color the pictures to create your own patterns.
Core Standard: 4.OA.C.5
Circus Patterns
Identify each pattern, then write ABC, ABAB, AAB or ABB to describe the pattern.
Core Standard: 4.OA.C.5
Patterns
Identify and describe the repeating patterns.
Core Standard: 4.OA.C.5
Identifying Rules and Patterns - I
Follow the given rule and complete the input out put tables by applying those rules. identify the correct equation that represents the rule.
Core Standard: 4.OA.C.5
Patterns
Predict and extend repeating patterns.
Core Standard: 4.OA.C.5
Patterns
Create and describe repeating patterns.
Core Standard: 4.OA.C.5
Patterns
Identify the pattern unit to ï¬nd the missing piece.
Core Standard: 4.OA.C.5
Repeating Geometric Patterns
Identify and extend the geometric patterns.
Core Standard: 4.OA.C.5
Growing Geometric Patterns
Draw the missing figure in each growing pattern.
Core Standard: 4.OA.C.5
Growing and Repeating Patterns
Students will identify and extend the repeating / growing geometric patterns.
Core Standard: 4.OA.C.5
Patterns in Geometry - I
When you repeat geometrical shapes, they make a pattern. Learn to identify and extend patterns by predicting the next occurrence/s. Write the rules for the pattern you find.
Core Standard: 4.OA.C.5
Patterns in Geometry - II
Look at a geometrical pattern and identify the rule that the pattern follows. Read a rule and practice making patterns based on the rule.
Core Standard: 4.OA.C.5
Numbers Patterns - I
Looking at a number sequence, identify the rule. Then use the same rule to extend the number patterns. Rules can be based on any of the four operations.
Core Standard: 4.OA.C.5
Numbers Patterns - II
Find the missing number in a number pattern once you are able to identify the rule that the pattern follows. There are some really tricky problems in this worksheet based on rules and patterns.
Core Standard: 4.OA.C.5
Identifying Rules and Patterns - III
Using the given rule, extend the pattern. Practice forming equation using the given rule. This will help you in building equation to represent and solve a problem.
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Thermodynamics filled in class notes_Part_42
# Thermodynamics filled in class notes_Part_42 - 91 3.3 IDEAL...
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Unformatted text preview: 91 3.3. IDEAL MIXTURES OF IDEAL GASES Example 3.5 Find the expression for mixture entropy of the ideal gas. si = ci si = ci si = N s= i=1 ˆ cP i (T ) ˆ Pi , dT − Ri ln ˆ Po T 298 T ˆ cP i (T ) ˆ Pi ci so ,i + ci , dT − ci Ri ln 298 ˆ Po T 298 N N N T ˆ cP i (T ) ˆ o ci ci s298,i + ci Ri ln dT − ˆ T 298 i=1 i=1 i=1 T so ,i + 298 N ci so ,i + 298 = i=1 = so + 298 T 298 N T 298 i=1 ˆ ci cP i (T ) ˆ dT − ˆ T N ˆ cP (T ) ˆ dT − ˆ T i=1 (3.211) Pi Po ci Ri ln i=1 , (3.212) Pi Po N Pi Po ci Ri ln (3.210) , (3.213) . (3.214) All except the last term are natural extensions of the property for a single material. Consider now the last term involving pressure ratios. N − ci Ri ln i=1 Pi Po N = − i=1 N = −R ci i=1 Ri ln R N = = −R Pi Po ci Ri ln i=1 N −R i=1 −R −R i=1 i=1 = = −R ln −R ln i=1 Po P P N (3.216) P P − ln Po Po , (3.217) (3.218) P P + ln − ln , Po Po , (3.219) P P + ln Po Po , (3.220) yi + ln + ln PN Po i=1 P Po , yi (Pi ) i=1 (3.221) , (3.222) N 1 yi Po P + ln P Po yi Pi Po i=1 i=1 , − ln Pi Po Po PN (3.215) P P − ln Po Po yi Pi Po ln + ln , + ln Pi Po yi ln −R ln Pi Po =y i N = Pi Po ln ln N j =1 cj /Mj N = P P − ln Po Po + ln ci /Mi N = Pi Po R/Mi N j =1 cj R/Mj ci P P − R ln Po Po + R ln yi + ln P Po , (3.223) CC BY-NC-ND. 18 November 2011, J. M. Powers. 92 CHAPTER 3. GAS MIXTURES N = −R ln i=1 N = −R ln yi Pi P + ln yi P P i=1 yi N y yi i = −R ln P Po + ln i=1 + ln P Po , P Po (3.224) , (3.225) . (3.226) So the mixture entropy becomes s= = so + 298 so + 298 T 298 ˆ cP (T ) ˆ dT − R ln ˆ T T 298 N y yi i + ln i=1 ˆ P cP (T ) ˆ dT − R ln − R ln ˆ Po T classical entropy of a single body P Po , (3.227) N y yi i . (3.228) i=1 non−Truesdellian The extra entropy is not found in the theory for a single material, and in fact is not in the form suggested by Truesdell’s postulates. While it is in fact possible to redefine the constituent entropy definition in such a fashion that the mixture entropy in fact takes on the classical form of a single material via the T (ˆ Pi ˆ definition si = so ,i + 298 cP iˆ T ) dT − Ri ln Po + Ri ln yi , this has the disadvantage of predicting 298 T no entropy change after mixing two pure substances. Such a theory would suggest that this obviously irreversible process is in fact reversible. On a molar basis, one has the equivalents N ρ= ρi = i=1 1 v= N 1 i=1 v i ρ n = , V M = V 1 = = vM, n ρ (3.229) (3.230) N u= yi ui = uM, (3.231) yi hi = hM, (3.232) i=1 N h= i=1 N cv = yi cvi = cv M, (3.233) i=1 cv = cP − R, (3.234) N yi cP i = cP M, cP = i=1 CC BY-NC-ND. 18 November 2011, J. M. Powers. (3.235) ...
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > supxrre Structured version Visualization version GIF version
Theorem supxrre 12715
Description: The real and extended real suprema match when the real supremum exists. (Contributed by NM, 18-Oct-2005.) (Proof shortened by Mario Carneiro, 7-Sep-2014.)
Assertion
Ref Expression
supxrre ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) = sup(𝐴, ℝ, < ))
Distinct variable group: 𝑥,𝑦,𝐴
Proof of Theorem supxrre
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 simp1 1130 . . . 4 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → 𝐴 ⊆ ℝ)
2 ressxr 10679 . . . 4 ℝ ⊆ ℝ*
31, 2sstrdi 3983 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → 𝐴 ⊆ ℝ*)
4 supxrcl 12703 . . 3 (𝐴 ⊆ ℝ* → sup(𝐴, ℝ*, < ) ∈ ℝ*)
53, 4syl 17 . 2 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) ∈ ℝ*)
6 suprcl 11595 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ, < ) ∈ ℝ)
76rexrd 10685 . 2 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ, < ) ∈ ℝ*)
86leidd 11200 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ, < ))
9 suprleub 11601 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ sup(𝐴, ℝ, < ) ∈ ℝ) → (sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ, < ) ↔ ∀𝑧𝐴 𝑧 ≤ sup(𝐴, ℝ, < )))
106, 9mpdan 683 . . . 4 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ, < ) ↔ ∀𝑧𝐴 𝑧 ≤ sup(𝐴, ℝ, < )))
11 supxrleub 12714 . . . . 5 ((𝐴 ⊆ ℝ* ∧ sup(𝐴, ℝ, < ) ∈ ℝ*) → (sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ, < ) ↔ ∀𝑧𝐴 𝑧 ≤ sup(𝐴, ℝ, < )))
123, 7, 11syl2anc 584 . . . 4 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ, < ) ↔ ∀𝑧𝐴 𝑧 ≤ sup(𝐴, ℝ, < )))
1310, 12bitr4d 283 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ, < ) ↔ sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ, < )))
148, 13mpbid 233 . 2 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ, < ))
155xrleidd 12540 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ*, < ))
16 supxrleub 12714 . . . . 5 ((𝐴 ⊆ ℝ* ∧ sup(𝐴, ℝ*, < ) ∈ ℝ*) → (sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ*, < ) ↔ ∀𝑥𝐴 𝑥 ≤ sup(𝐴, ℝ*, < )))
173, 5, 16syl2anc 584 . . . 4 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ*, < ) ↔ ∀𝑥𝐴 𝑥 ≤ sup(𝐴, ℝ*, < )))
18 simp2 1131 . . . . . . . 8 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → 𝐴 ≠ ∅)
19 n0 4314 . . . . . . . 8 (𝐴 ≠ ∅ ↔ ∃𝑧 𝑧𝐴)
2018, 19sylib 219 . . . . . . 7 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → ∃𝑧 𝑧𝐴)
21 mnfxr 10692 . . . . . . . . 9 -∞ ∈ ℝ*
2221a1i 11 . . . . . . . 8 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → -∞ ∈ ℝ*)
231sselda 3971 . . . . . . . . 9 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → 𝑧 ∈ ℝ)
2423rexrd 10685 . . . . . . . 8 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → 𝑧 ∈ ℝ*)
255adantr 481 . . . . . . . 8 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → sup(𝐴, ℝ*, < ) ∈ ℝ*)
2623mnfltd 12514 . . . . . . . 8 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → -∞ < 𝑧)
27 supxrub 12712 . . . . . . . . 9 ((𝐴 ⊆ ℝ*𝑧𝐴) → 𝑧 ≤ sup(𝐴, ℝ*, < ))
283, 27sylan 580 . . . . . . . 8 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → 𝑧 ≤ sup(𝐴, ℝ*, < ))
2922, 24, 25, 26, 28xrltletrd 12549 . . . . . . 7 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ 𝑧𝐴) → -∞ < sup(𝐴, ℝ*, < ))
3020, 29exlimddv 1929 . . . . . 6 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → -∞ < sup(𝐴, ℝ*, < ))
31 xrre 12557 . . . . . 6 (((sup(𝐴, ℝ*, < ) ∈ ℝ* ∧ sup(𝐴, ℝ, < ) ∈ ℝ) ∧ (-∞ < sup(𝐴, ℝ*, < ) ∧ sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ, < ))) → sup(𝐴, ℝ*, < ) ∈ ℝ)
325, 6, 30, 14, 31syl22anc 836 . . . . 5 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) ∈ ℝ)
33 suprleub 11601 . . . . 5 (((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) ∧ sup(𝐴, ℝ*, < ) ∈ ℝ) → (sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ*, < ) ↔ ∀𝑥𝐴 𝑥 ≤ sup(𝐴, ℝ*, < )))
3432, 33mpdan 683 . . . 4 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ*, < ) ↔ ∀𝑥𝐴 𝑥 ≤ sup(𝐴, ℝ*, < )))
3517, 34bitr4d 283 . . 3 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → (sup(𝐴, ℝ*, < ) ≤ sup(𝐴, ℝ*, < ) ↔ sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ*, < )))
3615, 35mpbid 233 . 2 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ, < ) ≤ sup(𝐴, ℝ*, < ))
375, 7, 14, 36xrletrid 12543 1 ((𝐴 ⊆ ℝ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦𝑥) → sup(𝐴, ℝ*, < ) = sup(𝐴, ℝ, < ))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 207 ∧ wa 396 ∧ w3a 1081 = wceq 1530 ∃wex 1773 ∈ wcel 2107 ≠ wne 3021 ∀wral 3143 ∃wrex 3144 ⊆ wss 3940 ∅c0 4295 class class class wbr 5063 supcsup 8898 ℝcr 10530 -∞cmnf 10667 ℝ*cxr 10668 < clt 10669 ≤ cle 10670 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1904 ax-6 1963 ax-7 2008 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2153 ax-12 2169 ax-ext 2798 ax-sep 5200 ax-nul 5207 ax-pow 5263 ax-pr 5326 ax-un 7455 ax-cnex 10587 ax-resscn 10588 ax-1cn 10589 ax-icn 10590 ax-addcl 10591 ax-addrcl 10592 ax-mulcl 10593 ax-mulrcl 10594 ax-mulcom 10595 ax-addass 10596 ax-mulass 10597 ax-distr 10598 ax-i2m1 10599 ax-1ne0 10600 ax-1rid 10601 ax-rnegex 10602 ax-rrecex 10603 ax-cnre 10604 ax-pre-lttri 10605 ax-pre-lttrn 10606 ax-pre-ltadd 10607 ax-pre-mulgt0 10608 ax-pre-sup 10609 This theorem depends on definitions: df-bi 208 df-an 397 df-or 844 df-3or 1082 df-3an 1083 df-tru 1533 df-ex 1774 df-nf 1778 df-sb 2063 df-mo 2620 df-eu 2652 df-clab 2805 df-cleq 2819 df-clel 2898 df-nfc 2968 df-ne 3022 df-nel 3129 df-ral 3148 df-rex 3149 df-reu 3150 df-rmo 3151 df-rab 3152 df-v 3502 df-sbc 3777 df-csb 3888 df-dif 3943 df-un 3945 df-in 3947 df-ss 3956 df-nul 4296 df-if 4471 df-pw 4544 df-sn 4565 df-pr 4567 df-op 4571 df-uni 4838 df-br 5064 df-opab 5126 df-mpt 5144 df-id 5459 df-po 5473 df-so 5474 df-xp 5560 df-rel 5561 df-cnv 5562 df-co 5563 df-dm 5564 df-rn 5565 df-res 5566 df-ima 5567 df-iota 6313 df-fun 6356 df-fn 6357 df-f 6358 df-f1 6359 df-fo 6360 df-f1o 6361 df-fv 6362 df-riota 7108 df-ov 7153 df-oprab 7154 df-mpo 7155 df-er 8284 df-en 8504 df-dom 8505 df-sdom 8506 df-sup 8900 df-pnf 10671 df-mnf 10672 df-xr 10673 df-ltxr 10674 df-le 10675 df-sub 10866 df-neg 10867 This theorem is referenced by: supxrbnd 12716 ovoliunlem1 24037 ovoliun2 24041 ioombl1lem4 24096 uniioombllem2 24118 uniioombllem6 24123 itg1climres 24249 itg2monolem1 24285 itg2i1fseq2 24291 nmcexi 29736 itg2addnc 34832 supxrrernmpt 41579 supminfxr 41624 sge0supre 42556 sge0reuzb 42615
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# 2.14: Facilitated Diffusion
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Can you help me move?
What is one of the questions no one likes to be asked? Sometimes the cell needs help moving things as well, or facilitating the diffusion process. And this would be the job of a special type of protein.
## Facilitated Diffusion
What happens if a substance needs assistance to move across or through the plasma membrane? Facilitated diffusion is the diffusion of solutes through transport proteins in the plasma membrane. Facilitated diffusion is a type of passive transport. Even though facilitated diffusion involves transport proteins, it is still passive transport because the solute is moving down the concentration gradient.
Small nonpolar molecules can easily diffuse across the cell membrane. However, due to the hydrophobic nature of the lipids that make up cell membranes, polar molecules (such as water) and ions cannot do so. Instead, they diffuse across the membrane through transport proteins. A transport protein completely spans the membrane, and allows certain molecules or ions to diffuse across the membrane. Channel proteins, gated channel proteins, and carrier proteins are three types of transport proteins that are involved in facilitated diffusion.
A channel protein, a type of transport protein, acts like a pore in the membrane that lets water molecules or small ions through quickly. Water channel proteins (aquaporins) allow water to diffuse across the membrane at a very fast rate. Ion channel proteins allow ions to diffuse across the membrane.
A gated channel protein is a transport protein that opens a "gate," allowing a molecule to pass through the membrane. Gated channels have a binding site that is specific for a given molecule or ion. A stimulus causes the "gate" to open or shut. The stimulus may be chemical or electrical signals, temperature, or mechanical force, depending on the type of gated channel. For example, the sodium gated channels of a nerve cell are stimulated by a chemical signal which causes them to open and allow sodium ions into the cell. Glucose molecules are too big to diffuse through the plasma membrane easily, so they are moved across the membrane through gated channels. In this way glucose diffuses very quickly across a cell membrane, which is important because many cells depend on glucose for energy.
A carrier protein is a transport protein that is specific for an ion, molecule, or group of substances. Carrier proteins "carry" the ion or molecule across the membrane by changing shape after the binding of the ion or molecule. Carrier proteins are involved in passive and active transport. A model of a channel protein and carrier proteins is shown in Figure below.
Facilitated diffusion through the cell membrane. Channel proteins and carrier proteins are shown (but not a gated-channel protein). Water molecules and ions move through channel proteins. Other ions or molecules are also carried across the cell membrane by carrier proteins. The ion or molecule binds to the active site of a carrier protein. The carrier protein changes shape, and releases the ion or molecule on the other side of the membrane. The carrier protein then returns to its original shape.
An animation depicting facilitated diffusion can be viewed at http://www.youtube.com/watch?v=OV4PgZDRTQw (1:36).
### Ion Channels
Ions such as sodium (Na+), potassium (K+), calcium (Ca2+), and chloride (Cl-), are important for many cell functions. Because they are charged (polar), these ions do not diffuse through the membrane. Instead they move through ion channel proteins where they are protected from the hydrophobic interior of the membrane. Ion channels allow the formation of a concentration gradient between the extracellular fluid and the cytosol. Ion channels are very specific, as they allow only certain ions through the cell membrane. Some ion channels are always open, others are "gated" and can be opened or closed. Gated ion channels can open or close in response to different types of stimuli, such as electrical or chemical signals.
## Summary
• Facilitated diffusion is the diffusion of solutes through transport proteins in the plasma membrane. Channel proteins, gated channel proteins, and carrier proteins are three types of transport proteins that are involved in facilitated diffusion.
## Explore More
### Explore More I
Use this resource to answer the questions that follow.
• Facilitated Diffusion atwww.physiologyweb.com/lecture_notes/membrane_transport/facilitated_diffusion.html.
1. Define facilitative diffusion.
2. Describe the alternating access model.
3. What is meant by the occluded state?
4. What occurs after the occluded state?
## Review
1. What is facilitated diffusion?
2. What is a transport protein? Give three examples.
3. Assume a molecule must cross the plasma membrane into a cell. The molecule is very large. How will it be transported into the cell?
4. Explain how carrier proteins function?
5. Explain the role of ion channels. Why are ion channels necessary?
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# What shall be the number of zeros at the right-end if 200!
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What shall be the number of zeros at the right-end if 200! [#permalink]
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04 Oct 2017, 03:41
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What shall be the number of zeros at the right-end if 200! is expanded. [n!= n x (n-1) x (n-2)...x 1, where n is a positive integer]
A. 20
B. 22
C. 40
D. 48
E. 49
Courtsey: Experts' Global
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Re: What shall be the number of zeros at the right-end if 200! [#permalink]
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04 Oct 2017, 04:02
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sanjana90 wrote:
What shall be the number of zeros at the right-end if 200! is expanded. [n!= n x (n-1) x (n-2)...x 1, where n is a positive integer]
A. 20
B. 22
C. 40
D. 48
E. 49
Courtsey: Experts' Global
Discussed here: how-many-terminating-zeroes-does-200-have-125105.html
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# A comparison operation for lists, based on P(a > b)
Consider two lists $A$ and $B$, both containing $n$ numbers. A comparison operation can be defined based on the probability of uniformly selecting an element $a$ from list $A$ and an element $b$ from list $B$ such that $a>b$.
More precisely, define $$Q(A,B) = \sum_{i,j} \text{sign}(a_i-b_j)$$ If $Q(A,B) = 0$, then we can say the lists "tie". If $Q(A,B)>0$ then $A$ "beats" $B$, and likewise $Q(A,B)<0$ can be said as $A$ "loses" to $B$.
The naive way of calculating this comparison of lists involves checking each pair of elements and thus takes $n^2$ time.
Is there a faster way to calculate this?
Note, that somewhat non-intuitively, $A$ can still beat $B$ even if the average value of $A$ is less than $B$. (Also, even more non-intuitive, this comparison operator is non-transitive. Known as "intransitive dice", you can have dice such that A beats B, B beats C, C beats A.)
I believe if we take the $O(n \log n)$ hit to sort $A$ and $B$ first, then there should be a better way to calculate the comparison. But I'm having trouble filling in the pieces. Hopefully this is a useful hint instead of a red-herring.
Let $A \cup B$ denote the union of $A$ and $B$. Sort $A \cup B$ into increasing order, keeping track for each element which list it came from. Now for each element of this sorted list, annotate it with the number of elements of $A$ that are smaller than it and the number of elements of $B$ that are smaller than it. Those annotations can be filled in with a single linear scan (starting from the start of the list, and scanning left-to-right, and keeping track of the number of elements of $A$ and $B$ seen so far). Then you can compute the sum from those annotations: for each $i$, you can compute $\sum_j \text{sign}(a_i-b_j)$ directly from the annotation.
This algorithm takes $O(n \log n)$ time: $O(n \log n)$ time to sort, plus $O(n)$ for the scan, plus $O(n)$ time to add up the sums. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://cs.stackexchange.com/questions/75346/a-comparison-operation-for-lists-based-on-pa-b/75356#75356", "fetch_time": 1632471469000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-39/segments/1631780057508.83/warc/CC-MAIN-20210924080328-20210924110328-00291.warc.gz", "warc_record_offset": 255590852, "warc_record_length": 37742, "token_count": 538, "char_count": 1969, "score": 3.875, "int_score": 4, "crawl": "CC-MAIN-2021-39", "snapshot_type": "latest", "language": "en", "language_score": 0.8757650256156921, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"} |
# How to test an identity of rational numbers?
Hi,
Why is there a difference in output between:
print 5-1 == 0
print 3^(1/3) - 3^(1/5) == 0
False
3^(1/3) - 3^(1/5) == 0
What can I do if $x,y,s,t \in \mathbb Q$, and I want to test $x^y == s^t$?
Roland
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There is also a difference between the level of evaluation of the two expressions. The sage (man) sees 5-1, and thinks mayby it is better to write a 4 immediately, then seeing 4 == 0... and in the other case thinks, let it be as it comes...
If a True or a False is needed, then we have to require in the second case an explicit evaluation.
sage: print 3^(1/3) - 3^(1/5) == 0
3^(1/3) - 3^(1/5) == 0
sage: print bool( 3^(1/3) - 3^(1/5) == 0 )
False
Note also how the sage interpreter reshapes expressions in the following cases:
sage: print 4 == 0
False
sage: print 3^(1/5) - 3^(1/5) == 0
0 == 0
sage: print bool( 3^(1/5) - 3^(1/5) ) == 0
True
sage: print type( 3^(1/5) - 3^(1/5) )
<type 'sage.symbolic.expression.Expression'>
sage: print type(0)
<type 'sage.rings.integer.Integer'>
sage: print 98^(1/2) - 8^(1/6) == 5184^(1/4)
7*sqrt(2) - 8^(1/6) == 6*4^(1/4)
sage: print bool( 98^(1/2) - 8^(1/6) == 5184^(1/4) )
True
Note: This answer differs only psychologically from the above one. (Please let it above.)
more
The difference you noticed comes from the fact that 5 and 3^(1/3) have different natures
sage: parent(5)
Integer Ring
sage: parent(3^(1/3))
Symbolic Ring
Comparisons between integers (and more generally rationals) is very different from comparisons between symbolic elements. In order to test x^y == s^t you would better assume that bot y and t are integers (that you can always do by taken appropriate powers). Assuming that you only care about positive numbers you can use
def test_equality(x, y, s, t):
m = lcm(y.denominator(), t.denominator())
return x^(y*m) == s^(t*m)
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+0
# A square region with perimeter 60 inches is made with square inch tiles. Bob removes one tile from the square and rearranges the remaining t
0
108
1
A square region with perimeter 60 inches is made with square inch tiles. Bob removes one tile from the square and rearranges the remaining tiles without any overlap to make a rectangular region with minimum perimeter. How many inches are in the perimeter?
Guest Aug 5, 2017
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### 1+0 Answers
#1
+76929
+2
If the original perimeter was 60 inches, there must have been 15 tiles per side and 15 * 15 tiles = 225 tiles in all for an area of 225 in^2
Removing one tile will give us an area of 224 in^2 with 224 tiles
It can be shown that the perimeter will be minimized whenever the difference between the length and width is minimized
The divisors of 224 are : 1 | 2 | 4 | 7 | 8 | 14 | 16 | 28 | 32 | 56 | 112 | 224
And the possible lengths and widhts giving and area of 224 in^2 with the associated perimeters, P, is as follows :
1 , 224 P = 450
2 , 112 P = 228
4 , 56 P = 120
7, 32 P = 78
8, 28 P = 72
14, 16 P = 60
So......14 x 16 tiles minimizes the perimeter = 60 inches [ oddly..... the same perimeter before we removed a tile !!! ]
Note......this will always happen when we remove exactly one tile from any square as in this situation..... if s is the side of the square....4s is the perimeter.....if we remove one tile, the new length and width that minimizes the perimeter is
(s - 1), (s +1)....but the perimeter is also 2 [(s - 1) + (s +1)] = 2 [ 2s ] = 4s !!!!
CPhill Aug 5, 2017
edited by CPhill Aug 5, 2017
edited by CPhill Aug 5, 2017
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Multiplication of Integers
Chapter 1 Class 7 Integers
Concept wise
### Transcript
Question 5 Replace the blank with an integer to make it a true statement. (a) (–3) × _____ = 27 (−3) × _____ = 27 We know that 3 × 9 = 27 And −1 × −1 = 1 So, (−3) × (−9) = 27 Question 5 Replace the blank with an integer to make it a true statement. (b) 5 × _____ = –35 5 × _____ = −35 We know that 5 × 7 = 35 And 1 × −1 = −1 So, 5 × (−7) = −35 Question 5 Replace the blank with an integer to make it a true statement. (c) _____ × (– 8) = –56 _____ × (−8) = −56 We know that 7 × 8 = 56 And 1 × −1 = −1 So, 7 × (−8) = −56 Question 5 Replace the blank with an integer to make it a true statement. (d) _____ × (–12) = 132 _____ × (−12) = 132 We know that 11 × 12 = 132 And −1 × −1 = 1 So, (−11) × (−12) = 132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.teachoo.com/7741/2376/Ex-1.3--9/category/Multiplication-of-Integers/", "fetch_time": 1723416256000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00783.warc.gz", "warc_record_offset": 795057991, "warc_record_length": 21190, "token_count": 299, "char_count": 789, "score": 4.125, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.7804420590400696, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00058-of-00064.parquet"} |
# Evaluate 2st2 – 52 for s = 4 and t = 8
• Last Updated : 15 Jan, 2022
The basic concept of algebra taught us how to express an unknown value using letters such as x, y, z, etc. These letters are termed here as variables. this expression can be a combination of both variables and constants. Any value that is placed before and multiplied by a variable is termed a coefficient. An idea of expressing numbers using letters or alphabets without specifying their actual values is defined as an algebraic expression.
### Algebraic Expression
An expression that is made up of variables and constants along with algebraic operations such as addition, subtraction, etc. is termed an algebraic expression. These Expressions are made up of terms. Algebraic expressions are the equations when the operations such as addition, subtraction, multiplication, division, etc. are operated upon any variable. A combination of terms by the operations such as addition, subtraction, multiplication, division, etc is termed as an algebraic expression (or) a variable expression. Examples: 2x + 4y – 7, 3x – 10, etc.
The above expressions are represented with the help of unknown variables, constants, and coefficients. The combination of these three terms is termed as an expression. Unlike the algebraic equation, It has no sides or ‘equals to’ sign.
### Types of Algebraic expression
There are three types of algebraic expressions based on the number of terms present in them. They are monomial algebraic expressions, binomial algebraic expressions, and polynomial algebraic expressions.
• Monomial Expression: An expression that has only one term is termed a Monomial expression. Examples of monomial expressions include 5x4, 2xy, 2x, 8y, etc.
• Binomial Expression: An algebraic expression which is having two terms and unlike are termed as a binomial expression. Examples of binomial include 5xy + 8, xyz + x2, etc.
• Polynomial Expression: An expression that has more than one term with non-negative integral exponents of a variable is termed a polynomial expression. Examples of polynomial expression include ax + by + ca, 3x3 + 5x + 3, etc.
Other Types of Expression
Apart from monomial, binomial, and polynomial types of expressions, there are other types of expressions as well that are numeric expressions, variable expressions.
• Numeric Expression: An expression that consists of only numbers and operations, but never includes any variable is termed a numeric expression. Some of the examples of numeric expressions are 11 + 5, 14 ÷ 2, etc.
• Variable Expression: An expression that contains variables along with numbers and operations to define an expression is termed A variable expression. Some examples of a variable expression include 5x + y, 4ab + 33, etc.
Some algebraic formulae
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a – b) = a2 – b2
(x + a)(x + b) = x2 + x(a + b) + ab
(a + b)3 = a3 + b3 + 3ab(a + b)
(a – b)3 = a3 – b3 – 3ab(a – b)
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
### Evaluate 2st2 – 52 for s = 4 and t = 8
Solution:
Given expression: s = 4 and t = 8
= 2st2 – 52
Now put value of s and t
= 2 (4)(8)2 – 52
= 8(64) – 25
= 512 – 25
= 487
### Some Questions
Question 1: Solve for x: 5x – 50 = x + 3x
Solution:
5x – 50 = x + 3x
5x – 50 = 4x
5x – 4x = 50
x = 50
Question 2: Simplify (4x – 5) – (5x + 1)
Solution:
Given that, (4x – 5) – (8x + 1)
Step 1: Remove parentheses and apply the signs carefully.
= 4x – 5 – 8x – 1
Step 2: Bring like terms together
= 4x – 8x – 5 – 1
Step 3: Now add or subtract the like terms
= -4x – 6
= -2(2x + 3)
So the final result is -2(2x + 3)
Question 3: Solve for the value of t: 31 + t = 4 (t – 3) + 22.
Solution:
Given: 31 + t = 4 (t – 3) +22
31 + t = 4 (t – 3) + 22
31 + t = 4t – 12 + 22
31 + t = 4t + 10
31 – 10 = 4t – t
21 = 3t
t = 21/3
t = 7
So, the value of t is 7
Question 4: Solve for y in the equation: -1/y = -0.25x2 – 7.5
Solution:
Given: -1/y = -0.25x2 – 7.5
Multiply both sides by y
⇒ (-1/y)(y) = y(-0.25x2 – 7.5)
⇒ -1 = -0.25x2y – 7.5y
⇒ -1 = y(-0.25x2 – 7.5)
⇒ -1 = -y(-0.25x2 – 7.5)
⇒ y = 1/ (-0.25x2 – 7.5)
My Personal Notes arrow_drop_up | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.geeksforgeeks.org/evaluate-2st2-52-for-s-4-and-t-8/?ref=rp", "fetch_time": 1653187237000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00447.warc.gz", "warc_record_offset": 890099428, "warc_record_length": 24483, "token_count": 1325, "char_count": 4208, "score": 4.65625, "int_score": 5, "crawl": "CC-MAIN-2022-21", "snapshot_type": "latest", "language": "en", "language_score": 0.9299800395965576, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00051-of-00064.parquet"} |
# Understanding the definition of dense sets
I am recently confused about the definition of dense sets. What I learned is as the following
In topology and related areas of mathematics, a subset $A$ of a topological space $X$ is called dense (in $X$) if any point $x$ in $X$ belongs to $A$ or is a limit point of $A$.
The point is that when we say "a set $A$ is dense in a topological space $X$" we should first have the fact that $A$ is a subset of $X$. However, there is a notion that "a set $E$ is dense in a ball $X$" as I mentioned in this question. In this notion, $E$ is not necessarily a subset of $X$.
[Edit: How should I understand the definition of dense sets?] Can anybody tell me a reference for the most general definition for the dense sets?
-
It isn’t so much a matter of needing another definition as of getting the hang of how mathematical language evolves and is actually used ‘in the wild’. In the example in question, ‘$E$ is dense in the ball $B$’ can be thought of as verbal shorthand for ‘$E \cap B$ is a dense subset of the ball $B$’ $-$ which, when you think about it, is just about the only thing that it could reasonably mean.
How does such shorthand arise? Let’s say that you have the idea of a set $D$ in a topological space $X$ being dense in $X$ according to the definition that you gave. Any subset $Y$ of $X$ can be viewed as a space in its own right, with the subspace topology inherited from $X$, so you can certainly talk about a subset of $Y$ being dense in $Y$. But somewhere along the line you prove that $D$ is dense in $X$ iff $\operatorname{cl}D = X$; this characterization is far too useful to be omitted. What happens when you try to apply this definition to the subspace $Y$? It’s certainly true that a set $D \subseteq Y$ is dense in $Y$ iff $\operatorname{cl}_Y D = Y$, but if you’re thinking of $Y$ as a subset of $X$ rather than as a space in its own right, it’s more convenient to look at $\operatorname{cl}_X D$, which isn’t necessarily $Y$: it may be a proper superset of $Y$. (Take $X$ to be $E^1$, $Y$ to be $(0,1)$, and $D$ to be $\mathbb{Q} \cap (0,1)$.)
Thus, if we don’t want to bother with different closures in different sets, it’s much more convenient to say simply that a subset $D$ of a set $Y$ is dense in $Y$ iff $\operatorname{cl}D \supseteq Y$, where the closure operator is $\operatorname{cl}_X$, the closure in the whole space $X$. And once you get to that point, it’s sometimes convenient to separate the notion of dense in $Y$ from the notion of subset of $Y$ altogther. That is, we might as well say that $D$ is dense in $Y$ iff $\operatorname{cl}D \supseteq Y$, irrespective of whether $D \subseteq Y$. Then, for instance, we can say simply that the rationals are dense in $(0,1)$; we don’t have to say explicitly ‘the rationals in $(0,1)$’.
Many textbooks, including some very good ones (Munkres, Dugundji, Willard) define dense subset of a space but (so far as I can quickly tell) silently leave it to the reader to make the generalization to dense subset of a set in a space and the further generalization to dense in a set in a space, or to pick them up from context. Of course, one can define dense more generally. For example, John Greever’s Theory and Examples of Point-Set Topology gives this definition:
(2.14) Definition. If $(X,\mathscr{T})$ is a topological space and $A,B \subset X$, then $A$ is dense in $B$ if and only if $\overline{A} \supset B$.
Notice that he doesn’t require $A$ to be a subset of $B$: right from the start he’s using the most general of the three forms that I considered above. Had you learned the concept from Greever, you’d not have had a problem with the dense in a ball language. But whether with this concept or with another, you’re eventually going to encounter standard usages that require a little interpretation of or extrapolation from the ones that you’ve been taught.
-
Dear Brian, I'm a bit confused by the assertion that cl $D \supseteq Y$ iff cl $D\cap Y \supseteq Y$. Couldn't $D \cap Y = \emptyset$, e.g. if $X = \mathbb R$, $D = \mathbb Q$, and $Y = \mathbb R\setminus \mathbb Q$? Or have I gotten things muddled? Best wishes, – Matt E Mar 11 '14 at 4:07
Brian, this is actually the same question as above. If we take the reals with the standard topology and put $Y=\{1\}$ and $D=(0,1)$, then $Y\subseteq\mathrm{cl}\,D$, yet $\mathrm{cl}\,(D\cap Y)=\emptyset$. – Mad Hatter Aug 17 '15 at 9:01
Indeed, if $Y\not\subset \overline{\operatorname{int}_X Y}$, then $Y\subset \overline{D}$ does not imply $Y\subset \overline{Y\cap D}$. – Daniel Fischer Aug 17 '15 at 14:56
In the second paragraph, $cl_Y(D)=cl_X(D)\cap Y$. Hence $cl_Y(D)=Y$ iff $cl_X(D)\cap Y=Y$ iff $cl_X(D)\supset Y$. I think this might be a reason why we don't need to bother with different closures in different sets. – Jack Sep 30 '15 at 14:36
@MadHatter: You and Matt E are quite right; fixed now. (I’m not sure how I missed yours before.) – Brian M. Scott Sep 30 '15 at 14:57 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/58709/understanding-the-definition-of-dense-sets", "fetch_time": 1462133036000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-18/segments/1461860116886.38/warc/CC-MAIN-20160428161516-00148-ip-10-239-7-51.ec2.internal.warc.gz", "warc_record_offset": 183187637, "warc_record_length": 19016, "token_count": 1411, "char_count": 4981, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2016-18", "snapshot_type": "longest", "language": "en", "language_score": 0.9528929591178894, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"} |
# 5-Minute Fillers: Compound Words and More
Volume 1
Sticky-Note Compounds
Builds vocabulary, spelling skills
Prepare sticky notes in advance by writing on them the individual words that together form compound words. (For example, write foot on one sticky note and ball on another.) Put one note on each student's desk. As students come into the classroom, challenge them to find a classmate who has a sticky note that when paired with the one on their desk will form a compound word. Set a time limit on the activity.
Builds math computation and thinking skills
Provide for students several columns of numbers. A sum appears at the bottom of each column. But the numbers don't add up to the sum. They will add up, however, if one of the numbers is removed. Challenge students to figure out which number in each column does not belong. For example, the number 14 does not belong in this column.
26
19
14
22
+18
85
Pose the following question to students to start a lively discussion or use is as a prompt for a quick journal-writing activity: What if you could meet one famous person, either alive or dead? Who would that person be?
Words Up!
Builds spelling skills
Write a long word on the board. Tell students they have 3 minutes to write down as many small words as they can find in that long word. Students can use the letters in any sequence. They might work independently or in pairs. You might allow them to use a dictionary to check their work. For older students, limit words to those with four letters or more or reduce the time allotted to complete the activity.
Anagram Puzzles
Anagrams are a terrific tool for stimulating students to think critically.
Write the four phrases below on a board or chart. The letters in each phrase can be rearranged to spell a word. The words all have something in common. Challenge students to figure out the four words and what the words have in common.
Adapt the activity for younger students: To make the activity easier, tell students what the words have in common or arrange students in pairs to solve the anagram puzzles.
• I PLOT
• TRAITS
• DENTS IT
• RANGER ED
Answers: pilot, artist, dentist, and gardener are all jobs | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.educationworld.com/a_lesson/fillers/fillers001.shtml", "fetch_time": 1723688564000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00412.warc.gz", "warc_record_offset": 585309176, "warc_record_length": 33998, "token_count": 482, "char_count": 2186, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.9435920715332031, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"} |
## Re: Arkham Horror probability problem
So you want to know the probability of getting three or more rolls of 4+ on nine dice?
This is easier stated as 1 minus the probability of only getting 0,1 or 2 such successes. The binomial theorem gives us
1 way to get 0 successes
9 ways to get only 1 success
9*8/2=36 ways to get only 2 successes
p=1-((1*(3/6)^9)+(9*(3/6)*(3/6)^8)+(36*(3/6)^2*(3/6)^7))
=1-(46*(1/2)^9)
that's a 91 percent chance of success if I've understood the question. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.warseer.com/forums/showthread.php?411792-Arkham-Horror-probability-problem&s=ce928652fa56f12a4f3ea6c5c307a366&p=7527819", "fetch_time": 1611617196000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-04/segments/1610704792131.69/warc/CC-MAIN-20210125220722-20210126010722-00391.warc.gz", "warc_record_offset": 1082845984, "warc_record_length": 14559, "token_count": 159, "char_count": 485, "score": 3.65625, "int_score": 4, "crawl": "CC-MAIN-2021-04", "snapshot_type": "latest", "language": "en", "language_score": 0.8473238348960876, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00030-of-00064.parquet"} |
# “why is expected return considered forward-looking? what are the
EXPECTED RETURN AND MARKET RISK PROBLEM SOLUTION
(***** 100% Correct With Calculation *****)
Unit 4Assignment 2
Instructions
Answer the following questions and complete the following problems, as applicable:
You may solve the following problems algebraically, or you may use a financial calculator or Excel spreadsheet. If you choose to solve the problems algebraically, be sure to show your computations. If you use a financial calculator, show your input values. If you use an Excel spreadsheet, show your input values and formulas.
1. “Why is expected return considered forward-looking? What are the challenges for practitioners to utilize expected return” (Cornett, Adair, & Nofsinger, 2014, p. 250)?
2. “Describe how different allocations between the risk-free security and the market portfolio can achieve any level of market risk desired” (Cornett, Adair, & Nofsinger, 2014).
3. Refer to the table below to complete this question. “Compute the expected return given these three economic states, their likelihoods, and the potential returns” (Cornett, Adair, & Nofsinger, 2014).
4. “If the risk-free rate is 6 percent and the risk premium is 5 percent, what is the required return” (Cornett, Adair, & Nofsinger, 2014, p. 251)?
5. “The average annual return on the Standard and Poor’s 500 Index from 1986 to 1995 was 15.8 percent. The average annual T-bill yield during the same period was 5.6 percent. What was the market risk premium during these 10 years” (Cornett, Adair, & Nofsinger, 2014, p. 251)?
6. “Hastings Entertainment has a beta of 0.24. If the market return is expected to be 11 percent and the risk-free rate is 4 percent, what is Hastings’ required return” (Cornett, Adair, & Nofsinger, 2014, p. 251)?
7. Use the capital asset pricing model to calculate Hastings’ required return.
8. Calculate the beta of your portfolio, which comprises the following items: (a) Olympic Steel stock, which has a beta of 2.9 and comprises 25 percent of your portfolio, (b) Rent-a-Center stock, which has a beta of 1.5 and comprises 35 percent of your portfolio, and (c) Lincoln Electric stock, which has a beta of 0.2 and comprises 40 percent of your portfolio (Cornett, Adair, & Nofsinger, 2014).
Economic State Probability Return Fast Growth 0.30 40% Slow Growth 0.50 10% Recession 0.20 −25%
Submit your completed assignment as an attachment in the assignment area. You may use either a Word document or an Excel spreadsheet for your work, but not both. Prior to submitting your assignment, review the Estimating Risk and Return Scoring Guide to ensure you have met all of the requirements and as a self-assessment of your work.
Reference
Cornett, M. M., Adair, T. A., & Nofsinger J. (2014). M: Finance (2nd ed.). New York, NY: McGraw-Hill.
U Can Also Purchase Unit 4 Assignment 1 (***** Just Click On Below Link *****)
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# Statistics Study Guide
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Introduction
Statistics Study Guide Data are observations that have been collected (EX. Measurements, gender, survey responses) Statistics are a collection of methods for planning experiments, obtaining data, & then organizing, summarizing, presenting, analyzing, interpreting, and drawing conclusion based on the data. Stats are represented by a bell curve represents any population of anything. Census is the collection of data from every member of the population. Observational study- we observe and measure specific characteristics, but we don't attempt to modify the subjects being studied . Population is the complete collection of all elements to be studied. Parameter is a numerical measurement describing some characteristic of a sample. What is your data set composed of? Qualitative Data: Data are separated into different categories that are distinguished by same non-numerical characteristic. Quantitative Data - consist of numbers representing counts or measurements. Discrete data result when the number of possible values is either a finite number or a "countable" number (0, 1, 2, 3, etc.) Continuous (numerical) data results from infinitely many possible values that correspond to some continuous scale that covers a range of values without gaps, interruptions, or jumps. Ways to classify data Nominal data- characterized by data that consists of names labels, or categories. The data cannot be arranged in an ordering scheme (high to low, best to worst, Male or female) Ordinal- Can be arranged in same order, but difference between data values either cannot be determine or are meaningless. (1st, 2nd, 3rd....Likert scales: 1disagree...2...3...4...5...6 agree) Interval- distance between values becomes meaningful. However there is no natural zero starting point and ratios are meaningless. (none of the quantities are present) (Temperature 100-80=20... no mult or div) Ratio- Just like interval data, but add a natural zero staring point. For the values at this level, differences and ratios are both meaningful. (Can mult & div...no neg numbers) (Height, weight, prices, and distance traveled, miles per hr.) ...read more.
Middle
Variance- of a set of values is a measure of variation equal to the square of the standard deviation. Sample variance: Square of the standard deviation s. S= 7.0 min then, sample variance=s2=7.0 sq'd=49.0 min sq'd Population variance: Square of the population standard deviation. For estimating a value of the standard deviation s: to roughly estimate the standard deviation, use S=range/4 where range= (highest value)-(lowest value) Example.......... Lowest is 0 and highest is 491 so.......S= range/4 491/4=122.75=123 For interpting a known value of the standard deviation S: If the standard deviation s is known, use it to find rough estimates of the minimum and maximum "usual" sample values by using Min "usual" value= (mean) - 2 multiplied (standard deviation) Max "usual" value = (mean) + 2 multiplied (standard deviation) Example......the mean is 40.05cm and the standard deviation of 1.64cm Min= (mean) - 2 mult (standard deviation) 40.05 - 2(1.64) = 36.77 cm Max= (mean) + 2 mult (standard deviation) 40.05 + 2(1.64) = 43.33 cm A standard score, or z score, is the number of standard deviation that a given value x is above or below the mean. It is found using the following expressions: Sample Population Z= x- x/s Z= x - /o (round z to 2 decimal places) Percentile of a value x = number of values less than x/total number of value mult 100 L= (k/100)n where.....n= number of values......k= percentile in question Example......L= 68/100 multiplied by 40 = 27.2 then is it a whole # ? No ......change L by rounding it up to the next larger whole #...The Pk is the Lth value, counting from the lowest. Yes......The value of the Kth percentile is midway between the Lth value and the next value in the sorted set of data. Find Pk by adding the Lth value and the next value and diving the total by 2. Boxplot is a graph of a data set that constist of a line extending from the min value to the max value, and a box with lines drawn at the first quartile, q1; the median; and the third quartile, Q3. ...read more.
Conclusion
Complements: The probability of "at least one" is equivalent to "one or more" *The complement of getting at least one item of a particular type is that you get no items of that type. Conditional probability of an event is a probability obtained with the additional information that some other event has already occurred.P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred, and it can be found by dividing the probability of events A and B both occurring by the probability of the event A: P(B|A) = P(A and B)/P(A) Intuitive approach to conditional Probability The conditional probability of B given A can be found by assuming that event A has occurred and, working under that assumotion, calculating the proability that event B will occur. Simulation of a procedure is a process that behaves the same way as the ptocedure, so that similar results are produced. Fundamental Counting Rule For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m X n ways. Notation The factorial symbol! Denotes the product of decreasing positive whole numbers. For example, 4! = 4 x 3 x 2 x 1 = 24. By special definition, 0! =1. ( Many calculators have a factorial key.) Factorial Rule A collection of n different items can be arranged in order n! different ways. (This factoral rule reflects the fact that the frist item may be selected n different ways, the second item may be selected n - 1 ways, and so on.) Permutations Rule (when items are all different) The number of permutations (for sequences) of r items selected from n available items (without replacement) is nPr = n!/(n -r)! Permutations Rule (when some items are identical to others) If there are n items with n1 alike, n2 alike,.....nk alike, the number of permutations of all n items is n!/ n1!n2!....nk! Combinations Rule The number of combinations of r items selected from n different items is nCr = n!/(n -r)!r! ...read more.
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# SOLUTION: this rectangle will be scaled on graph paper.the scale is 1cm =5feet what will the longer side measure and the short side ? its a rectangle with 25 feet on the left side and 50 fee
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Click here to see ALL problems on Bodies-in-space Question 246078: this rectangle will be scaled on graph paper.the scale is 1cm =5feet what will the longer side measure and the short side ? its a rectangle with 25 feet on the left side and 50 feet on the bottom ?Answer by richwmiller(9144) (Show Source): You can put this solution on YOUR website!this rectangle will be scaled on graph paper.the scale is 1cm =5feet what will the longer side measure and the short side ? its a rectangle with 25 feet on the left side and 50 feet on the bottom ? 1cm=5ft so for every five feet you will use 1 cm. so the width will be 5 cm. 25/5=5 and the length 10 cm. 50/5=10 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.algebra.com/algebra/homework/Bodies-in-space/Bodies-in-space.faq.question.246078.html", "fetch_time": 1371590298000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368707188217/warc/CC-MAIN-20130516122628-00063-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 299104160, "warc_record_length": 4523, "token_count": 352, "char_count": 1392, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2013-20", "snapshot_type": "latest", "language": "en", "language_score": 0.8572105169296265, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00018-of-00064.parquet"} |
A quadratic form is a function which assigns a number to each vector, in such a way that it is sum of products of two coordinates, e.g. . The square of the norm is also an example of a quadratic form ().
In other words, a quadratic form can be described as , where is a bilinear form. We can always take a symmetric bilinear form, if the characteristic of the field is not equal to , and we are going to make such an assumption from now on.
### Positive and negative definite forms
We can classify forms with respect to possible sign of results:
• form is positively definite, if for all , we get .
• form is negatively definite, if for all , we get .
• form is positively semidefinite, if for all , we get .
• form is negatively semidefinite, if for all , we get .
Obviously a form may not fall in any of those categories, if for some we have and . Such forms are called indefinite.
### Matrix of a form
The matrix of a quadratic form with respect to basis is the matrix , where is a symmetric bilinear form such that . E.g., let , then:
so notice, that the coefficients are divided by outside the diagonal, because the same expression is generated twice.
### Sylvester’s criterion
Sylvester’s criterion determines whether a form is positively definite or negatively definite. Notice that it does not tell anything about the categories with semidefinite forms!
How does it work? We study determinants of minors: let be the matrix of size in the left upper corner of the matrix of a form we study. Let be the size of the matrix of this form. Sylvester’s criterion consists of the two following facts:
• if for any we have , the form is positively definite,
• if for any we have for even , and , for odd , then the form is negatively definite.
E.g. let , so its matrix: , so and , therefore and , so form is positively definite.
E.g., let , so its matrix: , so and , also , therefore and and , so form is negatively definite.
Finally let , its matrix: , so and , therefore and , so form is neither positively definite nor negatively definite.
But to check everything (including semi definiteness), we have to diagonalize the form, i.e. find a basis in which its matrix is diagonal (a diagonal congruent matrix). Then, obviously if:
• it has only positive entrees on the diagonal, then is positive definite,
• it has only negative entrees on the diagonal, then is negative definite,
• it has only nonnegative entrees on the diagonal, then is positive definite,
• it has only nonpositive entrees on the diagonal, then is negative semi definite,
• it has a positive and a negative entree on the diagonal, then is nondefinite.
It can be done it the tree following methods
### Diagonalization of a form: complementing to squares
We may complement a formula of a form to squares making sure to use all expressions with the first variable first, and then all with the second one, and so on.
E.g.
where , i , so the form is non-definite. The basis , in which the formula is expressed is , because
### Diagonalization of a form: orthogobal basis
We may also find an orthogonal basis with respect to the symmetrical bilinear form related to the considered quadratic form. Then the entrees on the diagonal are the values of the form on the vectors from this basis.
### Diagonalization of a form: eigenvalues
Finally, we shall remind ourselves that there exists a basis consisting of eigenvectors of a self-adjoint endomorphism described by the same matrix, which is orthogonal with respect to the symmetrical bilinear form related to the considered quadratic form. Then the entrees on the diagonal are the eigenvalues of the matrix.
E.g.: let , the matrix: , so its characteristic polynomial: has zeroes in and , so it has eigenvalues of both signs, so is is indefinite. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.mimuw.edu.pl/~m_korch/12-quadratic-forms/", "fetch_time": 1606954053000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-50/segments/1606141717601.66/warc/CC-MAIN-20201203000447-20201203030447-00302.warc.gz", "warc_record_offset": 604690900, "warc_record_length": 10098, "token_count": 835, "char_count": 3785, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2020-50", "snapshot_type": "latest", "language": "en", "language_score": 0.921297013759613, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00001-of-00064.parquet"} |
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# 1 Lecture Notes on Fluid Dynamics (1.63J/2.21J) by Chiang C.
Mei, MIT
CHAPTER 1: BASICS
1-1-LagEul.tex Lecture 1
1.1
## Methods of Describing Fluid Motion
To describe the uid motion, one needs to know the variations of physical quantities such as , density, velocity, pressure, temperature, stresses, etc., as functions of time, everywhere within a certain spatial region. There are two ways to describe the uid motion. One is called Lagrangian, where one follows all uid particles and describes the variations around each uid particle along its trajectory. The other is Eulerian, where the variations are described at all xed stations as a function of time. In the second, dierent particles pass the same station at dierent times. Consider rst the Lagrangian description.
1.1.1
Lagrangian description
A particle is identied by its initial position at time t0 , x0 = (x0 , y0 , z0 ) Let x = x (x0 , t) (1.1.2) be the position of the same particle at the later time t. By denition, at t = t0 , x (x0 , t0 ) = x0 . Instead of the initial position x0 , one can use any three quantities a, b, c, which are uniquely related to x0 : x0 = x0 (a) a (a, b, c) (1.1.3) to identify the particle. Thus the path of a particle identied by a is given by : x = x(a, t). (1.1.4) (1.1.1)
Note that a, t are the independent variables, and x = (x, y, z) are dependent variables. From the particle position we can calculate the particle velocity: q= x , t a (1.1.5)
## 2 as well as the particle acceleration: q t =
a
2x . t2 a
(1.1.6)
Other physical quantities such as density and pressure p can also be expressed in terms of a, t, e.g., = (a, t) , p = p (a, t) . (1.1.7)
1.1.2
Eulerian Description
Here we specify the uid properties (density, velocity, pressure...) at a xed point and a chosen time. Thus we are dealing with the time records of a property at a xed measuring probe. q (x, t) , (x, t) , p (x, t) , etc. (1.1.8) Once a physical property of all particles are known at all times, how do we get the property at all xed points at all times? In other words, how are the Eulerian and Lagrangian systems related? Suppose rst that the Lagrangian information of the velocity of all particles are known : q = q (a, t) (1.1.9) then we can integrate the denition x = q (a, t) t for to get the position of a particle, x = x (a, t) so that x(a, 0) = a. The result can be inverted in principle to get a = a (x, t) as long as the Jacobian of transformation (x, y, z) = J= (a, b, c)
x a y a z a x b y b z b x c y c z c
(1.1.10)
(1.1.11)
(1.1.12)
= 0.
(1.1.13)
does not vanish. Once the position of a particle at time t is known, any physical property of a particle can be translated to the property at certain position in space. For example, by substituting (1.1.12) into (1.1.9), we get q = q (x, t) , (1.1.14)
3 which gives the Eulerian velocity, i.e., the velocity eld at all xed points. Similarly from the Lagrangian pressure p((a, t) we get the Eulerian pressure p((a(x, t), t) = p(x, t) Consider next that the Eulerian velocity eld is known everywhere: q = q (x, t) (1.1.15) Since at t the particle a is at x (a, t); the Lagrangian velocity of particle a is the same as the Eulerian velocity. Therefore, x q (x, t) = , (1.1.16) t This equation is a system of nonlinear ordinary dierential equations. If they can be solved for x = x (a, t) subjected to the initial condition x = x0 (a) , t = t0 , we then have x = q [x (a, t) , t] t qL (a, t) qLagrangian For the rate of variations we need to calculate derivatives of some physical quantity A (x, t) whose Eulerian information is given. The rate of variation following a uid particle is A , t
a A = A + xi t (Lagrangian) A t a xi t a
A + qi xi (Eulerian)
A t
(1.1.17)
D = + qi (1.1.18) Dt t xi has dierent names: the substantial, or total, or material derivative. The second term is referred to as the convective rate of variation since it arises because the uid particle moves to new places. In particular, the acceleration of a uid particle is 2 x(a, t) q Dq = +q q = . (1.1.19) 2 t t Dt A more physical interpretation of Df /Dt can be derived by using the Eulerian picture directly. Consider f (x, t). After t, f (x, t) is changed to f (x + qt, t + t). Therefore, the total change of f is f = f [x + qt, t + t] f (x, t) f Df = f (x, t) + q f + t f (x, t) = t. t Dt It follows that f t =
a
The operator
Df f = +q Dt t
(1.1.20)
where
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Get Full Access to Calculus - 8 Edition - Chapter 4 - Problem 19
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# Let fsxd a1 sin x 1 a2 sin 2x 1 1 an sin nx, where a1, a2,
ISBN: 9781285740621 127
## Solution for problem 19 Chapter 4
Calculus | 8th Edition
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Problem 19
Let fsxd a1 sin x 1 a2 sin 2x 1 1 an sin nx, where a1, a2, . . . , an are real numbers and n is a positive integer. If it is given that | fsxd | < | sin x | for all x, show that | a1 1 2a2 1 1 nan | < 1
Step-by-Step Solution:
Step 1 of 3
1.2 Finding limits Graphically To find the limit graphically of a function you have to follow the function from both the positive side and from the negative side of the coordinate system towards the number x approaches. For example, For this function as x approaches -2, if you follow the function from both sides the limit equals 4. 1.3 Finding limits Numerically To find the limit of a function numerically there are different step you have to take: 1. To start finding the limit of a function you have to substitute the number that approaches x in the limit. For example, lim (3 + 2) = 3(-3) + 2 = -7 ▯→▯▯ 2. In the case that the limit is unsolvable by substitution you have to simplify the function by
Step 2 of 3
Step 3 of 3
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Year 5
Year 5
# Converting between miles and km
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## Lesson details
### Key learning points
1. In this lesson, we will introduce the distance measurements miles and kilometre and learn how to convert between them.
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This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
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## Starter quiz
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### 5 Questions
Q1.
A rectangle has a width of 4 m and a length of 20 m. What is it's perimeter?
24 m
44 m
80 m
Q2.
If a rectangle has a width of 4 metres and a length of 700cm, what is the perimeter of the shape?
0.2 m
20 m
22 cm
Q3.
A living room has a perimeter of 26 m in total. Which of the below show its possible dimensions?
width = 2.6 m and length = 6.2 m
width = 20 m and length = 6 m
Correct answer: width = 3.5 m and length = 9.5 m
width = 4 m and length = 6.5 m
Q4.
A games room has a perimeter of 36 m in total. Which of the options below cannot be the dimensions of the game room?
width of 4 m and length of 14 m
Correct answer: width of 6 m and length of 14 m
width of 7 m and length of 11 m
width of 8 m and length of 10 m
Q5.
Which of the dimensions below does not have the same perimeter as the rest?
Width = 2.5 m and Length = 3.6 m
Correct answer: Width = 3.6 m and Length = 3.5 m
Width = 4.5 m and Length = 1220 cm
Width = 480 cm and Length = 260 cm
## Exit quiz
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### 5 Questions
Q1.
What is the name of the sequence which can be used to roughly convert between miles and kilometres?
Fibonni
Harmonic
Triangular
Q2.
What is the name of the imperial unit officially used in the UK?
feet
inches
pounds
Q3.
Approximately how many miles are there in 8 km?
10 miles
12 miles
16 miles | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.thenational.academy/teachers/programmes/maths-primary-ks2-l/units/converting-units-of-measure-1475/lessons/converting-between-miles-and-km-c4rk4t", "fetch_time": 1719011676000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00240.warc.gz", "warc_record_offset": 899974677, "warc_record_length": 35835, "token_count": 552, "char_count": 1962, "score": 3.921875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9082452058792114, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00049-of-00064.parquet"} |
# Area of a Circle Calculator
To use the area of a circle calculator, fill the required input boxes and hit the calculate button
## Area of a circle Calculator
Circle area calculator is used to find the area of a circle with the help of a given radius of that circle. This area of a circle calculator also computes the diameter and circumference of the circle.
## What is a Circle?
circle is a round plane figure whose boundary or circumference consists of points equidistant from a fixed center. A Circle is a set of points that are formed in a closed loop, every point on which is a fixed distance from a center point.
The radius of a circle is the distance from the center of a circle to any point on the circle. The distance across a circle through the center is called the diameter.
### Formulas
Area of Circle = πr²
Diameter of Circle = 2r
Circumference of Circle = 2πr = πd
where,
• d = diameter
• π = 3.14
## Properties of a Circle:
Center - A point inside the circle. All points on the circle are equidistant (same distance) from the center point.
RadiusThe radius is the distance from the center to any point on the circle. It is half the diameter. See the Radius of a circle.
Diameter - The distance across the circle. The length of any chord passing through the center. It is twice the radius. See the Diameter of a circle.
Circumference - The circumference is the distance around the circle.
## How to find the area, diameter, and circumference of a circle?
Examples:
Find the area, diameter, and circumference of a circle with the given radius 4.
Solution
Step 1: Find the area.
Area = πr² = 3.14 * 4² = 3.14 * 16 = 50.24
Step 2: Find the diameter.
Diameter = 2r = 2 *4 = 8
Step 3Find the circumference.
Circumference = πd = 3.14 *8 = 25.12 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.meracalculator.com/area/circle.php", "fetch_time": 1720919350000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00028.warc.gz", "warc_record_offset": 767921397, "warc_record_length": 8998, "token_count": 451, "char_count": 1783, "score": 4.375, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8414873480796814, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"} |
# Word
What is the probability that a random word composed of chars T, H, A, M will be MATH?
Result
p = 4.17 %
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Math student
this is weird
Math student
good math
#### To solve this example are needed these knowledge from mathematics:
Would you like to compute count of combinations? See also our permutations calculator.
## Next similar examples:
1. One green
In the container are 45 white and 15 balls. We randomly select 5 balls. What is the probability that it will be a maximum one green?
2. Balls
From the urn in which are 7 white balls and 17 red, gradually drag 3-times without replacement. What is the probability that pulls balls are in order: red red red?
3. Cards
The player gets 8 cards of 32. What is the probability that it gets a) all 4 aces b) at least 1 ace
4. Hearts
5 cards are chosen from a standard deck of 52 playing cards (13 hearts) with replacement. What is the probability of choosing 5 hearts in a row?
5. Raffle
There are 200 draws in the raffle, but only 20 of them win. What is the probability of at least 4 winnings for a group of people who have bought 5 tickets together?
6. Theorem prove
We want to prove the sentense: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?
7. A book
A book contains 524 pages. If it is known that a person will select any one page between the pages numbered 125 and 384, find the probability of choosing the page numbered 252 or 253.
8. Classroom
Of the 26 pupils in the classroom, 12 boys and 14 girls, four representatives are picked to the odds of being: a) all the girls b) three girls and one boy c) there will be at least two boys
9. A pizza
A pizza place offers 14 different toppings. How many different three topping pizzas can you order?
10. Division
Division has 18 members: 10 girls and 6 boys, 2 leaders. How many different patrols can be created, if one patrol is 2 boys, 3 girls and 1 leader?
11. Permutations without repetition
From how many elements we can create 720 permutations without repetition?
12. Calculation of CN
Calculate: ?
13. A student
A student is to answer 8 out of 10 questions on the exam. a) find the number n of ways the student can choose 8 out of 10 questions b) find n if the student must answer the first three questions c) How many if he must answer at least 4 of the first 5 que
14. The confectionery
The confectionery sold 5 kinds of ice cream. In how many ways can I buy 3 kinds if order of ice creams does not matter?
15. Trinity
How many different triads can be selected from the group 43 students?
16. Fish tank
A fish tank at a pet store has 8 zebra fish. In how many different ways can George choose 2 zebra fish to buy?
17. Three workplaces
How many ways can we divide nine workers into three workplaces if they need four workers in the first workplace, 3 in the second workplace and 2 in the third? | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.hackmath.net/en/example/278", "fetch_time": 1555647542000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-18/segments/1555578527135.18/warc/CC-MAIN-20190419041415-20190419063415-00472.warc.gz", "warc_record_offset": 708346325, "warc_record_length": 6991, "token_count": 744, "char_count": 2986, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2019-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9309855103492737, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"} |
Lecture 2. Basics / law of small numbers | Stochastic Analysis Seminar Lecture 2. Basics / law of small numbers – Stochastic Analysis Seminar
## Lecture 2. Basics / law of small numbers
Due to scheduling considerations, we postpone the proof of the entropic central limit theorem. In this lecture, we discuss basic properties of the entropy and illustrate them by proving a simple version of the law of small numbers (Poisson limit theorem). The next lecture will be devoted to Sanov’s theorem. We will return to the entropic central limit theorem in Lecture 4.
Conditional entropy and mutual information
We begin by introducing two definitions related to entropy. The first definition is a notion of entropy under conditioning.
Definition. If and are two discrete random variables with probability mass functions and , then the conditional entropy of given is defined as
where is the conditional probability mass function of given .
Remark. If and are absolutely continuous random variables, the conditional differential entropy is defined analogously (where the probability mass functions are replaced by the corresponding probability densities with respect to Lebesgue measure).
Note that
That is, the conditional entropy is precisely the expectation (with respect to the law of ) of the entropy of the conditional distribution of given .
We now turn to the second definition, the mutual information. It describes the degree of dependence between two random variables.
Definition. The mutual information between two random variables and is defined as
where , and denote the distributions of , and .
Conditional entropy and mutual information are closely related. For example, suppose that has density with respect to the Lebesgue measure, then
In particular, since is always positive (because it is a relative entropy), we have just shown that , that is, conditioning reduces entropy. The same result holds for discrete random variables when we replace by .
Chain rules
Chain rules are formulas that relate the entropy of multiple random variables to the conditional entropies of these random variables. The most basic version is the following.
Chain rule for entropy. . In particular, .
Proof. Note that
Thus,
Taking the expectation on both sides under the distribution gives the desired result.
Corollary. Entropy is sub-additive, that is, .
Proof. Combine the chain rule with .
There is also a chain rule for relative entropy.
Chain rule for relative entropy.
The following identity will be useful later.
Lemma.
Proof. Note that
Data processing and convexity
Two important properties of the relative entropy can be obtained as consequences of the chain rule.
Data processing inequality. Let and be two probability measures on and suppose is measurable. Then , where is the distribution of when .
The data processing inequality tells us that if we process the data (which might come from one of the two distributions and ), then the relative entropy decreases. In other words, it becomes harder to identify the source distribution after processing the data. The same result (with the same proof) holds also if and are transformed by a transition kernel, rather than by a function.
Proof. Denote by and the joint laws of and when and . By the chain rule and nonnegativity of relative entropy
On the other hand, using again the chain rule,
where we used . Putting these together completes the proof.
Convexity of relative entropy. is jointly convex in its arguments, that is, if , , , are probability measures and , then
Proof. Let be a random variable that takes value with probability and with probability . Conditionally on , draw and . Then and . Using the chain rule twice, we obtain
and the right hand side is precisely .
Corollary. The entropy function is concave.
Proof for a finite alphabet. When the alphabet is finite, the corollary can be proven by noting that .
Relative entropy and total variation distance
Consider the hypothesis testing problem of testing the null hypothesis against the alternative hypothesis . A test is a measurable function . Under the constraint , it can be shown that the optimal rate of decay of as a function of the sample size is of the order of . This means that is the measure of how well one can distinguish between and on the basis of data.
We will not prove this fact, but only introduce it to motivate that the relative entropy is, in some sense, like a measure of distance between probability measures. However, it is not a metric since and the triangle inequality does not hold. So in what sense does the relative entropy represent a distance? In fact, it controls several bona fide metrics on the space of probability measures. One example of such metric is the total variation distance.
Definition. Let and be probability measures on . The total variation distance is defined as .
The following are some simple facts about the total variation distance.
1. .
2. If and have probability density functions and with respect to some common probability measure , then . To see this, define . Then
3. .
The following inequality shows that total variance distance is controlled by the relative entropy. This shows that the relative entropy is a strong notion of distance.
Pinsker’s inequality. .
Proof. Without loss of generality, we can assume that and have probability density functions and with respect to some common probability measure on . Let and .
Step 1: Prove this inequality by simple calculation in the case when contains at most elements.
Step 2: Note that and are defined on the space . So Pinsker’s inequality applies to and . Thus,
Law of small numbers
As a first illustration of an application of entropy to probability, let us prove a simple quantitative law of small numbers. An example of the law of small numbers is the well known fact that in distribution as goes to infinity. More generally, if are Bernoulli random variables with , if are weakly dependent, and if none of the dominates the rest, then where . This idea can be quantified easily using relative entropy.
Theorem. If and may be dependent, then
where and .
Proof. Let be independent random variables with . Then . We have
To conclude, it is enough to note that
Remark. If and are independent, then the inequality in the theorem becomes . However, this rate of convergence is not optimal. One can show that under the same condition, , using tools similar to those that will be used later to prove the entropic central limit theorem. Note that it is much harder to prove in the entropic central limit theorem, even without rate of convergence!
Lecture by Mokshay Madiman | Scribed by Che-yu Liu
03. October 2013 by Ramon van Handel
Categories: Information theoretic methods | Comments Off on Lecture 2. Basics / law of small numbers | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://blogs.princeton.edu/sas/2013/10/03/lecture-2-basic-properties-law-of-small-numbers/", "fetch_time": 1500800061000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00011.warc.gz", "warc_record_offset": 45494101, "warc_record_length": 19700, "token_count": 1352, "char_count": 6825, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2017-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8979364633560181, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"} |
# Strategies and tricks of the french variation
When you know this game as popular as French roulette, you need to find that formula that will help you win. Keep in mind that roulette is a game of chance whether in an online or physical casino. This means that none of the systems to win at roulette is fail-safe. This does not mean that you should not know the rules, the different bets and strategies of French roulette. Therefore, if you want to know more about these methods, continue reading here.
## MARTINGALE STRATEGY
It is one of the best known roulette strategies that is applied both in French roulette and in other table games. This is based on betting a fixed amount on the first bet. If you lose, you double the amount bet until you win the bet, thus guaranteeing the initial bet. We clarify that this method is only applied to simple bets. Let’s give you an example.
If you bet €1 and win, you bet the same again. If you lose, you bet €2, if you win, you bet the initial amount, if you lose, you bet €4 (double your previous bet). So on.
## D’ALEMBERT STRATEGY
This strategy is based on adding a unit to the bet if you lose, and subtracting that unit if you win. Here we understand as a unit, a token. This strategy is less risky than the Martingale if you want to maintain low losses and fixed bets. Let’s exemplify this method.
Your initial bet is €1, if you lose, you bet 1 more unit which would be €2. If you lose again you add one more unit and this time you would bet €3. If you win in the next round, then you subtract one unit and bet €2.
D’Alembert’s objective is that when you level the number of wins and losses, the amount is in your favor. All this in order that you can retire with profits.
## FIBONACCI STRATEGY
It is the famous sequence of numbers, where each number is the sum of the two previous ones. This is like this: 1-1-2-3-5-8-13-21-34-55-89-144-233 and so on infinitely. This means that if you lose, you bet according to the sequence. Then, if you win, you go back two bets, and that amount is what you are going to bet. We give you an example so that you better understand the dynamics of this method.
You bet €1 to start, if you lose, you bet €1 again. If you win you balance the loss. If you lose then you bet €2, if you lose again you bet €3. If you continue to lose, you have to bet €5. If you win, you go back and bet €2.
## JAMES BOND STRATEGY
This system owes its name to its origin in the movies of our beloved agent 007. It arises from the movie “Casino Royale” where the agent is an expert player. James Bond played French roulette just before facing his enemy at Baccarat. That simple fact prompted his roulette-savvy fans to devise a strategy in honor of him.
Now yes, let’s go to the topic to explain what this method is about. First, you must have 20 chips of the same value for each betting round. The profit goes up or down depending on the amount of the unit. Let’s see an example.
You will place three simple bets at the same time. First you put 14 units to the high group (19 to 36). Then, 5 units to the row from 13 to 18. Later, 1 unit to zero. If the ball lands in the high group, you will earn 8 units. If it falls on one of the lines, your profit will be 10 units, and if it goes to zero, you will get 16. The other numbers that are not included in this technique will cause you to lose 20 units.
As you can see, it is a highly risky strategy in French roulette. But if you win, it will turn you into a James Bond. Additionally, we recommend that you pay attention to your budget, we want you to have fun and have a great experience. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "miraraiexchange.org", "fetch_time": 1726603142000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00642.warc.gz", "warc_record_offset": 360236638, "warc_record_length": 13975, "token_count": 889, "char_count": 3614, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.934441864490509, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"} |
Q. 165.0( 1 Vote )
# If x = a (1 + cos
Idea of parametric form of differentiation:
If y = f (θ) and x = g(θ) i.e. y is a function of θ and x is also some other function of θ.
Then dy/dθ = f’(θ) and dx/dθ = g’(θ)
We can write :
Given,
y = a (θ + sin θ) ……equation 1
x = a (1+ cos θ) ……equation 2
to prove : .
We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.
Let’s find
As,
So, lets first find dy/dx using parametric form and differentiate it again.
…..equation 3
Similarly,
……equation 4
[
…..equation 5
Differentiating again w.r.t x :
Using product rule and chain rule of differentiation together:
[using equation 4]
As we have to find
put θ = π/2 in above equation:
=
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# How do i solve this trig question? identities etc?
1. Mar 20, 2009
### Dell
How do i solve this trig question? identities etc?
4*[cos(B) + 3*sin(B)]=1+ 2*[3*sin(39) - cos(39)]
i can get it to
cosB + 3sinB=0.08054076 using basic algebra, but how do i find a value for B
2. Mar 20, 2009
### danago
Can you think of a way to perhaps combine the sine and cosine terms of the left into a single trig function?
3. Mar 20, 2009
### Dell
no, thats what i need to somehow do
4. Mar 20, 2009
### danago
One common method is to use the 'auxiliary angle method', which is to combine the sine and cosine terms into a single sine (or cosine) term of the form:
R sin (B+a).
The reason why we can write it in such a form is because this is equivalent to: (using an appropriate trig identity)
R sin(B)cos(a) + R cos(B)sin(a)
So by comparing coefficients, you can see that:
R cos a = 12
R sin a = 4
From which you can solve to find a and R. See if you can continue from there.
5. Mar 20, 2009
### Dell
so we have
R cos a = 4
R sin a = 12
tan a = 3
a=71.565
R=12.65
am i meant to be saying that
R sin (B+a) = R sin(B)cos(a) + R cos(B)sin(a) ??????????
therefore
R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
sin(B+a)=0.25467
sin(c)=0.25467
c=14.75
c=B+71.565=14.75
B= (-56.81)
but this doesnt come right
cos(-56.81) + 3sin(-56.81)= -1.963
where i need
cosB + 3sinB=0.8054076
6. Mar 20, 2009
### danago
Yea the reason is because i made a mistake in my post but have since edited it out, and i think you have carried through that same mistake.
Comparing the two forms:
R sin(B)cos(a) + R cos(B)sin(a) = 4cos(B) + 12sin(B)
On the left hand side, the coefficient of cos(B) is R sin(a), and on the right hand side it is 4, and so we must have:
R sin(a) = 4
And by similar reasoning,:
R cos(a) = 12
So i had originally typed the 4 and 12 in the wrong places. I guess you read my post before i got a chance to edit it out (or perhaps coincidentally made the exact same mistake haha? :P)
7. Mar 20, 2009
### Dell
so...
R sin(a) = 4
R cos(a) = 12
tan(a)=1/3
(a)=18.34
R=12.65
R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
sin(B+a)=0.25467
sin(c)=0.25467
c=14.75
c=B+18.34=14.75
B= (-3.59)
cos(-3.59) + 3sin(-3.59)= -0.81
thanks!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/how-do-i-solve-this-trig-question-identities-etc.301173/", "fetch_time": 1508200460000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00585.warc.gz", "warc_record_offset": 1171231466, "warc_record_length": 15712, "token_count": 823, "char_count": 2342, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.8960604667663574, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00025-of-00064.parquet"} |
# Binary Search Tree Search and Insertion
Difficulty Level Easy
Frequently asked in Amazon DBOI Fanatics GE Healthcare MAQ Microsoft UHG Optum
Binary Search Tree Binary Tree Theory Tree
## Problem Statement
Write an algorithm to perform searching and insertion in Binary Search Tree. So what we are going to do is insert some of the elements from input into a binary search tree. Whenever asked to search a particular element, we’ll be searching it among the elements in BST(short for Binary Search Tree).
## Example
```Insert 10
Insert 15
Search 5
Insert 5
Insert 18
Search 5
Insert 12
Search 10```
```false
true
true```
## What is Binary Search Tree?
A Binary Search Tree is a special kind of Binary Tree that follows the following properties,
1. All the nodes smaller than the current node are present in its left sub-tree.
2. All the nodes greater than the current node are present in its right sub-tree.
3. Left and right sub-tree of a node is also Binary Search Tree and there are no repeated elements.
## Searching
### Algorithm
Binary Search Tree stores the data in a sorted way(as its in-order traversal leads to sorted data). So, searching in BST is quite simple and easy.
We check if the root equals to the target node, if yes, return true, else if the target is smaller than the root’s value we search it in the left sub-tree else we search it in the right sub-tree.
```1. Check if root equals to the target node, if yes, return true, else go to step 2.
2. If the target is smaller than the root's value, search the target in the left sub-tree.
3. Else search the target in the right sub-tree.```
### Time Complexity = O(n)
Since we are going to traverse the whole tree in the worst case. A worst-case can be we have a skewed tree and have our target value as the leaf of the tree. By the way, both searching and insertion in Binary Search Tree have same time complexity.
Given a binary tree, how do you remove all the half nodes?
### Space Complexity = O(1)
Since we are not using an array, or storing values for nodes during the algorithm. Thus, searching occurs in O(1) space complexity. The same goes for space complexity, both searching and insertion in Binary Search Tree are O(1) space complexity algorithms.
## Insertion
Inserting a new node into BST is similar to searching. We search for the first empty position in the BST, by fulfilling the properties of BST and insert the new Node at that place.
### Algorithm
```1. Allocate space for new Node, let it be node.
2. If root is null, make root as node and return.
3. If the value of new node is smaller than the root's value, insert the new node in the left sub-tree.
4. Else insert the new node in the right sub-tree.```
### Time Complexity = O(n)
Here again, we have a case we are provided elements either in increasing or decreasing order, or such that we may end up having a skewed tree. Then in that case, if the element to be inserted is such that it is going to become a leaf. We’ll have to traverse the whole of the tree. Thus contributing to O(n) time complexity.
### Space Complexity = O(1)
Here since we did not store any value corresponding to each node. We have constant space complexity.
## Code
### JAVA Code for searching and insertion in Binary Search Tree
```class BSTSearchAndInsert {
// class to represent node of a binary tree
static class Node {
int data;
Node left, right;
public Node(int data) {
this.data = data;
}
}
private static Node insertToBST(Node root, Node node) {
// if root is null, then make root as node and return
if (root == null) {
root = node;
return root;
}
// if node's value is less than root, insert it to left subtree
if (node.data < root.data) {
root.left = insertToBST(root.left, node);
}
// else insert it to right subtree
else {
root.right = insertToBST(root.right, node);
}
// return the updated root
return root;
}
private static Node insert(Node root, int value) {
// allocate memory for new node
Node node = new Node(value);
// insert the new node to tree
return insertToBST(root, node);
}
private static boolean search(Node root, int val) {
// if root is null, return false
if (root == null) {
return false;
}
// if root is equals to target, return true
if (root.data == val) {
return true;
}
// else if val is less than root, search in left subtree
else if (val < root.data) {
return search(root.left, val);
}
// else search in right subtree
else {
return search(root.right, val);
}
}
public static void main(String[] args) {
// Example
Node root = null;
root = insert(root, 10);
root = insert(root, 15);
System.out.println(search(root, 5));
root = insert(root, 5);
root = insert(root, 18);
System.out.println(search(root, 5));
root = insert(root, 12);
System.out.println(search(root, 10));
}
}```
```false
true
true```
### C++ Code for searching and insertion in Binary Search Tree
```#include <bits/stdc++.h>
using namespace std;
// class representing node of a binary tree
class Node {
public:
int data;
Node *left;
Node *right;
Node(int d) {
data = d;
left = right = NULL;
}
};
Node* insertToBST(Node *root, Node *node) {
// if root is null, then make root as node and return
if (root == NULL) {
root = node;
return root;
}
// if node's value is less than root, insert it to left subtree
if (node->data < root->data) {
root->left = insertToBST(root->left, node);
}
// else insert it to right subtree
else {
root->right = insertToBST(root->right, node);
}
// return the updated root
return root;
}
Node* insert(Node *root, int value) {
// allocate memory for new node
Node *node = new Node(value);
// insert the new node to tree
return insertToBST(root, node);
}
bool search(Node *root, int value) {
// if root is null, return false
if (root == NULL) {
return false;
}
// if root is equals to target, return true
if (root->data == value) {
return true;
}
// else if val is less than root, search in left subtree
else if (value < root->data) {
return search(root->left, value);
}
// else search in right subtree
else {
return search(root->right, value);
}
}
int main() {
// Example
Node *root = NULL;
root = insert(root, 10);
root = insert(root, 15);
if (search(root, 5)) {
cout<<"true"<<endl;
} else {
cout<<"false"<<endl;
}
root = insert(root, 5);
root = insert(root, 18);
if (search(root, 5)) {
cout<<"true"<<endl;
} else {
cout<<"false"<<endl;
}
root = insert(root, 12);
if (search(root, 10)) {
cout<<"true"<<endl;
} else {
cout<<"false"<<endl;
}
return 0;
}```
```false
true
true``` | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.tutorialcup.com/interview/tree/binary-search-tree-search-and-insertion.htm", "fetch_time": 1638933953000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00445.warc.gz", "warc_record_offset": 1068608209, "warc_record_length": 38992, "token_count": 1614, "char_count": 6469, "score": 4.1875, "int_score": 4, "crawl": "CC-MAIN-2021-49", "snapshot_type": "longest", "language": "en", "language_score": 0.8336973786354065, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00053-of-00064.parquet"} |
# Evaluating $\int_0^{+\infty}(\frac{\arctan t}{t})^2dt$
Is it possible to calculate $$\int_0^{+\infty}\Big(\frac{\arctan t}{t}\Big)^2 dt$$ without using complex analysis?
I found this on a calculus I book and I don't know how to solve it.
I tried to set $t = \tan u$ but it didn't help.
• I am sorry. I don't know how to format the expressions correctly. Thank you for your help. – und3rd06012 Dec 27 '15 at 17:30
• This question is a bit similar. – mickep Dec 27 '15 at 21:07
One may first integrate by parts, \begin{align} \int_0^{+\infty}\!\! \left(\frac{\arctan t}{t}\right)^2\!\!dt=\color{#365A9E}{\left.-\frac1{t}\left(\arctan t\right)^2\right|_0^{+\infty}}\!+2\!\!\int_0^{+\infty} \!\frac{\arctan t}{t(1+t^2)}dt=\color{#365A9E}{0}+2\!\!\int_0^{+\infty}\! \frac{\arctan t}{t(1+t^2)}dt \end{align} then making the change of variable $x=\arctan t$, $dx=\dfrac{dt}{1+t^2}$, in the latter integral gives \begin{align} 2\!\int_0^{+\infty}\! \frac{\arctan t}{t(1+t^2)}dt=2\!\int_0^{\pi/2} \!\!x\:\frac{\cos x}{\sin x}dx=\left.2x\log(\sin x)\right|_0^{\pi/2}-2\!\int_0^{\pi/2} \!\log(\sin x)dx=\pi\log 2 \end{align} where we have used the classic result:$\color{#365A9E}{\displaystyle \int_0^{\pi/2} \!\log(\sin x)\:dx=\!\int_0^{\pi/2} \!\log(\cos x)\:dx=-\frac{\pi}2\log 2.}$
Finally,
$$\int_0^{+\infty}\! \left(\frac{\arctan t}{t}\right)^2dt=\pi\log 2.$$
• That is actually very good! Thanks Olivier. – und3rd06012 Dec 27 '15 at 18:24
• Solid answer. +1 - Mark – Mark Viola Dec 27 '15 at 18:38
Consider
$$I(a,b) = \int_0^{\infty} dt \frac{\arctan{a t}}{t} \frac{\arctan{b t}}{t}$$
Then
\begin{align}\frac{\partial^2 I}{\partial a \, \partial b} &= \int_0^{\infty} \frac{dt}{(1+a^2 t^2)(1+b^2 t^2)}\\ &= \frac{\pi}{2 (a+b)} \end{align}
This result may be derived using Parseval's theorem from Fourier transforms.
$$\implies \frac{\partial I}{\partial a} = \frac{\pi}{2}\log{(a+b)} + f(a)$$ $$\frac{\partial I}{\partial b} = \frac{\pi}{2}\log{(a+b)} + g(b)$$
$$I_a(a,0)=0 \implies f(a)=-\frac{\pi}{2} \log{a}$$ $$I_b(0,b)=0 \implies g(b)=-\frac{\pi}{2} \log{b}$$
Then
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - (a+b) - a \log{a} + a \right ] + F(b)$$
but
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - (a+b) - b \log{b} + b \right ] + G(a)$$
Accordingly, $F(b) = -\frac{\pi}{2}(b \log{b}-b)$ and $G(a) = -\frac{\pi}{2}(a \log{a}-a)$ and therefore
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - a \log{a} - b \log{b} \right ]$$
and the integral we want is
$$I(1,1) = \int_0^{\infty} dt \frac{\arctan^2{t}}{t^2} = \pi \log{2}$$
• Wow! I would have never thought of that.Thanks Ron. – und3rd06012 Dec 27 '15 at 18:16
• @rongordon Happy Holidays! In the development, I believe you wanted to write $$\left.\frac{\partial I}{\partial a}\right|_{b=0}=0\implies f(a)=-\frac{\pi}{2}\log(a)$$and likewise for the partial of $I$ with respect to $b$ to find $g$. +1 for the answer - Mark – Mark Viola Dec 27 '15 at 18:37
• @Dr.MV: yes, thanks. And HH to you as well. – Ron Gordon Dec 27 '15 at 18:38
The answer can be derived from a tailor-made version of Parseval's theorem, as already suggested by Ron Gordon. The Fourier sine series of $x$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is given by: $$x = \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}\,\sin(2k x) \tag{1}$$ and if $k,j\in\mathbb{N}^+$ we have: $$\int_{0}^{\pi/2}\frac{\sin(2kt)}{\sin(t)}\cdot\frac{\sin(2jt)}{\sin(t)}\,dt = \pi\cdot\min(j,k).\tag{2}$$ That gives:
$$\int_{0}^{\pi/2}\frac{x^2}{\sin^2 x}\,dx = \pi \sum_{j,k\geq 1}\frac{(-1)^{j+k}\min(j,k)}{jk}\tag{3}$$ and by reindexing over $j+k$, then using partial fraction decomposition, it is not difficult to check that the RHS of $(3)$ equals $\color{red}{\pi\log 2}$. On the other hand, the LHS of $(3)$ equals $\int_{0}^{+\infty}\frac{\arctan^2 t}{t^2}\,dt$ through the obvious substitution.
$J=\displaystyle\int_0^{\infty} \dfrac{dt}{(1+a^2 t^2)(1+b^2 t^2)}=\int_0^{\infty}\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}dt-\int_0^{\infty}\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}dt$
If $a>0$, one obtains (change of variable $y=at$):
$\displaystyle \int_0^{\infty}\dfrac{a}{1+a^2t^2}dx=\dfrac{\pi}{2}$
Thus, $\displaystyle J=\dfrac{\pi}{2}\left(\dfrac{b}{b^2-a^2}-\dfrac{a}{b^2-a^2}\right)=\dfrac{\pi}{2(a+b)}$
• I am sorry I don't quite understand what you did. Is it possible for you to further explain please? – und3rd06012 Dec 29 '15 at 7:51
• $\dfrac{1}{(1+a^2t^2)(1+b^2t^2)}=\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}-\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}$ – FDP Dec 29 '15 at 10:10
• Oh ok. I completely missed the numerator in the integral. Thank you for the explanation. – und3rd06012 Dec 29 '15 at 13:54 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1590772/evaluating-int-0-infty-frac-arctan-tt2dt", "fetch_time": 1561021513000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00299.warc.gz", "warc_record_offset": 520969065, "warc_record_length": 38154, "token_count": 1980, "char_count": 4633, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2019-26", "snapshot_type": "longest", "language": "en", "language_score": 0.584212601184845, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"} |
# Uniformly convex space
Uniformly convex spaces are a special class of normalized spaces considered in mathematics . These spaces were introduced in 1936 by James A. Clarkson by means of a geometric property of the unit sphere. The uniformly convex Banach spaces are reflexive and have an important property for approximation theory .
## Motivation and Definition
The midpoint between and lies in the case of the Euclidean norm, not in the case of the sum norm.${\ displaystyle e_ {1}}$${\ displaystyle e_ {2}}$
Since the unit sphere of a normalized space is convex , the midpoint between two vectors and the unit sphere is again in the unit sphere. We investigate the distance of such a center from the edge of the unit sphere. ${\ displaystyle \ {x \ in E; \ | x \ | \ leq 1 \}}$${\ displaystyle E}$ ${\ displaystyle {\ tfrac {1} {2}} (x + y)}$${\ displaystyle x}$${\ displaystyle y}$
Considering on the Euclidean norm , then the unit sphere of the unit circle in the plane. If the center of two edge points is formed, then this center point lies further inside the circle, the further the two edge points are from each other. ${\ displaystyle {\ mathbb {R}} ^ {2}}$
If, on the other hand, one looks at the sum norm defined by , then the 'unit sphere' is a square. It applies to apparently , , and . Although the two edge points and are very far apart, their center point is still on the edge of the unit sphere. ${\ displaystyle {\ mathbb {R}} ^ {2}}$${\ displaystyle \ | (x, y) \ | _ {1}: = | x | + | y |}$${\ displaystyle e_ {1}: = (1,0), e_ {2}: = (0,1)}$${\ displaystyle \ | e_ {1} \ | _ {1} = 1}$${\ displaystyle \ | e_ {2} \ | _ {1} = 1}$${\ displaystyle \ | {\ tfrac {1} {2}} (e_ {1} + e_ {2}) \ | _ {1} = 1}$${\ displaystyle \ | e_ {1} -e_ {2} \ | = 2}$${\ displaystyle e_ {1}}$${\ displaystyle e_ {2}}$
It is therefore a special geometrical property that two vectors of the unit sphere must be close to each other if their center is close to the edge. Therefore one defines:
A normed space is uniformly convex , if for every one there, so the following applies: If using , and so follows . ${\ displaystyle E}$${\ displaystyle \ varepsilon> 0}$${\ displaystyle \ delta> 0}$${\ displaystyle x, y \ in E}$${\ displaystyle \ | x \ | \ leq 1}$${\ displaystyle \ | y \ | \ leq 1}$${\ displaystyle \ | {\ tfrac {1} {2}} (x + y) \ |> 1- \ delta}$${\ displaystyle \ | xy \ | <\ varepsilon}$
This is a property of the norm . If one goes over to an equivalent norm , this property can be lost, as the two examples considered at the beginning show.
## Examples
• YES Clarkson has this property for the Banach L p [0,1] , detected ( set of Clarkson ). A simpler proof emerged as a consequence of the Hanner inequalities proved by Olof Hanner in 1956 . This statement was substantially generalized in 1950 by EJ McShane . If the space is uniformly convex, any positive measure , then uniformly convex is also . It is the Banach space of equivalence classes of measurable functions with values in so .${\ displaystyle 1 ${\ displaystyle E}$${\ displaystyle \ mu}$${\ displaystyle 1 ${\ displaystyle L ^ {p} (\ mu, E)}$${\ displaystyle L ^ {p} (\ mu, E)}$${\ displaystyle f}$${\ displaystyle E}$${\ displaystyle \ int \ | f (\ cdot) \ | ^ {p} d \ mu <\ infty}$
• 1967 CA McCarthy has the uniform convexity for shadow classes with proven.${\ displaystyle 1
## Milman's Theorem
David Milman has demonstrated the following important property of uniformly convex spaces:
Milman's theorem : Uniformly convex Banach spaces are reflexive.
This result was also found independently of Milman by Billy James Pettis (1913–1979), which is why one sometimes speaks of the Milman-Pettis theorem . The class of uniformly convex spaces is really smaller than the class of reflexive spaces, because there are reflexive Banach spaces that are not isomorphic to uniformly convex spaces.
One can even show that uniformly convex Banach spaces have the Banach-Saks property (a theorem by S. Kakutani ), and that Banach spaces with the Banach-Saks property are reflexive (a theorem by T. Nishiura and D. Waterman ).
## The approximation theorem
The following statements, which are also known as the approximation theorem, show the importance of the uniformly convex spaces for approximation theory. Many approximation problems can be reformulated in such a way that a vector can be found in a convex set (e.g. in a subspace) which has the shortest distance to a given vector. The following statements for a real normed space apply , and a completed and convex subset : ${\ displaystyle E}$${\ displaystyle x \ in E}$${\ displaystyle Y \ subset E}$
• Uniqueness: If strictly convex , there is at most one with .${\ displaystyle E}$ ${\ displaystyle y_ {0} \ in Y}$${\ displaystyle \ | x-y_ {0} \ | = \ inf _ {y \ in Y} \ | xy \ |}$
• Existence: If there is a uniformly convex Banach space, then there is a (according to the above uniquely determined) with . (Note that evenly convex spaces are strictly convex.)${\ displaystyle E}$${\ displaystyle y (x) \ in Y}$${\ displaystyle \ | x-y_ {0} \ | = \ inf _ {y \ in Y} \ | xy \ |}$
• Aspect of continuity: If there is a uniformly convex Banach space and a normalized subspace in closed , then the proximum mapping that assigns the (previously described) to each is continuous .${\ displaystyle E}$${\ displaystyle Y \ subset E}$${\ displaystyle E}$ ${\ displaystyle x \ mapsto y (x)}$${\ displaystyle x \ in E}$${\ displaystyle y (x) \ in Y}$
## Convexity module
You bet for a number ${\ displaystyle 0 \ leq \ alpha \ leq 2}$
${\ displaystyle \ delta _ {E} (\ alpha): = \ inf \ {1 - {\ frac {1} {2}} \ | x + y \ |; \, x, y \ in E, \ | x \ | \ leq 1, \ | y \ | \ leq 1, \ | xy \ | = \ alpha \}}$
and calls the function defined thereby the convexity modulus of . For uniformly convex spaces, by definition, applies to all , and it can be shown that the convexity module is a monotonic function, even the mapping is monotonic. A theorem by MI Kadec represents a necessary condition for the unconditional convergence of series in uniformly convex spaces: ${\ displaystyle \ delta _ {E}: [0.2] \ rightarrow [0.1]}$${\ displaystyle E}$${\ displaystyle \ delta _ {E} (\ alpha)> 0}$${\ displaystyle \ alpha> 0}$${\ displaystyle \ alpha \ mapsto \ delta _ {E} (\ alpha) / \ alpha}$
If a sequence is in a uniformly convex space with for all and if the series is unconditionally convergent, then we have . ${\ displaystyle (x_ {n}) _ {n}}$${\ displaystyle E}$${\ displaystyle \ | x_ {n} \ | \ leq 2}$${\ displaystyle n \ in {\ mathbb {N}}}$${\ displaystyle \ textstyle \ sum _ {n \ in \ mathbb {N}} x_ {n}}$${\ displaystyle \ textstyle \ sum _ {n \ in \ mathbb {N}} \ delta _ {E} (\ | x_ {n} \ |) <\ infty}$
## Other room classes
The uniform convexity condition discussed here is the strongest among several convexity conditions , each of which leads to different room classes. In particular, it results that evenly convex spaces are strictly convex and strongly convex and have the Radon-Riesz property .
## Individual evidence
1. Friedrich Hirzebruch, Winfried Scharlau: Introduction to Functional Analysis. 1971, ISBN 3-860-25429-4 , definition 16.1
2. James A. Clarkson: Uniformly convex spaces. Transactions of the American Mathematical Society, Volume 40, 1936, pages 396-414.
3. CA McCarthy, C p , Israel Journal of Mathematics (1967), Volume 5, pages 249-271.
4. ^ D. Milman: On some criteria for the regularity of spaces of type (B). Comptes Rendus (Doklady) de l'Académie des Sciences de l'URSS, Volume 20, 1938, pages 243-246.
5. ^ BJ Pettis: A proof that every uniformly convex space is reflexive. Duke Math. J., Volume 5, 1939, pages 249-253.
6. Mahlon M. Day: Reflexive Banach spaces not isomorphic to uniformly convex spaces. Bulletin of the American Mathematical Society, Vol. 47, No. 4, 1941, pp. 313-317.
7. ^ Arnold Schönhage: Approximation theory. 1971, p. 15
8. ^ Joseph Diestel: Sequences and Series in Banach Spaces. 1984, ISBN 0-387-90859-5 , Chapter VIII, Theorem 2 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 63, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://de.zxc.wiki/wiki/Gleichm%C3%A4%C3%9Fig_konvexer_Raum", "fetch_time": 1713723244000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00230.warc.gz", "warc_record_offset": 166909085, "warc_record_length": 14898, "token_count": 2251, "char_count": 8032, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.8228403925895691, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00020-of-00064.parquet"} |
Homura said that she uses applied statistics in witch hunts
The cast of Mahou Shoujo Madoka Magica are students of Mitakihara Middle School. Even though it is merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:
## Episode 1
Episode one Math question 1
### Question 1
Any integer divided by 14 will have a remainder between 0 and 13. Given that $a$ has a remainder of 6 and $b$ has a remainder of 1 when divided by 14, what is the remainder of $x$ when divided by 14, given $x$ is an integer solution to $x^2-2ax+b=0$?
Solution:
This problem can be solved simply with modular arithmetic:
• $a \equiv 6 \pmod{14}$;
• $b \equiv 1 \pmod{14}$.
• $\displaystyle{x^2 - 2ax + b = 0}$ implies
• $x^2 - 12x + 1 \equiv 0 \pmod{14}$
• $x^2 + 2x + 1 \equiv 0 \pmod{14}$ (because $-12 \equiv 2 \pmod{14}$)
• $(x + 1)^2 \equiv 0 \pmod{14}$
• $x + 1 \equiv 0 \pmod{14}$ (because 14 is square-free)
• $x \equiv 13 \pmod{14}$ (because $-1 \equiv 13 \pmod{14}$).
Second Solution:
Homura used in Episode 1 a basic approach with usual integer arithmetic.
Let $\displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1}$.
Substitute into $\displaystyle{x^2 - 2ax + b = 0}$ to get after some calculations
$\displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}$
14 divides $14c$ and 0, hence it also divides $(r+1)^2$. This implies that $r+1$ is divisible by 14.
The question asks for a remainder $r$ between 0 and 13, so we obtain $r = 13$.
### Question 2
Episode one Math question 2
Assuming that $p$ is a prime number and $n$ is an arbitrary natural number, prove that $(1+n)^p - n^p - 1 \,$ is divisible by $p$.
Solution:
By Fermat's Little Theorem, for any prime $p$ and integer $a$,
$a^p \equiv a \pmod{p} \,$
Thus:
$(1+n)^p - n^p - 1 \pmod{p} \,$ is equivalent to
$(1+n) - n - 1 \pmod{p} \,$ is equivalent to
$0 \pmod{p} \,$
So the overall expression is divisible by $p$.
Second solution:
The problem can be solved with the binomial theorem:
For $a, b$ not equal to $0$ and nonnegative integer $p$ it holds that:
$\displaystyle (a + b)^p = \sum_{k=0}^{p} \binom{p}{k} a^{p-k} b^k$
where the binomial coefficient $\binom{p}{k}$ is the integer $\frac{p(p - 1) \cdots (p - k + 1)}{k(k-1) \cdots 1}$ for $0 \leq k \leq p$.
Therefore,
$\displaystyle (1 + n)^p = \sum_{k=0}^{p} \binom{p}{k} n^k$
and
$\displaystyle (1 + n)^p - n^p - 1 = \sum_{k = 1}^{p - 1} \binom{p}{k} n^k$
since $\binom{p}{0} = \binom{p}{p} = 1$.
It holds that $\binom{p}{1} = p$. Since $p$ is a prime number, the factor $p$ in $\binom{p}{k}$ is not divisible by $k, k-1,\ldots, 2$ for $2 \leq k \leq p-1$. Therefore, $p$ divides $\binom{p}{k}$ for $1 \leq k \leq p-1$.
Since each summand on the right side is divisible by $p$, the whole sum, i.e. the left side is also divisible by $p$.
### Question 3
Episode one Math question 3
Find the integer solutions $(a,b)$ with $(a^3 + a^2 - 1) - (a - 1)b = 0 \,$.
Solution:
Episode one Math solution to question 3
\begin{align} (&a^3 + a^2 - 1) - (a - 1)b \\ & = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\ & = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\ & = (a - 1)(a^2 + 2a - b + 2) + 1 = 0 \end{align}
Therefore,
$(a - 1)(a^2 + 2a - b + 2) = -1 \,$
$a, b$ are integers, therefore the factors $a - 1$ and $a^2 + 2a - b + 2 \,$ are integers too. Since the product is $-1$, one of the factors must be equal to $-1$, the other to $1$.
If $a - 1 = -1$ then $a = 0$ and from $a^2 + 2a - b + 2 = 1 \,$ we obtain $b = 1$.
If $a - 1 = 1$ then $a = 2$ and $a^2 + 2a - b + 2 = -1 \,$ implies that $b = 11$.
There are two integer solutions: $(a,b) = (0,1) \,$ or $(a,b) = (2,11) \,$.
Second Solution
The question can also be solved by polynomial division.
$(a^3 + a^2 - 1) - (a - 1)b = 0 \,$
$b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,$
As $b \in \mathbb{Z}$, $a^2 + 2a + 2 + \frac{1}{a - 1} \in \mathbb{Z}$, so $\frac{1}{a - 1} \in \mathbb{Z}$
As $1$ has only two factors: $1$ and $-1$,
$a-1 = 1 \,$ or $a-1 = -1 \,$
$a = 2 \,$ or $a = 0 \,$
Substituting the above values into the equation to find the corresponding $b$, we have
$(a, b) = (0, 1) \,$ or $(2, 11)$.
### Question 4
Episode one Math question 4
Given $\displaystyle F(x) = \frac{4x + \sqrt{4x^2 - 1}}{\sqrt{2x + 1} + \sqrt{2x-1}}$,
find the sum of $\displaystyle{F(1)+F(2)+F(3)+...+F(60)}$.
Solution:
Episode one Math, partial solution to question 4. Note that there are errors.
Simplying the fraction:
Notice that for any variables $a$ and $b$:
$\displaystyle{(a-b)(a+b) = a^2-b^2}$.
Let $\displaystyle{a = \sqrt{2x + 1}}$ and $\displaystyle{b = \sqrt{2x - 1}}$.
Multiply $\displaystyle{F(x)}$ by 1 or $\displaystyle{\frac{a-b}{a-b}}$ which is equal to $\displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}}$. The denominator becomes:
\begin{align} &(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\ &= (2x + 1) - (2x - 1)\\ &= 2 \end{align}
The numerator becomes:
\begin{align} &(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ &=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\ &=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (a-b)\\ &=(4x + a b ) (a - b)\\ &=4x (a - b) + a^2 b - a b^2\\ &=4x (a - b) + (2x + 1) b - a (2x - 1)\\ &=4x (a - b) - 2x (a - b) + a + b\\ &=2x ( a - b) + a + b\\ &=(2x + 1) a - (2x - 1) b\\ &=a^3 - b^3 \end{align}
Therefore:
$\displaystyle {F(x) = \frac{a^3 - b^3 }{2}}$
Note that when $\displaystyle{ x = 1, 2, 3, 4, \ldots, 60}$;
$\displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}}$;
$\displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} }$;
Taking the sum over $\displaystyle{F(x)}$ for $\displaystyle{x}$ between $1$ and $60$, observe that the majority of the terms in $\displaystyle{a^3}$ and $\displaystyle{b^3}$ cancel out, leaving:
\begin{align} \sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\ &=\frac{(11^2)^\frac{3}{2} - 1}{2}\\ &= 665 \end{align}
Thus, sum of $\displaystyle{F(x)}$ for $\displaystyle{x}$ between $1$ and $60$ is $665$. The solution can be generalized as equal to $\displaystyle{\frac{a(x_{\max}^3) - b(x_{\min}^3)}{2}}$. Where $\displaystyle{a}$ and $\displaystyle{b}$ are interpreted as functions of $\displaystyle{x}$.
## Episode 8
Homura predicted the location of Walpurgis Night to be the clock tower using statistics. The material Homura presents to Kyoko shows what looks like a sinusoidal regression, where a time series data is assumed to be following a sine curve over time and the properties (amplitude, frequency, phase) of the sine wave is predicted through statistics. For the purpose of predicting the location of Walpurgis Night, statistics done this way would use geographic coordinates (longitude and latitude) as observed data points and attempt to fit two different sine waves to data, one for longitude and one for latitude. The fact that a large pendulum exists in Homura's room may be viewed as supporting this interpretation that what Homura used was a sinusoidal regression, because the mechanical movement of a pendulum also follows a sinusoidal pattern.
Irrespective of what type of statistics was used, the fact that Walpurgis Night actually appeared at the predicted location indicates that the model fitted the data very well with only a negligible amount of error.
## Episode 9
Fibonacci sequence question that appeared in Episode 9
This advanced Algebra problem previously appeared in the first Tokyo University Entrance Exam:
### Problem
A number sequence $\displaystyle{ \{F(n)\} }$ that can be defined as $\displaystyle{F(1) = 1, F(2) = 1}$, $\displaystyle{F(n+2) = F(n) + F(n+1)}$ (where $\displaystyle{n \in \mathbb{N}}$) is called the Fibonacci sequence and its general solution is given by,
$\displaystyle{F(n) = \frac{\varphi^{n} - (-1/\varphi)^{n}}{\sqrt{5}}}$
$\displaystyle{\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887 \dots}$
Answer the following questions by using this fact if needed:
Define a sequence of natural numbers $\displaystyle{ \{X(n)\} }$ (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:
(i) $\displaystyle{X(1) = 1}$
(ii) We define $\displaystyle{X(n+1)}$ as a natural number, which can be obtained by replacing the digits of $\displaystyle{X(n)}$ with 1 if the digit is 0, and with 10 if the digit is 1.
For example, $\displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots}$
Anecdote: $\displaystyle{X(n)}$ can be considered a clever analogy to the show, where Episode 10 should replaces Episode 1 and Episode 1 replaces Episode 0 in the viewer's next viewing.
(1) Find $\displaystyle{A(n)}$, defined as the number of digits of $\displaystyle{X(n)}$.
(2) Find $\displaystyle{B(n)}$, defined as the numbers of times '01' appears in $\displaystyle{X(n)}$? For example, $\displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots}$
### Solution
#### Part 1
Let $\displaystyle{A(n)}$ equal to the number of digits in $X(n)$, which consists solely of 1s and 0s. Let's suppose $\displaystyle{x(n)}$ is the number of 0s in $\displaystyle{X(n)}$ at the n-th iteration (poor choice of variable by the student). Let's suppose $\displaystyle{y(n)}$ is the number of 1s in $\displaystyle{X(n)}$ in the n-th iteration, then $\displaystyle{x(n) + y(n) = A(n)}$. Since
• Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in $\displaystyle{X(n+1)}$.
• Every time a 1 appears, it is replaced with 10 at the next iteration, contributing to a single 1 and a single 0 in $X(n+1)$.
it follows that the number of 0s in the next iteration is equal to the number of 1s previously:
$\displaystyle{x(n+1)= y(n)}$ ;
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:
$y(n+1)= \displaystyle{x(n)+y(n)}$ .
Next, prove that $x(n)$ is a Fibonacci sequence, since we know that:
• $\displaystyle{y(n+1) = x(n) + y(n)}$ ;
• $\displaystyle{x(n+2) = y(n+1)}$ ;
• $\displaystyle{x(n+1) = y(n)}$ ;
by substition we can show,
\begin{align} &x(n+2)\\ &= y(n+1)\\ &= x(n) + y(n)\\ &= x(n) + x(n+1) \end{align}
Hence, $\displaystyle{x(n+2) = x(n+1) + x(n)}$.
Thus $\displaystyle{x(n)}$ fits the definition of a Fibonacci sequence. Since $\displaystyle{x(n)}$ is a Fibonacci sequence, it follows that $\displaystyle{y(n)}$ is also a Fibonacci sequence (given that $\displaystyle{x(n+1) = y(n)}$). Therefore, since it has already been shown that:
• $\displaystyle{x(n+2)=x(n+1)+x(n)}$
• $\displaystyle{y(n+2)=y(n+1)+y(n)}$
It follows that $\displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)}$.
Recall that $\displaystyle{A(n)=x(n)+y(n)}$, therefore
$\displaystyle{A(n+2)=A(n+1)+A(n)}$ .
Since, $\displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots}$
It follows that $\displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots}$
Recall by definition of the Fibonacci sequence: $\displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}}$. Therefore $\displaystyle{A(n) = F(n+1)}$, or
$\displaystyle{A(n) = F(n+1) = \frac{\varphi^{n+1} - (-1/\varphi)^{n+1}}{\sqrt{5}}}$
#### Part 2
Let $B(n)$ be the number of times the sequence $01$ appears in $X(n)$. Any two digits in $X(n)$ may be $00$, $01$, $10$, $11$, and the corresponding digits in the next iteration $X(n+1)$ will be
• $00 \rightarrow 11$
• $01 \rightarrow 110$
• $10 \rightarrow 101$
• $11 \rightarrow 1010$
Thus, any two digit sequence in $X(n)$ that begins with 1* will contribute to a $01$ sequence in the next iteration. In other words, $B(n+1)$ equals $y(n)$, except when unit digit of $X(n)$ is 1, or:
$B(n+1) = y(n) - odd(X(n))$
where
• $odd(n) = 1$ if $n$ is odd (unit digit is 1)
• $odd(n) = 0$ if $n$ is even (unit digit is 0)
It can be seen that $odd(n) = odd(X(n))$ for all $n$. To prove this inductively: observe that $odd(X(1)) = odd(1)$. Now let $odd(X(n)) = odd(n)$, then if $n$ is even, $X(n)$ is even, thus $X(n)$ ends with 0, which gets mapped to 1, making $X(n+1)$ odd and so $odd(X(n+1)) = odd(n+1)$. And if $n$ is odd, then $X(n)$ is odd, thus $X(n)$ ends with 1, which gets mapped to 10, making $X(n+1)$ even, and so $odd(X(n+1)) = odd(n+1)$.
$y(n)$ is a fibonacci sequence with starting values: $y(1) = 1$, $y(2) = 1$, $y(3) = 2$, thus $y(n) = F(n)$
$B(n+1) = F(n) - odd(n)$
$B(n) = F(n-1) - odd(n-1)$, or
$\displaystyle{B(n) = F(n-1) - odd(X(n-1)) = \frac{\varphi^{n-1} - (-1/\varphi)^{n-1}}{\sqrt{5}} - odd(X(n-1))}$
## Episode 10
Same as Episode 1.
## Episode 11
Pieces of papers floats by Homura as she faces Walpurgis Night. On it are calculations done by hand by Homura. Most likely, these are ballistics of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.
## Movie 3: Rebellion
### Problem A
The first problem posed, (a), was to calculate $\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx$
We can perform the substitution $u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}$ and so rewrite the integral as $\int u^3\,du$
This can be easily solved via basic integration formulas: $\int u^3\,du=\frac{1}{4}u^4+C$
Undoing the substitution we get the solution: $\frac{1}{4}\arcsin^4x + C$
### Problem B
The second problem, (b), was to calculate $\int x\ln(x^2+y)\,dx$ In order to solve it we'll assume $\displaystyle y$ to be a parameter independent from the value of $\displaystyle x$ and $\displaystyle\ln x$ to denote the natural logarithm of $\displaystyle x$.
Performing the substitution $u=x^2+y\longrightarrow\frac{du}{dx}=2x$ we can rewrite the integral as $\frac{1}{2}\int\ln u\,du$
This can be solved integrating by parts: $\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)$
And the final answer in obtained undoing the substitution: $\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C$
### Problem C
One of the problems posed, (c), was to calculate $\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx$.
The answer is to first simplify the function: $\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}$. Note that $x^2-x+1$ has only simple complex zeros.
It holds $\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C$.
The general formula $\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C$ can be used to calculate $\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C$.
For the last term, $\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C$ can be used (with some additional calculations) to obtain the solution
$\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&oldid=110413", "fetch_time": 1660307957000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00572.warc.gz", "warc_record_offset": 545664833, "warc_record_length": 13385, "token_count": 5610, "char_count": 14973, "score": 4.75, "int_score": 5, "crawl": "CC-MAIN-2022-33", "snapshot_type": "latest", "language": "en", "language_score": 0.8290334939956665, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"} |
# Reducing Spaces: Characterization
Given a Hilbert space $\mathcal{H}$.
Consider an operator: $$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}:=\mathcal{D}(T)$$
Regard a subspace: $$\mathcal{S}\leq\mathcal{H}:\quad\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^\perp$$
Then one has for reducibility: $$PT\subseteq TP\iff T\mathcal{S}^{(\perp)}\subseteq\mathcal{S}^{(\perp)}\quad(P\mathcal{D}\subseteq\mathcal{D})$$
In general inclusion will be strict: $$\mathcal{D}(PT)=\mathcal{D}\subsetneq P^{-1}\mathcal{D}=\mathcal{D}(TP)$$
How to prove this from scratch?
Preparation
Before it always holds: $$\mathcal{D}\supseteq\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$
Especially one has: $$P\mathcal{D}\subseteq\mathcal{D}\iff\mathcal{D}=\mathcal{D}\cap\mathcal{S}+\mathcal{D}\cap\mathcal{S}^\perp$$ (That will be crucial below!)
Characterization
On the one hand one has: $$PT\subseteq TP\implies P\mathcal{D}\subseteq\mathcal{D}$$
Concluding that they're invariant as: $$\varphi\in\mathcal{D}\cap\mathcal{S}:\quad T\varphi=TP\varphi=PT\varphi\in\mathcal{S}$$ $$\psi\in\mathcal{D}\cap\mathcal{S}^\perp:\quad T\psi=T(1-P)\psi=(1-P)T\psi\in\mathcal{S}^\perp$$
On the other hand one has: $$P\mathcal{D}\subseteq\mathcal{D}\implies\mathcal{D}(PT)\subseteq\mathcal{D}(TP)$$
Concluding that they commute as: $$\varphi\in\mathcal{D}:\quad PT\varphi=PT\varphi_\parallel+ PT\varphi_\perp=T\varphi_\parallel=TP\varphi$$ (Here equality above was crucial!)
Strictness
Consider an operator: $$T:\mathcal{D}(T)\to\mathcal{H}:\quad\mathcal{D}(T)\subsetneq\mathcal{H}$$
Regard any closed subspace: $$P^2=P=P^*:\quad\mathcal{R}(P)\subseteq\mathcal{D}(T)$$
Then extension will be strict: $$\mathcal{D}(PT)=\mathcal{D}(T)\subsetneq\mathcal{H}=\mathcal{D}(TP)$$ (Note that happens in most cases!) | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://math.stackexchange.com/questions/1187110/reducing-spaces-characterization?noredirect=1", "fetch_time": 1558655779000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00088.warc.gz", "warc_record_offset": 567995224, "warc_record_length": 31668, "token_count": 701, "char_count": 1802, "score": 3.78125, "int_score": 4, "crawl": "CC-MAIN-2019-22", "snapshot_type": "latest", "language": "en", "language_score": 0.4228285253047943, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"} |
As per usual, in this blog post we will practice solving algorithms problems. And today our leetcode problem is: Merge k Sorted Lists.
Problem definition:
Merge k sorted linked lists and return it as one sorted list.
Example:
``````Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6
``````
Two sorted lists
In case of the only two sorted lists we can use simple binary search with two pointers. Similarly like we did with Median of Two Sorted lists problem. But when we have 3 or more lists to iterate through it becomes a bit more complicated.
K Sorted lists
First let’s check how many lists we have in our input. It will help us determine how many pointers we need. In case of two lists we can just hardcode two pointers and it will work. But for an unknown number of lists we will need different way of keeping track of pointers and their positions.
Also we will have a problem growing complexity of comparison. In case of two lists we will need to do only one comparison operation, in case of 4 lists, 4 operations and so on. If we have 100 lists, it becomes a little less elegant to hard code conditions. We will need to come up with a smart way to compare all numbers at the same time. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://morioh.com/a/a2244057d885/algorithms-with-javascript-merge-k-sorted-lists", "fetch_time": 1713779418000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-18/segments/1712296818105.48/warc/CC-MAIN-20240422082202-20240422112202-00509.warc.gz", "warc_record_offset": 364540865, "warc_record_length": 15621, "token_count": 291, "char_count": 1215, "score": 3.578125, "int_score": 4, "crawl": "CC-MAIN-2024-18", "snapshot_type": "latest", "language": "en", "language_score": 0.9193704724311829, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"} |
# Convert 1 mm to fractional inches
1 mm = 3/64 in
Result: 1 millimeter is about 3/64 of an inch.
mm
## What is 1 mm in inches fraction?
The answer is 3/64 inch. The result is sometimes only an approximation of the real value because of rounding. But it's accurate enough for typical everyday measurements.
## How to Convert 1 mm to fractional inches?
To convert 1 millimeter to an inch fraction, you need to first convert 1 millimeter to inches. Since 1 inch is equal to 25.4 millimeters, divide 1 by 25.4 to get decimal inches:
1 mm ÷ 25.4 = 0.0393 in
Then, let's convert the decimal value of 0.0393 inches to a fraction.
To convert 0.0393 inches to a fraction, use a ruler or a decimal to fraction conversion table to find the nearest fraction with a reasonable denominator (fractional inches usually have denominators that are powers of 2, such as 2, 4, 8, 16, 32, 64).
In this example, use a ruler to find the closest fraction.
Using a ruler, you can see that 0.0393 inches is very close to 3/64 inch.
Therefore, 1 millimeter is approximately equal to 3/64 inch. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://inches.guru/1-mm-to-inches-fraction", "fetch_time": 1722933655000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00158.warc.gz", "warc_record_offset": 248442856, "warc_record_length": 4051, "token_count": 296, "char_count": 1080, "score": 3.953125, "int_score": 4, "crawl": "CC-MAIN-2024-33", "snapshot_type": "latest", "language": "en", "language_score": 0.9006896615028381, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00031-of-00064.parquet"} |
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You are watching: How much does a bag of sugar weigh
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## How many 1kg or 2lb Bags the Sugar room in a stone (and multiple Stones)
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Number that Stones Amount of 1kg (2.2lb) Bags the Sugar Amount of 2lb (0.9kg) Bags the Sugar 1 6.4 bags 7 bags 2 12.7 bags 14 bags 3 19.1 bags 21 bags 4 25.4 bags 28 bags 5 31.8 bags 35 bags 6 38.1 bags 42 bags 7 44.5 bags 49 bags 8 50.9 bags 56 bags 9 57.2 bags 63 bags 10 63.5 bags 70 bags
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## How many 500g or 1lb Bags that Sugar room in a stone (and lot of Stones)
The table listed below shows rock weights converted to one-pound street bags in addition to metric tantamount 500g bags.
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Number the Stones Amount that 500g (1.1lb) Bags of Sugar Amount the 1lb (0.45kg) Bags that Sugar 1 12.7 bags 14 bags 2 25.4 bags 28 bags 3 38.1 bags 42 bags 4 50.8 bags 56 bags 5 63.5 bags 70 bags 6 76.2 bags 84 bags 7 88.9 bags 98 bags 8 101.6 bags 112 bags 9 114.3 bags 126 bags 10 127 bags 140 bags
## More load Loss to sugar Bag Equivalents
If you’ve not rather made it to a rock yet, or you prefer to occupational in kilograms, then take a look at the charts listed below to see exactly what her weight ns looks like in terms of sugar bags.
See more: How Many Pictures Can 4Gb Hold Capacity (All Sizes), How Many Photos Per Gb
### Kilograms to street Bags
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Weight lose in Kilograms Amount of 500g (1.1lb) Bags that Sugar Amount the 1kg (2.2lb) Bags the Sugar 1 2 bags 0.5 bags 2 4 bags 1 bags 3 6 bags 1.5 bags 4 8 bags 2 bags 5 10 bags 2.5 bags 6 12 bags 3 bags 7 14 bags 3.5 bags 8 16 bags 4 bags 9 18 bags 4.5 bags 10 20 bags 5 bags 15 30 bags 7.5 bags 20 40 bags 10 bags 25 50 bags 12.5 bags
### Pounds to street Bags
If you favor to occupational weights the end in pounds, the table below will aid you to uncover the info you’re feather for:
Weight lose in Pounds Amount the 1lb (0.45kg) Bags of Sugar Amount the 2lb (0.9kg) Bags the Sugar 1 1 bags 0.5 bags 2 2 bags 1 bags 3 3 bags 1.5 bags 4 4 bags 2 bags 5 5 bags 2.5 bags 6 6 bags 3 bags 7 (half a stone) 7 bags 3.5 bags 8 8 bags 4 bags 9 9 bags 4.5 bags 10 10 bags 5 bags 11 11 bags 5.5 bags 12 12 bags 6 bags 13 13 bags 6.5 bags 14 (one stone) 14 bags 7 bags 15 15 bags 7.5 bags 20 20 bags 10 bags 25 25 bags 12.5 bags 30 30 bags 15 bags 35 35 bags 17.5 bags 40 40 bags 20 bags | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://bsci-ch.org/how-much-does-a-bag-of-sugar-weigh/", "fetch_time": 1653290465000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00176.warc.gz", "warc_record_offset": 197541007, "warc_record_length": 5602, "token_count": 1214, "char_count": 3925, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "latest", "language": "en", "language_score": 0.905315637588501, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00045-of-00064.parquet"} |
# Calculating motorcycle lean angle
motecben
I would like to calculate racing motorcycle lean angle but my trig is very rusty. Data I can supply for the calculation includes 3 axis of G, 3 axis of gyro (rate of change). The gyro drifts with time as all gyros do. I also have 2 wheel speeds and GPS speed available. There are maths functions available for integration and differentiation among others.There would obviously have to be a correction put in for lateral G as the sensor is fixed on the bike.
A further interesting addition would be a compensation for the banking of a track surface. Any help in this area would be greatly appreciated.
Last edited:
markjenn
motecben said:
I would like to calculate racing motorcycle lean angle but my trig is very rusty. Data I can supply for the calculation includes 3 axis of G, 3 axis of gyro (rate of change). The gyro drifts with time as all gyros do. I also have 2 wheel speeds and GPS speed available. There are maths functions available for integration and differentiation among others.There would obviously have to be a correction put in for lateral G as the sensor is fixed on the bike.
A further interesting addition would be a compensation for the banking of a track surface. Any help in this area would be greatly appreciated.
This is all back of the envelope, but it might be right:
Assuming a flat surface and no skidding, the lean angle of the motorcycle is related to it's speed and the cornering radius as follows:
theta = arctan ( V*V / r*g)
Where g is the gravitational constant.
Now, I think you can also show that the turning radius is related to speed and lateral acceleration as:
r = V*V / a
where a is the lateral acceleration.
Substitute r in the above and you get:
theta = arctan (a/g)
So to pull 1.0 g, then you need to be turning at a 45 degree lean angle. This is the theoretical limit assuming tires can't achieve coefficients of friction greater than 1.0.
So if you really have lateral acceleration, you don't really need anything else (I think) to get lean angle.
Where are you getting (or planning on getting) your lateral G's? Are there any small sensors out there suitable for motorcycles?
- Mark
Last edited:
motecben
Motorcycle lean angle
Hi Mark and thanks for responding. This was the first time I had used this forum and I posted the same question in 2 separate areas. Yours was the only reply after several weeks!
Yes we have small accelerometers that we can run on bikes from Crossbow in the US. They are very difficult to mount in such a way as to not get too much vibration interference but sometimes we get them to work OK. We can introduce filters that will assist somewhat. The problem that I am not sure whether you have addressed is that the lateral G sensor is fixed on the bike therefore the gravitational pull on the sensor changes direction as the bike leans so there has to be a correction put in somehow?
Thanks again for your thoughts. If you have any more, please continue to post them. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/calculating-motorcycle-lean-angle.88999/", "fetch_time": 1674809240000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00662.warc.gz", "warc_record_offset": 950567462, "warc_record_length": 14575, "token_count": 654, "char_count": 3009, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.9654711484909058, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00036-of-00064.parquet"} |
# The Math Patterns in a Rubik’s Cube
If you have never seen a Rubik’s cube, it is, in its original and basic form, a cube made out 27 tiny cubes that can be rotated. The goal of this 3D puzzle is to make every side of the cube a single color for that side. Sounds simple enough – if you start playing now, you will eventually come up with a solution to the puzzle, right? It’s not as simple as that.
## Combinations and Variations
For a classic Rubik’s cube, the number of combinations is staggering. Without just giving you the number, let’s go one step at a time. First, let us figure out how many pieces we can move and how many positions they can occupy. Since the centers of the cube are what holds the rotating center together, they don’t count, or rather, we count them separately since they always occupy the same position.
We have 8 corner pieces that can occupy 8 different positions. In other words 8! of combinations. We have 12 side pieces, which, again, are in 12 different positions – 12!. Next, we need to consider the fact that 8 corners each have three different ways to be placed once they are settled in their position. That’s 38 of combinations. Every side piece has two placements in their position, so that’s 212. We now multiply these numbers together. Pro tip: your pocket calculator probably won’t work here.
Unfortunately, this number allows us to come up with combinations that you will not normally find on a Rubik’s cube, save some tampering in the form of breaking a piece off or taping the cube differently.
The next step is to divide the total with the number of parities, or, more accurately parity errors, on the cube. Corner pieces have three positions, side pieces have two, and you could potentially pick a piece and swap it with another piece, which means two parity errors there. So it’s 2x3x2, or 12. If we divide the previous number with 12, we get a number that is slightly higher than 43 quintillions.
Interestingly enough, the permutations for the 2x2x2 cube come to ‘just’ 3,674,160. Using the pattern from above, you are more than welcome to try and calculate the number of permutations of a 5x5x5 cube. As you know, there are even harder cubes out there.
## God’s Number
God’s number is the smallest number of moves needed to solve a Rubik’s cube from any position except for the illegal ones that contain parity errors. The latest estimate is that you can solve the Rubik’s cube in 20 moves. In other words, if you know how, it should not take you more than twenty moves to solve the cube. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.21ct.org/the-math-patterns-in-a-rubiks-cube/", "fetch_time": 1596790095000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-34/segments/1596439737172.50/warc/CC-MAIN-20200807083754-20200807113754-00337.warc.gz", "warc_record_offset": 125490776, "warc_record_length": 9555, "token_count": 589, "char_count": 2548, "score": 4.71875, "int_score": 5, "crawl": "CC-MAIN-2020-34", "snapshot_type": "latest", "language": "en", "language_score": 0.9423507452011108, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00043-of-00064.parquet"} |
### Sample Problem
A rectangular swimming pool measures 30 m by 10 m by 2 m. How many cubic meters of water will be required to fill the pool completely?
The volume of water is m3.
#### Solution
The volume of rectangular swimming pool is:
30 × 10 × 2 = 600 m3
So the volume of the water is: 600 m3 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://braingenie.ck12.org/skills/102901/learn", "fetch_time": 1558652937000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00113.warc.gz", "warc_record_offset": 424542689, "warc_record_length": 1684, "token_count": 81, "char_count": 303, "score": 3.9375, "int_score": 4, "crawl": "CC-MAIN-2019-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9095262885093689, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00029-of-00064.parquet"} |
# Quantum mechanics, potential well
1. Oct 14, 2011
### fluidistic
1. The problem statement, all variables and given/known data
A particle of mass m is find to be inside a uni-dimensional potential well of the form: $V(x)=0$ for $x \leq -a$ and $a\leq x$ and $V(x)=-V_0$ for $-a <x<a$.
1)Write down the corresponding Schrödinger's equation.
2)Consider the case $-V_0<E<0$. Determine the contour conditions and conditions of continuity that an eigenfunction must satisfy inside this potential well.
3)Show that the eigenfunctions have a definite parity and that they are real for the "linked states". I'm not really sure about the word "linked state", this is my own translation of the original problem.
4)Show that the energy of the linked states only take discrete values. (Maybe linked states means eigenstates.)
5)Consider now that E>0. Find and graph the wave function of the particle. Calculate the transmission coefficient when the particle moves along the potential well.
2. Relevant equations
$-\frac{\hbar ^2 }{2m} \frac{d ^2 \Psi (x)}{d x^2}+[V(x)-E]\Psi (x)=0$.
3. The attempt at a solution
Part 1) is solved if you take the equation above and replace V(0) by its values, for each region of the potential well.
I won't put all my work for 2), since it's extremely long.
I reached that in the first region $(x \leq -a)$, $\Psi _I (x)=De^{\alpha x}[e^{x(\alpha+\beta )}-e^{x(\alpha - \beta)}]$.
For the region of the well, region II, $\Psi _{II} (x)=D(e^{-\beta x}-e^{\beta x})$.
For the third region, when $x \geq a$, $\Psi _{III}(x)=De^{-\alpha x} [e^{x(\alpha - \beta)}-e^{x(\alpha + \beta)}]$.
Where $\alpha = \frac{\sqrt {-2mE}}{\hbar}$ and $\beta = \frac{\sqrt{2m(-V_0-E)}}{\hbar}$.
I never met any complex exponential since despite its negative look, all the arguments I met under the square roots were positive.
For the contour condition, I assumed that Psi should not diverge when x tends to negative and positive infinity. From this, I could "drop" 2 constants, or more precisely, equal them to 0.
For the continuity, I assumed that the function Psi and its first derivative must be continuous over the whole 3 regions. This is how I could reduce the number of unknown constant to 1 (I started with A, B, C and D).
My question is now... how do I obtain D? Should I normalize something?
I don't think I'm done for part 2).
For the first part of part 3), it's easy. In previous exercise I showed that if V(x)=V(-x) then Psi is either odd or even. And this is the case in this exercise so by the previous exercise, the eigenfunctions have a definite parity. I'll have to show that they are real and take discrete values.
But first, I want to deal with D... but have no idea how to.
Any help will be appreciated! Thanks!
2. Oct 14, 2011
### vela
Staff Emeritus
Yes, you determine D by normalizing the wave function.
3. Oct 14, 2011
### fluidistic
Ah ok thanks! So I must do $\int _{-\infty} ^{\infty} |\Psi (x)|^2dx=1$, right? Of course I'll have to split the integral according to the 3 regions...
4. Oct 14, 2011
### vela
Staff Emeritus
Yes, exactly.
5. Oct 15, 2011
### fluidistic
I'm depressed. I made at least an error somewhere. If you look at my Psi II and Psi III, you can see that they are equal and this shouldn't be so... I realized this when trying to normalize the Psi function.
6. Oct 15, 2011
### vela
Staff Emeritus
Yeah, I was wondering why V0 makes an appearance in your solutions for regions I and III.
7. Oct 15, 2011
### fluidistic
Ok let's retake this.
Do you buy that the solution to S. equation for the region I is of the form $\Psi _I (x)=Ae^{\alpha x}+Be^{-\alpha x}$ (but B is worth 0 from the contour conditions), thus $\Psi ' _I (x)=\alpha A e^{\alpha x}$?
For the other 2 regions I get $\Psi _{II}(x)= C e^{\beta x}+De^{- \beta x}$, thus $\Psi ' _{II}(x)= \beta C e^{\beta x}-\beta D e^{- \beta x}$.
And $\Psi _{III}(x)=Fe^{\alpha x }+Ge^{-\alpha x}$ (G=0 for the later mentioned reason), thus $\Psi ' _{III}(x)=-\alpha G e^{-\alpha x}$.
Now I say that F and B are worth 0 as conditions of no divergence of Psi when x tends to infinities.
I get the continuity equations:
(1) $Ae^{- \alpha a}=Ce^{-\beta a}+D e^{\beta a}$
(2) $Ce^{\beta a}+D e^{-\beta a}=Ge^{-\alpha a}$
(3) $\alpha A e^{-\alpha a}=\beta C e^{-\beta a}-\beta D e^{\beta a}$
(4) $\beta C e^{\beta a}-\beta D e^{-\beta a}=G e^{-\alpha a}$.
Is what I've done so far okay?
After this, what I've done of my lengthy draft is playing with algebra to reach that, assuming C=-D, G must equal -A. I also isolated A to be worth $D[e^{a (\alpha + \beta)}-e^{a(\alpha - \beta)}]$.
So that I have A, G, and C in terms of D... the remaining unknown constant to be determined with the normalization of the Psi function.
8. Oct 15, 2011
### vela
Staff Emeritus
You should get sinusoidal solutions inside the well when E > -V0
Probably just a typo, but you're missing an alpha on the righthand side of (4). Your approach is fine.
Last edited: Oct 15, 2011
9. Oct 15, 2011
### fluidistic
Hmm ok, good to know.
A typo that costs me a lot! Originally a typo on my draft. I wrote $\Psi ' _{II}(a)= \Psi _{III}(a)$ instead of $\Psi ' _{II}(a)= \Psi' _{III}(a)$. I must redo the algebra.
By the way, is it ok to assume C=-D?
Thank you very much for your help!
10. Oct 15, 2011
### vela
Staff Emeritus
From the symmetry of the potential, you should expect to get even and odd solutions, which will correspond to C=D and C=-D or, respectively, cosines and sines. Those two conditions should come out of the math somehow.
11. Oct 15, 2011
### fluidistic
Ah I see. Well I'll try to reach this mathematically. I've tried to deal with matrices but after about 1 hour of algebra, I reach non sense so I made at least 1 error as usual. I even tried wolfram alpha (failed to calculate the coefficient A, C, D and G.)
Even mathematica 7:
Code (Text):
Solve[{A*e^(-k*a) - C*e^(-b*a) - D*e^(b*a) == 0,
C*e^(b*a) + D*e^(-b*a) - G*e^(-a*k) == 0,
k*A*e^(-a*k) - C*b*e^(-b*a) + b*D*e^(b*a) == 0,
b*C*e^(b*a) - b*D*e^(-b*a) - k*G*e^(-k*a) == 0}, {A, C, D, G}]
returns
Code (Text):
{{G -> 0, A -> 0, C -> 0, D -> 0}}
...
It's almost 2 am, time to sleep for me. I hope I'll wake up and attack this when I just wake up!
12. Oct 16, 2011
### vela
Staff Emeritus
Being lazy, I tried the same thing in Mathematica and got the same result. The problem is if you have four independent equations, the only solution is the trivial solution. You have to find the appropriate values for $\alpha$ and $\beta$ so the equations are no longer independent. These values happen to correspond to the allowed energies of the bound state.
I'm going to suggest you switch to sines and cosines for region II, just to simplify the algebra. Your four equations become:
\begin{align*}
C\cos k_2 a - D \sin k_2 a &= A e^{-k_1 a} \\
C \sin k_2 a + D \cos k_2 a &= \frac{k_1}{k_2} A e^{-k_1 a} \\
C\cos k_2 a + D \sin k_2 a &= G e^{-k_1 a} \\
C \sin k_2 a - D \cos k_2 a &= \frac{k_1}{k_2} G e^{-k_1 a}
\end{align*}where I replaced your $\alpha$ and $\beta$ by k1 and k2 because they're shorter to type.
If you add the first and third equations together, you get
$$2C\cos k_2 a = (A+G)e^{-k_1 a}$$Similarly, if you add the second and fourth equations together, you get
$$2C\sin k_2 a = \frac{k_1}{k_2}(A+G)e^{-k_1 a}$$If you divide these two equations, a bunch of factors cancel, and you get
$$\tan k_2 a = \frac{k_1}{k_2}$$This is one of the conditions you are looking for. You should be able to show that when this condition is satisfied, D has to vanish. Then it follows that A=G, and you have your wave function up to the normalization constant. If you do the same thing except this time isolating the terms with D in it, you'll get another condition, which leads to the rest of the solutions.
13. Oct 16, 2011
### fluidistic
Ok thank you very much vela. I do have a few questions though, I don't really get what you get.
From your first equation it seems like $e^{- \beta a}$ transforms into $\cos (\beta a)$ and $e^{\beta a}$ transforms into $-\sin (\beta a)$. However if I transform all equations according to this rule, I get different equations than yours, but for equation 1. So I didn't really understand how you simplified the exponential function for the region II.
14. Oct 16, 2011
### vela
Staff Emeritus
Your C and D aren't the same as my C and D. I should have made that clear. Remember that $\beta=ik_2$ will be complex, so $e^{\beta a} = \cos k_2a + i\sin k_2a$ and $e^{-\beta a} = \cos k_2a - i\sin k_2a$. Then you can collect terms and rename the constants. Alternately, you can write the solution to region II down as $\psi_\mathrm{II}(x) = C \cos k_2 x + D \sin k_2 x$ and then apply the boundary conditions.
15. Oct 16, 2011
### fluidistic
Ok, I'm going to retake all. I've tried to keep exponentials and reached $C=D \frac{\frac{\beta }{\alpha }+1}{1-\frac{\beta}{\alpha}}$ but I think it's wrong because $C \neq \pm D$ in this case I think.
So basically if I restart the whole exercise, I should define $\beta = \frac{\sqrt{2m(-V_0-E)}}{\hbar}= \frac{i \sqrt {2m (V_0+E)}}{\hbar}$ for convenience?
16. Oct 16, 2011
### vela
Staff Emeritus
I would actually pull the i out and define β as a real quantity, so your solution in region II would be
$$\psi(x) = C e^{i\beta x} + D e^{-i\beta x}$$or
$$\psi(x) = C \cos \beta x + D \sin \beta x$$I find it's simpler to work with real-valued variables if you can.
17. Oct 16, 2011
### fluidistic
Hmm I don't really understand. My beta was supposed to be real. It is/was worth $\frac{\sqrt {2m(-V_0-E)}}{\hbar}$ where both $V_0$ and $E$ are negative, so that $-V_0-E>0$. Thus the square root is real and so is beta.
If I factorizes by "i", I get a negative argument in the square root, which is worth a complex number (I can't define the new "beta" to be real in this case, which is what you've done I think. I don't really understand this part). In all cases beta is real valued.
18. Oct 16, 2011
### vela
Staff Emeritus
The way you originally stated the problem, the potential is -V0 at the bottom of the well, which implied V0>0 so that -V0<0. Since E<0 and it needs to be greater than -V0, you have E > -V0 so 0 > -V0-E. (These signs always mess me up too.)
19. Oct 16, 2011
### fluidistic
Oh... I totally missed this! Well thank you very much for pointing this out. I'm going to redo everything with this in mind!
20. Oct 16, 2011
### fluidistic
Okay your beta is worth $\frac{ \sqrt{2m(V_0+E)}}{\hbar}$ which is indeed real.
But I don't understand how you get such values for Psi.
$\Psi _{II}(x)=Ce^{i kx}+De^{-ikx}$ where my k is worth your beta.
Now, this is worth $C[\cos (kx)-i \sin (kx)]+D[\cos (kx)+i \sin (kx)]$. But from here, I don't undestand how you simplified to get $C \cos (kx)+D \sin (kx)$. Could you explain a bit more please?
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# Pls Help Solve These Math Questions
Hello Brilliantians
I do not know how to solve the following questions. I request the readers to kindly help me with solutions and the method of solving them.
$1. \quad The \quad 91st \quad term \quad of \quad the \quad sequence \quad 3, \quad 7, \quad 14, \quad 24, ... \quad is$
$2. \quad The \quad radius \quad of \quad a \quad circle \\ \quad is \quad 25cm. \quad The \quad radii \quad of \quad 3 \quad concentric \quad circles \quad drawn \quad in \quad such \quad \\ a \quad manner \quad that \quad the\quad whole \quad area \quad is \quad divided \quad into\\ \quad 4 \quad equal\quad parts \quad are:$
(a) 25, 25, 25
(b) 25, 50, 75
(c) 25, 30, 35
(d) 25. 25$\sqrt2$, 25$\sqrt3$
$3. \quad Number \quad of \quad real \quad roots \quad of \quad {x}^{2}+3|x|+2=0 \quad is$
$4. \quad If \quad one \quad root \quad a{x}^{2}+bx+c=0 is \quad the \quad square \quad of \quad the \quad other, \quad then$
(a) ${a}^{3}+{b}^{3}+{c}^{3}=3abc$
(b) ${a}^{2}+{b}^{2}+{c}^{2}=3abc$
(c) ${a}^{2}c+{b}^{3}+a{c}^{2}=3abc$
(d) ${a}^{2}c-{b}^{3}+a{c}^{2}=3abc$
$5. \quad If \quad the \quad ratio \quad of \\ \quad sum \quad to \quad 'n' \quad terms \quad of \\ \quad 2 A.P.s \quad (3n+4):(n+3), \quad then \quad the \quad ratio \quad of \quad \\ their \quad 'n'th \quad terms \quad is$
$6. \quad A \quad girl \quad writes \quad all \quad the \\ \quad natural \quad nos. \quad from \quad 100 \quad to \quad 999. \quad The \quad no. \quad \quad of \quad zeroes \quad \\ that \quad she \quad uses \quad is \quad x, \\ \quad the \quad no. \quad of \quad 5's \quad that \quad she \quad uses \quad is \quad y \quad and \quad \\ the \quad no.\quad of \quad 8's \quad that \quad \\ she \quad uses \quad is\quad z. \quad What \quad is \quad the \quad value \quad of \quad 2x+2z-3y$
$7. \quad The \quad remainder \quad when \quad {2}^{89} \quad is \quad divided \quad by \quad 89$
$8. \quad The \quad co-efficient \quad of \quad x \quad in \quad the \quad expansion \quad of \quad \\ (1+x)(1+2x)(1+3x)....(1+100x) \quad is$
Ritu Roy
Note by Ritu Roy
5 years, 2 months ago
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Method to solve first question :
1. However, there are many possibles sequences of such terms. But I solve it by assuming it to be second order A.P. which means second ordered difference is constant in the second order A.P.
i.e. $\quad \quad \quad \quad \quad 3, \quad \quad 7, \quad \quad 14, \quad \quad 24, ...........$
Ist order difference. $\quad 4,\quad \quad 7, \quad \quad 10, \quad \quad 13,.........$
2nd order difference $\quad \quad 3, \quad \quad 3, \quad \quad 3, ..............$
So here in this case, second order difference is constant.
if kth order difference is constant, then the general term is a polynomial expression of kth degree.
So, $t(n)$ is the general term of the given sequence, so $t(n)=an^2+bn+c$.
We know the values $t(1),t(2),t(3)$, from here find the values of $a,b,c$.
$t(n)=\dfrac{3n^2-n+4}{2}$. Now you can find whatever term you want by putting that value of $n$. @Ritu Roy
- 5 years, 2 months ago
Thanks a lot @Sandeep Bhardwaj .
- 5 years, 2 months ago
For 3,
$x^2\geq 0; 3|x|\geq 0; 2>0$
Adding them gives $x^2+3|x|+2>0$, so no solution
- 5 years, 2 months ago
For 8,
Coefficient of $x=1+2+...+100=\dfrac{100×101}{2}=5050$
Explanation: To get power of x as 1, you will have to take "nx" from one term and "1" from other 99. So it would become $x+2x+3x+...+100x$
- 5 years, 2 months ago
Probably, I can give you the answer of Q.7
According to Fermat's little theorem,
$a^{p-1}\equiv 1\pmod {p}$ where $p$ is a prime.
So, $2^{88}\equiv1\pmod{89}$ since $89$ is a prime .
Or, $2^{89}\equiv2\pmod{89}$
Hence the required remainder is $2$
- 5 years, 2 months ago
Good! I just think that there is a minor error in mentioning what the theorem is, but I see where you are going with this.
- 5 years, 2 months ago
$89$ is also a prime number. So, by one of my most favorite mod theorems, Euler's totient function, the expression will just reduce $2^{89} \equiv2^{89-88}\equiv 2 \pmod{89}$
- 5 years, 2 months ago
It works as well..
- 5 years, 2 months ago
Shouldn't it be $p-1$? I just realized it's just the same.. lol
- 5 years, 2 months ago
1. With the assumption that $a_1$ = 3, we can rewrite the sequence of the next term as:$a_{n}$ = $a_{n-1}+(3n-2)$ for $n \geq 2$. I found that with substitution, you can find the nth term of the sequence by $(a_{n}) = (a_1) + 3(1+2+...+(n-1)) + (n-1)$ for $n \geq 1$. So, for example, $a_{2}$ = $a_1$ + $3(2-1)$ + $2-1$ = $3 + 3 + 1 = 7$, $a_{3}$ = $a_1$ + $3(1+2)$ + $3-1$ = $3 + 9 + 2 = 14$, and $a_{4}$ = $a_1$ + $3(1+2+3)$ + $4-1$ = $3 + 18 + 3 = 24$, which are the second, third, and fourth terms of the sequence respectively. So, $a_{91}$ = $a_1$ + $3(1+2+...+89+90)$ + $91-1$ = $3 + 12285 + 90 = 12378$. Therefore, the 91st term of the sequence is 12378. Sorry for my bad typing of the solution. I'm pretty sure others can do better, but I hope this helps. I will probably clarify on how to get to find the nth term of the sequence in a separate note. EDIT: made a major/minor error in the calculation, fixed it now.
- 5 years, 2 months ago
Nice work! I will just put your idea in a better way. Just analyse the sequence in the following manner. Form a new sequence whose elements are the difference of the adjacent numbers in the given sequence. So the new sequence will read out as $4,7,10 \dots$. It is easy to realise that it is an arithmetic progression. So working backwards, the terms of the given sequence must be related to the sum of terms which form an AP. So for the AP, characterised by the initial term as $1$ and common difference as $3$ let $S(n)$ denote the sum upto $n$ terms. So just by observing it can be realised that the general term of the given sequence is $2 + S(n)$. Using standard formulae evaluate$S(n)$ and hence the required term.
- 5 years, 2 months ago
Thank you!
- 5 years, 2 months ago
@Ritu Roy The answer to the 4th question is (c)${ a }^{ 2 }c\quad +\quad { b }^{ 3 }\quad +\quad { c }^{ 2 }a\quad =\quad 3abc$
- 5 years, 2 months ago
Can you pls elaborate?
- 5 years, 2 months ago
I guess these are standard board examination questions...Right? Anyway, since no one has answered #5) , I'll post a solution to that one!.
Sum of an A.P upto N terms is :- $\dfrac{n}{2}*( {2a+(n-1)d})$ where the other symbols have their usual meanings. Substitute this with the ratios namely :- $(3n+4):(n+3)$ ..Taking the a's as a1 and a2 and d's as d1 and d2.: we get it as - \dfrac{2a_{1}+(n-1)*d_{1}}{2a_{2}+(n-1)*d_{2}= ratio given. Solve from here using the fact that $a+(n-1)*d$ is the $n$th term of an A.P
- 5 years, 2 months ago
A better way to say it is that sum of $2m +1$ terms of an AP is $2m +1$ times the $m^{\text{th}}$ term. So the ratio of $n^{\text{th}}$ terms is same as the ratio of their sums upto $2n+1$ terms.
- 5 years, 2 months ago
W.R.T #2) the question isn't stated clearly as to where the circles are drawn...
- 5 years, 2 months ago
@Krishna Ar That was how the question was stated.
- 5 years, 2 months ago
Is the answer of 5th question $\frac{3n-1}{n+1}$
- 5 years, 2 months ago
Pls help
- 5 years, 2 months ago
The only help I can offer you is this:
- 5 years, 2 months ago
Thanks to @Pranjal Jain @Anik Mandal , @Keshav Tiwari , @Sudeep Salgia , @Krishna Ar , @Rajdeep Dhingra , @Marc Vince Casimiro , @Sandeep Bhardwaj and others for your detailed solutions.
I also need the solutions for the ${2}^{nd}$, ${4}^{th}$ and ${7}^{th}$ questions. Thanks a lot once again.
- 5 years, 2 months ago
For number 4, assume that the roots of that equation were $s$ and $s^{2}$; thus, by Viete's formulas, $b$ = $-(s + s^{2})*a$ and $c$ = $(s)(s^{2})(a)$ = $(s^{3})*(a)$. Then you can substitute into each of the answer choices to get the answer: letter $(c)$
- 5 years, 2 months ago
@Ritu Roy None of the options in ques no 2 are right . The answer is $\frac { 25 }{ 2 } ,\frac { 25 }{ \sqrt { 2 } } ,\frac { 25\sqrt { 3 } }{ 2 }$
- 5 years, 2 months ago
Ritu Roy Here is the solution image equal all the area and you will get.the answer. Any problem just reply
- 5 years, 2 months ago
Thanks a lot.
- 5 years, 2 months ago
to improve your skill,analyse the questions..... understand the concept and analyse it
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# Andrew McLennan January 19, Winter Lecture 5. A. Two of the most fundamental notions of the dierential calculus (recall that
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1 Andrew McLennan January 19, 1999 Economics 5113 Introduction to Mathematical Economics Winter 1999 Lecture 5 Convergence, Continuity, Compactness I. Introduction A. Two of the most fundamental notions of the dierential calculus (recall that Leibniz and Newton are credited with being the creators of this) `limit' and `continuity,' were not successfully described with formal precision until the nineteenth century. 1. After a period of evolution, the mathematics profession settled on two principle frameworks, metric spaces and topological spaces, as the arenas in which these concepts are dened and discussed. 2. Let X be a set. A function d : X X! IR + is a metric if: a. for all x; y 2 X, d(x; y) = 0 if and only if x = y; b. for all x; y 2 X, d(x; y) =d(y; x); c. for all x; y; z 2 X, d(x; z) d(x; y) +d(y; z). A metric space is a pair (X; d) where X is a set and d is a metric on X. 3. The most fundamental geometric objects in X are the open balls: for x 2 X and >0, let B (x) =fx 0 2X:d(x; x 0 ) <g: 1
2 II. Sequences and Convergence A. Formally,asequence inxis a function from f1; 2; 3;:::g, or some similar index set, tox. 1. Informally, a sequence is akin to a set, and we denote it by fx n g (or sometimes x 1 ;x 2 ; :::) suggesting that we are concerned less with the particular ordering and more with the general tendency as n gets large. B. TRoughly speaking, the sequence fx n g converges to the limit x 2 X if it is eventually inside each B (x). 1. In more detail, for any >0 there is an integer N such that d(x n ;x) < for all n N. 2. The notation x n! x indicates that the sequence fx n g converges to x. 3. A sequence that converges to something is said to be convergent. Sequences that are not convergent are said to be divergent. 4. There are two ways that a sequence might be divergent: a. The sequence might just bounce around without ever settling down. This happens when, for some >0, it is possible, for any integer N, tondn; m N, such that d(x n ;x m ). b. A Cauchy sequence is a sequence such that for any >0there is an integer N such that d(x n ;x m ) <for all n; m N, i.e., a sequence that does not bounce around too much. c. If a Cauchy sequence fails to converge, the usual view is that this is not because the sequence is in any sense ill behaved. Instead, the fault lies with the given space. A metric space is said to be complete if every Cauchy sequence has a limit. C. Sequences in IR. 1. The basic reason that the real numbers are more useful than the rational numbers is as follows: 2
3 Theorem: IR is complete. Proof: Let ft n g be a Cauchy sequence in IR. Let S be the set of real numbers S such that t n >sfor all but at most nitely many n. It is straightforward to use the denition of a Cauchy sequence to show that S is nonempty and bounded above. Let s be the least uppoer bound of S. We claim that t n! s. In order to prove this by contradiction we suppose that it is not the case, which means that there is some >0 such that jt n, sj > for innitely many n. Suppose N is large enough that jt m, t n j <=2 for all n; m N, and choose M N such that jt M, sj >. Then jt n, t M j <=2 for all n M, soif t M >s+=2, then s + =2 is an element ofs, while if t M <s,=2, then s, =2 is an upper bound. In either case we have contradicted the assumption that s is the least uppoer bound of S. 2 It turns out that any metric space X can be \completed" by adding \the missing points." The formal procedure for doing this is to let X be the set of equivalence classes of Cauchy sequences in X, where two Cauchy sequences are equivalent if the distance between them goes toi zero in the limit. The details (dening the natural metric on X and showing that X is complete) are lengthy, so I will not discuss them, but that is not to say that it would be bad for you to think carefully about what is involved. 3 A sequence ft k )g in IR is increasing if t k t k+1 for all k, and it is strictly increasing if t k <t k+1 for all k. The sequence ft k g is bounded above if there is some t such that t k t for all k. Theorem: An increasing sequence ft k )g that is bounded above has a limit. Proof: We leave this as an exercise, since the ideas are very similar to those used to prove the last result. C. Sequences in IR n. 1. Unless we explicitly say otherwise, IR n is always endowed with the Eu- 3
4 clidean metric: d(x; y) = kx,yk. 2. Basically all you need to know about convergence in IR n is: Theorem: A sequence fx k =(x 1 k ;:::;xn k )gin IRn converges if and only if each component sequence fx i kg is convergent. Proof: If the sequence converges, say to x, then each component sequence converges because, for all i =1;:::n,jx i,x i k jkx,x kk. Conversely, suppose that each component sequence fx i k g converges to xi, and dene x to be the point (x 1 ;:::;x n ). For any y 2 IR n we have Consequently kyk 2 =jy 1 j 2 +:::+jy n j 2 (jy 1 j + :::+jy n j) 2 : kx k, xk jx 1 k,x 1 j+:::+jx n k,x 1 j!0: III. Open and Closed Sets A. A set C X is closed (or closed inxif some other containing space is possible) if it contains all its limit points. That is, whenever fx k g is a sequence in C that converges to some point x, x is an element ofc. 1. Example: IR n + is closed. B. A neighborhood of a point x 2 X is any set S X that contains B (x) for some >0. A set U X is open (or open in X) if it is a neighborhood of each of its points, so tha t for each x 2 U there is >0 for which B (x) U. 1. Example: IR n ++ is open. Theorem: For any x 2 X and >0, B (x) is open. Proof: For any y 2 B (x), the triangle inequality implies that B (,d(x;y)) (y) B (x). 3. Exercise: Prove that, for any x and, f y 2 X : d(x; y) >gis open. Theorem: A set U X is open if and only if its complement U c = XnU is closed. 4
5 Proof: Let C = U c. The assertion consists of two implications, the `if' and the `only if.' Suppose that C is closed, and x 2 U. If, for each natural number k, B 1=k (x) 6 U, we can choose a point inb 1=k (x) \ C, thereby constructing a sequence fx k g in C that converges to x. Since C is closed, this w ould imply that x 2 C, contrary to our assumption that x 2 U. Therefore B 1=k (x) U for large k, and since x was an arbitrary point ofu, wehave shown that U is open. Suppose that U is open, and that fx k g is a sequence in C that converges to x. If x 2 U, then B (x) U for some >0, and x k 2 B (x) for large k since x k! x, but this contradicts the a ssumption that x k 2 C. Thus C is closed. C. A topological space is a pair (T;) in which T is a set and is a collection of subsets of T, called the open sets of T, with the properties of the open subsets of X given by: Theorem: (a) ; and X are open sets. (b) The intersection of nitely many open sets is open. (c) The union of an arbitrary collection of open sets is open. Proof: This is all pretty obvious, so we will only mention that (b) is proved by noting that if U 1 ;:::;U p are open and B 1 (x) U 1 ;:::;B p (x) U p, then B minf1 ;::: p g(x) U 1 \ :::\U p : 1. Exercise: In a topological space a closed set is by denition a set whose complement is open. The collection of all closed subsets of T has properties that are similar to, and immediately derivable from, (a){(c). What are they? 2. The theory of topological spaces is much more complicated than the theory of metric spaces, essentially because any number of things can 5
6 go wrong. For starters, in a topological space there can be sets that contain all the limits of their convergent sequences, but are nonetheless not closed. 3. In the future, at least, I will try to be careful to give denitions that are valid for all topological spaces, not just metric spaces, and to be careful to indicate what properties of metric spaces might not be true more generally. However, we are basically going to just forget about general topological spaces. IV. Continuity A. Let (X; d X ) and (Y;d Y ) be metric spaces, and let f : X! Y be a function. 1. The function f is continuous if f,1 (V ):=fx2x:f(x)2vgis open whenever V Y is open. 2. This denition makes sense, and is correct, for general topological spaces, but has the unfortunate aspect of seeming strange on rst sight. Note, however, that if f is continuous, x 2 X, and >0, then f,1 (B (f(x))) is open, so that there exists some >0 such that B (x) f,1 (B (f(x))). 3. Exercise: The reverse implication also holds: (8x 2 X)(8 >0)(9 >0) B (x) f,1 (B (f(x))) implies that f is continuous. Prove this. B. The following test of continuity for functions between metric spaces is very useful. Theorem: f is continuous if and only if f(x n )! f(x) whenever fx n g is a sequence in X that converges to x. Proof: Suppose f is continuous. Let fx n g be a sequence in X with x n! x. For any > 0, f,1 (B (f(x))) is open and contains x, so there is some > 0 such that 6
7 B (x) f,1 (B (f(x))). Since x n! x, for all suciently large n we havex n 2B (x) and f(x n ) 2 B (f(x)). Since was arbitrary, wehave shown that f(x n )! f(x). Suppose that f(x n )! f(x) whenever x n! x. Let V Y be open. If f,1 (V )is not open, it must contain a point x such that for each natural number n we can choose x n 2 B 1=n (x) n f,1 (V ). Then x n! x. But V is open, so there is > 0 such that B (f(x)) V. For each n, f(x n ) =2 V, so that d(f(x n );f(x)) >and f(x n ) 6! f(x), contrary to assumption. Therefore f,1 (V ) is open, and since V was arbitrary, wehave shown that f is continuous. V. Compactness A. The concept of a compact set, developed in the rst part of this century, is now applied in most aspects of mathematics. B. The denition is far from intuitive. Let (X; d) be a metric space. 1. An open cover of a set C X is a collection fu g 2A of open sets that \covers" C in the sense that C 0 [ 2A U. 2. A subcover is a subset of fu g 2A that also covers C. 3 This set C is compact if every open cover has a nite subcover. C. It is easy to explain why such sets might be attractive from the point of view of optimization. Theorem: If f: X! IR is continuous, and C X is compact, then f attains its supremum on C (that is, arg max x2c f(x) is nonempty). Proof: For each x 2 C let U x = fy 2 X: f(y) <f(x)g=f,1 (,1;f(x)) Since f is continuous, U x is open. If f does not attain its maximum on C, then every point ofcis in some U x so fu x g x2c is a cover of C. Let U x1 ;:::;U xk be a nite subcover. 7
8 Reordering if we need to, suppose that f(x 1 ) f(x 2 ) f(x k ). Then U x1 U x2 U xk,sowehave CU x1 [[U xk = U xk : Since x k 2 C, this would yield f(x k ) <f(x k ). This contradiction completes the proof. D. What kinds of sets are compact? 1. Finite sets are obviously compact. Other examples are not so obvious. 2. A set C X is bounded if, for any x 2 C, there is M > 0 such that d(x; x 0 ) M for all x 0 2 C. Lemma: If C X is compact, it is bounded. Proof: For any x 2 C, the sets U k = fx 2 C: d(x; y) <kg (k=1;2;:::) constitute an open cover of C, and must have a nite subcover. Lemma: If C X is compact, it is closed. Proof: Suppose fx n g is a sequence in C that converges to x. Ifx=2C, the sets U k = y 2 C: d(x; y) > 1 (k =1;2;:::) k constitute an open cover of C which cannot have a nite subcover if there is a sequence in C converging to x. 3. A subsequence of a sequence fx n g is a sequence x n1 ;x n2 ;::: where n 1 < n 2 <. 4. An accumulation point of a sequence fx n g is a point x with the property that, for any >0, there are innitely many n such that x n 2 B (x). 8
9 a. Exercise: Prove that a sequence fx n g has a convergent subsequence if and only if it has an accumulation point. 5. A set C X is sequentially compact is every sequence in C has a convergent subsequence whose limit is in C. Theorem: A compact set C X is sequentially compact. Proof: By the exercise, it suces to show that a given sequence fx k g in C has an accumulation point. If not, for each x it is possible to nd an open set U x that contains x k for at most nitely many k. Since x 2 U x, for each x, fu x : x 2 Cg is an open cover of C. But the union of the elements of a nite subcover contains C, and contains x k for at most nitely many k. (The sum of nitely many nite numbers is nite.) Since this is impossible, the proof is complete. 6. The converse is also true for metric spaces (we will prove this shortly) but not for general topological spaces. E. Compact Subsets of Euclidean Space. Lemma: If fx k g is a bounded sequence in IR n, it has a convergent subsequence. Proof: We claim that it suces to prove this in the case n =1.Ifitisknown to be true for n = 1, then, in the general case, we can choose a subsequence such that the sequence of rst components is convergent, take a further subsequence of this subsequence for which the sequence of record components converges, and so on until, in the nth sequence, all sequences of components are convergent. From an earlier exercise we know that this is sucient for convergence. So, let ft k g be a bounded sequence in IR, with lower bound a 0 and upper bound b 0. We construct a sequence [a 1 ;b 1 ];[a 2 ;b 2 ];::: of closed intervals \inductively" by letting [a i ;b i ]= a i,1 ; a i,1+b i,1 2 9
10 if the right hand side contains t k for innitely many k. Otherwise we set [a i ;b i ]= ai,1 +b i,1 ;b i : 2 Arguing by induction, we can easily see that each [a i ;b i ] contains t k for innitely many k, since this is true for [a i,1 ;b i,1 ]. Now choose k 1 such that t k1 2 [a 1 ;b 1 ]. We construct fk i g inductively by choosing k i >k i,1 with t k 2 [a i ;b i ]. Since t ki ;t ki+1 ;::: are all contained in [a i ;b i ], which has length (b 0,a 0 ) 2 i, ft ki g is a Cauchy sequence, hence convergent since IR is complete. 1. All that remains is to bundle our ndings in a nice neat package. Theorem: A set C IR n is compact if and only if it is closed and bounded. Proof: We have already shown that a compact C is closed and bounded. Suppose that C is closed and bounded. Consider that any sequence in C has a convergent subsequence, by the limit, and the limit must be in C since C is closed. Thus C is sequentially compact, so it is compact since IR n is a metric space. V. Countable and Uncountable Sets A. Two sets X and Y have the same cardinality if there is a bijection f : X! Y. B. A set is said to be countable if it has the same cardinality as the natural numbers N := f1; 2; 3;:::g. 1. Some authors say that a set is countable if it is either nite or has the same cardinality asn, using the phrase \countably innite" to describe a set with exactly the same cardinality asn. 2. A set that is neither nite nor countable is said to be uncountable. C. Properties of Countable Sets. 1. Any innite subset of a countable set is countable. (Pf: If Y X = fx 1 ;x 2 ;:::g with Y innite, we map Y bijectively to N by letting f(x n ) be the number of i n such that x i 2 Y.) 10
11 2. The cartesian product of two countable sets is countable. 3. Examples: a. The rational numbers are countable since they can be put in one to one correspondence with a subset of N N. b. The real numbers are not countable. This is proved by producing a contradiction using Cantor's diagonalization procedure: if r 1 ;r 2 ;::: is an enumeration of the reals, create a number between 0 and 1 by choosing a rst digit (after the decimal point) that is dierent from the rst digit of r 1, a second digit dierent from the second digit of r 2, and so forth. All chosen digits can be dierent from 0 and 9, to avoid ambiguity about numbers such as 1 that may be written either 1:000 ::: of 0:999 :::.) The number constructed in this way is dierent from any number in the list. D. Let fu g 2A be a collection of open sets. Another collection of open sets fv g 2B is a renement of fu g 2A if S 2B V = S 2A U and, for each 2 B, there is some 2 A such that V U. Lemma: Any collection of fu g 2A of open subsets of IR n has a countable renement. Proof: Let B be the set of pairs (x; r) 2 IR n (0; 1) such that r and all components of x are rational numbers, and V (x;r) := B r (x) U for some. Then B is countable, by virtue of our remarks above, and we only need to show that S (x;r)2b V (x;r) = S 2A U. Choose 2 A and y 2 U arbitrarily. Then B (y) U for some >0. Choose a point x with rational coordinates in B =2 (y), and let r be a rational number between kx, yk and,kx,yk. Then y 2 B r (x) B (y) U : Lemma: If fv g 2B is a renement offu g 2A, and for some set S, nitely many elements of fv g 2B cover S, then there is a nite subcover of fu g 2A that covers S. 11
12 Proof: If V 1 ;:::;V K covers S, and for each k =1;:::;K, V K U k, then U 1 ;:::;U K covers S. Theorem: If C IR n is sequentially compact, then it is compact. Proof: Let fu g 2A be an open cover of C. This cover has a countable renement, and it suces to nd a nite subcover of the renement, which means that we only need to prove the result when A is countable, so we may assume that A = N. That is, we assume that the open cover is of the form fu 1 ;U 2 ;:::g. If there is no nite subcover, then for each k =1;2;::: wemaychoose x k 2 C n (U 1 [ :::[U k,1 ): By assumption fx k g 1 k=1 has a convergent subsequence whose limit x is in C. There is some K such that x 2 U K, and there is some >0 such that B (x) U K,sokx k,xkfor all but nitely many k, which is impossible if fx k g 1 k=1 has a subsequence converging to x. This contradiction completes the proof. 12
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Inferences for Population Proportions
Confidence Intervals for One Population Proportion
Statisticicans often need to determine the proportion (percentage) of a population that has a specific attribute. Some examples are:
• the percentage of U.S. adults who have health insurance
• the percentage of cars in the United States that are imports
• the percentage of U.S. adults who favor stricter clean air health standards
• the percentage of Canadian women in the labor force
In the first case, the population consists of all U.S. adults and the specified attribute is "has health insurance." For the second case, the population consists of all cars in the United States and the specific attribute is "is an import." The population in the third case is all U.S. adults and the specified attribute is "favors stricter clean air health standards." In the fourth case, the population consists of all Canadian women and the specified attribute is "is in the labor force."
We know that it is often impractical or impossible to take a census of a large population. In practice, therefore, we use data from a sample to make inferences about the population proportion.
A sample proportion, p^, is computed by using the formula
p^ = x / n
where x denotes the number of members in the sample that have the specified attribute and, as usual, n denotes the sample size. For convenience, we sometimes refer to x as the number of successes and to nx as the number of failures.
The Sampling Distribution of the Sample Proportion
To make inferences about a population mean, ๐, we must know the sampling distribution of the sample mean, that is, the distribution of the variable x(bar) (see detail for confidence interval for one population mean at thread "Statistic Procedure – Confidence Interval" http://www.tomhsiung.com/wordpress/2017/08/statistic-procedures-confidence-interval/). The same is true for proportions: To make inferences about a population proportion, p, we need to know the sampling distribution of the sample proportion, that is, the distribution of the variable p^. Because a proportin can always be regarded as a mean, we can use our knowledge of the sampling distribution of the sample mean to derive the sampling distribution of the sample proportion. In practice, the sample size usually is large, so we concentrate on that case.
The accuracy of the normal approximation depdends on n and p. If p is close to 0.5, the approximation is quite accurate, even for moderate n. The farther p is from 0.5, the larger n must be for the approximation to be accurate. As a rule of thumb, we use the normal approximation when np and n(1 – p) are both 5 or greater. Alternatively, another commonly used rule of thumb is that np and n(1 – p) are both 10 or greater; still another is that np(1 – p) is 25 or greater.
Below is the one-proportion z-interval procedure, which is also known as the one-sample z-interval procedure for a population proportion and the one-variable proportion interval procedure. Of note, as stated in Assumption 2 of Procedure 12.1, a condition for using that procedure is that "the number of successes, x, and the number of failures, nx, are both 5 or greater." We can restate this condition as "np^ and n(1 – p^) are both 5 or greather," which, for an unknown p, corresponds to the rule of thumb for using the normal approximation given after Key Fact 12.1.
Determining the Required Sample Size
If the margin of error (E) and confidence level are specified in advance, then we must determine the sample size required to meet those specifications. Solving for n in the formula for margin of error, we get
n = p^(1 – p^)(Z๐ผ/2 / E)2
This formula cannot be used to obtain the required sample size because the sample proportion, p^, is not known prior to sampling. There are two ways around this problem. To begin, we examine the graph of p^(1 – p^) versus p^ shown in Figure 12.1. The graph reveals that the largest p^(1 – p^) can be is 0.25, which occurs when p^ = 0.5. The farther p^ is from 0.5, the smaller will be the value of p^(1 – p^). Because the largest possible value of p^(1 – p^) is 0.25, the most conservative approach for determining sample size is to use that value in the above equation. The sample size obtained then will generally be larger than necessary and the margin of error less than required. Nonetheless, this approach guarantees that the specifications will at least be met. In the same vein, if we have in mind a likely range for the observed value of p^, then, in light of Figure 12.1, we should take as our educated guess for p^ the value in the range closest to 0.5. In either case, we should be aware that, if the observed value of p^ is closer to 0.5 than is our educated guess, the margin of error will be larger than desired.
Hypothesis Tests for One Population Proportion
Just earlier, we showed how to obtain confidence intervals for a population proportion. Now we show how to perform hypothesis tests for a population proportion. This procedure is actually a special case of the one-mean z-test. For Key Fact 12.1, we deduce that, for large n, the standardized version of p^,
has approximately the standard normal distribution. Consequently, to perform a large sample hypothesis test with null hypothesis H0: p = p0, we can use the variable
at the test statistic and obtain the critical value(s) or P-value from the standard normal table. We call this hypothesis-testing procedure the one-proportion z-test.
Hypothesis Tests for Two Population Proportions
For independent samples of sizes n1 and n2 from the two populations, we have Key Fact 12.2
Now we can develop a hypothesis-testing procedure for comparing two population proportions. Our immediate goal is to identify a variable that we can use as the test statistic. From Key Fact 12.2, we know that, for large, independent samples, the standardized variabvle z has approximately the standard normal distribution. They null hypothesis for a hypothesis test to compare two population proportions is H0: p1 = p2. If the null hypothesis is true, then p1 – p2 = 0, and, consequently, the bariable in
becomes
However, because p is unknown, we cannot use this variable as the test statistic. Consequently, we must estimate p by using sample information. The best estimate of p is obtained by pooling the data to get the proportion of successes in both samples combined; that is, we estimate p by
Where the p^p is called the pooled sample proportion. After replacing the p by p^p we get the final test statistic, which can be used as the test statsitic and has approximately the standard normal distribution for large samples if the null hypothesis is true. Hence we have Procedure 12.3, the two-proportions z-test. Also, it is known as the two-sample z-test for two population proportions and the two-variable proportions test.
It is very fortunate that the confidence intervals for the difference between two population proportions could be computed. As we can use Key Fact 12.2 to derive a confidence-interval procedure for the difference between two population proportions, called the two-proportions z-interval procedure. Note the following: 1) The two-proportions z-interval procedure is also known as the two-sample z-interval procedure for two population proportions and the two-variable proportions interval procedure. 2) Guidelines for interpreting confidence intervals for the difference, p1p2, between two population proportions are similar to those for interpreting confidence intervals for the difference, ๐1 – ๐2, between two population means, as describe in other relative threads.
Update on Oct 2 2017
Supplemental Data – Confidence Intervals of Odds Ratio (OR) and Relative Risk (RR)
OR
The sampling distribution of the odds ratio is positively skewed. However, it is approximately normally distributed on the natural log scale. After finding the limits on the LN scale, use the EXP function to find the limits on the original scale. The standard deviation of LN(OR) is
SD of LN(OR) = square root of (1/a + 1/b + 1/c + 1/d)
Now we know the distribution of LN(OR) and the standard deviation (mean and variation) of LN(OR), and the z-proportion procedure could be conducted to compute the confidence intervals of LN(OR).
RR
Similar with OR, the sampling distribution of the relative risk is positively skewed but is approximately normally distributed on the natural log scale. Constructing a confidence interval for the relative risk is similar to constructing a CI for the odds ratio except that there is a different formula for the SD.
SD of LN(RR) = square root of [ b/a(a+b) + d/c(c+d) ]
Statistic Procedures – Confidence Interval
Confidence Intervals for One Population Mean
A common problem in statistics is to obtain information about the mean, μ, of a population. One way to obtain information about a population mean μ without taking a census is to estimate it by a sample mean x(bar). So, a point estimate of a parameter is the value of a statistic used to estimate the parameter. More generally, a statistic is called an unbiased estimator of a parameter if the mean of all its possible values equals the parameter; otherwise, the statistic is called a biased estimator of the parameter. Ideally, we want our statistic to be unbiased and have small standard error. In that case, chances are good that our point estimate (the value of the statistic) will be close to the parameter.
However, it is not uncommon that a sample mean is usually not equal to the population mean, especially when the standard error is not small as stated previously. Therefore, we should accompany any point estimate of μ with information that indicates the accuracy of that estimate. This information is called a confidence-interval estimate for μ. By definition, the confidence interval (CI) is an interval of numbers obtain from a point estimate of a parameter. The confidence level is the confidence we have that the parameter lies in the confidence interval. And the confidence-interval estimate is the confidence level and confidence interval. An confidence interval for a population mean depends on the sample mean, x(bar), which in turn depdends on the sample selected.
Margin of error E indicates how accurate the sample mean of x(bar) is as an estimate for the value of the unknown parameter of μ. With the point estimate and confidence-interval estimate (of 95% confidence interval), we can be 95% confident that the μ is within E of the sample mean. Simply, it means that the μ = point estimate +- E.
Summary
• Point estimate
• Confidence-interval estimate
• Margin of error
Computing the Confidence-Interval for One Population Mean (σ known)
We not develop a step-by-step procedure to obtain a confidence interval for a population mean when the population standard deviation is known. In doing so, we assume that the variable under consideration is normallhy distributed. Because of the central limit theorem, however, the procedure will also work to obtain an approximately correct confidence interval when the sample size is large, regardless of the distribution of the variable. The basis of our confidence-interval procedure is the sampling distribution of the sample mean for a normally distributed variable: Suppose that a variable x of a population is normally distributed with mean μ and standard deviation σ. Then, for samples of size n, the variable x(bar) is also normally distributed and has mean μ and standard deviation σ/√n. As a consequence, we have the procedure to compute the confidence-interval.
PS: The one-mean z-interval procedure is also known as the one-sample z-interval procedure and the one-variable z-interval procedure. We prefer "one-mean" because it makes clear the parameter being estimated.
PS: By saying that the confidence interval is exact, we mean that the true confidence level equals 1 – α; by saying that the confidence that the confidence interval is approximately correct, we mean that the true confidence level only approximately equals 1 – α.
Before applying Procedure 8.1, we need to make several comments about it and the assumptions for its use, including:
• We use the term normal population as an abbreviation for "the variable under consideration is normally distributed."
• The z-interval procedure works reasonably well even when the variable is not normally distributed and the sample size is small or moderate, provided the variable is not too far from being normally distributed. Thus we say that the z-interval procedure is robust to moderate violations of the normality assumption.
• Watch for outlilers because their presence calls into question the normality assumption. Moreover, even for large samples, outliers can sometimes unduly affect a z-interval because the sample mean is not resistant to outliers.
• A statistical procedure that works reasonably well even when one of its assumptions is violated (or moderately violated) is called a robust procedure relative to that assumption.
Summary
Key Fact 8.1 makes it clear that you should conduct preliminary data analyses before applying the z-interval procedure. More generally, the following fundamental principle of data analysis is relevant to all inferential procedures: Before performing a statistical-inference procedure, examine the sample data. If any of the conditions required for using the procedure appear to be violated, do not apply the procedure. Instead use a different, more appropriate procedure, if one exists. Even for small samples, where graphical displays must be interpreted carefully, it is far better to examine the data than not to. Remember, though, to proceed cautiously when conducting graphical analyses of small samples, especially very small samples – say, of size 10 or less.
Sample Size Estimation
If the margin of error and confidence level are specified in advance, then we must determine the sample size needed to meet those specifications. To find the formula for the required sample, we solve the margin-of-error formula, E = zα/2 · σ/√n, for n. See the computing formula in Formula 8.2.
Computing the Confidence-Interval for One Population Mean (σ unknown)
So far, we have discussed how to obtain the confidence-interval estimate when the population standard deviation, σ, is known. What if, as is usual in practice, the population standard deviation is unknown? Then we cannot base our confidence-interval procedure on the standardized version of x(bar). The best we can do is estimate the population standard deviation, σ, by the sample standard deviation, s; in other words, we replace σ by s in Procedure 8.1 and base our confidence-interval procedure on the resulting variable t (studentized version of x(bar)). Unlike the standardize version, the studentized version of x(bar) does not have a normal distribution.
Suppose that a variable x of population is normally distributed with mean μ. Then, for samples of size n, the variable t has the t-distribution with n-1 degrees of freedom. A variable with a t-distribution has an associated curve, called a t-curve. Although there is a different t-curve for each number of degrees of freedom, all t-curves are similar and resemble the standard normal cruve. As the number of degrees of freedom becomes larger, t-curves look increasingly like the standard normal curve.
Having discussed t-distributions and t-curves, we can now develop a procedure for obtaining a confidence interval for a population mean when the population standard deviation is unknown. The procedure is called the one-mean t-interval procedure or, when no confusion can arise, simply the t-interval procedure.
Properties and guidelines for use of the t-interval procedure are the same as those for the z-interval procedure. In particular, the t-interval procedure is robust to moderate violations of the normality assumption but, even for large samples, can sometimes be unduly affected by outliers because the sample mean and sample standard deviation are not resistant to outliers.
What If the Assumptions Are Not Satisfied?
Suppose you want to obtain a confidence interval for a population mean based on a small sample, but preliminary data analyses indicate either the presence of outliers or that the variable under consideration is far from normally distributed. As neither the z-interval procedure nor the t-interval procedure is appropriate, what can you do? Under certain conditions, you can use a nonparametric method. Most nonparametric methods do not require even approximate normality, are resistant to outliers and other extreme values, and can be applied regardless of sample size. However, parametric methods, such as the z-interval and t-interval procedures, tend to give more accurate results than nonparametric methods when the normality assumption and other requirements for their use are met.
How to compute the expected 95% CI
The Random Sampling Distribution of Means
Imagine you have a hat containing 100 cards, numbered from 0 to 99. At random, you take out five cards, record the number written on each one, and find the mean of these five numbers. Then you put the cards back in the hat and draw another random sample, repeating the same process for about 10 minutes.
Do you expect that the means of each of these samples will be exactly the same? Of course not. Because of sampling error, they vary somewhat. If you plot all the means on a frequency distribution, the sample means form a distribution, called the random sampling distribution of means. If you actually try this, you will note that this distribution looks pretty much like a normal distribution. If you continued drawing samples and plotting their means ad infinitum, you would find that the distribution actually becomes a normal distribution! This holds true even if the underlying population was not all normally distributed: in our population of cards in the hat, there is just one card with each number, so the shape of the distribution is actually rectangular, yet its random sampling of means still tends to be normal.
These principles are stated by the central limit theorem, which states that the random sampling distribution of means will always tend to be normal, irrespective of the shape of the population distribution from which the samples were drawn. According to the theorem, the mean of the random sampling distribution of meansย is equal the mean of the original population.
Like all distributions, the random sampling distribution of means not only has a mean, but also has a standard deviation. This particular standard deviation, the standard deviation of the random sampling distribution of means is the standard deviation of the population of all the sample means. It has its own name: standard error, or standard error of the mean. It is a measure of the extent to which the sample means deviate from the true population mean.
When repeated random samples are drawn from a population, most of the means of those samples are going to cluster around the original population mean. If the samples each consisted of just two cards what would happen to the shape of the random sampling distribution of means? Clearly, with an n of just 2, there would be quite a high chance of any particular sample mean falling out toward the tails of the distribution, giving a broader, fatter shape to the curve, and hence a higher standard error. On the other hand, if the samples consisted of 25 cards each (n = 25), it would be very unlikely for many of their means to lie far from the center of the curve. Therefore, there would be a much thinner, narrower curve and a lower standard error.
So the shape of the random sampling distribution of means, as reflected by its standard error, is affected by the size of the samples. In fact, the standard error is equal to the population standard deviation (ฯ) divided by the square root of the size of the samples (n).
Using the Standard Error
Because the random sampling distribution of means is normal, so the z score could be expressed as follow. It is possible to find the limits between which 95% ย of all possible random sample means would be expected to fall (z score = 1.96).
Estimating the Mean of a Population
It has been shown that 95% of all possible members of the population (sample means) will lie within approximately +-2 (or, more exactly, +-1.96) standard errors of the population mean. The sample mean lies within +-1.96 standard errors of the population mean in 95% of the time; conversely, the population mean lies within +-1.96 standard errors of the sample mean 95% of the time. These limits of +-1.96 standard errors are called the confidence limits.
Therefore, 95% confidence limits are approximately equal to the sample mean plus or minus two standard errors. The difference between the upper and lower confidence limits is called the confidence interval – sometimes abbreviated as CI. Researchers obviously want the confidence interval to be as narrow as possible. The formula for confidence limits shows that to make the confidence interval narrower (for a given level of confidence, such as 95%), the standard error must be made smaller.
Estimating the Standard Error
According to the formula above, we cannot calculate standard error unless we know population standard deviation (ฯ). In practice,ย ฯ will not be known: researchers hardly ever know the standard deviation of the population (and if they did, they would probably not need to use inferential statistics anyway).
As a result, standard error cannot be calculated, and so z scores cannot be used. However, the standard error can be estimated using data that are available from the sample alone. The resulting statistic is the estimated standard error of the mean, usually called estimated standard error, as shown by formula below.
where S is the sample standard deviation.
t Scores
The estimated standard error is used to find a statistic, t, that can be used in place of z score. The t score, rather than the z score, must be used when making inferences about means that are based on estimates of population parameters rather than on the population parameters themselves. The t score is Studentโs t, which is calculated in much the same way as z score. But while z was expressed in terms of the number of standard errors by which a sample mean lies above or below the population mean, t is expressed in terms of the number of estimated standard errors by which the sample mean lies above or below the population mean.
Just as z score tables give the proportions of the normal distribution that lie above and below any given z score, t score tables provide the same information for any given t score. However, there is one difference: while the value of z for any given proportion of the distribution is constant, the value of t for any given proportion is not constant – it varies according to sample size. When the sample size is large (n >100), the value of t and z are similar, but as samples get smaller, t and z scores become increasingly different.
Degree of Freedom and t Tables
Table 2-1 (right-upper) is an abbreviated t score table that shows the values of t corresponding to different areas under the normal distribution for various sample sizes. Sample size (n) is not stated directly in t score tables; instead, the tables express sample size in terms of degrees of freedom (df). The mathematical concept behind degrees of freedom is complex and not needed for the purposes of USMLE or understanding statistics in medicine: for present purposes, df can be defined as simply equal to n – 1. Therefore, to determine the values of t that delineate the central 95% of the sampling distribution of means based on a sample size of 15, we would look in the table for the appropriate value of t for df = 14; this is sometimes written as t14. Table 2-1 shows that this value is 2.145.
As n becomes larger (100 or more), the values of t are very close to the corresponding values of z.
Evaluate The Article About Therapy (Randomized Trials)
Section 1 How Serious Is The Risk of Bias?
Consider the question of whether hospital care prolongs life. A study finds that more sick people die in the hospital than in the community. We would easily reject the naive conclusion that hospital care kills people because we recognize that hospitalized patients are sicker (worse prognosis) than patients in the community. Although the logic of prognostic balance is vividly clear in comparing hospitalized patients with those in the community, it may be less obvious in other contexts.
Were Patients Randomized?
The purpose of randomization is to create groups whose prognosis, with respect to the target outcomes, is similar. The reason that studies in which patient or physician preference determines whether a patient receives treatment or control (observational studies) often yield misleading results is that morbidity and mortality result from many causes. Treatment studies attempt to determine the impact of an intervention on events such as stroke, myocardial infarction, and death – occurrences that we call the trial's target outcomes. A patient's age, the underlying severity of illness, the presence of comorbidity, and a host of other factors typically determine the frequency with which a trial's target outcome occurs (prognostic factors or determinants of outcome). If prognostic factors – either those we know about or those we do not know about – prove unbalanced between a trial's treatment and control groups, the study's outcome will be biased, either underestimating or overestimating the treatment's effect. Because known prognostic factors often influence clinicians' recommendations and patients' decisions about taking treatment, observational studies often yield biased results that may get the magnitude or even the direction of the effect wrong.
Observational studies can theoretically match patients, either in the selection of patients for study or in the subsequent statistical analysis, for known prognostic factors. However, not all prognostic factors are easily measured or characterized, and in many diseases only a few are known. Therefore, even the most careful patient selection and statistical methods are unable to completely address the bias in the estimated treatment effect. The power of randomization is that treatment and control groups are more likely to have a balanced distribution of know and unknown prognostic factors. However, although randomization is a powerful technique, it does not always succeed in creating groups with similar prognosis. Investigators may make mistakes that compromise randomization, or randomization may fail because of chance – unlikely events sometimes happen.
Was Randomization Concealed?
When those enrolling patients are unware and cannot control the arm to which the patient is allocated, we refer to randomization as concealed. In unconcealed trials, those responsible for recruitment may systematically enroll sicker – or less sick – patients to either a treatment or control group. This behavior will compromise the purpose of randomization, and the study will yield a biased result (imbalance in prognosis).
Were Patients in the Treatment and Control Groups Similar With Respect to Known Prognostic Factors? (The Importance of Sample Size)
The purpose of randomization is to create groups whose prognosis, with respect to the target outcomes, is similar. Some times, through bad luck, randomization will fail to achieve this goal. The smaller the sample size, the more likely the trial will have prognostic imbalance.
Picture a trial testing a new treatment for heart failure that is enrolling patients classified as having New York Heart Association functional class III and class IV heart failure. Patients with class IV heart failure have a much worse prognosis than those with class III heart failure. The trial is small, with only 8 patients. One would not be surprised if all 4 patients with class III heart failure were allocated to the treatment group and all 4 patients with class IV heart failure were allocated to the control group. Such a result of the allocation process would seriously bias the study in favor of the treatment. Were the trial to enroll 800 patients, one would be startled if randomization placed all 400 patients with class III heart failure in the treatment arm. The larger the sample size, the more likely randomization will achieve its goal of prognostic balance.
The smaller the sample size, the more likely the trial will have prognostic imbalance. We can check how effectively randomization has balanced known prognostic factors by looking for a display of patient characteristics of the treatment and control groups at the study's commencement – the baseline or entry prognostic features. Although we will never know whether similarity exists for the unknown prognostic factors, we are reasssured when the known prognostic factors are well balanced. All is not lost if the treatment groups are not similar at baseline. Statistical techniques permit adjustment of the study result for baseline differences. When both adjusted analyses and unadjusted analyses generate the same conclusion, clinicians gain confidence that the risk of bias is not excessive.
Was Prognostic Balance Maintained as the Study Progressed?
To What Extent Was the Study Blinded?
If randomization succeeds, treatment and control groups begin wtih a similar prognosis. Randomization, however, provides no guarantees that the 2 groups will remain prognostically balanced. Blinding is the optimal strategy for maintaining prognostic balance. There are five groups that should, if possible, be blind to treatment assignment, including:
• Patients – to avoid placebo effects
• Clinicians – to prevent differential administration of therapies that affect the outcome of interest (cointervention)
• Data collectors – to prevent bias in data collection
• Adjudicators of outcome – to prevent bias in decisions about whether or not a patient has had an outcome of interest
• Data analysts – to avoid bias in decisions regarding data analysis
These 5 groups involved in clinical trials will remain unware of whether patients are receiving the experimental therapy or control therapy.
Were the Groups Prognostically Balanced at the Study's Completion?
It is possible for investigators to effectively conceal and blind treatment assignment and still fail to achieve an unbiased result.
Was Follow-up Complete?
Ideally, at the conclusion of a trial, investigators will know the status of each patient with respect to the target outcome. The greater the number of patients whose outcome is unknown – patients lost of follow-up – the more a study is potentially compromised. The reason is that patients who are retained – they may disappear because they have adverse outcomes or because they are doing well and so did not return for assessment. The magnitude of the bias may be substantial. See two examples in Pharmacy Profession Forum at http://forum.tomhsiung.com/pharmacy-practice/clinical-trials/852-example-how-lost-to-follow-up-affect-the-outcome-of-a-rct.html
Loss to follow-up may substantially increase the risk of bias. If assuming a worst-case scenario does not change the inferences arising from study results, then loss to follow-up is unlikely a problem. If such an assumption would significantly alter the results, the extent to which bias is introduced depends on how likely it is that treatment patients lost to follow-up fared badly, whereas control patients lost to follow-up fared well. That decision is a matter of judgement.
Was the Trial Stopped Too Early?
Stopping trial early (i.e., before enrolling the planned sample size) when one sees an apparent large benefit is risky and may compromise randomization. These stopped early trials run the risk of greatly overestimating the treatment effect.
A trial designed with too short a follow-up also may compromise crucial information that adequate length of follow-up would reveal. For example, consider a trial that randomly assigned patients with an abdominal aortic aneurysm to either an open surgical repair or a less invasive, endovascular repair technique. At the end of the 30-day follow-up, mortality was significantly lower in the endovascular technique group. The investigators followed up participants for an additional 2 years and found that there was no difference in mortality between groups after the first year. Had the trial ended earlier, the endovascular technique may have been considered substantially better than the open surgical techinique.
Were Patients Analyzed in the Groups to Which They Were Randomized?
Investigators will undermine the benefits of randomization if they omit from the analysis patients who do not receive their assigned treatment or, worst yet, count events that occur in nonadherent patients who were assigned to treatment against the controll group. Such analyses will bias the results if the reasons for nonadherence are related to prognosis. In a number of randomized trials, patients who did not adhere to their assigned drug regimens fared worse than those who took their medication as instructed, even after taking into account all known prognostic factors. When adherent patients are destined to have a better outcome, omitting those who do not receive assigned treatment undermines the unbiased comparison provided by randomization. Investigators prevent this bias when they follow the intention-to-treat principle and analyze all patients in the group to which they were randomized irrespective of what treatment they actually received. Following the intention-to-treat principle does not, however, reduce bias associated with loss to follow-up.
Section 2 What Are the Results?
How Large Was the Treatment Effect?
Most frequently, RCTs monitor dichotomous outcomes (e.g., "yes" or "no" classifications for cancer recurrence, myocardial infarction, or death). Patients either have such an event or they do not, and the article reports the proportion of patients who develop such events. Consider, for example, a study in which 20% of a control group died but only 15% of those receiving a new treatment died. How might one express these results?
One possibility is the absolute difference (known as the absolute risk reduction [ARR] or risk difference) between the proportion who died in the control group (control group risk [CGR]) and the proportion who died in the experimental group (experimental group risk [EGR]), or CGR – EGR = 0.20 – 0.15 = 0.05. Another way to express the impact of treatment is as the RR: the risk of events among patients receiving the new treatment relative to that risk among patients in the control group, or EGR/CGR = 0.15/0.20 = 0.75.
The most commonly reported measure of dichotomous treatment effects is the complement of the RR, the RRR. It is expressed as a percentage: 1 – (EGR/CGR) x 100% = (1 – 0.75) x 100% = 25%. An RRR of 25% means that of those who would have died had they been in the control group, 25% will not die if they receive treatment; the greater the RRR, the more effective the therapy. Investigators may compute the RR during a specific period, as in a survival analysis; the relative measure of effect in such a time-to-event analysis is called the hazard ratio. When people do not specify whether they are talking about RRR or ARR – for instance, "Drug X was 30% effective in reducing the risk of death" or "The efficacy of the vaccine was 92%" – they are almost invariably taking about RRR.
How Precise Was the Estimate of the Treatment Effect?
We can never be sure of the true risk reduction; the best estimate of the true treatment effect is what we observe in a well-designed randomized trial. This estimate is called a point estimate to remind us that, although the true value lies somewhere in its neighborhood, it is unlikely to be precisely correct. Investigators often tell us the neighborhood within which the true effect likely lies by calculating CIs, a range of values within which one can be confident the true effect lies.
We usually use the 95% CI. You can consider the 95% CI as defining the range that – assuming the study has low risk of bias – includes the true RRR 95% of the time. The true RRR will generally lie beyond these extremes only 5% of the time, a property of the CI that relates closely to the conventional level of statistical significance of P <0.05.
Example
If a trial randomized 100 patients each to experimental and control groups, and there were 20 deaths in the control group and 15 deaths in the experimental group, the authors would calculate a point estimate for the RRR of 25% [(1-0.15/0.20) x 100 = 25%]. You might guess, however, that the true RRR might be much smaller or much greater than 25%, based on a difference of only 5 deaths. In fact, you might surmise that the treatment might provide no benefit (an RRR of 0%) or might even do harm (a negative RRR). And you would be right; in fact, these results are consistent with both an RRR of -38% and and RRR of nearly 59%. In other words, the 95% CI on this RRR is -38% to 59%, and the trial really has not helped us decide whether or not to offer the new treatment.
If the trial enrolled 1000 patients per group rather than 100 patients per group, and the same event rates were observed as before. There were 200 deaths in the control group and 150 deaths in the experimental group. Again, the point estimate of the RRR is 25%. In this larger trial, you might think that our confidence that the true reduction in risk is close to 25% is much greater. Actually, in the larger trial the 95% CI on the RRR for this set of results is all on the positive side of 0 and runs from 9% to 41%.
These two examples show that the larger the sample size and higher the number of outcome events in a trial, the greater our confidence that the true RRR (or any other measure of effect) is close to what we observed. As one considers values farther and farther from the point estimate, they become less and less likely to represent the truth. By the time one crosses the upper or lower bundaries of the 95% CI, the values are unlikely to represnet the true RRR. All of this assumes the study is at low risk of bias.
Section 3 How Can I Apply the Results to Patient Care?
Were the Study Patients Similar to the Patient in My Practice?
If the patient before you would have qualified for enrollment in the study, you can apply the results with considerable confidence or consider the results generalizable. Often, your patient has different attributes or characteristics from those enrolled in the trial and would not have met a study's eligibility criteria. Patients may be older or younger, may be sicker or less sick, or may have comorbid disease that would have excluded them from participation in the study.
A study result probably applies even if, for example, adult patients are 2 years too old for enrollment in the study, had more severe disease, had previously been treated with a competing therapy, or had a comorbid condition. A better approach than rigidly applying the study inclusion and exclusion criteria is to ask whether there is some compelling reason why the results do not apply to the patient. You usually will not find a compelling reason, in which case you can generalize the results to your patient with confidence.
A related issue has to do with the extent to which we can generalize findings from a study using a particular drug to another closely (or not so closely) related agent. The issue of drug class effects and how conservative one should be in assuming class effects remains controversial. Generalizing findings of surgical treatment may be even riskier. Randomized trials of carotid endarterectomy, for instance, demonstrate much lower perioperative rates of stroke and death than one might expect in one's own community, which may reflect on either the patients or surgeons (and their relative expertise) selected to participate in randomized trials.
A final issue arises when a patient fits the features of a subgroup of patients in the trial report. We encourage you to be skeptical of subgroup analyses. The treatment is likely to benefit the subgroup more or less than the other patients only if the difference in the effects of treatment in the subgroups is large and unlikely to occur by chance. Even when these conditions apply, the results may be misleading, particularly when investigators did not specify their hypotheses before the study began, if they had a large number of hypotheses, or if other studies fail to replicate the finding.
Were All Patient-Important Outcomes Considered?
Treatments are indicated when they provide important benefits. Demonstrating that a bronchodilator produce small increments in forced expiratory volume in patients with chronic airflow limitation, that a vasodilator improves cardiac output in heart failure patients, or that a lipid-lowering agent improves lipid profiles does not provide sufficient justification for administering these drugs. In these instances, investigators have chosen substitute outcomes or surrogate outcomes rather than those that patients would consider important. What clinicians and patients require is evidence that treatments improve outcomes that are important to patients, such as reducing shortness of breath during the activities required for daily living, avoiding hospitalization for heart failure, or decreasing the risk of a major stroke.
Substitute/Surrogate Outcomes
Trial of the impact of antiarrhythmic drugs after myocardial infarction illustrate the danger of using substitute outcomes or end points. Because abnormal ventricular depolarizations were associated with a high risk of death and antiarrhythmic drugs demonstrated a reduction in abnormal ventricular depolarizations (the substitute end point), it made sense that they should reduce death. A group of investigators, performed randomized trials on 3 agents (encainide, flecainide, and moricizine) that were previously found to be effective in suppressing the substitute end point of abnormal ventricular depolarizations. The investigators had to stop the trials when they discovered that mortality was substantially higher in patients receiving antiarrhythmic treatment than in those receiving placebo. Clinicians replying on the substitue end point of arrhythmia suppression would have continued to administer the 3 drugs, to the considerable detriment of their patients.
Even when investigators report favorable effects of treatment on a patient-important outcome, you must consider whether there may be deleterious effects on other outcomes. For instance, cancer chemotherapy may lengthen life but decrease its quality. Randomized trials often fail to adequately document the toxicity or adverse effects of the experimental intervention.
Composite End Points
Composite end points represent a final dangerous trend in presenting outcomes. Like surrogate outcomes, composite end points are attractive for reducing sample size and decreasing length of follow-up. Unfortunately, they can mislead. For example, a trial that reduced a composite outcome of death, nonfatal myocardial infarction, and admission for an acute coronary syndrome actually demonstrated a trend toward increased mortality with the experimental therapy and covincing effects only on admission for an acute coronary syndrome. The composite outcome would most strongly reflect the treatment effect of the most common of the components, admission for an acute coronary syndrome, even though there is no convincing evidence the treatment reduces the risk of death or myocardial infarction.
Another long-neglected outcome is the resource implications of alternative management strategies. Health care systems face increasing resource constraints the mandate careful attention to economic analysis.
PS: Substitute/surrogate end points
In clinical trials, a surrogate endpoint (or marker) is a measure of effect of a specific treatment that may correlate with a real clinical endpoint but does not necessarily have a guaranteed relationship. The National Institutes of Health(USA) defines surrogate endpoint as "a biomarker intended to substitute for a clinical endpoint".[1][2]
Surrogate markers are used when the primary endpoint is undesired (e.g., death), or when the number of events is very small, thus making it impractical to conduct a clinical trial to gather a statistically significant number of endpoints. The FDA and other regulatory agencies will often accept evidence from clinical trials that show a direct clinical benefit to surrogate markers. [3]
A surrogate endpoint of a clinical trial is a laboratory measurement or a physical sign used as a substitute for a clinically meaningful endpoint that measures directly how a patient feels, functions or survives. Changes induced by a therapy on a surrogate endpoint are expected to reflect changes in a clinically meaningful endpoint. [6]
A commonly used example is cholesterol. While elevated cholesterol levels increase the likelihood for heart disease, the relationship is not linear – many people with normal cholesterol develop heart disease, and many with high cholesterol do not. "Death from heart disease" is the endpoint of interest, but "cholesterol" is the surrogate marker. A clinical trial may show that a particular drug (for example, simvastatin (Zocor)) is effective in reducing cholesterol, without showing directly that simvastatin prevents death.
Are the Likely Treatment Benefits Worth the Potential Harm and Costs?
If the results of a study apply to your patient and the outcomes are important to your patient, the next question concerns whether the probable treatment benefits are worth the associated risks, burdern, and resource requirements. A 25% reduction in the RR of death may sound impressive, but its impact on your patient may nevertheless be minimal. This notion is illustrated by using a concept called number needed to treat (NNT), the number of patients who must receive an intervention fo therapy during a specific period to prevent 1 adverse outcome or produce 1 positive outcome. See here for how to calcuate NNT: http://forum.tomhsiung.com/pharmacy-practice/pharmacy-informatics-and-drug-information/424-evidence-based-medicine-what-is-number-needed-to-treat-and-number-needed-to-harm.html
The impact of a treatment is related not only to its RRR but also to the risk of the adverse outcome it is designed to prevent. One large trial in myocardial infarction suggests that clopidogrel in addition to aspirin reduces the RR of death from a cardiovascular cause, nonfatal myocardial infarction, or stroke by approximately 20% in comparison to aspirin alone. Table 6-3 considers 2 patients presenting with acute myocardial infarction without elevation of ST segments on their electrocardiograms. Compared with aspirin alone, both patients have a RRR of approximately 20%, but the ARR is quite different between the two patients, which results in a siginifant different NNT.
A key element of the decision to start therapy, therefore, is to consider the patient's risk of the event if left untreated. For any given RRR, the higher the probability that a patient will experience an adverse outcome if we do not treat, the more likely the patient will benefit from treatment and the fewer such patients we need to treat to prevent 1 adverse outcome. Knowing the NNT assists clinicians in helping patients weigh the benefits and downsides associated with their management options. What if the siutation changes to the other end (Treatment usually will induces harm compared with control [adverse event is the nature of drugs], in this example, the harm is the increased risk of bleeding)? The answer is, for any given RRI (relative risk increasing), the higher the probability that a patient will experience an adverse outcome if we treat, the more likely the patient will get harm from treatment and the fewer such patients we need to treat to cause 1 adverse outcome.
Trading off benefits and risk also requires an accurate assessment of the adverse effects of treatment. Randomized trials with relatively small sample sizes are unsuitable for detecting rare but catastrophic adverse effects of therapy. Clinicians often must look to other sources of information – often characterized by higher risk of bias – to obtain an estimate of the adverse effects of therapy.
When determining the optimal treatment choice based on the relative benefits and harms of a therapy, the values and preferences of each individual patient must be considered. How best to communicate information to patients and how to incorporate their values into clinical decision making remain areas of active investigation in evidence-based medicine.
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# In Delta PQR, If angleR=angleP, QR=4 cm and PR = 5 cm .Then, the length of PQ is
Question from Class 9 Chapter Triangles
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Answer Text
4 cm 5 cm2 cm2.5 cm
Answer :
A
Solution :
Given Delta PQR such that anlge R=angleP,QR=4cm and PR = 5cm <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_IX_C07_S01_005_S01.png" width="80%"> <br> In Delta PQR, " " angleR=angleP <br> rArr " "PQ=QR [sides opposite to equla anlges are equal ] <br> rArr " " PQ=4 cm " "[because QR=4 cm] <br> Hence, the length of PQ is 4 cm. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://doubtnut.com/question-answer/in-delta-pqr-if-angleranglep-qr4-cm-and-pr-5-cm-then-the-length-of-pq-is-28219582", "fetch_time": 1591120399000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347425481.58/warc/CC-MAIN-20200602162157-20200602192157-00571.warc.gz", "warc_record_offset": 324645174, "warc_record_length": 51043, "token_count": 275, "char_count": 765, "score": 3.515625, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.6619424223899841, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00032-of-00064.parquet"} |
IsConnected - Maple Help
Hypergraphs
IsConnected
Check whether an hypergraph is connected or not
Calling Sequence IsRegular(H)
Parameters
H -
Description
• The command IsConnected(H) checks whether the hypergraph H is connected or not.
Terminology
• Connected : A hypergraph H is said connected whenever its vertex edge incidence graph is connected.
Examples
> $\mathrm{with}\left(\mathrm{Hypergraphs}\right):$$\mathrm{with}\left(\mathrm{GraphTheory}\right):$$\mathrm{with}\left(\mathrm{ExampleHypergraphs}\right):$
Create a hypergraph from its vertices and edges
> $H≔\mathrm{Hypergraph}\left(\left[1,2,3,4,5,6,7\right],\left[\left\{1,2,3\right\},\left\{2,3\right\},\left\{4\right\},\left\{3,5,6\right\}\right]\right)$
${H}{≔}{\mathrm{< a hypergraph on 7 vertices with 4 hyperedges >}}$ (1)
Print its vertices and edges
> $\mathrm{Hypergraphs}:-\mathrm{Vertices}\left(H\right);$$\mathrm{Hyperedges}\left(H\right)$
$\left[{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}{,}{7}\right]$
$\left[\left\{{4}\right\}{,}\left\{{2}{,}{3}\right\}{,}\left\{{1}{,}{2}{,}{3}\right\}{,}\left\{{3}{,}{5}{,}{6}\right\}\right]$ (2)
Draw a graphical representation of this hypergraph
> $\mathrm{Draw}\left(H\right)$
Check whether H is connected
> $\mathrm{Hypergraphs}:-\mathrm{IsConnected}\left(H\right)$
${\mathrm{false}}$ (3)
Check whether H is linear
> $\mathrm{IsLinear}\left(H\right)$
${\mathrm{false}}$ (4)
Construct the line graph L of H
> $L≔\mathrm{Hypergraphs}:-\mathrm{LineGraph}\left(H\right)$
${L}{≔}{\mathrm{Graph 1: an undirected graph with 4 vertices, 3 edge\left(s\right), and 4 self-loop\left(s\right)}}$ (5)
Draw a graphical representation of L
> $\mathrm{DrawGraph}\left(L\right)$
Construct the vertex-edge-incidence graph M of H
> $M≔\mathrm{VertexEdgeIncidenceGraph}\left(H\right)$
${M}{≔}{\mathrm{Graph 2: an undirected graph with 11 vertices and 9 edge\left(s\right)}}$ (6)
Draw a graphical representation of L
> $\mathrm{DrawGraph}\left(M\right)$
Create another hypergraph
> $H≔\mathrm{Lovasz}\left(3\right)$
${H}{≔}{\mathrm{< a hypergraph on 6 vertices with 10 hyperedges >}}$ (7)
Print its vertices and edges
> $\mathrm{Hypergraphs}:-\mathrm{Vertices}\left(H\right);$$\mathrm{Hyperedges}\left(H\right)$
$\left[{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}\right]$
$\left[\left\{{1}{,}{2}{,}{4}\right\}{,}\left\{{1}{,}{3}{,}{4}\right\}{,}\left\{{2}{,}{3}{,}{4}\right\}{,}\left\{{1}{,}{2}{,}{5}\right\}{,}\left\{{1}{,}{3}{,}{5}\right\}{,}\left\{{2}{,}{3}{,}{5}\right\}{,}\left\{{1}{,}{2}{,}{6}\right\}{,}\left\{{1}{,}{3}{,}{6}\right\}{,}\left\{{2}{,}{3}{,}{6}\right\}{,}\left\{{4}{,}{5}{,}{6}\right\}\right]$ (8)
Draw a graphical representation of this hypergraph
> $\mathrm{Draw}\left(H\right)$
Check whether H is connected
> $\mathrm{Hypergraphs}:-\mathrm{IsConnected}\left(H\right)$
${\mathrm{true}}$ (9)
Check whether H is linear
> $\mathrm{IsLinear}\left(H\right)$
${\mathrm{false}}$ (10)
Construct the line graph L of H
> $L≔\mathrm{Hypergraphs}:-\mathrm{LineGraph}\left(H\right)$
${L}{≔}{\mathrm{Graph 3: an undirected graph with 10 vertices, 45 edge\left(s\right), and 10 self-loop\left(s\right)}}$ (11)
Draw a graphical representation of L
> $\mathrm{DrawGraph}\left(L\right)$
Construct the vertex-edge-incidence graph M of H
> $M≔\mathrm{VertexEdgeIncidenceGraph}\left(H\right)$
${M}{≔}{\mathrm{Graph 4: an undirected graph with 16 vertices and 30 edge\left(s\right)}}$ (12)
Draw a graphical representation of L
> $\mathrm{DrawGraph}\left(M\right)$
References
Claude Berge. Hypergraphes. Combinatoires des ensembles finis. 1987, Paris, Gauthier-Villars, translated to English.
Claude Berge. Hypergraphs. Combinatorics of Finite Sets. 1989, Amsterdam, North-Holland Mathematical Library, Elsevier, translated from French.
Charles Leiserson, Liyun Li, Marc Moreno Maza and Yuzhen Xie " Parallel computation of the minimal elements of a poset." Proceedings of the 4th International Workshop on Parallel Symbolic Computation (PASCO) 2010: 53-62, ACM.
Compatibility
• The Hypergraphs[IsConnected] command was introduced in Maple 2024. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 37, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://jp.maplesoft.com/support/help/maple/view.aspx?path=Hypergraphs%2FIsConnected", "fetch_time": 1726208525000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00202.warc.gz", "warc_record_offset": 297398916, "warc_record_length": 24219, "token_count": 1392, "char_count": 4118, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.5239625573158264, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00027-of-00064.parquet"} |
# Linear Algebra: 2 eigenfunctions, one with eigenvalue zero
## Homework Statement
If I have two eigenfunctions of some operator, that are linearly indepdendent e.g ##F(x) , G(x)+16F(x) ## and ##F(x)## has eigenvalue ##0##, does this mean that ## G(x) ## must itself be an eigenfunction?
I thought for sure yes, but the way I particular question I just worked through went seemed to suggest it shouldn't be obvious, so perhaps not always guaranteed too.
## Homework Equations
So I have ## \hat{P} F(x) = 0 F(x) ##
##\hat{P}G(x)=16F(x)+G(x)##
## The Attempt at a Solution
[/B]
##=> \hat{P}(16F(x)+G(x))= \hat{P}(16F(x))+\hat{P}(G(x))=0+\hat{P}(16F(x)+G(x))## therefore ##16F(x)+G(x)## is an eigenfunction with eigenvalue ##1##
Intuition says ##G(x)## should be an eigenfunction, I can't think how to show it from the above however.
Thanks
Orodruin
Staff Emeritus
Homework Helper
Gold Member
A linear combination of eigenfunctions is generally not an eigenfunction. It is only an eigenfunction if the eigenvalues are the same.
MathematicalPhysicist
Gold Member
Perhaps you drank too much vodka... but P(16F(x)+G(x))= itself is not the definition of an eigenvalue equation.
A linear combination of eigenfunctions is generally not an eigenfunction. It is only an eigenfunction if the eigenvalues are the same.
Totally agree, didn't say that.
I thought if one has eigenvalue zero it may be a special case.
Perhaps you drank too much vodka... but P(16F(x)+G(x))= itself is not the definition of an eigenvalue equation.
Not this time my friend, unfortunately.
Typo =..0× F(x) + P(G(x))= 16F(x) + G(x)
PeroK
Homework Helper
Gold Member
2020 Award
## Homework Statement
If I have two eigenfunctions of some operator, that are linearly indepdendent e.g ##F(x) , G(x)+16F(x) ## and ##F(x)## has eigenvalue ##0##, does this mean that ## G(x) ## must itself be an eigenfunction?
I thought for sure yes, but the way I particular question I just worked through went seemed to suggest it shouldn't be obvious, so perhaps not always guaranteed too.
## Homework Equations
So I have ## \hat{P} F(x) = 0 F(x) ##
##\hat{P}G(x)=16F(x)+G(x)##
## The Attempt at a Solution
[/B]
##=> \hat{P}(16F(x)+G(x))= \hat{P}(16F(x))+\hat{P}(G(x))=0+\hat{P}(16F(x)+G(x))## therefore ##16F(x)+G(x)## is an eigenfunction with eigenvalue ##1##
Intuition says ##G(x)## should be an eigenfunction, I can't think how to show it from the above however.
Thanks
You've lost me here. In general, if ##F## and ##G## are eigenfunctions, then ##F+G## is an eigenfunction iff ##F## and ##G## have the same eigenvalue:
##\hat{P}(F + G) = \lambda F + \mu G = \lambda (F+G) + (\mu - \lambda) G##
Which is just what @Orodruin said in post #2.
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Totally agree, didn't say that.
I thought if one has eigenvalue zero it may be a special case.
Why would it be a special case? There is no reason for that. As has already been said, the only relevant information is if the eigenvalues are the same.
You've lost me here. In general, if ##F## and ##G## are eigenfunctions, then ##F+G## is an eigenfunction iff ##F## and ##G## have the same eigenvalue:
##\hat{P}(F + G) = \lambda F + \mu G = \lambda (F+G) + (\mu - \lambda) G##
Which is just what @Orodruin said in post #2.
I am doing the opposite of this.
I have that f and 16f +g are both eigenfunctions . F has eigenvalue 0.
Orodruin
Staff Emeritus
Homework Helper
Gold Member
I am doing the opposite of this.
I have that f and 16f +g are both eigenfunctions . F has eigenvalue 0.
No, you are really not doing the opposite. Let h = 16f + g and then what you are asking is if g = h - 16 f is an eigenvector if h and f are eigenvectors.
PeroK
PeroK | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/linear-algebra-2-eigenfunctions-one-with-eigenvalue-zero.945727/", "fetch_time": 1627471904000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00389.warc.gz", "warc_record_offset": 990220836, "warc_record_length": 17404, "token_count": 1109, "char_count": 3716, "score": 3.625, "int_score": 4, "crawl": "CC-MAIN-2021-31", "snapshot_type": "latest", "language": "en", "language_score": 0.8941150903701782, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00056-of-00064.parquet"} |
# Parallel Systems RBD
Parallel Systems RBD
This example validates the results for parallel systems in BlockSim's analytical and simulation diagrams.
Reference Case
The data set is from example 4.6 on page 73 in the book Life Cycle Reliability Engineering by Dr. Guangbin Yang, John Wiley & Sons, 2007.
Data
A lighting system uses three identical bulbs. The lifetimes of the bulbs are assumed to follow a 2-paremeter Weibull with parameters α = 35,800 hours (this is the eta value in BlockSim) and β = 1.35. We calculate the reliability of the system after 8760 hours of use, and the number of bulbs needed to be connected in parallel to achieve 99.99% reliability at that time.
Result
Since the lives of the bulbs are modeled with the Weibull distribution, the reliability of a single bulb is calculated as:
$R_{0} = exp \left [ -\left(\frac{t}{\lambda}\right)^{\beta} \right ] = exp\left [ -\left(\frac{8,760}{35,800}\right)^{1.35} \right ] = 0.8611\,\!$
Substituting the value of R0 into Equation 4.11 on page 72:
$R = 1 - (1-R_{0})^{n}\,\!$
where R is the system reliability and n is the number of identical components that form the system. The reliability of the system can be calculated as:
$R = 1 - (1-0.8611)^{3} = 0.9973\,\!$
The minimum number of bulbs required to achieve 99.99% reliability after 8760 hours of use can be calculated via Equation 4.12 on page 72.
$n = \frac{ln(1-R)}{ln(1-R_{0})} = \frac{ln(1-0.9999)}{ln(1-0.8611)} = 4.6658\,\!$
Therefore, a minimum of 5 bulbs in parallel configuration are needed to achieve 99.99% reliability.
Results in BlockSim
In BlockSim, the lighting system RBD is configured as shown below.
Each bulb is modeled with a 2-parameter Weibull distribution with parameters eta = 35,800 hours and beta = 1.35.
Analytical Proof
The reliability after 8760 hours of use is estimated to be 99.73%, which is the same result as calculated in the reference book.
The minimum number of bulbs required to achieve 99.99% reliability after 8760 hours of use can be estimated via the Allocation Analysis tool.
The analysis shows that the number of bulbs needs to be increased to 1.55506. Therefore, the total number of bulbs needed in parallel to achieve 99.99% reliability after 8760 hours of use is 4.66518 bulbs, which is the sum of the "Equivalent Parallel Units" column shown below. This number is almost exactly the same as the one calculated in the reference book.
Simulation Proof
We can also estimate the results by using the simulation tool in BlockSim. The simulation settings are shown below.
The point reliability after 8760 hours of use is estimated to be 99.77%. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://reliawiki.org/index.php?title=Parallel_Systems_RBD&printable=yes", "fetch_time": 1675789807000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00557.warc.gz", "warc_record_offset": 36587711, "warc_record_length": 7077, "token_count": 702, "char_count": 2639, "score": 3.796875, "int_score": 4, "crawl": "CC-MAIN-2023-06", "snapshot_type": "latest", "language": "en", "language_score": 0.8687605261802673, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00022-of-00064.parquet"} |
# Optimizing Box Volume with Marshmallows
The problem is a classic in algebra and calculus, and to be honest it really did not make sense to me until I built the boxes, when I was 35 years old. Here is the problem:
• Given a standard 8 ½ inch x 11 inch piece of paper, determine a function which gives the volume of a box (without a lid) made by cutting squares from each of the corners and folding up the sides. Let x be the length of a side of the square, and write the volume as a function of x. What are the units of x and v(x)? What is the maximum volume of the box?
Granted this is an expanded version of the problem. You couldn’t ask all of these questions on a standardized test. But this problem is full of concepts like: domain, range, polynomials, graphing, roots, and volume just to name a few. My classrooms have students at varying degrees of experiences in math. For instance I will have students that are ready for calculus working alongside students who really don’t understand what volume means. In fact if you ask some of them, they will reply, “L X W X H”. If that is all students think about when they are asked to find volume, then they probably do not understand the concept of volume.
I have a bunch of 5X8 note cards, so I ask the kids to imagine cutting equal squares out of the corner of the note card and filling the resulting tray with marshmallows. How many marshmallows could we contain in this resulting tray if we don’t pack them and we do not heap them above the box? I let them fiddle around with the marshmallows. I then give them 1 minute and tell them to write their name and their guess on a piece of small yellow post it note. We then arrange them on the board in a makeshift line graph. I do this so that students buy into the problem and practice estimation. This guess cements their curiosity. They start asking themselves. “Am I right?” ” Is my answer the most?” “Is my answer even possible?” Some students will try and build the box so that it matches their answer.
As they are building these boxes some students will ask, “How big should my squares be?” I respond, “How big can your squares be?” I notice that the older I get the more I sound like a psychiatrist. But here I am trying to lead students to the domain. Eventually we will formalize the domain and the range for this problem. But first we have to see how many marshmallows we can stuff in these boxes.
What we are finding so far is that unsurprisingly the number of marshmallows divided by 8 is just under the actual volume of each box in cubic inches. Here we can talk about the gaps between mallows and ask questions like, does smashing them give us closer estimates? How about piling them? (You are on your own here. The kids ask this but I redirect them to the GeoGerba applet below:)
Finally, we talk about the function that represents the volume of the box for any square of length x cut from the corners of the 5 X 8 note card. In this case V(x) = x(5 – 2x)(8-2x) (I will let you do the distribution to find the simplified version of that polynomial;). And about the best place to explore the function is in Ted Coe’s GeoGebra worksheet at http://www.tedcoe.com( His site is incredible!!!!! BTW). Here is the html applet. I hope it works because it is awesome!!!
The power of this project comes in the form of having students tape their creations to a projected GeoGebra display of the function the class developed. Every time I do this it brings a small tear to my eyes at its beauty. To me it shows the true power of technology in our classrooms. The first time I saw this problem was Algebra II when I was 16 years old and our internet was floppy discs. Along with the fact that we only used computers when instruction was complete. I am sure my teacher at the time tried to show us the solution, but I did not register any understanding. It was one of those problems that the teacher assigns and then when no one tries, he/she solves it for them on the board while all students nod their heads that they understand. I saw it again in Calculus and I still don’t think I truly understood anything outside of (the algorithm) the fact that when I applied the derivative to the function and then set the resulting function to zero and solved I would get my maximum and minimum values. The third time I saw this problem was in my masters program at the University of Nebraska-Lincoln. Here is where I came up with the idea that maybe when we solve this problem we should actually construct the boxes. I am pretty sure that constructing the boxes actually makes this problem a Jo Boaler Low floor high ceiling problem. (YOUCUBED.ORG has changed my very perception of mathematics education). I give this problem to all of my students 9 – 12 and they all love it. Probably because they get to eat marshmallows, but I hope that at the very least some finally see the connections between stuffing marshmallows in a box and volume are the same thing. If we get to the connections between algebra, calculus, and geometry…. that would be bonus for those that are ready to make those connections. I am pretty sure I wish I lived right now. Education is awesome!!!
# Who thought about that problem differently?
The question above should be asked by math teachers after almost every problem. During my bell ringer today we had unexpected, crazy, rich discussion over two problems that I never thought of as rich in mathematical ideas. The problems were :
In a certain school, 45 percent of the students purchased a yearbook. If 540 students purchased yearbooks, how many students did not buy a yearbook?
and
If a machine produces 240 thingamabobs per hour, how many minutes are needed for the machine to produce 30 thingamabobs?
Solve these and see what you come up with. Keep track of what you do. In my classroom, I randomly choose students to come up and show their methods. Sometimes they are wrong and we use incorrect answers as an opportunity to show common mistakes and I do everything in my power to show that these mistakes are helping everyone in class grow their brain (Thanks youcubed.org!!!) In this first problem, many students were blurting 243. In this case students were multiplying 0.45 and 540 or switching up their proportional reasoning. To handle this I told them to slow down and read the question again. Does that answer make sense? Also, please stop blurting answers! You are stealing someones opportunity to think.
Problem 1 method 1 The first student came up and told me that you first take 540 students divided by 45 percent to find the number of students per percent. Then take that answer (12) times 55 percent for 660. I asked them how they came up with 55 percent and they were able to tell me that 100 – 45 = 55 which is the percent of students who did not buy the yearbook. Then I asked, “Who thought of this differently?” Surprisingly, many hands flew up.
Problem 1 Method 2 The next method was the straight up algebra. If x is the total number of students, then 0.45x = 540. Solve this and 540/.45 = 1200. Since there are 1200 students we know that 1200-540 = 660 student that did not buy the yearbook. Each time a new method popped up I would give it the name of the student who produced it.
Problem 1 Method 3 In the third method a student divided 45 and 540 by 9. This student then proceeded to tell me that because of this every 5 percent is 60 students, so 11 times 60 is 660. Here the 11 felt like witchcraft for some of the kids, so I had the student explain where the 11 came from and he was able to tell everyone that 55% has 11 groups of 5. This method is similar to method 1 and this student may just have wanted to come to the board, but that is ok. Kudos for thinking deeply.
Problem 1 Method 4 A student kept blurting out FISH. At this students previous school they had been taught that proportional reasoning was a method called fish. When you set up the proportion 45/540 = 55/x , where x is the total number of students. We then proceed to cross multiply and divide. When you trace out the pattern of this the Christian fish symbol is produced. .
In all cases I was able to show students the connections between their solutions and use the language of mathematics. In the word percent, per is divide and cent is 100. Using labels helps you keep everything organized. I had one student use 45/100 = x/540 to come up with 243. Again, when I added the labels for each numbers 45( purchasers)/100 (entire class) = x (entire class)/540 (purchasers) several students in the class had an aha moment.
Problem 2 method 1 By the time we made it to the second problem, I had students practically running up to the front of the class to show what they did. So I didn’t have to randomly pick them. The first student told me it takes 8 boxes of 30 thingamabobs to package 60 minutes worth of product, so 60/8 means that it takes 7.5 minutes to produce one box.
Problem 2 method 2 On the second method a student told us that if 240 thingamabobs are produced in 60 minutes, then 120 are produced in 30 minutes. Following this reasoning then 60 are produced in 15 minutes and 30 are produced 7.5 minutes.
Problem 2 method 3 It surprised me that the algorithm we teach is always the last method shown. My personal opinion is that when we follow these “efficient” algorithms it slows down thinking . In this method the student lets x = minutes for 30 thingamabobs. Then 240(thingamabobs)/60(minutes) = 30(thingamabobs)/x(minutes). Then students cross multiply and divide to get 7.5 minutes. 240x=1800 then 1800/240 = 7.5 I think “cross multiply and divide” is a “trick” that students don’t really understand. I try and always show that it is really just common denominators and the multiplication of the entire equation by that denominator in order to simplify the fraction.
```Hans Rosling and his site gapminder.org are revolutionary and should be changing the way all of us look at local data! I am surprised that his concept of the motion graph has not trickled down into more local applications. Especially where these motion graphs can be made with Google spread sheets for free!!!!! (visualization may not work on phones or tablets) I have been part of negotiations since 2000. Communicating to other teachers, community members, and school board members what is going on statistically in the negotiations process has always been our biggest problem. When I first saw Hans Rosling's 2006 TED talk, I knew that we needed this visualization in local education.```
```First off, play around with the graphic above. Notice how dynamically all the variables can be changed. Check the box marked "Average". This is the average for each year. Compare it to other districts. Are district above or below the average? Change the variables. This is data taken directly from district arrays. What about teachers who are MA+18. How do the districts compare when we change this variable? Change the x-axis to Approximate base cost per contract day (12 X monthly insurance + base)/(contract days). Change the y-axis to ultimate max. Now you are comparing what minimum costs districts pay per contract day and the most salary a district will pay to a new teacher. Notice that the two large districts seem to spend less per contract day, but teachers can ultimately make higher salaries. Does this suggest that larger districts are more efficient? The data from Scottsbluff public schools and Gering public schools, skews the average. But I threw those two schools in just to prove that we could create one of these visualizations that housed all Nebraska School districts. This data is negotiations and compensation data from https://www.nsea.org/compensation. I have spent well over 100 hours typing all of this data into a google spread sheet. There are mistakes. Mainly because this is for instructional purposes only!!! PLEASE do not use this to make decisions if you are one of the schools in this visualization!! Although it is public data that can be accessed by anyone, I know that it is full of mistakes. At the same time, I know that something like this would be useful in communicating data to local communities all across Nebraska. So I cannot keep it secret any longer! It must be shared. Imagine what these visuals could do for local communities!!!! ``` | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://searchingformath.com/2015/10/", "fetch_time": 1493471746000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-17/segments/1492917123491.79/warc/CC-MAIN-20170423031203-00224-ip-10-145-167-34.ec2.internal.warc.gz", "warc_record_offset": 357567847, "warc_record_length": 15513, "token_count": 2857, "char_count": 12433, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2017-17", "snapshot_type": "longest", "language": "en", "language_score": 0.9393317103385925, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00013-of-00064.parquet"} |
# 95862 (number)
95,862 (ninety-five thousand eight hundred sixty-two) is an even five-digits composite number following 95861 and preceding 95863. In scientific notation, it is written as 9.5862 × 104. The sum of its digits is 30. It has a total of 4 prime factors and 16 positive divisors. There are 29,472 positive integers (up to 95862) that are relatively prime to 95862.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 30
• Digital Root 3
## Name
Short name 95 thousand 862 ninety-five thousand eight hundred sixty-two
## Notation
Scientific notation 9.5862 × 104 95.862 × 103
## Prime Factorization of 95862
Prime Factorization 2 × 3 × 13 × 1229
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 95862 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 95,862 is 2 × 3 × 13 × 1229. Since it has a total of 4 prime factors, 95,862 is a composite number.
## Divisors of 95862
1, 2, 3, 6, 13, 26, 39, 78, 1229, 2458, 3687, 7374, 15977, 31954, 47931, 95862
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 206640 Sum of all the positive divisors of n s(n) 110778 Sum of the proper positive divisors of n A(n) 12915 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 309.616 Returns the nth root of the product of n divisors H(n) 7.42253 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 95,862 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 95,862) is 206,640, the average is 12,915.
## Other Arithmetic Functions (n = 95862)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 29472 Total number of positive integers not greater than n that are coprime to n λ(n) 3684 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9222 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 29,472 positive integers (less than 95,862) that are coprime with 95,862. And there are approximately 9,222 prime numbers less than or equal to 95,862.
## Divisibility of 95862
m n mod m 2 3 4 5 6 7 8 9 0 0 2 2 0 4 6 3
The number 95,862 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (95862)
Base System Value
2 Binary 10111011001110110
3 Ternary 11212111110
4 Quaternary 113121312
5 Quinary 11031422
6 Senary 2015450
8 Octal 273166
10 Decimal 95862
12 Duodecimal 47586
20 Vigesimal bjd2
36 Base36 21yu
## Basic calculations (n = 95862)
### Multiplication
n×y
n×2 191724 287586 383448 479310
### Division
n÷y
n÷2 47931 31954 23965.5 19172.4
### Exponentiation
ny
n2 9189523044 880926058043928 84447333776207025936 8095290310454757920276832
### Nth Root
y√n
2√n 309.616 45.7666 17.5959 9.91584
## 95862 as geometric shapes
### Circle
Diameter 191724 602319 2.88697e+10
### Sphere
Volume 3.69001e+15 1.15479e+11 602319
### Square
Length = n
Perimeter 383448 9.18952e+09 135569
### Cube
Length = n
Surface area 5.51371e+10 8.80926e+14 166038
### Equilateral Triangle
Length = n
Perimeter 287586 3.97918e+09 83018.9
### Triangular Pyramid
Length = n
Surface area 1.59167e+10 1.03818e+14 78271
## Cryptographic Hash Functions
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# A puck slides a total of 12m: Problem relating kinteic friction to Newton's law
1. Nov 1, 2009
### Kat11111
1. The problem statement, all variables and given/known data
In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient of kinetic friction between the puck and board is 0.28, what was the initial speed of the puck?
2. Relevant equations
f=ma
fk=U*N
a=delta v/t
3. The attempt at a solution
I tried to find the acceleration using the equation above but since I don't know the initial speed, I get 2 unknown. I tried to substitute a by delat v/t but I don't have time either. I don't have the mass of the puck either so I can't find the normal force. How can I start?
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Nov 1, 2009
### Staff: Mentor
Use Newton's 2nd law to find the acceleration. (You don't need the actual mass--just call it m.)
3. Nov 1, 2009
### ApexOfDE
From Newton's law: you can find accleration of puck.
P + Ff + N = ma
After find a, you will use this eq:
$$v^2 - v_0^2= 2as$$
Because puck is at rest after going distance = 12m, v = 0. Plug a, then you get answer.
4. Nov 1, 2009
### Kat11111
Thanks to both of you, I got it! I didn't see the masses canceled in the first equation. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/a-puck-slides-a-total-of-12m-problem-relating-kinteic-friction-to-newtons-law.350793/", "fetch_time": 1521779045000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-13/segments/1521257648177.88/warc/CC-MAIN-20180323024544-20180323044544-00340.warc.gz", "warc_record_offset": 860572646, "warc_record_length": 14710, "token_count": 389, "char_count": 1336, "score": 3.5625, "int_score": 4, "crawl": "CC-MAIN-2018-13", "snapshot_type": "latest", "language": "en", "language_score": 0.9120234251022339, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"} |
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Multi-Step Word Problems Rational Numbers
20 Q
6th - 7th
## Explore printable Mixed Multiplication and Division worksheets for 6th Grade
Mixed Multiplication and Division worksheets for Grade 6 are essential resources for teachers looking to enhance their students' math skills in a structured and engaging manner. These worksheets provide a variety of problems that challenge students to apply their knowledge of multiplication and division concepts to solve real-world scenarios. By incorporating mixed operations, students are encouraged to think critically and develop a deeper understanding of the relationships between numbers. Teachers can easily integrate these worksheets into their lesson plans, ensuring that their Grade 6 students receive ample practice in mastering these crucial math skills. With a wide range of difficulty levels and problem types, these Mixed Multiplication and Division worksheets for Grade 6 cater to the diverse learning needs of students, making them an invaluable tool for any math classroom.
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# Physics
A uniform half metre rule is freely pivoted at the 15cm Mark and it balanced horizontally when a body of mass 40g is hung from the 2cm Mark. calculate the mass of the rule
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2. 👎
3. 👁
1. Draw a clear force diagram of the arrangements
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2. 👎
2. the center of mass of the rule is at the 25 cm mark
40 g * (15 cm - 2 cm) = M * (25 cm - 15 cm)
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2. 👎
3. Since a meter is 1m long which z equal to 100cm,,the half meter rule(simple calculation) is 0.5m or 50cm. Hence,we will b working with the 50cm rule.
Now, the center of mass of a 50cm rule is at 25cm.furthermore,from the data we're told that the object is pivoted at 15cm and the object of 40g is hung at 2cm from the beggining of the rule, hence using the simple formula we will relate the two scenarios which are;
Distance from 2cm to where the pivot is multiplied by mass( 40g*13) must b equal to the distance from the pivot to the center of mass* mass(10*xg).
Doing the simple calculation,the answer has to be 52g
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2. 👎
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.A uniform metre rule balances horizontally on a knife edge placed at the 58 cm mark when a weight of 20gf is suspended from one end. what is the weight of the rule? * | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.jiskha.com/questions/1827169/a-uniform-half-metre-rule-is-freely-pivoted-at-the-15cm-mark-and-it-balanced-horizontally", "fetch_time": 1623854757000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00568.warc.gz", "warc_record_offset": 742256161, "warc_record_length": 5020, "token_count": 985, "char_count": 3513, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2021-25", "snapshot_type": "latest", "language": "en", "language_score": 0.8822897672653198, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00002-of-00064.parquet"} |
# Calculator first order differential equation
## Bernoulli differential equation
The Bernoulli differential equation is a special form of the first order nonlinear differential equation and is:
y′(x) = f(x) ⋅ y(x) + g(x) ⋅ yn(x)
with the initial values
y(x0) = y0
where y is a sought function of x and f(x) and g(x) are continuous functions on an interval I and n is a real number not equal to one. The Bernoulli differential equation occurs frequently in physics and engineering, especially in fluid mechanics and aerodynamics.
### Calculator for the initial value problem of the Bernoulli equation with the initial values x0, y0
The solution of the Bernoulli differential equation is solved numerically. The used method can be selected. Three Runge-Kutta methods are available: Heun, Euler and RK4. The initial value can be varied by dragging the red point on the solution curve. In the input fields for the functions f and g, up to three parameters a, b and c are used which can be varied by means of the slider in the graphics.
↹#.000
🔍↔
🔍↕
Method:
Steps:
Exponent n =
Grid points:
Scale grid:
Curve:
Grid:
f(x):
g(x):
Axes ranges
x-min=
x-max=
y-min=
y-max=
Initial values
x0=
y0=
Parameter value
a=
b=
c=
Parameter ranges
a-min=
b-min=
c-min=
a-max=
b-max=
c-max=
f(x) =
g(x) =
cl
ok
Pos1
End
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x
4
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a
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c
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π
(
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0
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+
sin
cos
tan
ex
ln
xa
a/x
^
asin
acos
atan
x2
√x
ax
a/(x+b)
|x|
sinh
cosh
a⋅x+c / b⋅x+c
a+x / b+x
x2-a2/ x2+b2
a / x+b
1+√x / 1-√y
exsin(x)cos(x)
x+a
ea⋅x
a⋅x2+b⋅x+c
FunctionDescription
sin(x)Sine of x
cos(x)Cosine of x
tan(x)Tangent of x
asin(x)arcsine
acos(x)arccosine of x
atan(x)arctangent of x
atan2(y, x)Returns the arctangent of the quotient of its arguments.
cosh(x)Hyperbolic cosine of x
sinh(x)Hyperbolic sine of x
pow(a, b)Power ab
sqrt(x)Square root of x
exp(x)e-function
log(x), ln(x)Natural logarithm
log(x, b)Logarithm to base b
log2(x), lb(x)Logarithm to base 2
log10(x), ld(x)Logarithm to base 10
more ...
### Screenshot of the graph
Print or save the image via right mouse click.
## Releated sites
Here is a list of of further useful sites: | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://elsenaju.eu/Integral/ODE-Bernoulli-equation.htm", "fetch_time": 1718550112000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00752.warc.gz", "warc_record_offset": 201016421, "warc_record_length": 7692, "token_count": 719, "char_count": 2125, "score": 3.671875, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.7296463251113892, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00038-of-00064.parquet"} |
# A problem with a trigonometric equation
I'm trying to solve this problem but I can't figure how. Can you help me?
$$A=\frac{\sin \alpha+\cos(3\pi/2-\alpha)+\tan(5\pi+\alpha)}{\csc(2\pi-\alpha)+\sin(5\pi/2+\alpha)}$$
If $\tan \alpha=-2/3$ and $\alpha \in$ IV quadrant, calculate the value of the expression.
Thanks.
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Pardon me, but what are "sen" and "tg"? Are these abbreviations for "sine" and "tangent"? if so, the only standard abbreviations in English are "sin" and "tan". – MJD Mar 26 '12 at 2:58
Fixed. Sorry for that, the english is not my native language. – ignaces Mar 26 '12 at 3:01
No problem; that's why I asked. – MJD Mar 26 '12 at 3:03
Can somebody help? – ignaces Mar 26 '12 at 23:33
You can find $\cos(\alpha)$ from the identity $\sec^2(\alpha) = \tan^2(\alpha) + 1$ and the fact that $\alpha$ is in QIV. Then find the sine. After that use angle sum/difference formulas to simplify terms that involve a trig function of a sum or difference. For example, $\cos(3\pi/2 - \alpha) = \cos(3\pi/2)\cos(\alpha) + \sin(3\pi/2)\sin(\alpha) = -\sin(\alpha)$. It's a bit messy but I don't see an easier solution to the problem. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/124488/a-problem-with-a-trigonometric-equation", "fetch_time": 1469529278000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00127-ip-10-185-27-174.ec2.internal.warc.gz", "warc_record_offset": 166770892, "warc_record_length": 17041, "token_count": 378, "char_count": 1141, "score": 3.828125, "int_score": 4, "crawl": "CC-MAIN-2016-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8352841734886169, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00061-of-00064.parquet"} |
# How many trial picks expectedly sufficient to cover a sample space?
Consider a sequence of independent events where an $r$ element subset of an $n$ element set is picked uniformly randomly (ie. any of the $\begin{pmatrix}n\newline r\end{pmatrix}$ possibilities being equally likely).
What is the expected number of subsets one has pick to cover the whole set?
Here the terminology means: a sequence of picks $A_1,A_2,\ldots,A_n$ covers the whole set if $|A_1 \cup \cdots \cup A_n| = n$. A sequence $A_1, A_2,\ldots$ succeeds to cover the whole set in $n$ steps, if $A_1,\ldots,A_n$ covers the whole set but $A_1,\ldots, A_{n-1}$ does not.
The expected numbers seems to be much higher than one would imagine. But I could not quite come up with a closed form. But chances are, its always a rational number.
• You should edit your question to not use $n$ for both the cardinality of the set and the number of steps to cover the set. Also, interesting question! I look forward to seeing what people come up with for this one. – Zev Chonoles Jan 27 '10 at 20:15
• As you expected, it is always rational. If you let F(n,k,r) denote the expected number of additional sets you need when you already have covered k elements of your n, then you can set up a linear recurrence for F(n,k,r) in terms of F(n, k-1, r), F(n, k-2, r), ..., F(n, k-r, r) by looking at how many elements are covered by your next set. Combined with the boundary condition F(n,0,r)=0, you could in theory solve to get F(n,0,r) as a rational number. This is what is done in the "coupon collector" problem referenced by Tal K (the case r=1), but is impractical for, say, n/r bounded. – Kevin P. Costello Jan 27 '10 at 21:21
## 3 Answers
This process will cover the set faster than making $r$ random selections of a single element at each step ("sampling with replacement", producing a multiset of $r$ not-necessarily-distinct elements instead of a set of $r$ distinct elements). The latter is taking $r$ steps at a time in the Coupon Collector process which takes $n * log(n)$ steps. So we need at least $(n/r) * log(n)$ steps on average. This should be a close approximation when $n/r$ is large and within a bounded (not necessarily constant) factor of the truth when $n/r$ is bounded. The case when $n=2r$ is close to the "20 questions" problem of Erdos and Renyi.
• Do you mean "at most (n/r) log(n) steps"? – Reid Barton Jan 27 '10 at 20:48
• Yes, "at most", meaning that the slower coverage process takes (n/r)*log(n). Thanks for catching that. Also, when I say "a close approximation" I suppose that the asymptotic difference between the with- and without- replacement expected times (in the case when n/r is large) would be an additive difference of O(log n), not a multiplicative difference of a constant factor in the larger main term. In the n/r bounded case there could well be some log-periodic function as the "constant", as in the Erdos-Renyi problem. It would take a more detailed calculation to find out. – Tal K Jan 27 '10 at 20:59
• It seems like even when n/r is bounded that n/r log n should be the right answer up to a (1+o(1)) multiplicative factor. If we considered an alternative model where each element is included in a set INDEPENDENTLY with some probability p, then it's easy to see (e.g. by computing the second moment of the number of omitted elements) that the threshold is log n/p sets. But the threshold is monotone in p, and you can sandwich the original problem with r=cn in between p=c-o(1) and p=c+o(1) with high probability. – Kevin P. Costello Jan 27 '10 at 21:26
• If you fill cartons of r distinct coupons, it takes an average of (n/n + n/(n-1) + ... + n/(n-r+1)) coupons to fill a carton. So, a random r selection is like taking that many steps in the coupon collector process. – Douglas Zare Jan 27 '10 at 21:42
• Kevin, your alternative model with p=1/2 is the Erdos-Renyi "20 Questions" problem, and the expected coverage time in that case involves both the [base 2] log(n) and some function of the fractional part of log(n). That's not necessarily inconsistent with your remark (for instance the log-periodic term could be additive, not multiplicative, I don't have the reference handy to check which it is). – Tal K Jan 27 '10 at 21:47
The expected number of picks needed equals the sum of the probabilities that at least $t$ picks are needed, which means that $t-1$ subsets left at least one value uncovered. We can use inclusion-exclusion to get the probability that at least one value is uncovered.
The probability that a particular set of $k$ values is uncovered after $t-1$ subsets are chosen is
$$\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1}$$
So, by inclusion-exclusion, the probability that at least one value is uncovered is
$$\sum_{k=1}^n {n \choose k}(-1)^{k-1}\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg) ^{t-1}$$
And then the expected number of subsets needed to cover everything is
$$\sum_{t=1}^\infty \sum_{k=1}^n {n \choose k}(-1)^{k-1} \Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1}$$
Change the order of summation and use $s=t-1$:
$$\sum_{k=1}^n {n \choose k}(-1)^{k-1} \sum_{s=0}^\infty \Bigg( \frac{n-k \choose r}{n \choose r}\Bigg)^s$$
The inner sum is a geometric series.
$$\sum_{k=1}^n {n \choose k} (-1)^{k-1}\frac{n \choose r}{{n \choose r}-{n-k \choose r}}$$
$${n \choose r} \sum_{k=1}^n (-1)^{k-1}\frac{n \choose k}{{n \choose r}-{n-k \choose r}}$$
I'm sure that should simplify further, but at least now it's a simple sum. I've checked that this agrees with the coupon collection problem for $r=1$.
Interestingly, Mathematica "simplifies" this sum for particular values of $r$, although what it returns even for the next case is too complicated to repeat, involving EulerGamma, the gamma function at half-integer values, and PolyGamma[0,1+n].
• Maple doesn't give a simpler form even for r=2, although there's no guarantee that there's not some trick it doesn't see. Also, your answer seems to agree with Tal K's asymptotics below. – Michael Lugo Jan 28 '10 at 1:04
• If you plug r=1 into that formula, it still takes some manipulation to convert the alternating sum of (n choose k)/k to (1 + 1/2 + 1/3 + ... + 1/n). Can one express it as a similar sum of positive decreasing terms? – Douglas Zare Jan 28 '10 at 4:47
• For r=2, here is what Mathematica reports: n/(4^n (1-2n)^2 Sqrt[Pi]) * (Gamma[1/2-n](-2n^2 Gamma[n] + Gamma[1+n])) + n(n-1)/(1-2n)^2*(1+EulerGamma(-1+2n)+(-1+2n)PolyGamma[0,1+n]). I've tried to simplify this by cancelling a few terms, and I hope I haven't introduced any errors. Note that 2n-1 or n-1/2 shows up in many places. For r=3, the formula involves many occurrences of Sqrt[1+6n-3n^2] which is imaginary for n \ge 3. – Douglas Zare Jan 28 '10 at 11:38
EDIT: While the $r=1$ case is the easiest, I thought it would be helpful to work it out anyway. I get that the expected number of picks necessary for $r=1$ is $nH_n$, where $H_n$ is the $n$th harmonic number, which is in line with Tal K's answer since $H_n\approx\ln(n)$.
Suppose the total number of elements covered by our picks so far is $k$. If we calculate the expected number of picks it will take to get to $k+1$, then we simply take the sum of our result from $k=0$ to $k=n-1$. There are $n-k$ elements we still need to hit, so there is an $\frac{n-k}{n}$ probability of having $k+1$ covered after 1 pick, $\frac{n-k}{n}(\frac{k}{n})$ probability of having $k+1$ covered after exactly 2 picks, and in general $\frac{n-k}{n}(\frac{k}{n})^j$ probability of going to $k+1$ after exactly $j$ picks. Thus, the expected number of picks to go from $k$ covered to $k+1$ covered is $(\frac{n-k}{n})\sum_{j=1}^\infty k(\frac{k}{n})^{k-1}$, which by the standard derivative trick we know is $(\frac{n-k}{n})\frac{1}{(1-\frac{k}{n})^2}=\frac{n}{n-k}$. Thus the expected number of picks of 1 element subsets necessary to cover an $n$ element set is $\sum_{k=0}^{n-1}\frac{n}{n-k}=n\sum_{k=1}^n\frac{1}{k}=nH_n$. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://mathoverflow.net/questions/13171/how-many-trial-picks-expectedly-sufficient-to-cover-a-sample-space", "fetch_time": 1571537629000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00331.warc.gz", "warc_record_offset": 607535098, "warc_record_length": 29732, "token_count": 2353, "char_count": 7961, "score": 3.734375, "int_score": 4, "crawl": "CC-MAIN-2019-43", "snapshot_type": "latest", "language": "en", "language_score": 0.9231845140457153, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00033-of-00064.parquet"} |
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# CBSE Sample Paper-05
SUMMATIVE ASSESSMENT II
MATHEMATICS
Class X
Time allowed: 3 hours
Maximum Marks: 90
General Instructions:
a) All questions are compulsory.
b) The question paper consists of 31 questions divided into four sections A, B, C and D.
c) Section A contains 4 questions of 1 mark each which are multiple choice questions, Section B
contains 6 questions of 2 marks each, Section C contains 10 questions of 3 marks each and
Section D contains 11 questions of 4 marks each.
d) Use of calculator is not permitted.
Section A
1.
2.
3.
4.
Cards each marked with one of the numbers 6, 7, 8, ., 15 are placed in a box and mixed
thoroughly. One card is drawn at random from the box. The probability of getting a card with a
number less than 10 is:
1
3
2
4
(a)
(b)
(c)
(d)
5
5
5
5
The value of x for which the distance between the points A ( 2, 3) and B ( x,5 ) is 10 units is:
(a) 2
(b) 4
(c) 6
(d) 8
The sum of first five multiples of 4 is:
(a) 30
(b) 40
(c) 50
(d) 6
The angle of depression and the angle of elevation from an object on the ground to an object in
the air are related as:
(a) greater than
(b) less than
(c) equal
(d) all of them
Section B
5.
8.
Find the radius of the circle whose circumference is equal to the sum of circumferences of the
two circles of diameter 30 cm and 24 cm.
A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the
shape of a right circular cone mounted on a hemisphere of radius 3 cm. If the height of the toy
is 12 cm, find the number of toys so formed.
Water flows through a circular pipe, whose internal diameter is 2 cm, at the rate of 0.7 m per
second into a cylindrical tank, the radius of whose base is 40 cm. By how much will the level of
water in the cylindrical tank use in half an hour?
For what value of k , are the roots of the equation 3 x 2 + 2 kx + 27 = 0 are real and equal?
9.
Two APs have the same common difference. The first two terms of one of these is 3 and not
6.
7.
## of the other is 7. Find the difference between their 4th terms.
10. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If PCA =
100, then find CBA.
Section C
11. If A ( 5, 1) , B ( 3, 2 ) and C ( 1,8) are the vertices of triangle ABC, find the length of median
through A and the coordinates of centroid.
12. If the point ( x, y ) is equidistant from the points ( a + b, b a ) and ( a b, a + b ) , then prove that
bx = ay.
13. A chord AB of a circle of radius 14 cm makes a right angle at the centre (O) of the circle. Find
22
7
22
## face of the clock between 6 a.m. and 6.05 a.m. Use =
7
15. Find the number of coins 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right
circular cylinder of height 10 cm and diameter 4.5 cm.
3
1
16. Solve the quadratic equation by using quadratic formula: 2 x 2
x+
=0
2
2
17. Find the middle term of the AP 10, 7, 4,......., ( 62 ) .
18. ABCD is a quadrilateral such that D = 90 . A circle C (O, r ) touches the sides AB, BC, CD and
DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, then find r.
19. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height
5 m. From a point on the plane the angles of elevation of the bottomand top of the flagstaff of
height 5 m. From a point on the plane the angles of elevation of the bottom and top of the
flagstaff are respectively 30 and 60 . Find the height of the tower.
20. Find the probability that a number selected at random from the numbers 1, 2, 3, ., 35 is:
(i)
a prime number.
(ii)
multiple of 7.
(iii) multiple of 3 or 5.
Section D
21. Draw any quadrilateral ABCD. Construct another quadrilateral ABCD similar to the
4
quadrilateral ABCD with each side equal to th of the corresponding side of quadrilateral
5
ABCD. Write the steps of construction also.
22. A tree breaks due to the storm and the broken part bends so that the top of the tree touches
the ground making an angle of 30 with the ground. The distance from the foot of the tree to
the point where the top touches the ground is 10 meters. Find the height of the tree.
23. A box contains 19 balls bearing numbers 1, 2, 3, .., 19. A ball is drawn at random from the
box. Find the probability that the number on the ball is:
(i) aprime number
(ii) divisible by 3 or 5
(iii) neither divisible by 5 nor by 10
(iv) an even number
a 2a
24. If P and Q are two points whose coordinates are ( at 2 , 2at ) and 2 ,
respectively and S is
t
t
1
1
+
is independent of t.
the point ( a, 0 ) , then show that
SP SQ
25. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid
cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume
of:
(i) water displaced out of the cylindrical vessel.
22
## (ii) water left in the cylindrical vessel. Use =
26. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and
of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two
places of decimals. Also find the total surface area of the remaining solid. ( Use = 3.1416 )
1
1
11
= ; x 4, 7
x + 4 x 7 30
Two years ago, a mans age was three times the square of his sons age. Three years hence, his
age will be four times his sons age. Find their present ages.
Ram and Shyam have been given to find out the number of two digit numbers in between 6 and
102 which are divisible by 6. Ram calculated it by using AP while Shyam calculated it directly.
Read the above passage and answer the following questions:
(i) How many two digits number are there in between 6 and 102 which are divisible by 6??
(ii) What value is depicted by Ram?
The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided
by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle.
With the vertices of a triangle ABC as centre, three circles are described, each touching the
other two externally. If the sides of the triangle are 9 cm, 7 cm and 6 cm, then find the radii of
the circle.
28.
29.
30.
31. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://ja.scribd.com/document/310926276/2016-10-Mathematics-Sample-Paper-Sa2-05", "fetch_time": 1563383898000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183540-00006.warc.gz", "warc_record_offset": 437848138, "warc_record_length": 59855, "token_count": 1771, "char_count": 6137, "score": 3.765625, "int_score": 4, "crawl": "CC-MAIN-2019-30", "snapshot_type": "latest", "language": "en", "language_score": 0.903156578540802, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00052-of-00064.parquet"} |
## SHM p.9
### Oscillation in a Plane (2-d)
Consider a particle held to the origin with a spring, but allowed to move in a (x-y) plane. Schrödinger's equation reads:
Proceeding exactly as before we have the dimensionless form
Since the total Hamiltonian can be written as the sum of Hamiltonians (one just in x' the other just in y', each exactly the same form as in the 1-d oscillator), the solution is just the product of 1-d oscillator wavefunctions (one just in x' the other just in y'):
with E'= (2nx+1 + 2ny+1), i.e., the energy is just the sum of the "x" E' and the "y" E'. Note one new feature in the solution: there are different wavefunctions with the same energy (e.g., nx=3 & ny=0 ; nx=2 & ny=1 ; nx=1 & ny=2 ; nx=0 & ny=3 all have E'=8). This seemingly small difference turns out to be quite important; it's called degeneracy.
Here's what the probability density looks like for nx=16 and ny=4:
In this display the bright parts of the image are where the particle is likely to be found. Here is a expanded version of the same:
Note the 4 nodal lines perpendicular to y and the 16 nodal lines perpendicular to x.
Since the problem is rotationally symmetric rather than square-symmetric, it makes more sense to use a coordinate system that reflects the symmetry in the problem. Hence, I'd like to solve this problem is polar coordinates (r'-).
Changing coordinates changes the Schrödinger's equation to:
Seeking a solution where the dependence on r' factors from the dependence on ("separation of variables"), we find:
The solution to the equation is easy:
Our "boundary condition" is that if we go round the origin exactly once, the value of the wavefunction is exactly the same, i.e., the wavefunction must be the same at and +2:
Hence m must be an integer.
This leaves us with the r' differential equation (given below in three equivalent forms):
We have to work harder to solve this differential equation! Start by seeing how the the equation must work for large r'. The only term that has a chance of matching the ever growing r'2 is the two-derivative term. Just as in the 1-d oscillator, an approximate cancellation requires exp(-½r'2) behavior.
For small r' only the derivative terms have a chance to match the ever growing m2/r'2 term. If we try a power-law solution r'n
we find n=|m|.
Factoring out all the required behavior for large and small r', we hope to find a simple function (polynomial with luck) G(r') that contains the behavior for intermediate r'.
=r'|m| exp(-½r'2) G(r')
So the next step is to rewrite Schrödinger's equation for G(r').
So:
(In the above we've used m where we actually mean |m|.)
If we now try to write G as a polynomial:
we can plug the polynomial form into the differential equation:
The result is a two term recursion relation (i.e., the result has just two as so, for example, given a0 we can calculate a2, from which we can calculate a4, etc. until we're done.)
Note that the the recursion relation connects even k to even k. Since a0 may not be zero (as if it were we'd factor out r', thus increasing the r'm term to r'm+1), the polynomials will have only even terms. Thus we write our even k=2i, where i is an integer.
It turns out that the polynomial must end if the wavefunction is to be normalizable. One can show that the non-terminating sum gets at least as big as exp(+r'2), so instead of going to zero for large r' gets big like: exp(+r'2/2). The only way the sum can end is if the numerator of the recursion relation is zero, i.e., if
Thus if E'=2(2nr+|m|+1) then ai+1=0 for i=nr (and hence all further as are zero) and the highest power of r'2 in the polynomial is nr.
Note that if we had tried for a polynomial solution for itself (rather than factoring out the r'm exp(-r'2/2) first), we would get an (unsolvable) three term recursion relation.
The polynomials we have been calling G are Laguerre polynomials (see for example, Abramowitz & Stegun, §22 p. 771 or Szegö Ch.V).
Here is the overall solution:
Here is what the probability density looks like for nr=10, m=0 (again, the bright areas are places where the particle is likely to be found, the dark where it is unlikely to be found)
Note the solution is circle-symmetric and has 10 nodal circles.
Here is what the probability density looks like for nr=5, m=10 (all of these density-plot solutions have E'=42)
Note the solution is circle-symmetric and has 5 nodal circles, and that the particle is unlikely to be found exactly at the origin.
Note that in producing these circle-symmetric solutions, we have produced complex rather than real wavefunctions.... that factor of exp(im). This factor cancels out of our * probability density (so the probability density is circle-symmetric), but if we were to look at just the real part of we would find oscillation as a function of . Oscillation indicates momentum, so these solutions have angular momentum...in fact m angular momentum in the "z" direction (i.e., perpendicular to the plane).
Because the angular dependence is "simple" we can usefully plot the wavefunction just as a function of r'. Here are "stacked wavefunction" plots for m=0,1,3:
The red line is the classical "effective potential"= m2/r'2+r'2 which includes the "centrifugal barrier" m2/r'2 for a particle with z angular momentum m. Because of the centrifugal barrier non-zero m wavefunctions are excluded from the origin. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://physics.csbsju.edu/QM/shm.09.html", "fetch_time": 1618595432000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00234.warc.gz", "warc_record_offset": 76172232, "warc_record_length": 3537, "token_count": 1336, "char_count": 5400, "score": 3.71875, "int_score": 4, "crawl": "CC-MAIN-2021-17", "snapshot_type": "latest", "language": "en", "language_score": 0.9299441576004028, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00018-of-00064.parquet"} |
# Question #da2fb
Oct 31, 2016
$2. 9$
$4. 10$
#### Explanation:
I'll do questions $2$ and $4$.
$2.$ We need to factor the numbers within the square roots.
$\implies 5 \sqrt{4 \times 3} - 2 \sqrt{25 \times 3} + 3 \sqrt{9 \times 3}$
$\implies 5 \left(2\right) \sqrt{3} - 2 \left(5\right) \sqrt{3} + 3 \left(3\right) \sqrt{3}$
$\implies 10 \sqrt{3} - 10 \sqrt{3} + 9 \sqrt{3}$
$\implies 9 \sqrt{3}$
The numerical coefficient is $9$.
$4.$ This is a trinomial factoring problem. When we want to factor a trinomial of the form $y = a {x}^{2} + b x + c , a \ne 1 , 0$, we need to find two numbers that multiply to $a c$ and that add to $b$.
$\implies 10 {r}^{2} - 8 r + 35 r - 28$
$\implies 2 r \left(5 r - 4\right) + 7 \left(5 r - 4\right)$
$\implies \left(2 r + 7\right) \left(5 r - 4\right)$
The value of $a + b + c + d$ is $2 + 7 + 5 - 4 = 10$
Hopefully this helps! | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 19, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/581778cab72cff294b7da2fb", "fetch_time": 1575728362000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-51/segments/1575540499439.6/warc/CC-MAIN-20191207132817-20191207160817-00304.warc.gz", "warc_record_offset": 551852463, "warc_record_length": 6072, "token_count": 374, "char_count": 878, "score": 4.53125, "int_score": 5, "crawl": "CC-MAIN-2019-51", "snapshot_type": "latest", "language": "en", "language_score": 0.5876309275627136, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
# Project Euler # Triangular, pentagonal and hexagonal in Python
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
• Triangle: $$\T_n=\frac{n(n+1)}{2}\$$ 1, 3, 6, 10, 15, …
• Pentagonal: $$\P_n=\frac{n(3n−1)}{2}\$$ 1, 5, 12, 22, 35, …
• Hexagonal: $$\H_n=n(2n−1)\$$ 1, 6, 15, 28, 45, …
It can be verified that $$\T_{285} = P_{165} = H_{143} = 40755\$$.
Find the next triangle number that is also pentagonal and hexagonal.
from time import time
def tri_pent_hex(n):
"""yields triangles that are also pentagons and hexagons in range n."""
triangles = set((num * (num + 1)) // 2 for num in range(1, n))
pentagons = set((num * (3 * num - 1)) // 2 for num in range(1, n))
hexagons = set(num * (2 * num - 1) for num in range(1, n))
for triangle in triangles:
if triangle in pentagons and triangle in hexagons:
yield triangle
if __name__ == '__main__':
start_time = time()
print(max(tri_pent_hex(1000000)))
print(f'Time: {time() - start_time} seconds.')
Instead of generating 3 million numbers and store them in memory, you only need to generate 3 of them at a time (one triangular, one pentagonal and one hexagonal) and compare them. The lowest should be advanced to the next of its kind and the processus repeated. That way you only generate the amount of numbers you need to achieve your goal and you don't put that much pressure on your memory.
You can use 3 simples functions to generate each kind of number:
def triangular_numbers():
n = 0
while True:
n += 1
yield (n * (n + 1)) // 2
def pentagonal_numbers():
n = 0
while True:
n += 1
yield (n * (3 * n - 1)) // 2
def hexagonal_numbers():
n = 0
while True:
n += 1
yield (n * (2 * n - 1))
And combine them to generate each number that are the 3 at the same time:
def triangular_pentagonal_and_hexagonal_numbers():
"""yields triangles that are also pentagons and hexagons."""
triangles = triangular_numbers()
pentagons = pentagonal_numbers()
hexagons = hexagonal_numbers()
t = next(triangles)
p = next(pentagons)
h = next(hexagons)
while True:
if t == p == h:
yield t
m = min(t, p, h)
if m == t:
t = next(triangles)
elif m == p:
p = next(pentagons)
else:
h = next(hexagons)
You now only need to ask this function for the number you’re after:
def main():
numbers = triangular_pentagonal_and_hexagonal_numbers()
one = next(numbers)
assert one == 1
given = next(numbers)
assert given == 40755
return next(numbers)
if __name__ == '__main__':
print(main())
Now that it works, you can simplify the first 3 functions using itertools.count:
def triangular_numbers():
yield from ((n * (n + 1)) // 2 for n in count(start=1))
def pentagonal_numbers():
yield from ((n * (3 * n - 1)) // 2 for n in count(start=1))
def hexagonal_numbers():
yield from (n * (2 * n - 1) for n in count(start=1))
$python emadboctor.py 1533776805 Time: 1.0859220027923584 seconds. Mine is around 20 times faster: $ python -m timeit 'import tri_pen_hexa; tri_pen_hexa.main()'
5 loops, best of 5: 50.4 msec per loop
• @ Mathias that approach you suggested, I did it before this one but i failed to go any further with it because I did not think about next-ing the trio, i made 3 functions with 3 generators and then i thought I need all numbers at once in a container for comparison because i don't know for what value of triangle will == both other numbers. Anyway i'll try your approach, thanks for the feedback.
– user203258
Jul 22, 2019 at 11:13
• @ Mathias There is another problem I applied this mechanism it's called 'pentagon numbers' or something similar. You might check it if you want and tell me what you think.
– user203258
Jul 22, 2019 at 11:16
The way this was written, it was meant to steer you in the direction you did.
Find the next triangle number that is also pentagonal and hexagonal.
You loop the triangles to find matches in the others:
for triangle in triangles:
if triangle in pentagons and triangle in hexagons:
yield triangle
But if you think about it, there are more triangles than pentagons and more pentagons than hexagons. And since you need to find occurences where all 3 types match, wouldn't it be better to loop the hexagons, and find a match on the others from there?
My point is:
Don't let the way a question is asked influence the way you implement an algorithm.
Disclaimer: This is not a proper review. I just want to give a taste of how the Project Euler problems shall be addressed.
Given a number $$\x\$$, it is triangular if there exist $$\n\$$ such that
$$\\dfrac{n(3n-1)}{2} = x\$$
Solving for $$\n\$$, gives $$\n = \dfrac{1 + \sqrt{1 + 24x}}{6}\$$
Similarly, for it to be pentagonal, there must be $$\k\$$ such that
$$\\dfrac{k(k+1)}{2} = k\$$
Solving for $$\k\$$ gives $$\k = \dfrac{-1 + \sqrt{1 + 8k}}{2}\$$
It means that $$\1 + 8x\$$ and $$\1 + 24x\$$ must simultaneously be perfect squares. We may stop here. Generate hexagonal numbers and test them against a condition above. It is already better than the solution above (we only need to generate one sequence vs three), but just by a linear factor; besides, testing for something being a perfect square does not feel comfy. Let's continue.
We are after integer $$\u\$$ and $$\v\$$ such that $$\1 + 8x = u^2\$$ and $$\1 + 24x = v^2\$$. Excluding $$\x\$$ results in $$\v^2 - 3u^2 = -2\$$. This is a variation of a famous Pell's equation, and it is where a real efficiency comes from: the solutions of Pell's equation grow exponentially fast.
Now, solve the Pell equation (it is trivial) for $$\u_i, v_i\$$. Recreate $$\x_i\$$. Test it to be a hexagonal number (yet another quadratic).
• Unless I am mistaken, you mixed up the formulas for triangular and pentagonal numbers. Note also that $1 + 24x$ being a perfect square is a necessary, but not a sufficient condition for a pentagonal number (example: $x = 15$). Jul 23, 2019 at 7:00 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://codereview.stackexchange.com/questions/224656/project-euler-triangular-pentagonal-and-hexagonal-in-python", "fetch_time": 1657034591000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00008.warc.gz", "warc_record_offset": 216530926, "warc_record_length": 68175, "token_count": 1699, "char_count": 5829, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2022-27", "snapshot_type": "latest", "language": "en", "language_score": 0.8168730139732361, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00044-of-00064.parquet"} |
What are the factors of 243?
243 is a multiple of 3 and 3 is the only prime number that is a factor of 243, that's to say 3 is the unique prime factor of 243. Since 243 is not a prime number, 243 has more than two factors. 243 is a composite number with 6 factors, here are the 6 factors of 243: 1; 3; 9; 27; 81; 243.
A factor of 243 is a whole number that we can multiply by another whole number to retrieve 243.
Example: 27 is a multiple of 243 because we can retrieve 243 by multiplying 27 with another positive integer that is equal to 9. 27*9=243. Since 27 multiplied by 9 gives 243 then 27 and 9 are both factors of 243.
Remark: If we divide 243 by one of its factors we get as quotient another factor of 243.
Example: 81 is a factor of 243, so 243 divided by 81 gives as quotient 3 that is another factor of 243 and the remainder of the division is equal to zero.
Here are all the multiplications of two integers that the product is equal to 243: 1*243=243; 3*81=243; 9*27=243.
The prime factorization of 243 is: 3^5.
243 is not a multiple of 2 then 243 is an odd number.
Factors of 243:
Justify why 1; 3; 9; 27; 81 and 243 are factors of 243?
Note: A factor of 243 is a whole number that can divide 243 by giving an integer quotient and a remainder that is equal to zero or it's a whole number that we can multiply with another whole number to get 243 as product.
1 is a factor of 243 because 243 divided by 1 is equal to 243 and the remainder of the division is equal to zero.
3 is a factor of 243 because 243 divided by 3 gives 81 as quotient and the remainder of the division is equal to zero.
9 is a factor of 243 because 243 divided by 9 gives 27 as quotient and the remainder of the division is equal to zero.
27 is a factor of 243 because the multiplication of 27 and 9 gives 243 as product.
81 is a factor of 243 because the multiplication of 81 by 3 gives 243 as product.
243 is a factor of 243 because the multiplication of 243 by 1 gives 243 as product.
Multiples of 243:
Justify if the following natural numbers are multiples of 243: 234; 500; 1215; 486; 729; 2430; 1701; 3159; 3162; 5346; 5103; 5113; 7290; 1944; 1948; 3645; 4131; 2673; 3402; 3408; 8019.
Note: A positive integer is a multiple of 243 if its division by 243 gives a whole number as quotient and a remainder that is equal to zero.
If a natural number is a multiple of 243 then 243 is its factor.
243 is an odd number so 243 can have even numbers or odd numbers as multiples.
If we divide 243 by 243 we get an integer quotient that is equal to 1 and the remainder of the division is equal to zero so 243 is a multiple of 243 (243 and 1 are factors of 243).
500 is not a multiple of 243 because 500 divided by 243 gives 2 as quotient and the remainder of the division is different from zero so 243 is not a factor of 500.
By dividing 1215 by 243 we get 5 as quotient and the remainder of that division is equal to zero so 1215 is a multiple of 243 and 5 (243 and 5 are factors of 1215).
486 is divisible by 243 because 486 divided by 243 gives 2 as quotient and zero remainder so 243 is a factor or divisor of 486.
729 is the product of the multiplication of 243 by 3 so 729 is a common multiple of 243 and 3 (243 and 3 are factors of 729).
243 is a factor of 2430 because the division of 2430 by 243 gives 10 as quotient and the remainder of the division is equal to zero so 2430 is a multiple of 243.
By multiplying 243 by 7 we get 1701 as product so 1701 is a common multiple of 243 and 7 (243 and 7 are divisors or factors of 1701).
3159 is a multiple of 243 and 243 is a factor of 3159 because the quotient of the division of 3159 by 243 is equal to 13 and the remainder of that division is equal to zero.
243 is not a factor of 3162 because 3162 is not divisible by 243 so 3162 is not a multiple of 243.
5346 divided by 243 gives 22 as quotient and a remainder that is equal to zero so 5346 is a common multiple of 243 and 22 (243 and 22 are factors of 5346).
The multiplication of 243 by 21 gives 5103 as product so 5103 is a commun multiple of 243 and 21 (243 and 21 are divisors of 5103).
5113 is not a factor of 243 because the division of 5113 by 243 gives 21 as quotient and a remainder that is not equal to zero.
If we multiply 243 by 30 we get 7290 so 7290 is a commun multiple of 243 and 30 (243 and 30 are factors of 7290).
1944 divided by 243 is equal to 8 and the remainder of the division is equal to zero so 1944 is a multiple of 243 and 243 is a factor of 1944.
The division of 1948 by 243 gives 8 as quotient and a remainder that is equal to 4 so 1948 is not a multiple of 243 and 243 is not a factor of 1948.
3645 is a common multiple of 243 and 15 because the multiplication of 243 and 15 gives 3645 as product so 243 and 15 are factors of 3645.
By multiplying 243 by 17 we get 4131 as product so 4131 is a multiple of 243 and 17 (243 and 17 are factors of 4131).
By dividing 2673 by 243 we get 11 as quotient and zero remainder so 2673 is a multiple of 243 and 11 (243 and 11 are factors of 2673).
3402 is divisible by 243 because the division of 3402 by 243 gives 14 as quotient and a remainder that is equal to zero so 3402 is a common multiple of 243 and 14 (243 and 14 are factors of 3402).
3408 is not divisible by 243 because we cannot find a positive integer that we can multiply by 243 to get 3408 so 243 is not a factor of 3408 and 3408 is not a multiple of 243.
8019 divided by 243 gives 33 as quotient and the remainder of the division is equal to zero so 8019 is a multiple of 243 and 243 is a factor of 8019.
Arithmetic and algebra expressions:
Positive integer: Positive integers or natural numbers are numbers that are written only by using the combination of the ten existing digits (0; 1; 2; 3; 4; 5; 6; 7; 8; 9). The natural numbers are infinite, we cannot list them all.
The first number that is a natural number is 0.
Rational number: Rational numbers are numbers we can write them as fractions that the numerators are relative integers and the denominators are whole numbers different from zero.
Fraction: A fraction is a division where the dividend is called the numerator and the divisor is called the denominator.
Factors : A factor of a whole number is a positive integer that can divide that whole number, each whole number has at least two factors.
Common factors: If a natural number divides two numbers by giving an integer quotient and zero remainder then that natural number is a common factor or common divisor of the two numbers.
Prime factors: A prime factor of a whole number is its factor that is a prime number.
Composite numbers: a composite number is a natural number that is not a prime number.
Even numbers: An even number is a positive integer that is divisible by 2.
Odd numbers: An odd number is a positive integer that is not divisible by 2.
Decimals : decimals are all numbers that can be written with a point or not, all natural number are decimals but all decimal numbers are not natural numbers.
Substraction : a subtraction is one of the four arithmetic operations that allows you to tell the difference between two numbers.
Divisibility rules : a rule of divisibility helps to know if a natural number is divisible by another natural number or not.
find the lcm and the gcf of two integers
find the first n multiples of an integer
find the multiples of an integer inside an interval
retrieve prime numbers inside an interval
retrieve the irreducible value of a fraction | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://chrislink.net/codeSource/factors-of-243.php", "fetch_time": 1718403273000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00450.warc.gz", "warc_record_offset": 143079043, "warc_record_length": 4313, "token_count": 2006, "char_count": 7438, "score": 4.625, "int_score": 5, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.9453163743019104, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00035-of-00064.parquet"} |
## Thursday, 23 November 2017
### Mathematics - Chapter - 5 (Logical Reasoning)
Coding & Decoding
Coding is a method of transmitting a message between the sender and the receiver that no third person can understand it. The coding and decoding one’s ability of deciphering the rule and breaking the code to decipher the message will be tested to know. Some words are stand for some another words which is known code of the word. Process of replacing some word by code word is known as Coding. Decoding is reverse of Coding. Coding and Decoding test is set up to judge candidates’ ability to decipher the rule.
Approach
1. Observe alphabets or numbers given in the code keenly.
2. Find the sequence it follows whether it is ascending or descending.
3. Detect the rule in which the alphabets/numbers/words follow.
4. Fill the appropriate letter/number/word in the blank given.
Letter coding
Alphabets in a word are replaced by other alphabets according to a specific rule to know its code. So the common rule should be detected first. Some examples are given below.
1. ‘ZYXW’ as coded as ‘ABCD’ then ‘STUV’ is coded as...........
Answer: Z – A, Y– B, X – C, W – D
V – E, U – F, T – G, S – H
STUV = HGFE
Rule= front alphabet= back alphabet.
2. ‘bcd’ is coded as ‘def’ then ‘True’ is coded as..........
Answer: b – d (+2)
c – e (+2)
d – f (+2)
+2 letters are considered in this code.
True – Vtwg
3. ‘Hyderabad’ is coded as ‘Ixedszcze’ then ‘Chennai’ is coded as..............
Answer: H – I (+1)
Y – X (1–)
D – E (1+)
E – D (1-)
R – S (1+)
A – Z (1–)
B – C (1+)
A – Z (1– )
D – E (1+)
Here if we observe alternatively the letter increasing and one letter decreasing.
Chennai =dgfmozj
Number Coding
In this each alphabets or words are assigned to the numeric values we should observe the given letters and the assigned values and use the same rule to find the value to of given code. Some examples are given below:
1. Apple is coded as 25563, Rung is coded as 7148. Then purple is coded as
PURPLE – 517563
2. In a language A is coded as 1, B is coded as 2, ….. then FACE is coded as
Then FACE = 6135
3. PUSH is coded as 1234, ROUGH is coded as 65274. Then SOUP is coded as
SOUP = 3521
Substitution
In this section an object names are substituted with different object names. We should carefully trace the substitution and answer given question. Some examples are given below.
1. ‘book’ is coded as ‘pencil’, ‘pencil’ is coded as ‘mirror’, ‘mirror is coded as ‘book’. Then what is useful to write on a paper?
Answer: Pencil is coded as mirror
2. ‘Man’ is coded as ‘woman’, woman is coded as ‘girl’, and ‘girl’ is coded as boy ’,‘ boy is coded as ‘worker’ then 6 years female is known as?
Answer: 6 years female = girl, but ‘girl’ is coded as ‘boy’.
3. ‘Reds’ are ‘blues’, ‘blues’ are ‘whites’, ‘whites’ are ‘yellows’, ‘yellows’ are ‘oranges’, ‘oranges’ are ‘pinks’, then what is the colour of the sky?
Answer: Sky is blue, but blues are whites
Exercise questions
4. In a certain code, COMPUTER is written as RFUVQNPC. How is MEDICINE written in the same code ?
a) MFEDJJOE b) EOJDEJFM
c) MFEJDJOE d)EOJDJEFM
Ans: Option d
The letters of the word are written in reverse order and expect the first and the last letter all other letters are
move one step forward
5. In a code language, A is written as B, B is written as C, C is written as D and so on, then how will SMART be written in that code language ?
a) TLBSU b)SHBSU
c)TNBSU d)SNBRU
Ans: Option c
The letters are coded by moving them 1 step forward.
6. In a certain code , RIPPLE is written as 613382 and LIFE is written as 8192. How is PILLER written in that code?
a) 318826 b)776655
c)786543 d)156724
Ans: Option a:
Word : R I P P L E L I F E P I L L E R
Code : 6 1 3 3 8 2 8 1 9 2 3 1 8 8 2 6
7. In a certain code FLOWER is coded as 36 and SUNFLOWER is coded as 81, then how to code FOLLOWS?
a) 42 b)49
c)63 d) 36
Ans: Option b
The word FLOWER has 6 letters . 62 is 36
The word SUNFLOWER has 9 letters. 92 is 81
Like FOLLOWS has 7 letters. So 72 is 49
8. In a certain code ,'il be pee' means 'roses are blue','sik hee' means 'red flowers' and 'pee mit hee' means 'flowers are vegetables', How is 'red' written in that code?
a) hee
b) sik
c) be
d) cannot be determined
e) none
Ans: Option b
Code Sentence
Il be pee roses are blue
Sik hee red flowers
Pee mit hee flowers are vegetables
In II and III code ‘hee’ stands for ‘flowers’. So ‘sik’ stands for ‘red’
9. In a certain code language : ‘dugo hui mul zo’ stans for ‘work is very hard’ ‘hui dugo ba ki’ for ‘Bingo is very smart’; ‘nano mul dugo’ for ‘cake is hard’; and ‘mul ki gu’ for ‘smart and hard’ Which of the following word stand for Bingo ?
a) Jalu
b) Dugo
c) Ki
d) Ba
Ans: Option d
Code Sentence
1.dugo hui mul zo work is very hard
2.hui dugo ba ki bingo is very smart
3.nano mul dugo cake is hard
4.mul ki gu smart and hard
From second code and its sentence neither ‘ba’ is repeated nor is ‘Bingo.
10. If rain is called water, water is called air, air is called cloud, cloud is called sky, sky is called sea, sea is called road, where do the aeroplanes fly ?
a) Water
c) Sea
d) Cloud
Ans: Option c
Aeroplanes fly in sky and as per given codes sky is sea
11. If Orange is called Lemon, Lemon is called Flower, Flower is called Fish, Fish is called Tail and Tail is called Pen, what is Rose ?
a) Pen
b) Lemon
c) Flower
d) Fish
Ans: Option d
Rose is a flower and as per given codes flower is fish.
12. In a certain code language \$#* means ‘Shirt is clean’, @ D# means ‘Clean and neat’ and @ ? means ‘neat boy’, then what is the code for ‘and’ in that language.
a) #
b) D
c) @
Ans: Option b
Code sentence
\$#* ‘Shirt is clean’,
@ D# ‘Clean and neat’
@ ? ‘neat boy’
Here # stands for clean and @ stands for neat. D stands for ‘and’
13. If A stands for +, B stands for -, C stands for x, what is the value of (10C4)(A) (4C4)B6 ?
a) 60
b) 50
c) 56
d) 46
Ans: Option b
(10C4)(A) (4C4)B6 = (10 * 4) + (4*4) –6 = 50
Mathematical Operations
This section deals with questions on simple mathematical operations. Here, the four fundamental operations -- addition, subtraction, multiplication and division and also statements such as 'less than', 'greater than', 'equal to', 'not equal to', etc. are represented by symbols, different from the usual ones. The questions involving these operations are set using artificial symbols. The candidate has to substitute the real signs and solve the questions accordingly, to get the answer.
1. E.g. In following alphabet series, one term missing as shown by question mark. Choose missing term from options.
Z, U, Q, ?, L
Option:
A. I
B. K
C. M
D. N
Answer: D . N
Justification:
The first, second, third,... letters of the series are respectively moved one, two, three,... steps forward to obtain the successive terms.
2. In following alphabet series , one term missing as shown by question mark . Choose missing term from options.
A, C, F, H, ?, M
Option:
A. L
B. K
C. J
D. I
Answer: B . K
Justification:
The letters are alternately moved two and three steps forward to obtain the successive terms.
3. In following alphabet series, one term missing as shown by question mark. Choose missing term from options.
A, Z, X, B, V, T, C, R, ?, ?
Option:
A. P, D
B. E, O
C. Q, E
D. O, Q
E. Q, O
Answer: A. P, D
Justification:
The first, fourth and seventh letters are in alphabetical order.
So, tenth letter would be the letter after C i.e. D.
Also, the second and third letters are alternate and in reverse order and so are the fifth and sixth letters and the eighth and ninth letters.
4. In following alphabet series, one term missing as shown by question mark . Choose missing term from options.
R, M,?, F, D, ?
Option:
A. C, B
B. J, H
C. B, H
D. H, C
E. I, C
Answer: E. I, C
Justification:
Letter is in reverse order in which from the last 0, 1, 2, 3 and 4 letters are missing between two consecutive letters.
Number Coding
In Number coding type of questions, either numerical code values are assigned to a word or alphabetical code letters are assigned to the numbers. The candidate is required to analyse the code as per the directions.
1. If ROSE is coded as 6821, CHAIR is coded as 73456 and PREACH is coded as 961473, what will be the code for SEARCH?
Option:
A. 246173
B. 214673
C. 214763
D. 216473
Justifications:
The alphabets are coded as shown:
R O S E C H A I P 6 8 2 1 7 3 4 5 9
So, in SEARCH,
S ia coded as 2,
E as 1,
A as 4,
R as 6,
C as 7,
H as 3.
Thus, the code for SEARCH is 214673.
2. If the letters in PRABA are coded as 27595, and THILAK are coded as 368451, how can BHARATHI be coded?
Option:
A. 37536689
B. 57686535
C. 96575368
D. 96855368
Answer: C . 96575368
Justification:
The alphabets are coded as shown:
P R A B T H I L K 2 7 5 9 3 6 8 4 1
So, B is coded as 9,
H as 6,
A as 5,
R as 7,
T as 3 and
I as 8.
Thus, the code for BHARATHI is 96575368.
Logical Venn Diagram
Introduction
The main aim of this section is to test your ability about the relation between some items of a group by diagrams. In these questions some figures of circles and some words are given. You have to choose a figure which represents the given words.
Some critical examples are given below:
Example 1:
If all the words are of different groups, then they will be shown by the diagram as given below.
Dog, Cow, Horse
All these three are animals but of different groups, there is no relation between them. Hence they will be represented by three different circles.
Example 2:
If the first word is related to second word and second word is related to third word. Then they will be shown by diagram as given below.
Unit, Tens, Hundreds
Ten units together make one Tens or in one tens, whole unit is available and ten tens together make one hundreds.
Example 3:
If two different items are completely related to third item, they will be shown as below.
Pen, Pencil, Stationery
Example 4:
If there is some relation between two items and these two items are completely related to a third item they will be shown as given below.
Women, Sisters, Mothers
Some sisters may be mothers and vice-versa. Similarly some mothers may not be sisters and vice-versa. But all the sisters and all the mothers belong to women group.
Example 5:
Two items are related to a third item to some extent but not completely and first two items totally different.
Students, Boys, Girls
The boys and girls are different items while some boys may be students. Similarly among girls some may be students.
Example 6:
All the three items are related to one another but to some extent not completely.
Boys, Students, Athletes
Some boys may be students and vice-versa. Similarly some boys may be athletes and vice-versa. Some students may be athletes and vice-versa.
Example 7:
Two items are related to each other completely and third item is entirely different from first two.
Lions, Carnivorous, Cows
All the lions are carnivorous but no cow is lion or carnivorous.
Example 8:
First item is completely related to second and third item is partially related to first and second item.
Dogs, Animals, Flesh-eaters
All the dogs are belonging to animals but some dogs are flesh eater but not all.
Example 9:
First item is partially related to second but third is entirely different from the first two.
Dogs, Flesh-eaters, Cows
Some dogs are flesh-eaters but not all while any dog or any flesh-eater cannot be cow.
1. In a country three persons A, B and C live. They are three different persons. This information can be represented as:
Here, we can see that A, B and C are different elements so they’ve represented by different circles.
2. If we were to represent information in which two elements are intermingled while the third one is different we’ll do that a bit differently.
For example: Hindu, Indian and Parrot. Now, logically we know that Some Hindus are Indian (as some Hindu might be living abroad and be Australian or any other country’s citizen) and also, no Parrot is Hindu (as animals have no religion) also no Parrot is Indian (as no animal has ethnicity). This information of Hindu, Indian, Parrot can be represented as follows:
Here, the shaded area shows those Hindus who are Indians at the same time. Parrot are represented in a different circle.
3. Suppose, we need to convey this: Dog, Animal and Cow.
Now we know that all dogs are animals (clearly no dog is human) so the circle of ‘dog’ will have to be completely surrounded by circle of animal though circle of animal can have some spare space aside from dog as dog isn’t the only animal. Similarly, all ‘cows’ are animals so the circle of ‘cow’ will have to be completely surrounded by circle of animal though circle of animal can have some spare space aside from cow as cow isn’t the only animal. This information can be represented as:
Here, we can see that ‘Animal’ has been represented by a big circle which encompasses the circles for both ‘cow’ and ‘dog. Notice, the circle for ‘animal’ has some spare space as this can contain other types of animals because ‘cow’ and ‘dog’ aren’t the only type of animal.
Types of questions asked in Competition Exam
Let’s have a look at the type of questions asked specifically in SSC exam. There are basically two types of questions:
Finding relationship
To solve these kinds of questions, we need to have a strong grip on common relationships that exist in the world around us. Like to define the relationship between Catholics & Christian we need to know that Catholics are the type of Christians hence we can easily conclude that all Catholics are Christian but some Christians will not be Catholics as they will be the other type of Christians. This information can be represented as:
A typical question might look like this: Dean, Painter, Singer.
We live in a diverse world where people can be multi-talented also people possess just one talent so this info can be represented by 7 categories of people:
a) Who are only Dean
b) Who are only Painter
c) Who are only Singer
d) Who are both Dean & Painter
e) Who are both Painter & Singer
f) Who are both Singer & Dean
g) Who are all Dean, Painter & Singer
This information can be represented by Venn - diagram as follow: (for reader’s convenience, the different regions have been labeled as named above but in exams, questions aren’t marked this way) | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "bsau1.blogspot.com", "fetch_time": 1558582775000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-22/segments/1558232257002.33/warc/CC-MAIN-20190523023545-20190523045545-00471.warc.gz", "warc_record_offset": 31117573, "warc_record_length": 19511, "token_count": 3964, "char_count": 14510, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2019-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8822889924049377, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00000-of-00064.parquet"} |
# Does this limit exist?
1. Aug 12, 2009
### nietzsche
1. The problem statement, all variables and given/known data
A population develops according to logistic equation:
$$\frac{dy}{dt} = y(0.5-0.01y)$$
Determine the following:
$$\lim_{t\rightarrow\infty} y(t)$$
3. The attempt at a solution
By finding an equilibrium solution to the differential equation, we see that dy/dt = 0 when y = 0 or y = 50.
But does the limit exist? What if y = 0 initially?
2. Aug 12, 2009
### Cyosis
You should try to solve the differential equation first.
3. Aug 12, 2009
### nietzsche
But do you have to do the differential to determine the limit? Because we can see that if y=0 and t goes to infinity, y will still equal 0. and if y does not equal 0 initially, then y will go to 50.
4. Aug 12, 2009
### Cyosis
You have picked two fixed values for y which indeed satisfy the differential equation, however these values aren't the only solutions. The more interesting solutions are the ones that still depend on t.
5. Aug 12, 2009
### nietzsche
I'm sorry, I don't understand.
I solved the differential equation:
y(t) = 50 / (1+Ae^(-0.5t)
where A = (50-y(0)) / y(0)
so isn't it still dependent on the value of y(0)?
Last edited: Aug 12, 2009
6. Aug 12, 2009
### nietzsche
OH. so now if we take the limit, as t approaches infinity, then A will go to 0, no matter what. is that correct?
7. Aug 12, 2009
### Fightfish
Yea, more correctly, Ae^(-0.5t) goes to zero
8. Aug 12, 2009
### nietzsche
ah okay, thanks.
i still don't get the concept though. how can the limit of the equation be 50 if the initial population is 0?
9. Aug 12, 2009
### Fightfish
Since the equation is modeling the development of a population, that would imply that y(0) cannot be zero; otherwise you have no population to begin with.
10. Aug 12, 2009
### nietzsche
haha, i suppose. thanks very much cyosis and fightfish.
11. Aug 12, 2009
### Staff: Mentor
No, it's not. As fightfish said, Ae-.5t goes to 0 as t approaches infinity. This, however, does not imply that A is 0 or is approaching zero. A is a constant in the problem.
12. Aug 12, 2009
### nietzsche
sorry, i'll be more specific next time...
13. Aug 12, 2009
### g_edgar
>> how can the limit of the equation be 50 if the initial population is 0?
You need to object to nietsche, who said the solution is:
y(t) = 50 / (1+Ae^(-0.5t)
that is NOT the solution if y(0)=0... | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/does-this-limit-exist.330320/", "fetch_time": 1508669381000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00770.warc.gz", "warc_record_offset": 997608072, "warc_record_length": 16460, "token_count": 731, "char_count": 2425, "score": 3.96875, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "longest", "language": "en", "language_score": 0.8891645669937134, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00057-of-00064.parquet"} |
# All the letters of the word 'EAMCET' are arranged in different possible ways.
Question:
All the letters of the word 'EAMCET' are arranged in different possible ways. The number of such arrangements in which two vowels are adjacent to each other, is
(a) 360
(b) 144
(c) 72
(d) 54
Solution:
Total number of words using letters of EAMCET is
$\frac{6 !}{2}=\frac{6 \times 5 \times 4 \times 3 \times 1}{2}$ (Since $\mathrm{E}$ is respected twice)
= 360
Total number of words with no two vowels together is :-
Since vowels are $E, A, E=3$ and consonant is $M_{1} C_{1} T=3$.
i.e number of ways to arrange consonant is 3!
– C – C – C –
i.e vowel can take any of 4 places
i.e number of vowels arrangement $=3 ! \times{ }^{4} P_{3} \times \frac{1}{2}(\because E$ is repeated twice $)$
i.e $3 ! \times \frac{4 !}{1 !} \times \frac{1}{2}=72$
$\therefore$ Number of words arrangement so that two vowels are adjacent $=\frac{360-72}{2}=144$ | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.esaral.com/q/all-the-letters-of-the-word-eamcet-are-arranged-in-different-possible-ways-79972", "fetch_time": 1721147705000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00619.warc.gz", "warc_record_offset": 659915984, "warc_record_length": 11918, "token_count": 307, "char_count": 948, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.7536155581474304, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00028-of-00064.parquet"} |
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# ICSE Class 9 Mathematics Revision Notes Chapter 18 - Statistics
Last updated date: 15th Jul 2024
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## Revision Notes for ICSE Class 9 Mathematics Chapter 18 - Free PDF Download
“Statistics” is the method of assortment and arrangement of data in such a manner that it can be understood and showcased. We use statistics to make our littlest decisions. To get in-depth of this chapter, one must have a good grasp of the basics of this chapter when it eventually broadens in older classes to excel.
For starters, you learn about what statistics is, the variable which means any quantity that varies and its types. Then, gradually it tells us about tabulation of data,i.e., representation of data in tabular form.
The segment proceeds on to topics like frequency and how it's distributed into different types. To understand this, students should know about class intervals and class limits, and subtopics like adjustment, class size, and class-mark. The first section comes to an end with the concepts of cumulative frequency and cumulative frequency table.
The upcoming section is about the graphical representation of data and studying it with the help of a histogram and frequency polygon. There is a slight difference between a histogram and frequency polygon. A histogram resembles a bar graph, with width definite spaces between every single data. But, in a frequency polygon, we join the upper middle points in case of making it with the help of a histogram) or without histogram (but for that, we will need to calculate the class mark and use it accordingly).
Competitive Exams after 12th Science
## Sections of Chapter 18 - Statistics
18.1 Introduction
18.2 Variable
18.3 Tabulation of data
18.4 Frequency
18.5 Frequency distribution
18.6 Types of frequency distributions
18.7 Class intervals and class limits
18.8 Cumulative frequency and cumulative frequency table
18.9 Graphical representation of data
18.10 Graphical representation of continuous frequency distribution
### List of Exercises in Chapter 18 - Statistics
Exercise 18(A) - 13 questions
Exercise 18(B) - 5 questions
### Revision Notes for ICSE Class 9 Mathematics Chapter 18 - Free PDF Downloads
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2. Which concepts in Statistics Chapter 18 are hard to deal with?
One may find it hard at first to implement the fundamentals of this chapter and calculate the values but as you and we have heard time and again, practice makes perfect! Try doing regular questions based on frequency, cumulative frequency, class intervals, and the tabular representations required. Students should also start using these concepts in real life which helps them to connect better with the concepts. Through this one is sure to get a good understanding of these concepts as well as this chapter.
3. What is the best way to prepare for Class 9 Maths Chapter 18?
Revise the concepts of Class 9 Maths individually and at the end of every topic do questions related to it, then move on to other topics and do the same, when all the topics and subtopics are covered then, in the end, take one whole practice test. This will help you to realize your mistakes and your weak concepts. Then revise again and continue the cycle of practice tests, highlight your mistakes and work on them. Slowly and steadily, this will enhance your Maths skills.
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5. How can we calculate Adjustment in Chapter 18?
For calculating the adjustment or adjustment factor, we need to follow two steps. First, we have to find the difference between the upper limit of one class and the lower limit of the class ahead. The second step is to divide the obtained value by 2. The value thus obtained is known as the adjustment factor which will be further deducted from the class limits and this is how we get actual class limits or true class limits. These are also known as class boundaries. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.vedantu.com/icse/class-9-mathematics-revision-notes-chapter-18", "fetch_time": 1721091025000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00522.warc.gz", "warc_record_offset": 935232670, "warc_record_length": 31308, "token_count": 1179, "char_count": 5718, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2024-30", "snapshot_type": "latest", "language": "en", "language_score": 0.8833989500999451, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00062-of-00064.parquet"} |
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Closed 5 year ago.
You are watching: 1,2,6,24,120
I to be playing with No Man"s Sky once I ran right into a collection of numbers and was asked what the next number would certainly be.
\$\$1, 2, 6, 24, 120\$\$
This is for a terminal assess password in the video game no man sky. The 3 options they offer are; 720, 620, 180
The following number is \$840\$. The \$n\$th ax in the sequence is the the smallest number through \$2^n\$ divisors.
Er ... The following number is \$6\$. The \$n\$th ax is the the very least factorial many of \$n\$.
No ... Wait ... It"s \$45\$. The \$n\$th term is the best fourth-power-free divisor of \$n!\$.
Hold on ... :)
Probably the price they"re looking for, though, is \$6! = 720\$. But there are lots of other justifiable answers!
After some trial and error I found that this numbers space being multiply by their matching number in the sequence.
For example:
1 x 2 = 22 x 3 = 66 x 4 = 2424 x 5 = 120Which would typical the next number in the sequence would be
120 x 6 = 720and for this reason on and also so forth.
Edit: thanks to
GEdgar in the comments because that helping me make pretty cool discovery about these numbers. The totals are also made increase of multiplying each number approximately that existing count.
For Example:
2! = 2 x 1 = 23! = 3 x 2 x 1 = 64! = 4 x 3 x 2 x 1 = 245! = 5 x 4 x 3 x 2 x 1 = 1206! = 6 x 5 x 4 x 3 x 2 x 1 = 720
The following number is 720.
The sequence is the factorials:
1 2 6 24 120 = 1! 2! 3! 4! 5!
6! = 720.
(Another way to think of that is each term is the term before times the next counting number.
See more: Which Artist Made The Concept Of Collage Into A Form Of Art In 1912?
T0 = 1; T1 = T0 * 2 = 2; T2 = T1 * 3 = 6; T3 = T2 * 4 = 24; T4 = T3 * 5 = 120; T5 = T4 * 6 = 720.
\$egingroup\$ it's yet done. You re welcome find one more answer , a little bit initial :) maybe with the amount of the number ? note also that it starts with 1 2 and ends through 120. Possibly its an possibility to concatenate and add zeroes. An excellent luck \$endgroup\$
## Not the price you're spring for? Browse various other questions tagged sequences-and-series or asking your own question.
Is there any kind of discernible pattern for the number perform \$frac73, frac54, 1, frac89, ...\$
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# Weyl method
in number theory
A method for obtaining non-trivial estimates of trigonometric sums (cf. Trigonometric sum) of the form
$$\tag{1 } S( f ) = \sum _ {1 \leq x \leq P } e ^ {2 \pi i f( x) } ,$$
where
$$f( x) = \alpha _ {n} x ^ {n} + \dots + \alpha _ {1} x$$
and $\alpha _ {n} \dots \alpha _ {1}$ are arbitrary real numbers. Developed by H. Weyl [1] to establish criteria for uniform distribution (cf. Weyl criterion).
The method may be described as follows. The sum (1) is raised to the power $\rho _ {0} = 2 ^ {n-} 1$ by successive squaring operations in order to reduce the degree of the polynomial $f( x)$. Thus, in the first stage
$$S ^ {2} ( f ) = \sum _ {\lambda _ {1} \neq 0 } \sum _ { x } e ^ {2 \pi i ( f( x+ \lambda _ {1} )- f( x)) } + O( P),$$
where the summations are performed over intervals of lengths $\ll P$; now
$$f( x+ \lambda _ {1} ) - f( x) = \ n \alpha _ {n} \lambda _ {1} x ^ {n-} 1 + \dots$$
is a polynomial of degree $n - 1$ in $x$( the symbols $O( P)$, $\ll P$ denote magnitudes of order $P$). At the $( n - 1)$- st step one obtains inner sum
$$\tag{2 } S( \alpha )= \sum _ {x= a+ 1 } ^ { {a+ } P _ {1} } e ^ {2 \pi i \alpha x } ,$$
where $P _ {1} \leq P$, $\alpha = n! \lambda _ {1} {} \dots \lambda _ {n-} 1 \alpha _ {n}$, $\lambda _ {i} \neq 0$. Sums of the form (2) are estimated using the inequality
$$| S( \alpha ) | \leq \min \left ( P _ {1} , \frac{1}{| \sin \pi \alpha | } \right ) ,$$
and the resulting estimate is:
$$\tag{3 } | S( f ) | ^ {\rho _ {0} } \leq P ^ {\rho _ {0} - 1 } +$$
$$+ P ^ {\rho _ {0} {- n + \epsilon } } \sum _ {0 < y \leq P ^ {n-} 1 } \min \left ( P _ {1} , \frac{1}{| \sin \pi y \alpha _ {n} | } \right ) .$$
The inequality (3) yields different estimates for the sum (1) in case $1/ | \sin \pi y \alpha _ {n} |$ is small in comparison to $P$. These estimates depend on the accuracy with which the coefficient $\alpha _ {n}$ of the polynomial $f( x)$ can be approximated by rational fractions.
Example. Let
$$\left | \alpha _ {n} - \frac{p}{q} \right | \leq \frac{1}{q ^ {2} } , \ ( a, q) = 1.$$
Then
$$| S( f ) | \ll P ^ {1 + \epsilon } q ^ \epsilon \left ( \frac{1}{P} + \frac{1}{q} + \frac{q}{P ^ {n} } \right ) ^ {\rho _ {0} } .$$
In particular, if
$$P \leq q \leq P ^ {n-} 1 ,$$
then
$$| S( f) | \ll P ^ {1- \rho _ {0} + \epsilon } .$$
Weyl's method gives solutions, to a first approximation, of several important problems in number theory. The estimate (3) and its corollaries were used to study the distribution modulo one of the polynomial $f( x)$. G.H. Hardy and J.E. Littlewood (1919) gave a solution of the Waring problem which was based on estimating the sums (1) by Weyl's method. They could thus estimate the values of $r= r( k)$ for which the equation
$$N = x _ {1} ^ {k} + \dots + x _ {r} ^ {k} ,$$
where $N > 0$ is an integer and $x _ {i}$ are integers, is solvable, and even gave an asymptotic formula for the number of solutions. A generalization of the estimate (3) to the case of functions $f( x)$ which are not polynomials but are in a certain sense close to polynomials, resulted in the improvement of certain theorems in the theory of the distribution of prime numbers (an estimate of the difference between two successive prime numbers and an estimate of the residual term in the asymptotic formula for the number $\pi ( N)$ of prime numbers not exceeding $N$).
The insufficient strength of the estimates obtained by Weyl's method is due to the high power $\rho _ {0}$ to which the sum $S( f )$ is raised. J. van der Corput proposed a somewhat improved method for estimating the sums (1). The Vinogradov method yields a very accurate upper bound for the integral
$$\int\limits _ { 0 } ^ { 1 } \dots \int\limits _ { 0 } ^ { 1 } | S( f ) | ^ {2b} d \alpha _ {n} \dots d \alpha _ {1}$$
already for $b \geq cn ^ {2}$( $c > 0$ a constant, $n \geq 2$). This estimate (cf. Vinogradov theorem about the average) may be used to deduce essentially new estimates of Weyl sums (1) (with reduction factor $P ^ {- \rho }$, $\rho = c _ {1} n ^ {2} \mathop{\rm ln} n$; $c _ {1} > 0$ a constant), which cannot be attained by Weyl's method.
#### References
[1] H. Weyl, "Ueber die Gleichverteilung von Zahlen mod. Eins" Math. Ann. , 77 (1916) pp. 313–352 [2] I.M. Vinogradov, "The method of trigonometric sums in the theory of numbers" , Interscience (1954) (Translated from Russian)
How to Cite This Entry:
Weyl method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Weyl_method&oldid=49204
This article was adapted from an original article by B.M. Bredikhin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://encyclopediaofmath.org/wiki/Weyl_method", "fetch_time": 1652891498000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00010.warc.gz", "warc_record_offset": 292817612, "warc_record_length": 7317, "token_count": 1588, "char_count": 4702, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2022-21", "snapshot_type": "longest", "language": "en", "language_score": 0.6683695316314697, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00050-of-00064.parquet"} |
# Year 12 Sound Problem (Involves resonance in a closed pipe).
1. Feb 26, 2009
### caitie_72
1. The problem statement, all variables and given/known data
Hi :)
This question is from a test i had today. For some stupid reason we get to finish it off tomorrow. I figured i do what everyone else will, and sort out the what i didn't understand. Luckily this was it. Anyway, the question is typed from memory so it may seem a bit odd or inaccurate.
A pipe is in a tub of water. As it is moved down (or up, can't remember!) it resonates at 288 Hz and 512 Hz. The distance between these two points is 13.9 cm (0.139 m). What is the velocity of sound in the tube?
I'm not sure of the order of frequencies occurring is. I think it was probably the lowest frequency first. I don't really need an answer, just an understanding of how to do a problem like this. All i know is that the pipe resonates at those two frequencies, and the distance between the points where it resonates is 13.9m. If you've ever seen a question like this ... then you'd probably get what i mean, hopefully.
Do I used simultaneous questions? I'm not sure.
2. Relevant equations
(w=wavelength)
v=F*w
3. The attempt at a solution
What i did first, in a fit of irrational confusion, was find the wavelength. I assumed that resonance was occurring at either nodes or antinodes, half a wavelength apart (???)
So w = 2 * 13.9.
Then i got confused, because i know the fundamental wavelength of a closed pipe is w=4L.
Then i did something stupid i knew wasn't right. Found the difference between the frequencies and then multiplied it with the wavelength. Stupid stupid, i know. Makes no sense.
I'm thinking simultaneous equations?
f=v/4L.
512=3v/4L 288=v/4L
Resonance must be occuring at first and second modes, right?
I'm just confused about what the length is! Does length, in this case, mean the length of the whole pipe or the length between the points of resonance? Does resonance occur at half wavelengths or quarter wavelengths? Don't understand!
Any help or direction at ALL is appreciated. Thanks :)
2. Feb 26, 2009
### rl.bhat
Since there is no other information, you have to assume that they are resonating in the fundamental mode.
So write 512=3v/4L 288=v/4L as 512=v/4L1,and 288=v/4L2. Write these equation in terms of L1 and L2. L1 - L2 is given. Solve for v.
3. Feb 26, 2009
### caitie_72
Thank you so much! You are seriously my favourite person ever right now.
Just checking, am i doing it right?
L1=3v/2048
L2=v/1152
0.139 = 3v/2048 - v/1152
v = 232.9 m/s
I assume so. Thanks again :)
4. Feb 26, 2009
### rl.bhat
L1=3v/2048
This is wrong. It should be L1=v/2048. Find L2-L1. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/year-12-sound-problem-involves-resonance-in-a-closed-pipe.295383/", "fetch_time": 1553293324000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-13/segments/1552912202698.22/warc/CC-MAIN-20190322220357-20190323002357-00282.warc.gz", "warc_record_offset": 878957802, "warc_record_length": 13138, "token_count": 733, "char_count": 2675, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2019-13", "snapshot_type": "latest", "language": "en", "language_score": 0.9666056036949158, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00010-of-00064.parquet"} |
Puzzle makers have been creating problems for people to solve for centuries. Some of the earliest puzzles date back to the time of the ancient Greeks and Romans. The Chinese have a ring puzzle that is thought to have been developed during the second century A.D. This was first described by Italian mathematician Girolamo Carolano (Cardan) in 1550. When the printing press was invented, complete books of mathematical and mechanical problems designed specifically for recreation were circulated. Rubiks Build It Solve It
Now you need to orient these pieces. Refer to the next picture. As you can see, the orange piece matches the orange centre. Look at the edges on your puzzle. You could have none matching, two matching or all matching. If you have all four edges matching the centres, your cross is solved. If you have none matching, perform a U move, then look around the cube again. You want to have at least two matching. If none of them match, do another U move. Repeat until you have either two or four edges matching their centres.
The Rubik's cube (sometimes misspelled rubix cube) is a mechanical 3D puzzle, invented more than 30 years ago and still considered as the best-selling toy of all times! Yet, solving the Rubik's Cube is considered a nearly-impossible task, which requires an IQ of 160... Is that really so hard? Definitely not!! Just follow this simple step by step solving guide and you'll shortly find out that you can solve the Rubik's cube as well… Let's get to work!
Constructing the Cube is a superb way to exercise those fine motor skills, visual and spatial comprehension and cognitive thinking from children. When the block is placed together, it is going to challenge the small ones to use their spatial and visual understanding as they know to spin the tiles. The block also helps kids learn about colours and fitting them. Rubix Building Products
If you're reading this, you're probably holding a cube in your hand and already feeling bad about yourself for needing to look up the solution. But don't worry! In fact, most of the “super-human-intelligence beings” (a common misconception) who have solved the cube thousands of times in their lifetimes were sitting as you are now. Whether you want to learn it to impress a girl, because your friends bet you couldn't, or just to close the book on the biggest time waste of your childhood by finally defeating it, this guide will take you through the simplest way to conquer the puzzle.
When it comes to building the Rubik’s Cube, it’s not as hard as it looks. In all actuality, it will take about fifteen minutes and the instructions are easy to follow. When it comes to placing the colored tiles, make sure you pay attention to where you’re supposed to place them, because if you snap them in the wrong place, you won’t be able to remove them. Yes, you will still be able to use the Rubik’s Cube, but you won’t be able to follow along with the instruction guide on solving the puzzle.
1 When production is initiated, the plastic pellets are transformed into Rubik's cube parts through injection molding. In this process, the pellets are put into the hopper of an injection molding machine. They are melted when they are passed through a hydraulically controlled screw. As the screw turns, the melted plastic is shuttled through a nozzle and physically forced, or injected, into the mold. Just prior to the arrival of the molten plastic, the two halves of the mold are brought together to create a cavity that has the identical shape of the Rubik's cube part. This could be an edge, a corner, or the center piece. Inside the mold, the plastic is held under pressure for a specific amount of time and then allowed to cool. While cooling, the plastic hardens inside the mold. After enough time passes, the mold halves are opened and the cube pieces are ejected. The mold then closes again and the process begins again. Each time the machine moulds a set of parts is one cycle of the machine. The Rubik's cube cycle time is around 20 seconds. Rubiks Build It Solve It Instructions
If there are no more edges left on the top layer, then they are probably either inserted in the right place but flipped, or inserted in the wrong place. To get an edge out of somewhere it shouldn't be, just insert one of the yellow edges into that slot. This should get the edge out and on the top layer, ready for you to use the above instructions to insert correctly.
# Just because this kit gives you a behind-the-scenes look as to how a Rubik's Cube is made along with tips for how to solve it doesn't mean that you'll be solving it like a pro within seconds. Even the solution booklet itself says that the first step will take practice and trial by error. So this is definitely going to be more fun for kids or adults who enjoy the puzzlement of a Rubik's Cube and have the patience to build it and practice using the solving tips. But once you finally do solve it, you'll be pretty proud of yourself, and your friends and family will be impressed.
Puzzle makers have been creating problems for people to solve for centuries. Some of the earliest puzzles date back to the time of the ancient Greeks and Romans. The Chinese have a ring puzzle that is thought to have been developed during the second century A.D. This was first described by Italian mathematician Girolamo Carolano (Cardan) in 1550. When the printing press was invented, complete books of mathematical and mechanical problems designed specifically for recreation were circulated. Rubix Build | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://newtoyslist.com/rubik/how-to-make-a-diy-rubiks-cube-how-to-build-a-rubiks-cube-2nd-layer.html", "fetch_time": 1566137276000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027313889.29/warc/CC-MAIN-20190818124516-20190818150516-00454.warc.gz", "warc_record_offset": 134870784, "warc_record_length": 4604, "token_count": 1163, "char_count": 5525, "score": 4.0, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "latest", "language": "en", "language_score": 0.9622994661331177, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00042-of-00064.parquet"} |
# How to write the probability of ending up in a state where qubit $A$ is in the state $|0\rangle_A$ when $A$ is measured, in the bra-ket notation?
I need some help with the bra-ket notation. Suppose we have a normalized wavefunction for a two-qubit system (where $A$ and $B$ denote the two qubits respectively), like: $$|\psi\rangle_{A B} = a(|0\rangle_A \otimes |0\rangle_B) + b (|0\rangle_A \otimes |1\rangle_B) + c (|1\rangle_A \otimes |0\rangle_B) + d (|1\rangle_A \otimes |1\rangle_B)$$
Then what is the correct notation for writing the probability of wavefunction to collapse to a state where the state of qubit $A$ is $|0\rangle_A$ ? Is it $|_A\langle 0|\psi\rangle_{AB}|^2$ (which should be $|a|^2 + |b|^2$) ?
Also it would be very helpful if someone could link me a webpage or resource which discusses the properties of this operation i.e. multiplying a bra with the tensor product of two ket vectors. Does multiplying $\langle 0|_A$ with $|0\rangle_A \otimes |0\rangle_B$ or $|0\rangle_A \otimes |1\rangle_B$ equal $1$ ? And does multiplying $\langle 1|_A$ with $|1\rangle_A \otimes |0\rangle_B$ or $|1\rangle_A \otimes |1\rangle_B$ equal $0$ ? I'm not very sure about this.
I will first answer the second question. In a two particle system the basis is made up of tensor products of the form $| \psi\rangle_A \otimes | \phi \rangle_B$ that we can write $| \psi \phi \rangle$ remembering that position $1$ refers to particle $A$ and postion $2$ refers to particle $B$. The bra-ket multiplication of two such states is $\langle \psi_1 \phi_1 | \psi_2 \phi_2\rangle = \langle \psi_1 | \psi_2 \rangle \cdot \langle \phi_1 | \phi_2 \rangle$. In this sense it is meaningless to think about the multiplication of ${}_A\langle 0 |$ with anything, because this is not a possible state of the system.
And yes, that is the probability. One of the postulates of quantum mechanics is the following. The measurement of an observable $A$ on a state $| \psi \rangle$ can only give as an answer an eigenvalue $a_n$ of $A$, with probability
$P (a_n) = \sum\limits_{i=1}^{g_n} \big| \langle u_n^i | \psi \rangle \big|^2$
where $\{ | u_n^i \rangle, i= 1, ..., g_n \}$ is the subspace of eigenvectors of $A$ with eigenvalue $a_n$, and degeneracy $g_n$.
In your case you may think of the observable "state of A" $S_A$ with some eigenvalues $\{y, y, n, n\}$ in the basis $\{ |00\rangle, | 01 \rangle, |10\rangle , |11\rangle\}$. You want to know the probability of measuring the eigenvalue $y$ ("$A$ being in state $|0\rangle$"), with associate subspace $\{ |00\rangle, | 01 \rangle\}$. Therefore
$P(y) = \big|\langle 00 |\psi \rangle_{AB} \big|^2 + \big|\langle 01 |\psi \rangle_{AB} \big|^2 = \big| a \big|^2 + \big| b\big|^2$
This is how I understand it, but maybe there is a simpler way to do it.
• Good answer! Could you please also have a look at these two questions of mine: physics.stackexchange.com/questions/382397/… and math.stackexchange.com/questions/2622115/… ? Jan 26, 2018 at 14:31
• Sorry but I'm not familiar with those terms. In the first question I don't know what you mean by $\langle \psi \rangle$. I only understand the expectation value for an observable, not for a state. And I don't understand either what you mean by measurement operators. Maybe what you are trying to do is determining the state of one of the particles. But this is what I wrote in the answer. Jan 26, 2018 at 14:55
• And in the second question, as I said, I'm not familiar with writing 2-qubits states as matrices. I would write them as vectors $|00\rangle = [ 1, 0, 0, 0]^T$, $|01\rangle = [ 0,1,0,0]^T$, and so on. This way the projection onto say $|10\rangle$ would be $|10\rangle \langle 10| = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. But again maybe there is a way to do it writting the states as $2 \times 2$ matrices. Jan 26, 2018 at 15:02
• You mean there is nothing like expectation value of the whole system's quantum state (if it includes multiple particles)? Just expectation values for observables for single particle is valid? In your answer isn't "simultaneous state of A AND B", an observable in itself? Jan 26, 2018 at 15:16 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://physics.stackexchange.com/questions/382388/how-to-write-the-probability-of-ending-up-in-a-state-where-qubit-a-is-in-the-s", "fetch_time": 1716858941000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00494.warc.gz", "warc_record_offset": 384012297, "warc_record_length": 39853, "token_count": 1299, "char_count": 4191, "score": 3.640625, "int_score": 4, "crawl": "CC-MAIN-2024-22", "snapshot_type": "longest", "language": "en", "language_score": 0.8588745594024658, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00026-of-00064.parquet"} |
# Search by Topic
#### Resources tagged with Multiplication & division similar to Rolling Along the Trail:
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### There are 163 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Largest Number
##### Stage: 3 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Diggits
##### Stage: 3 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### Long Multiplication
##### Stage: 3 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
### Two Many
##### Stage: 3 Challenge Level:
What is the least square number which commences with six two's?
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Ordering Cards
##### Stage: 1 and 2 Challenge Level:
This problem is designed to help children to learn, and to use, the two and three times tables.
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### A One in Seven Chance
##### Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### Round and Round and Round
##### Stage: 3 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help?
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### What's My Weight?
##### Stage: 2 Short Challenge Level:
There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?
### Difficulties with Division
##### Stage: 1 and 2
This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Being Determined - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require determination.
### Book Codes
##### Stage: 2 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Clever Santa
##### Stage: 2 Challenge Level:
All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Tom's Number
##### Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Fair Feast
##### Stage: 2 Challenge Level:
Here is a picnic that Petros and Michael are going to share equally. Can you tell us what each of them will have?
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Four Goodness Sake
##### Stage: 2 Challenge Level:
Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Numbers Numbers Everywhere!
##### Stage: 1 and 2
Bernard Bagnall recommends some primary school problems which use numbers from the environment around us, from clocks to house numbers.
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### The 24 Game
##### Stage: 2 Challenge Level:
There are over sixty different ways of making 24 by adding, subtracting, multiplying and dividing all four numbers 4, 6, 6 and 8 (using each number only once). How many can you find? | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=7183", "fetch_time": 1496142504000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-22/segments/1495463615093.77/warc/CC-MAIN-20170530105157-20170530125157-00037.warc.gz", "warc_record_offset": 341517769, "warc_record_length": 10317, "token_count": 2340, "char_count": 9673, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2017-22", "snapshot_type": "longest", "language": "en", "language_score": 0.80329829454422, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"} |
Explore BrainMass
# Probability
### Complement Probability
If P(A) = .02, P(A or B)= .40 and P(AB) =0.1, what is the probability P(~B)?
### Ways to Select a Committee with Specific Gender Requirements
The school board of a public school system must select a finance committee of 5 from its 15 man, 8 woman membership. In how many ways can this committee be selected so that it consists of 3 men and 2 women?
### Binomial Distribution and NC Army National Guard
In the military, everything works of numbers. Recruiting is a good example of a binomial distribution. I'll use NC Army National Guard for an example. The NCARNG must have a set number of enlistments every year and they get the requirement for the year during October. This year the number is: 2000 (so n = 2000), the 2 ou
### Determine the portion of time the staff person is busy and how long a guest must wait for his or her request to be addressed.
The Delacroix Inn in Alexandria is a small exclusive hotel with 20 rooms. Guests can call housekeeping from 8:00 am to midnight for any of their service needs. Housekeeping keeps one person on duty during this time to respond to guest calls. Each room averages 0.7 call per day to housekeeping (Poisson Distributed), and a gues
### Binomial Probability Distribution and Probability
1. Suppose that the number of cars, X, that pass through a car wash between 4:00pm and 5:00pm on any sunny Friday has the following probability distribution: X 4 5 6 7 8 9 P(X=x) 1/12 1/12 ¼ ¼ 1/6 1/6 What is the probability that at least 7 car
### Probability and Discrete Probability Distributions Model
In paragraph form, create an example in everyday life that uses the binomial distribution as a model. Make sure that you explain how your example fits the requirements for a binomial distribution.
### Prove that no matter what the posted odds
Based on the betting model in the Two-horse race_01.xls spreadsheet, prove that no matter what the posted odds and the bets received are, the expected loss/gain is always zero. How can the bookmaker make sure he can earn a certain amount that is above the expected zero income from the betting? See attached file for ful
### Normal approximation to binomial
A multiple-choice examination consists of 90 questions, each having possible choices a, b, c, d, and e. Approximate the probability that a student will get at least 21 answers correct if she randomly guesses at each answer. (Note that, if she randomly guesses at each answer, then the probability that she gets any one answer corr
### Probability of normal distribution
A cola-dispensing machine is set to dispense on average 7.00 ounces of cola per cup. The standard deviation is 0.10 ounces. The distribution of amounts dispensed follows a normal distribution. a. What is the probability that the machine will dispense between 7.10 and 7.25 ounces of cola? b. What is the probability that t
### Statistical Process Control Help
1. A medical supplies company buys its supplies in bulk and redistributes them to doctor's offices and clinics. They receive thermometers in lots of 500 from the vendor. They are considering a sampling plan of n=50 and c=1. a. Develop a OC curve for this sampling plan. (Use Poisson Tables) b. If an incoming lot has 5% defectiv
### Automobile mileage for company's sales force
Distances driven by the sales force average 5,650 miles a month, and has a Standard Deviation of 120. Lower limit for calculation is 5,500. What percentage drove 5,500 miles or more in a month? Would like to see in Excel format to identify relevant formula.
### find the mean and standard deviation
The below table lists the last digits of the 73 published distances (in feet) of the 73 home runs hit by Barry Bonds in 2001 when he set the record for the most home runs in a season (based on data form USA today) The last digit of a data set can sometimes be used to determine whether the data have been measured or simply report
### Returned checks at a business
Issue: Returned checks at a business. 12% are returned for insufficient funds. 50% of this 12% the customer received cash back. 10% of all customers ask to receive cash back at the end of a purchase. For 1000 customer visits calculate the following: a) Insufficient funds? b) Cash back? c) Both insufficient funds an
### Probability
Express the indicated degree of likelihood as a probability value. *"it will definitely turn dark tonight." a. 0.5 b. 1 c. 0.30 d. 0.67 *Which of the following can not be a probability? a. 1 b. 0 c. -1 d. 2 *Answer the question,considering t
### Quantitative Methods: Finding Probability
Potato chip are sold in 12 oz bags. The bag weights are expected to be normally distributed about this standard. If the standard deviation of the bags is 0.58 oz would it be unusual to find a bag containing 11 oz. The Frito Lay CEO wants to know if it will be unusual for a bag to be 10% or more short.
### Quantitative Methods - Simulation to Demonstrate Probability
As you discussed earlier, four students called in to say they would miss their exam due to a flat tire. On the make-up exam the instructor as each student to list the tire that was flat. We concluded that there was only a 1/64 chance that all four would randomly agree if they actually had not had a flat. Simulate the situatio
### Quantitative Methods - should a plane go commercial probability
Management is considering going live with a new product that research believes is promising. Engineering estimates that if a \$300,000 pilot plant is build there is a 0.75 chance of a high yield versus the alternative of a low yield. If the pilot plant shows a high yield they believe there is a probability of 0.8 that the comm
### Quantitative Methods - probability of colored marbles
What is the probability of getting at most 3 white marbles when drawing 5 marbles from a bucket containing 5 white marbles and 5 black marbles?
### Determining Probability: Duration of Drug Effectiveness
Researchers at a pharmaceutical company have found that the effective time duration of a safe dosage of a pain relief drug is normally distributed with mean 2 hours and standard deviation 0.3 hour. For a patient selected at random: a) What is the probability that the drug will be effective for 2 hours or less? b) What i
### Calculating the Probability of Occurrence
Roger has read a report that the weights of adult mail Siberian tigers have a distribution which is approximately normal with mean = 390 lb and standard deviation = 65 lb. a) Find the probability that an individual male Siberian tiger will weigh more than 450 lb. b) Find the probability that a random sample of 4 male Siberia
### Probability: Calculating Combinations
Laura is training for a week-long mountain cycling tour. She has 12 short hilly routes from which to choose mid-week rides. a) How many ways can she choose 4 different rides from the list for the first week's training if order matters? b) How many ways can she choose 4 different rides if order does not matter? c) If sh
### Variance of Poisson random variable
The number of false alarms a building fire alarm system registers has a Poisson distribution with a mean of 6 false alarms per year. The variance of the number of false alarms is what?
### Compute a probability of identical balls
An urn contains 8 balls identical in every respect except color. There are 4 blue balls, 3 red balls, and 1 white ball. a) If you draw one ball from the urn what is the probability that it is blue or white? b) If you draw two balls without replacing the first one, what is the probability that the first ball is red and the
### Waiter's Probability for Commission
John has two jobs. For daytime work at a jewelry store he is paid \$200 per month plus a commission. His monthly commission is normally distributed with a mean of \$600 and a standard deviation of \$40. At night he works as a waiter, for which his monthly income is normally distributed with a mean of \$100 and a standard deviation o
### probability a randomly selected patron
Suppose that patrons of a restaurant were asked whether they preferred beer or whether they preferred wine. 70% said that they preferred beer. 60% of the patrons were male. 80% of the males preferred beer. What is the probability a randomly selected patron prefers wine is? What is the probability a randomly selected patro
### Finding probability of sample mean
The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the probability that the sample mean will be between 39 and 48 minutes?
### Normal approximation to the binomial distribution
A. Tossing a Coin Simulate 6 independent tosses of a fair coin. Repeat the procedure to get 200 sets of 6 coin tosses. (That's 1200 total tosses.) List the results. Explain carefully your method of simulating the tosses, whether it involves a computer, a table of random digits, or even flipping several coins at once. Addre
### Select the best stock level.
Given Buy at \$4.00, Sell at \$7.00 Demand 0 P d .1 Demand 1 P d .2 Demand 2 p d .2 Demand 3 P d .3 Demand 4, P d .2 a.Select the best stock level. b. Repeat with salvage value of \$1
### Multiple choice questions on Probability & Hypothesis
See attached files for full problem description.
### Standard distribution
Assume that the population of heights of male college students is approximately normally distributed with mean m of 68 inches and standard deviation s of 3.75 inches. A random sample of 16 heights is obtained. d. Find the mean and standard error of the xbar distribution. e. Find P (xbar > 70) f. Find P ( xbar < 67) I kno | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://brainmass.com/statistics/probability/pg71", "fetch_time": 1600762366000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00053.warc.gz", "warc_record_offset": 314081041, "warc_record_length": 14520, "token_count": 2253, "char_count": 9780, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2020-40", "snapshot_type": "latest", "language": "en", "language_score": 0.9288236498832703, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00048-of-00064.parquet"} |
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# Triangle Inequality
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The triangle inequality theorem states that the sum of any two sides of a triangle must be greater than the length of the third side. To find a range of values for the third side when given two lengths, write two inequalities: one inequality that assumes the larger value given is the longest side in the triangle and one inequality that assumes that the third side is the longest side in the triangle. Combine the two inequalities for the final answer.
Let's say I gave you 3 pieces of spaghetti and you try to make a triangle out of them, so I'm going to say that this piece of spaghetti is 10 inches long, this piece right here is 3 and this piece right here is 6. Does it look like we're going to be able to make a triangle out of this? It looks like we don't quite have enough to close that arch, let's say however instead of 3 and 6 let's say I gave you pieces that were 3 and 7, then we would just have a straight line. Because there's no way for us to form an angle there.
What we're talking about here are triangle inequalities which means if you look at a triangle there's a special relationship between those 3 sides and it kind of make sense if you think about it. If you started at point a what's going to be the fastest way to go to point c? Is it going to be to walk to point b and then point c? or is it just going to be faster to go from point a to point c? Well pretty clearly you're just going to walk straight to point c. But how does this look in terms of relationships? Well we can say that line segment ab or side ab plus side bc must be greater than your third side ac, so what's that saying is that going straight from one point to another is going to be shorter or less than the sum of the other 2 sides. You can write 2 other inequalities, we could start with ab and go with our side ac and say ac has to be greater than your third side which is bc. At last you could write your last inequality which we'd say side ac plus side bc has to be greater than your side ab. So in essence you always have to make sure that two of your sides summed will be more than the third side which is why we couldn't make a triangle because 3 plus 6 is 9 which is less than 10. 3+7=10 but notice you do not have an equal to part to your inequality all you have is this greater than symbol.
## Find Videos Using Your Textbook
Enjoy 3,000 videos just like this one. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://brightstorm.com/math/geometry/triangles/triangle-side-inequalities/", "fetch_time": 1371706229000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2013-20/segments/1368710313659/warc/CC-MAIN-20130516131833-00013-ip-10-60-113-184.ec2.internal.warc.gz", "warc_record_offset": 39346547, "warc_record_length": 11235, "token_count": 568, "char_count": 2491, "score": 4.625, "int_score": 5, "crawl": "CC-MAIN-2013-20", "snapshot_type": "longest", "language": "en", "language_score": 0.9720413088798523, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00009-of-00064.parquet"} |
## The ancient Greek alphabetical numeric system
```The ancient Greek alphabetical numeric system:
This chart illustrates both the ancient Greek acrophonic and alphabetical numeric systems. However, the acrophonic system, used primarily in Classical Athens ca. 500 – 400 BCE, came much later than the alphabetical system. So in effect we must resort to the only Greek numeric system we can use to represent numbers in Mycenaean Greek numbers, i.e. the alphabetical system. The alphabetical numbers are displayed in the second column after the modern numbers, 1 – 100,000 in the following chart. Here are some examples of alphabetic numbers representing Mycenaean numbers:
```
## Derivation [D] of Linear B Numerics: 1-10 20 100 & 200 [Click to ENLARGE]:
Derivation [D] of Linear B Numerics: 1-10 20 100 & 200 [Click to ENLARGE]:
Derivation [D] of Linear B Numerics: 1-10 20 100 & 200 as reconstructed in Progressive Linear B Grammar:
Explanation of the Table of Numerics:
The Principle of Derivation [D] as applied to the reconstruction of the orthography of numerics in Linear B. Even though there are very few attested [A] examples of numbers spelled out on Linear B tablets, I believe that we can safely derive them with a considerable degree of confidence, if we strictly apply the spelling or orthographic conventions of Linear B, which are available online here:
NOTES:
[1] These numbers are all spelled identically in Linear B and in ancient alphabetic Greek.
[2] The linear B number QETORO [4] is obviously a variant of the Latin “quattro”. There is nothing unusual whatsoever about the parallelism between Linear B & Latin orthography, since linguistically speaking, Q or QU are interchangeable with T. First, we have the spellings of 4 in Greek (te/ssarej te/ttarej). Though it will come as a surprise to many of you that the Linear B spelling for 4 QETORO would eventually morph into (te/ssarej te/ttarej) in ancient Greek & then back to “quattro” in Latin, there is a perfectly logical explanation for this phenomenon. This is why you see a ? to the left of the Greek for 4, because it is all too easy to fall into the trap of erroneously concluding that Linear B QETORO cannot have been a ancestor of the ancient Greek spellings for 4, when in fact it is. In order to put this all into proper perspective, Latin spells 4 as “quattro” for the simple reason that “tattro” would simply not do. All this boils down to a single common denominator, the principle of euphonics, meaning the alteration of speech sounds, hence orthography, such that any word in any language sounds pleasing to the ear to native speakers of that language (not to non-speakers, i.e. anyone who cannot speak the language in question). Every language has its own elemental principles of euphony, but some languages place far more stress on it than others. Greek is notorious for insisting on euphony at all times.
? The masculine for the number 1 is missing for the exact same reason that the 2nd. person singular is missing from my paradigm for the present tense of Linear B verbs. I find it impossible to accurately reconstruct any Greek word ending in eij for the simple reason that it not possible for any word to end in a consonant in Linear B. So why bother? This handicap will return to haunt us over and over in the Regressive reconstruction of all the conjugations and declensions and parts of speech in Linear B Progressive grammar, leaving gaping holes all over the place.
Orthography of Numerics In Linear B:
At a glance, we can instantly see that the spelling of several numerals in Linear B is identical to that of their later Greek alphabetic counterparts, with the exception of marking initial aspirates & non-aspirates, which Linear B was unable to express with just one exception, the homophone for HA. This speaks volumes to the uniformity of the spelling of numerics over vast expanses of time, as in this case, from ca. 1450 BCE (Linear B) – ca. 100 AD and well beyond. In fact, some numbers are still spelled exactly the same way in modern Greek, meaning that the orthography of the Greek numeric system has remained virtually intact for over 3,400 years!
Linear B’s Conspicuous Reliance on Shorthand:
Since the Mycenaeans used Linear B primarily for accounting, agricultural, manufacturing & economic purposes, they almost never spelled out numbers, for the obvious reason that they needed to save space on tablets that were small, because they were baked, hence fragile, and not only that, destroyed at the end of every “fiscal” year, to make room for the following year’s statistics. Their habitual use of logograms for numbers was in fact, (a type of) shorthand. Keep this in mind, because Linear B makes use of shorthand for various annotative purposes – not just for numerals. This is hardly surprising, considering that the tablets were used almost exclusively for accounting. So if we think shorthand is a modern invention, we are sorely mistaken. The Mycenaeans & Minoans were extremely adept in the liberal use of shorthand to cram as much information, or if you like, data on what were, after all, small tablets. But there is more: every extant single tablet is a record of the agricultural, manufacturing & economic activities for one single year. It also means that there is no way for us to know which, if any, of the tablets from one site, for instance, Knossos, where the vast majority of tablets have been found, represent accounting statistics for exactly the same year as any of the others. So we have to be extremely careful in interpreting the discrepancies in the quasi-chronology and even reverse-chronology of the data found on the tablets. There are several other reasons why we need to exercise extreme caution in interpreting the data on extant tablets, but since these are of a different nature, I shall not address them here.
Richard
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Artistic Reconstruction and Original Translation From Homer's "Iliad" by Kathleen Vail | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://linearbknossosmycenae.com/tag/counting/", "fetch_time": 1590413267000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00196.warc.gz", "warc_record_offset": 405405364, "warc_record_length": 31410, "token_count": 2030, "char_count": 8639, "score": 3.609375, "int_score": 4, "crawl": "CC-MAIN-2020-24", "snapshot_type": "latest", "language": "en", "language_score": 0.8700069189071655, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00054-of-00064.parquet"} |
## Chem – College: Conversion Between Kc and Kp
How do you convert between Kc and Kp?
Another problem you might come across is how to convert between Kp and Kc. That is to say how to convert between the equilibrium constant from pressure units and the equilibrium constant from concentration units. The formula is below.
Kp = Kc (RT)Δ n
Above the symbols represent:
Kp is the equilibrium constant for pressure
Kc is the equilibrium constant for molar concentration
R is our old friend the gas constant from the. R = 0.0821 when using atm as pressure units.
T is temperature in Kelvin.
Δ n is the change in moles. This the exponent of R and T multiplied together. If you need to know how to calculate Δ n then go back to the section on
VIDEO Converting Between Kc and Kp Demonstrated Example 1: If the Kc for the chemical equation below is 0.5 at a temperature of 300K, then what is the Kp?
2 OF2(g) + 2 NH3(g) <—> N2F4(g) + O2(g) + 3 H2(g)
What information does the problem give you?
Kc = 0.5
T = 300K
R = 0.0821 (you should always have this constant)
What is the Δ n for this chemical equation?
Answer: 5 – 4 = 1
Set up the math equation.
Kp = Kc * (R* T)Δ n 1
Put in all the values (numbers).
Kp = 0.5 * (0.0821 * 300K)1 1
Multiply what is in the parenthesis.
Kp = 0.5 * (24.63)1 1
Distribute the exponent inside the parenthesis.
Kp = 0.5 * 24.63 1
Multiply the rest together.
Kp = 12.3 1
VIDEO Converting Between Kc and Kp Demonstrated Example 2: If the Kp for the chemical equation below is 8.7 * 10-3 at a temperature of 280K, then what is the Kc?
2 C4H10(l) + 13 O2(g) <—-> 8 CO2(g) + 10 H2O(g)
What information does the problem give you?
Kp = 8.7 * 10-3
T = 280K
R = 0.0821 (you should always have this constant)
What is the Δ n for this chemical equation?
Answer: 18 – 13 = 5 don’t count the liquid
Set up the math equation.
Kp = Kc * (R* T)Δ n 1
Put in all the values (numbers).
8.7 * 10-3 = Kc * (0.0821 * 280K)5 1
Multiply what is in the parenthesis.
8.7 * 10-3 = Kc * (23)5 1
Distribute the exponent inside the parenthesis.
8.7 * 10-3 = Kc * (6.44 * 106) 1
Divide both sides by 6.44 * 106
8.7 * 10-3 = Kc * (6.44 * 106) 6.44 * 106 6.44 * 106
Cancel out the right side.
8.7 * 10-3 = Kc * (6.44 * 106) 6.44 * 106 6.44 * 106
Simplify
8.7 * 10-3 = Kc 6.44 * 106
Divide
1.35 * 10-9 = Kc 1
COMPLETE ANSWER: Kc = 1.35 * 10-9
PRACTICE PROBLEMS: Solve the question below involving Kp and Kc.
If the Kc for the chemical equation below is 25 at a temperature of 400K, then what is the Kp?
N2(g) + 3 H2(g) <—-> 2 NH3(g)
Answer: Kp = 9.05 * 10-4
If the Kc for the chemical equation below is 6.9 * 10-4 at a temperature of 250K, then what is the Kp?
2 N2O3(g) <——> 2 N2(g) + 3 O2(g)
Answer: Kp = 8.6 * 103
If the Kp for the chemical equation below is 5.7 * 10-2 at a temperature of 273K, then what is the Kc?
Ba(OH)2(aq) + 2 HCl(aq) <—-> 2 H2O(l) + BaCl2(aq) | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://scientifictutor.org/1531/chem-college-conversion-between-kc-and-kp/", "fetch_time": 1719144050000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00881.warc.gz", "warc_record_offset": 27171570, "warc_record_length": 7966, "token_count": 1054, "char_count": 2931, "score": 3.984375, "int_score": 4, "crawl": "CC-MAIN-2024-26", "snapshot_type": "latest", "language": "en", "language_score": 0.865972638130188, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00046-of-00064.parquet"} |
OML-IZ Search
## Fractions - Mixed Number Multiplication
Calculate and enter the answers. Enter fractions in reduced, mixed number form, with a space separating the whole from the fractional part, and a slash separating the numerator and the denominator, omitting the whole or fraction part if they don't apply. E.g. "1 1/2", "3/4", or "5" without the quotes. You may use the TAB key to move to the next question. When you are done, click Submit.
1. $$5 \dfrac{ 9 }{ 10 } \times \dfrac{ 1 }{ 10 } =$$
2. $$3 \dfrac{ 3 }{ 7 } \times 5 \dfrac{ 5 }{ 7 } =$$
3. $$3 \dfrac{ 7 }{ 10 } \times 3 \dfrac{ 5 }{ 7 } =$$
4. $$5 \dfrac{ 1 }{ 8 } \times 5 \dfrac{ 2 }{ 3 } =$$
5. $$\dfrac{ 7 }{ 9 } \times 3 \dfrac{ 5 }{ 7 } =$$
6. $$2 \dfrac{ 3 }{ 7 } \times 2 \dfrac{ 1 }{ 3 } =$$
7. $$3 \dfrac{ 3 }{ 7 } \times 5 \dfrac{ 1 }{ 10 } =$$
8. $$5 \dfrac{ 1 }{ 8 } \times 2 \dfrac{ 1 }{ 7 } =$$
9. $$5 \dfrac{ 1 }{ 6 } \times 1 \dfrac{ 3 }{ 4 } =$$
10. $$3 \dfrac{ 2 }{ 5 } \times \dfrac{ 7 }{ 10 } =$$
We hope that the free math worksheets have been helpful. We encourage parents and teachers to adjust the worksheets according to the needs of the child. For more difficult questions, the child may be encouraged to work out the problem on a piece of paper before entering the solution. We hope that the kids will also love the fun stuff and puzzles. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://interactive.onlinemathlearning.com/fraction_mixed_multiplication.php?action=generate&numProblems=10", "fetch_time": 1508265639000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-43/segments/1508187822480.15/warc/CC-MAIN-20171017181947-20171017201947-00800.warc.gz", "warc_record_offset": 171220623, "warc_record_length": 5672, "token_count": 501, "char_count": 1339, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2017-43", "snapshot_type": "latest", "language": "en", "language_score": 0.6679339408874512, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00052-of-00064.parquet"} |
# Work-Energy Theorem locomotive Question
1. Dec 25, 2013
### Speedking96
1. The problem statement, all variables and given/known data
A locomotive exerts a 20 000 Newton force upon a train. It propels four cards, each with a 15 tonne mass. The locomotive's mass is 40 tonnes and friction is considered negligible. Initially at rest, the train accelerates over a distance of one kilometer.
a) What is the work done by the locomotive on the train?
b) What is the total work?
c) What is the velocity at the end of the 1 kilometer?
2. Relevant equations
F=ma
Ek = (m v^2)/2
W = F * d
3. The attempt at a solution
(a)
w = f * d = (20 000)(1000) = 20 000 000 Joules
(b) How is that different from question (a)?
(c) 20 000 000 = 0.5 * 100 000 * v^2
v = 20 m/s
2. Dec 25, 2013
### PhanthomJay
Looks good .
3. Dec 25, 2013
### Speedking96
I just have a quick question. Suppose there was friction. And they asked for the total work done... would I have to subtract the work by friction from the total amount of Joules? Do we add "work" like vectors?
Thank you.
4. Dec 25, 2013
### haruspex
Work is a scalar.
5. Dec 25, 2013
### PhanthomJay
If there were friction, the total work done would be the algebraic sum of the work done by the train and the work done by friction .
Work , like energy, is a scalar quantity, not a vector quantity. It can have a positive or negative value, but it has no direction. In your example, yes, you would have to subtract the work done by friction from the work done by the engine to get the total, or net, work done on the train .
6. Dec 26, 2013
### Speedking96
For example, if the work done by friction was 100 000 Joules, then the total work would just be:
20 000 000 Joules + 100 000 Joules = 20 100 000 Joules.
However, if I wanted to find the total work done on the train, I would have to subtract the 100 000 Joules as you had said.
7. Dec 26, 2013
### haruspex
I find the question somewhat ambiguous. Does 'train' include the locomotive? These two excerpts make me think it's just the four cars:
8. Dec 27, 2013
### PhanthomJay
No!
Let's start from scratch with a different problem, hopefully clear of ambiguities. Suppose that a horizontal force F was applied to an object of mass M sitting on a horizontal friction-less surface.
If F = 20 000 N and M = 1 000 kg ,
a) How much work is done by the Force F on the object after it has accelerated over a distance of 1 km?
b) What is the total work done on the object?
For part a), you reason that the work done by F is F.d = 20 000 000 J.
For part b), the total work is the algebraic sum of the work done by all the forces acting on the object. The other forces acting on the object are the objects weight and the normal force of the floor on the object, neither of which does any work on the object (convince yourself of this). Thus, the total work is just the the work done by F, or 20 000 000 J, same answer as part a. I think you understand this already.
Now, same problem, same distance traveled, but lets add friction, assume a small 100 N friction force opposes the motion. The work done by F, the applied force, is still 20 000 000 J. The work done by friction is -100 000 J. Please note and understand the use of the minus sign here. The total work done on the object is thus 19 900 000 J (just add them up algebraically). No way no how can you ever say that the total work done on the object is 20 100 000 J. | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://www.physicsforums.com/threads/work-energy-theorem-locomotive-question.729975/", "fetch_time": 1511300351000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-47/segments/1510934806426.72/warc/CC-MAIN-20171121204652-20171121224652-00444.warc.gz", "warc_record_offset": 859504015, "warc_record_length": 18011, "token_count": 950, "char_count": 3441, "score": 3.84375, "int_score": 4, "crawl": "CC-MAIN-2017-47", "snapshot_type": "latest", "language": "en", "language_score": 0.9470523595809937, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00031-of-00064.parquet"} |
# Chemistry
posted by on .
OMG! HELP!
1. How many grams of CaCl2 would be required to produce a 3.5 M (molar) solution with a volume of 2.0 L?
• Chemistry - ,
How many mols do you want? That's mols = M x L = ?
Then mol = grams/molar mass. You know molar mass and mols, solve for grams.
• Chemistry - ,
3.5x2.0=7
7/3.5=2
Is that right? /:
• Chemistry - ,
The first part is right. The second is not.
Yes, M x L = mols = 3.5 x 2L = 7 mols
Then mol = g/molar mass
7 = grams/molar mass CaCl2.
grams = 7*molar mass CaCl2 = estimated 7*111 = about 777 grams.
• Chemistry - ,
I still don't get it. How or where'd you get 111 from? /:
• Chemistry - ,
Dr. Bob222 gave you the setup to calculate the grams of CaCl2.
But I will go into a little bit more detail and maybe this will help:
Molarity=Moles/Volume (L)
The question tells you that you need need a total of 2L, but you don't know how many moles are in a 3.5M solution with a volume of 2L, but you can find that out: how do you calculate the moles needed to make a 3.5M solution that has a volume of 2L?
3.5M=x moles/2L
Solve for X:
3.5*2L=x moles
***You did this correctly, but you have to convert moles to grams because the question asks you for grams.
Stoichiometry allows you to do this, so how do you convert moles to grams?
x moles * molecular weight
x moles * 111g of CaCl2/mole=???????
I'll let you take care of it from here.
• Chemistry - ,
111 is the approximate molar mass of CaCl2.
Ca = about 40 from the periodic table.
71+40 = about 111. You can look up the numbers and obtain a more accurate answer. You should be able to look at the math equation I wrote above (mols = g/molar mass CaCl2 and g = mols x molar mass CaCl2) to see that I substituted 7 for mols from the first part and that I substituted 111 for molar mass. Then molar mass x mols = 111 x 7 = about 777. What I'm saying is that you should be able to tell that the 111 MUST be for the molar mass of CaCl2.
• Chemistry - ,
Oh.Oh okay. I get it. Thanks! | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://www.jiskha.com/display.cgi?id=1376773394", "fetch_time": 1496129374000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2017-22/segments/1495463614615.14/warc/CC-MAIN-20170530070611-20170530090611-00084.warc.gz", "warc_record_offset": 669909236, "warc_record_length": 4661, "token_count": 613, "char_count": 2005, "score": 3.703125, "int_score": 4, "crawl": "CC-MAIN-2017-22", "snapshot_type": "latest", "language": "en", "language_score": 0.9309675693511963, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00023-of-00064.parquet"} |
# 9.3: Outcomes and the Type I and Type II Errors
When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis $$H_0$$ and the decision to reject or not. The outcomes are summarized in the following table:
Table 9.2
$$\textbf{Statistical Decision}$$ $$\bf{H_0} \textbf{ is actually...}$$
True False
Cannot accept $$H_0$$ Type I error Correct outcome
Cannot reject $$H_0$$ Correct outcome Type II error
The four possible outcomes in the table are:
1. The decision is cannot reject $$\bf{H_0}$$ when $$\bf{H_0}$$ is true (correct decision).
2. The decision is cannot accept $$\bf{H_0}$$ when $$\bf{H_0}$$ is true (incorrect decision known as a Type I error). This case is described as "rejecting a good null". As we will see later, it is this type of error that we will guard against by setting the probability of making such an error. The goal is to NOT take an action that is an error.
3. The decision is cannot reject $$\bf{H_0}$$ when, in fact, $$\bf{H_0}$$ is false (incorrect decision known as a Type II error). This is called "accepting a false null". In this situation you have allowed the status quo to remain in force when it should be overturned. As we will see, the null hypothesis has the advantage in competition with the alternative.
4. The decision is cannot accept $$\bf{H_0}$$ when $$\bf{H_0}$$ is false (correct decision).
Each of the errors occurs with a particular probability. The Greek letters $$\alpha$$ and $$\beta$$ represent the probabilities.
• $$\alpha$$ = probability of a Type I error = $$\bf{P}$$(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true: rejecting a good null.
• $$\beta$$ = probability of a Type II error = $$\bf{P}$$(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. ($$1 − \beta$$) is called the Power of the Test.
$$\alpha$$ and $$\beta$$ should be as small as possible because they are probabilities of errors.
Statistics allows us to set the probability that we are making a Type I error. The probability of making a Type I error is $$\alpha$$. Recall that the confidence intervals in the last unit were set by choosing a value called $$Z_{\alpha}$$ (or $$t_{\alpha}$$) and the alpha value determined the confidence level of the estimate because it was the probability of the interval failing to capture the true mean (or proportion parameter $$p$$). This alpha and that one are the same.
The easiest way to see the relationship between the alpha error and the level of confidence is with the following figure.
In the center of Figure 9.2 is a normally distributed sampling distribution marked $$H_0$$. This is a sampling distribution of $$\overline X$$ and by the Central Limit Theorem it is normally distributed. The distribution in the center is marked $$H_0$$ and represents the distribution for the null hypotheses $$H_0$$: $$\mu = 100$$. This is the value that is being tested. The formal statements of the null and alternative hypotheses are listed below the figure.
The distributions on either side of the $$H_0$$ distribution represent distributions that would be true if $$H_0$$ is false, under the alternative hypothesis listed as Ha. We do not know which is true, and will never know. There are, in fact, an infinite number of distributions from which the data could have been drawn if Ha is true, but only two of them are on Figure 9.2 representing all of the others.
To test a hypothesis we take a sample from the population and determine if it could have come from the hypothesized distribution with an acceptable level of significance. This level of significance is the alpha error and is marked on Figure 9.2 as the shaded areas in each tail of the $$H_0$$ distribution. (Each area is actually \alpha/2 because the distribution is symmetrical and the alternative hypothesis allows for the possibility for the value to be either greater than or less than the hypothesized value--called a two-tailed test).
If the sample mean marked as $$\overline{X}_{1}$$ is in the tail of the distribution of $$H_0$$, we conclude that the probability that it could have come from the $$H_0$$ distribution is less than alpha. We consequently state, "the null hypothesis cannot be accepted with (\alpha) level of significance". The truth may be that this $$\overline{X}_{1}$$ did come from the $$H_0$$ distribution, but from out in the tail. If this is so then we have falsely rejected a true null hypothesis and have made a Type I error. What statistics has done is provide an estimate about what we know, and what we control, and that is the probability of us being wrong, $$\alpha$$.
We can also see in Figure 9.2 that the sample mean could be really from an Ha distribution, but within the boundary set by the alpha level. Such a case is marked as $$\overline{X}_{2}$$. There is a probability that $$\overline{X}_{2}$$ actually came from Ha but shows up in the range of $$H_0$$ between the two tails. This probability is the beta error, the probability of accepting a false null.
Our problem is that we can only set the alpha error because there are an infinite number of alternative distributions from which the mean could have come that are not equal to $$H_0$$. As a result, the statistician places the burden of proof on the alternative hypothesis. That is, we will not reject a null hypothesis unless there is a greater than 90, or 95, or even 99 percent probability that the null is false: the burden of proof lies with the alternative hypothesis. This is why we called this the tyranny of the status quo earlier.
By way of example, the American judicial system begins with the concept that a defendant is "presumed innocent". This is the status quo and is the null hypothesis. The judge will tell the jury that they can not find the defendant guilty unless the evidence indicates guilt beyond a "reasonable doubt" which is usually defined in criminal cases as 95% certainty of guilt. If the jury cannot accept the null, innocent, then action will be taken, jail time. The burden of proof always lies with the alternative hypothesis. (In civil cases, the jury needs only to be more than 50% certain of wrongdoing to find culpability, called "a preponderance of the evidence").
The example above was for a test of a mean, but the same logic applies to tests of hypotheses for all statistical parameters one may wish to test.
The following are examples of Type I and Type II errors.
Example 9.4
Suppose the null hypothesis, $$H_0$$, is: Frank's rock climbing equipment is safe.
Type I error: Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe.
Type II error: Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.
$$\bf{\alpha =}$$ probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe. $$\bf{\beta =}$$ probability that Frank thinks his rock climbing equipment may be safe when, in fact, it is not safe.
Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)
This is a situation described as "accepting a false null".
Example 9.5
Suppose the null hypothesis, $$H_0$$, is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital. This is the status quo and requires no action if it is true. If the null hypothesis cannot be accepted then action is required and the hospital will begin appropriate procedures.
Type I error: The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error: The emergency crew does not know if the victim is alive when, in fact, the victim is dead.
$$\bf{\alpha =}$$ probability that the emergency crew thinks the victim is dead when, in fact, he is really alive = P(Type I error). $$\bf{\beta =}$$ probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead = P(Type II error).
The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)
Exercise 9.5
Suppose the null hypothesis, $$H_0$$, is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?
Example 9.6
It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, $$H_0$$, is: It’s a Boy Genetic Labs has no effect on gender outcome. The status quo is that the claim is false. The burden of proof always falls to the person making the claim, in this case the Genetics Lab.
Type I error: This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, \alpha.
Type II error: This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, \beta.
The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy.
Exercise 9.6
“Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.
Example 9.7
A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious?
Type I: A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%.
Type II: A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%.
In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option. | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://stats.libretexts.org/Bookshelves/Applied_Statistics/Book%3A_Business_Statistics_(OpenStax)/09%3A_Hypothesis_Testing_with_One_Sample/9.03%3A_Outcomes_and_the_Type_I_and_Type_II_Errors", "fetch_time": 1582130987000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00552.warc.gz", "warc_record_offset": 536547361, "warc_record_length": 24477, "token_count": 2542, "char_count": 11067, "score": 4.21875, "int_score": 4, "crawl": "CC-MAIN-2020-10", "snapshot_type": "latest", "language": "en", "language_score": 0.898666501045227, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00004-of-00064.parquet"} |
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# While on a straight road, car X and car Y are traveling at
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While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?
(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour
(2) Three minutes ago car X was 1/2 mile ahead of car X.
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16 Feb 2007, 04:03
devilmirror wrote:
While on a straight road, car X and car Y are traveling at different constant rates. If car X is now 1 mile ahead of car Y, how many minutes from now will car X be 2 miles ahead of car Y?
(1) Car X is traveling at 50 miles per hour and car Y is traveling at 40 miles per hour
(2) Three minutes ago car X was 1/2 mile ahead of car X.
Here is my explanation:
Let's say now is at time t = T mins
Cax X's speed = S1
Car Y's speed = S2
Distance travelled by X = d1
Distance travelled by Y = d2
We know, at t = T mins d1 = d2 + 1
We want to find after how many mins d1 = d2 + 2
Condition 1:
Cax X's speed = 50 miles/hr
Car Y's speed = 40 miles/hr
Therefore at t = T mins Car X travels T5/6 miles and Car Y travels T4/6 miles.
we know d1 = d2 + 1 at time t = Tmins
Hence, T5/6 = T4/6 + 1 this give T = 6mins
Now we need to find t = M mins when d1 = d2 + 2
Therefore at t = M mins Car X travels M5/6 miles and Car Y travels M4/6 miles.
we know d1 = d2 + 2 at t = Mmins
Hence M5/6 = M4/6 + 2 this give M = 12 so after 6 min Car X will be 2 miles ahead of Car Y
Condition 1 is sufficient.
Condition 2:
at time t = T mins we know d1 = d2 + 1
Car X will travel d1 = TS1/60 and Car Y will travel d2 = d1 + 1
Therefore, d2 = TS1/60 + 1 ---I
From condition 2 we know,
at time t = T-3 min d1 = d2 + 1/2
at time t= T-3
Car X will travel d1 = (T-3)*S1/60 and Car Y will travel d2 = d1 + 1/2
Therefore, d2 = (T-3)*S1/60 + 1/2 ---II
From I and II we can find out S1 = 10 miles/hr
We need find t = T + x when d1 = d2 + 2
at time t= T+x
Car X will travel d1 = (T+x)*S1/60 and Car Y will travel d2 = d1 + 2
Therefore, d2 = (T+x)*S1/60 + 2 ---III
From II and III we can find x and hence sufficient.
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Last edited by amorpheus on 16 Feb 2007, 05:24, edited 1 time in total.
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16 Feb 2007, 06:23
Amorpheus, I did the exact same thing when I tried to solve this problem.
However, the explanation of this answer can be extremely short. I want to post the question first and see who can come up with the shortest explanation.
When I read the short explanation, I learned a lot from it. This question is very creative so I want to share it with you guys.
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16 Feb 2007, 06:24
devilmirror wrote:
Amorpheus, I did the exact same thing when I tried to solve this problem.
However, the explanation of this answer can be extremely short. I want to post the question first and see who can come up with the shortest explanation.
When I read the short explanation, I learned a lot from it. This question is very creative so I want to share it with you guys.
Devilmirror: Would you mind sharing the shortest explanation please?
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"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi
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16 Feb 2007, 07:04
amorpheus wrote:
Devilmirror: Would you mind sharing the shortest explanation please?
Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.
(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.
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16 Feb 2007, 07:06
devilmirror wrote:
amorpheus wrote:
Devilmirror: Would you mind sharing the shortest explanation please?
Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.
(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.
You the man.... I like that sometimes I feel stupid, this actually happens to me a lot I always end up solving problems using the longest possible method. What you do to get such short cuts? Any special tricks?
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"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi
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16 Feb 2007, 08:00
amorpheus wrote:
devilmirror wrote:
amorpheus wrote:
Devilmirror: Would you mind sharing the shortest explanation please?
Ok, here the spoiler.
(1) At their constant rates (Vx = 50 mph and Vy = 40 mph), car X would increase its distance from car Y by 10 miles every hour or, equivalently, 1 mile every 6 minutes. suff.
(2) This state that car X increases its distance from car Y by 0.5 mile every 3 minutes, or alternately 1 mile every 6 minutes; suff.
You the man.... I like that sometimes I feel stupid, this actually happens to me a lot I always end up solving problems using the longest possible method. What you do to get such short cuts? Any special tricks?
No, no, no my friends. I used the longer method at first. But the OE is a lot shorter than I expected.
The explanation is from OE not from me. If I showed you my method, it would be 1 full A4 paper.
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16 Feb 2007, 09:33
I look at this problem and I understand that we need to have the information to understand the difference in speed...
The first gives us this 50 mph and 40 mph 10 mph difference suff
second gives us the time it took for the cars to create a 1/2 mile distance suff since I know I can derive the difference in speed from this info.
No math is necessary to solve this problem
16 Feb 2007, 09:33
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