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Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| $(x^2+x)(y^2-1)=240$ implies that
$y = \sqrt{1 + \dfrac{240}{x(x+1)}}$
Clearly both $x$ and $x+1$ need to be divisors of $240$.
These are the divisors of $240$
\begin{align}
1 &\quad 240 \\
2 &\quad 120 \\
3 &\quad 80 \\
4 &\quad 60 \\
5 &\quad 48 \\
6 &\quad 40 \\
8 &\quad 30 \\
10 &\quad 24 \\
12 &\quad 20 \\
15 &\quad 16
\end{align}
For $x = 1,2,3,4,5,15$ we get
$y = 11, \sqrt{41}, \sqrt{21}, \sqrt{13}, 3, \sqrt 2$
So the minimum value of $x-y$ is $1-11 = -10$
and the maximum value of $x-y$ is $5-3 = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Calculate the rest of the division of $(4^{103} + 2(5^{104}))^{102}$ by $13$ I already know some techniques to solve big exponents such as, Euler/Fermat Theorem,Euler/Carmichael,successive squaring.
But these problems seem to be more dificult as they involve operations insted of a single number.
How could I solve them ?
*
*$(4^{103} + 2(5^{104}))^{102}$ by $13$
*$53^{103}+103^{53}$ by $39$
| Variants :
First congruence :
Observe first that $2$ has order $12 \bmod13$, so $4$ has order $6$, and that as $5^2\equiv -1\mod 13$, $5$ has order $4$. Thus
$$4^{103}+2\cdot 5^{104}\equiv 4^{103\bmod 6}2\cdot 5^{104\bmod 4}=4^1+2\cdot 5^0=6.$$
Now $6=2\cdot 3$ and we know $2$ has order $12 \bmod 13$, whereas $3$ has order $3$. We conclude that
$$6^{102}\equiv 2^{102\bmod 12}\cdot3^{102\bmod 3}=2^6\cdot3^0\equiv -1\mod 13.$$
Second congruence :
Using the Chinese remainder theorem, it is enough to calculate the expression modulo $3$ and modulo $13$.
*
*Modulo $3$: $\;53\equiv 2$, which has order $2\bmod 3$, and $103\equiv 1$, so
$$53^{103}+103^{53}\equiv 2^{103\bmod 2}+1^{53}\equiv 0\mod 3.$$
*Modulo $13$: $\;53\equiv 1$ and $103\equiv -1$, so $$53^{103}+103^{53}\equiv 1+(-1)^{53}\equiv 0\mod 13.$$
The Chinese remainder theorem asserts the canonical map
$$\mathbf Z/39\mathbf Z\longrightarrow\mathbf Z/3\mathbf Z\times\mathbf Z/13\mathbf Z $$
is an isomorphism, so we conclude that $\;53^{103}+103^{53}\equiv 0\mod 39$.
| {
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"timestamp": "2023-03-29T00:00:00",
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To find all natural triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$ Find all positive integer triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$.
I have no important result on this one!
| Appetizers from the comments:
*
*$c=1$ is left as an exercise
*$c=2$ works when both $a$ and $b$ are even
*$c=3$ works when $a,b$ and $0$ are pairwise non-congruent modulo $3$.
Then the main course:
Claim. There are no solutions with $c>3$.
Proof. We have trivially that if $a\equiv a'\pmod c$, then
$$
2^a\equiv2^{a'}\pmod{2^c-1}.
$$
Therefore without loss of generality we can assume that $0\le a,b<c$.
But in that case there are no solutions for larger $c$:
*
*If $a\neq b$ are distinct, then $2^a+2^b+2^0<2^c-1$ (observe that $2^c-1$ is the sum of $c>3$ distinct powers of two).
*If $a=b$, then it won't work. Either the sum $2^a+2^b+1$ is still $<2^c-1$, or $a=b=c-1$ and $2^a+2^b+1=2^c+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find an unknown power in this formula? $$ \frac{1}{2^a} = 3.0988$$
How do I solve for $a$?
The original equation was:
$$\frac{1}{i^a}-\left(B-\frac{NE}{P_1-E}\right) = 0$$
I know that:
*
*$B=10000$
*$N = 50$
*$E = 15$
*$P_1 = 3000$
$$\frac1{i^a}-\left(10000-\frac{50(15)}{3000-15}\right) = 0$$
$$\frac1{i^a} - 3.0988 = 0$$
so if $i = 2$ how do I solve for $a$?
| Using $\frac{1}{2^a} = 3.0988$,
$$\begin{align}
\ln\left(\frac {1}{2^a}\right) &= \ln\left(3.0988\right) \\
\ln\left(1\right) - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\
0 - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\
\ln\left(2^a\right) &= - \ln\left(3.0988\right) \\
a\ln\left(2\right) &= -\ln\left(3.0988\right) \\
a &= \frac{- \ln\left(3.0988\right)} {\ln\left(2\right)}
\end{align}$$
Or you simply work in $\log_2$:
$$\begin{align}
\frac{1}{2^a} &= 3.0988 \\
2^0 &= 2^a 2^{\log_2 3.0988 } \\
0 &= a+ \log_2 3.0988 \\
a&=-\log_2 3.0988
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Max and min of $f(x)=\sin^2{x}+\cos{x}+2$. My first try was to look for when $\cos{x}$ and $\sin{x}$ attain their min and max respectively, which is easy using the unit circle. So the minimum of $\sin{x}+\cos{x}$ has maximum at $\sqrt{2}$ and minimum at $-\sqrt{2}.$ But if the sine-term is squared, how do I find the minimum of $\sin^2{x}+\cos{x}$?
I then tried the differentiation method:
$$f'(x)=2\sin{x}\cos{x}-\sin{x}=(2\cos{x}-1)\sin{x}=0,$$
which in the interval $(0,2\pi)$, gave me the the roots $\left(\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3},\frac{5\pi}{3}\right)$. So
$$f(\pi/2)=3, \quad f(3\pi/2)=3, \quad f(\pi/3)=\frac{13}{4}, \quad f(5\pi/3)=\frac{13}{4}.$$
So $f_{\text{max}}=\frac{13}{4}$ and $f_{\text{min}}=3.$ But the minimum should be $1$.
| HINT: write $f(x)$ as
$$f(x)=1-\cos(x)^2+\cos(x)+2 \\=-\cos(x)^2+\cos(x)+3\\=-\left(\cos(x)^2-\cos(x)+\frac{1}{4}\right)+\frac{13}{4}\\=-\left(\cos(x)-\frac{1}{2}\right)^2+\frac{13}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Possible values of $a+b^2+c^3$? If $a, b$ and $c$ are rational numbers satisfying the equation $x^3+ax^2+bx+c=0$ then find the possible values of $a+b^2+c^3$.
I have found one of the possible values as 0 using some vieta's rules and substituting $c$ for $x$. But can't find other possible values. The question specifically asks $3$ possible values.
| Let $f(x)=x^3+ax^2+bx+c$. We are given that $f(a)=f(b)=f(c)=0$. Therefore
\begin{align*}
a+b+c & = -a\\
ab+bc+ca & =b\\
abc & = -c.
\end{align*}
The last equation suggests either $c=0$ or $ab=-1$.
If $c=0$, then from the first two equations we get $2a+b=0$ and $b(a-1)=0$. This gives us two possibilities $a=1, b=-2$ OR $a=0,b=0$. Consequently the possible values for $a+b^2+c^3=5 \text{ or } 0$.
Now consider the case $ab=-1$. Can you proceed from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the set of points in $\mathbb{C}$ s.t. $|z| = Rez + 1$ I was given the above problem as a homework question and I've arrived at a solution but I am not sure if my approach is the best one or even a valid one. Any help would be great!
Consider $z = a + bi : a, b \in \mathbb{R}$ with $a = Rez, b = Imz$. Then we would like $|z| = \sqrt{a^2 + b^2} = a + 1 = Rez + 1.$ To arrive at this, we require $|z| = \sqrt{(a+1)^2} = \sqrt{a^2 + 2a + 1}$, which means $b = \sqrt{2a+1}$ so that $|z| = \sqrt{a^2 + b^2} = \sqrt{a^2 + \sqrt{2a+1}^2} = \sqrt{a^2 + 2a + 1} = \sqrt{(a+1)^2} = a + 1 = Rez + 1$. We also require $2a+1 > 0$ or $a > \frac{-1}{2}.$
This gives us the set $\{a + bi : a > \frac{-1}{2}, b = \sqrt{2a + 1}\}$
| Note that $\sqrt{a^2 + b^2} = \sqrt{a^2 + 2a + 1} \implies \color{red}{b^2 = 2a + 1} \implies b=\color{red}{\pm} \sqrt{2a+1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
Evaluate
$$ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$
I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
| Dividing numerator and denominator by $\sqrt{2}$
$$\lim_{x\to \frac{π}{2}} \frac {\sqrt{\frac{1+\cos 2x}{2}}}{\sqrt {\frac{π}{2}} -\sqrt { x}}$$
$$$$
Multiplying and dividing by $(\sqrt {\frac{π}{2} } + \sqrt {x})$
$$=\lim_{x\to \frac {π}{2}}(\sqrt {\frac{π}{2}} + \sqrt x)(\frac {\sqrt {\frac {1+ \cos 2x}{2}}}{\frac {π}{2} -x})$$
$$$$
As $\sqrt {\frac {1+\cos 2x}{2}} = \cos x$
$$=(\lim_{x\to \frac{π}{2}} \sqrt {\frac{π}{2}} +\sqrt{x})(\lim _{x\to \frac{π}{2}}\frac {\cos x}{\frac {π}{2} - x})$$
$$= \sqrt {2π} \lim_{x\to \frac{π}{2}} \frac {\sin (\frac {π}{2} - x)}{\frac {π}{2} - x}$$
$$$$
Let $t = \frac {π}{2} -x$
$$$$
As $x\to \frac{π}{2}$ then $t\to 0$
$$ =\sqrt{2π} \lim_{t\to 0} \frac {\sin t}{t}$$
$$\lim_{x\to \frac{π}{2}} \frac {\sqrt {1+\cos 2x}}{\sqrt{π}-\sqrt 2x}=\sqrt {2π}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $2\tan{2x}\leq3\tan{x}.$ Problem:
$$\text{Solve} \quad 2\tan{2x}\leq3\tan{x}.$$
A problem of this character will yield 5 points on an exam. However, having the correct answer does not suffice to get all the 5 points. Full stringency and mathematical accuracy, on top of a correct final answer, warrants full house. I've decided to present a (partial) solution here and I want you to help me really to comb through it and search for possible logical loopholes.
Solution attempt:
Application of the double angle formula for $\tan{x}$ on LHS yields:
$$\text{LHS}=2\cdot\frac{\sin{2x}}{\cos{2x}}=2\cdot\frac{2\sin{x}\cos{x}}{\cos^2{x}-\sin^2{x}}=[\text{Divide by} \ \cos^2{x}\neq0]=\frac{4\tan{x}}{1-\tan^2{x}}.$$
Setting $t=\tan{x}$ and moving the RHS over and subtracting gives the equivalent inequality:
$$\frac{4t}{1-t^2}-3t=\frac{4t-3t(1-t^2)}{1-t^2}=\frac{t(1+3t^2)}{(1+t)(1-t)}=p(t)\leq0.$$
Since the factor $(1+3t^2)>0, \ \forall x\in \mathbb{R}, $ it suffices to examine the signs of the factors $t, \ (1+t), \ (1-t)$ and the entire expression that I denoted $p(t).$ The following table emerges:
\begin{array}
{|l|cr}
t= & -\infty & -1 & \ & 0 & \ & 1 & +\infty\\
\hline
1+t & - & 0 & + & & + & & +\\
\hline
t & - & & - & 0 & + & & +\\
\hline
1-t & + & & + & & + & 0 & -\\
\hline
p(t)& + & \varnothing & - & 0 & + & \varnothing & -\\
\end{array}
This indicates that the solutionset of $p(t)\leq0$ is given by $t\in(-1,0]\cup(1,\infty).$ Hereafter I'm stuck, I don't know how to revert to $x$. How do I do this in an effective way?
| Duplication formula for tangent
$\dfrac{4 \tan x}{1-\tan ^2 x}\leq 3 \tan x$
Bring all in the LHS
$\dfrac{4 \tan x}{1-\tan ^2 x}-3 \tan x\leq 0$
add together
$-\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\leq 0$
which is better as
$\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\geq 0$
The parenthesis $\left(3 \tan ^2 x+1\right)$ is positive for any $x$ because is the sum of two squares so we need to see where
$\tan x \geq 0$ verified in $0\leq x <\dfrac{\pi}{2}\lor \pi\leq x <\dfrac{3\pi}{2}$
and $\tan ^2 x-1>0$ verified when $\tan x<-1\lor \tan x>1$
which is $\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{\pi }{2}<x<\dfrac{3 \pi }{4}\lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{3 \pi }{2}<x<\dfrac{7 \pi }{4}$
The solution is then
$\dfrac{\pi }{4}<x<\dfrac{\pi }{2}\lor \dfrac{3 \pi }{4}<x\leq\pi \lor \dfrac{5 \pi }{4}<x<\dfrac{3 \pi }{2}\lor \dfrac{7 \pi }{4}<x\leq2 \pi$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that for $n\to \infty, u_n - \sqrt{n} - \frac{1}{2} \sim \frac{1}{8\sqrt{n}}$ $\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ with $u_1 = 1$.
I've already shown that $ u_n \sim \sqrt{n}$ and that $\underset{n\rightarrow \infty}{\lim} u_n - \sqrt{n} = \dfrac{1}{2}$
How to show that $n\rightarrow \infty, u_n - \sqrt{n} - \dfrac{1}{2} \sim \dfrac{1}{8\sqrt{n}}$ ? (Or it might not be this, maybe something else ? I dont really know... I intuited this result with the asymptotic exampsion of $\sqrt{n}$.
I know that I have to use the asymptotic expansion, but I don't know how to use it correctly (I've some difficulties to keep the $o(1)$).
What I've done :
$u_n = \sqrt{n+u_{n-1}}= \sqrt{n}(1+\dfrac{1}{2\sqrt{n}} + o(1)) = \sqrt{n} + \dfrac{1}{2} + o(1)$
| Letting $u_n = \sqrt{n}+\frac{1}{2}+\varepsilon_n$ with $\lim_{n\to\infty} \varepsilon_n=0$, you have
$$
\sqrt{n}+\frac{1}{2}+\varepsilon_n = \sqrt{n+\sqrt{n}+\frac{1}{2}+\varepsilon_{n-1}}
$$
for all $n$. Doing a Taylor expansion of the RHS yields, using the fact that $\varepsilon_{n-1}=o(1)$:
$$\begin{align}
\sqrt{n}+\frac{1}{2}+\varepsilon_n &= \sqrt{n}\sqrt{1+\frac{1}{\sqrt{n}}+\frac{1}{2n}+o\left(\frac{1}{n}\right)}
\\
&= \sqrt{n}\left(1+\frac{1}{2\sqrt{n}}+\frac{1}{4n}+o\left(\frac{1}{n}\right)-\frac{1}{8n}\right)
\\
&= \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)
\end{align}$$
and so $\varepsilon_n = \frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)$, giving the result:
$$
\boxed{u_n = \sqrt{n}+\frac{1}{2}+\frac{1}{8\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)}
$$
| {
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Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$
I am unsure on how to continue and would really appreciate some guidance.
| In this answer we continue where the OP left off, making one last 'normalization' change:
$\tag 1 \frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4} = \frac{-(x-1)(3(x-1)+4)}{2(x-1)^2 + 4(x-1) +4}$
Denominator:
If $|x - 1| \lt 1/2$ then
$\quad |2(x-1)^2 + 4(x-1) +4| \gt |4(x-1) +4| \gt 2 $
Note that $4(x-1) \gt -2$.
$\text{ }$
Numerator:
If $|x - 1| \lt 1/2$ then
$\quad \; | -(x-1)\;(3\,(x-1)+4) | = |x-1| \, |3(x-1)+4| $
$ \qquad \quad \le | x-1|\;( 3 \,|x-1| + 4) \lt 5.5 \, |x-1| $
Use the triangle inequality.
Now the absolute value of the ratio expression (1) gets bigger as the numerator gets larger and the denominator gets smaller. So if $|x - 1| \lt 1/2$ and since $\frac{5.5}{2} = \frac{11}{4}$,
$\tag 2 \text{Absolute value of (1) } \lt \frac{11}{4} |x -1|$
We are now ready! Let the $\varepsilon \gt 0$ challenge be given and set
$\qquad \delta = min(\frac{1}{2},\frac{4 \varepsilon}{11})$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it possible to have a 3rd number system based on division by zero? Is it possible in mathematics to use a third number line based on division by zero; in addition to the real and imaginary number lines?
This is because some solutions blow up when there is a division by zero. Would it be possible to solve them with this new number line?
$\therefore$ on the z axis we would have $\frac{1}{0}$ , $\frac{2}{0}$ , $\frac{3}{0}$ , etc. where, $p = \frac{1}{0}$ and $p \cdot 0 = 1$ .
division by zero graph
Is this a viable number system?
A similar question to this one: Is there a third dimension of numbers?
| $\textbf {Rules of the p number system using 1/0 and $\infty$ }$
If $p = \frac{1}{0}$ and $p \cdot 0 = 1$. Then I suggest the following rules.
$$p - p = \frac{1}{0} - \frac{1}{0} = \frac{0 \cdot 1 - 0 \cdot 1}{0 \cdot 0} = \frac{0 \cdot (1 - 1)}{0 \cdot 0} = \frac{0}{0} = 1$$
problem lies here (if it is a problem?) $$\frac{0}{0} \neq \frac{0}{\delta x} = 0$$
where $\delta x \to 0$ (see previous answer), going by the rule of $\frac{0}{0} = 1$ .
$\underline{ RULES \, that \,appear\, to \,work}$
$$\frac{2}{3} + \frac{0}{0} = \frac{2 \cdot 0 + 3 \cdot 0}{3 \cdot 0} = \frac{5 \cdot 0}{3 \cdot 0} = \frac{5}{3}$$
$$p \cdot p = \frac{1}{0} \cdot \frac{1}{0} = \frac{1}{0^2} = p^2 = \infty \cdot \infty = \infty^2$$
$$p + p = \frac{1}{0} + \frac{1}{0} = \frac{0 \cdot 1 + 0 \cdot 1}{0 \cdot 0} = \frac{0 \cdot (1 + 1)}{0 \cdot 0} = \frac{2}{0} = 2p$$
$$p \cdot 0 = 1 \neq p \cdot (0 + 0)$$
$$p \cdot (0 + 0) = p \cdot 2 \cdot 0 = \frac{1}{0} \cdot 2 \cdot 0 = 2$$
$$0 + 0 + 0 \,+ ... = 3 \cdot 0 \,+ ...$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $k\mid 12$ with $2^k=1\bmod 13$ Let $k$ be the smallest positive integer such that $2^k=1\bmod 13$. Show that $k\mid 12$.
I'm not very good at proofs and I'm confused as how to prove this. I started by saying $2^k-1=13n$. But I don't know where to go from there.
| $$
\begin{array}{c|c}
\hline
i & \text{ $2^i $ mod 13} \\
\hline
1&2 \\
2&4 \\
3&8 \\
4&3 \\
5&6 \\
6&12 \\
7&11 \\
8&9 \\
9&5 \\
10&10 \\
11&7 \\
12&1 \\
\hline
\end{array}
$$
First, for $0 \le k \le 12$, only $2^{12} \equiv 2^0 \equiv 1 \quad mod(13)$
Assume $k=12i+j$, where $0 \le j < 12$, then
$$2^k \equiv 2^{12i+j} \equiv 2^{12i} \cdot 2^j \equiv 1\cdot 2^j \equiv 1 \quad mod(13)$$
So $2^j=1$, $j=0$, and $k=12i, k|12$
(smallest positive integer of k is 12, obviously)
| {
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Solve the following recurrence $f(n) = 2^n f(n - 1) + 2^n$ given $f(0)=1$. Solve the recurrence $f(n) = 2^n f(n - 1) + 2^n$ given $f(0)=1$.
The pattern I can come up with is
$2^{2n-1} f(n - 2) + 2^{3n-1}$
$2^{3n-3} f(n - 3) + 2^{6n-4}$
$2^{4n-6} f(n - 4) + 2^{10n-10}$
How do I solve the recurrence relation from this pattern? Thanks.
| We have
$f(n) = 2^n f(n - 1) + 2^n
$.
Since
$\sum_{k=1}^n k
= n(n+1)/2
$,
divide the recurrence by
$2^{ n(n+1)/2}$.
It becomes
$\dfrac{f(n)}{2^{ n(n+1)/2}}
= \dfrac{f(n - 1)}{2^{ n(n+1)/2-n}} + \dfrac1{{2^{ n(n+1)/2-n}}}
= \dfrac{f(n - 1)}{2^{ n(n-1)/2}} + \dfrac1{{2^{ n(n-1)/2}}}
$.
Now,
let
$g(n)
=\dfrac{f(n)}{2^{ n(n+1)/2}}
$.
Then
$g(n-1)
=\dfrac{f(n-1)}{2^{ (n-1)n/2}}
$
so the recurrence becomes
$g(n)
= g(n-1) + \dfrac1{{2^{ n(n-1)/2}}}
$,
or
$g(n)- g(n-1)
= \dfrac1{{2^{ n(n-1)/2}}}
$.
Summing,
$g(n)-g(0)
=\sum_{k=1}^n (g(k_)-g(k-1))
=\sum_{k=1}^n \dfrac1{{2^{ k(k-1)/2}}}
$,
or
$\dfrac{f(n)}{2^{ n(n+1)/2}} -f(0)
=\sum_{k=1}^n \dfrac1{{2^{ k(k-1)/2}}}
$
or
$f(n)
=2^{ n(n+1)/2}+2^{ n(n+1)/2}\sum_{k=1}^n \dfrac1{{2^{ k(k-1)/2}}}
$.
Solving this involves summing
$\sum_{n=1}^m \dfrac1{{2^{ n(n-1)/2}}}
$.
I don't know how to sum this,
so I'll leave it here.
(added later)
Following up on my thought
that this looked
like a theta function,
I looked in good old
Whittaker and Watson.
In section 21.11,
page 464,
I find this
(with a script theta that I don't know
how to enter):
$\theta_2(z, q)
=2\sum_{n=0}^{\infty} q^{(n+1/2)^2}\cos((2n+1)z)
$.
Therefore
$\begin{array}\\
\theta_2(0, q)
&=2\sum_{n=0}^{\infty} q^{(n+1/2)^2}\\
&=2q^{1/4}\sum_{n=0}^{\infty} q^{n^2+n}\\
\text{so}\\
\theta_2(0, r^{1/2})
&=2r^{1/8}\sum_{n=0}^{\infty} r^{(n^2+n)/2}
\qquad q = r^{1/2}\\
\text{so}\\
\theta_2(0, (1/2)^{1/2})
&=2(1/2)^{1/8}\sum_{n=0}^{\infty} (1/2)^{(n^2+n)/2}\\
&=2^{7/8}\sum_{n=0}^{\infty} (1/2)^{(n^2+n)/2}
\qquad r = 1/2\\
\end{array}
$
Therefore
$\sum_{n=0}^{\infty} \dfrac1{2^{(n^2+n)/2}}
=2^{-7/8}\theta_2(0, (1/2)^{1/2})
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \dfrac{1}{6}<\sin x<1\tag*{}$
$\implies 5-2\sin x\geq36\sin^2x+1-12\sin x\implies 18\sin^2x-5\sin x-2\leq0$
$\implies(2\sin x-1)(9\sin x+2)\leq0$
$\implies\sin x\ \epsilon\ \bigg(\dfrac{1}{6},\dfrac{1}{2}\bigg]$
All of above implies $\sin x\ \epsilon\ \bigg[-1,\dfrac{1}{2}\bigg]$.
Answer is given in the form: $\bigg[\dfrac{\pi(12n-7)}{6},\dfrac{\pi(12n+1)}{6}\bigg]\ (n\epsilon Z)$
How do I reach the form given in options? I even don't know what I've is correct or not.
Please help.
| Your solution is right.
You need only to write the answer, for which
just take on the $y$-axis the point $\frac{1}{2}$ and you need $\sin{x}\leq\frac{1}{2}$, which gives $y\leq\frac{1}{2}$ and the arc
$$\left[-\frac{7\pi}{6},\frac{\pi}{6}\right]$$
on the trigonometric circle and add $2\pi n$ in the both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$ the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$.
I know that the $x$-intercepts are $x=1$ and $x=-3$, and that the vertex is $(-1,4)$. I'm trying to use the distance formula by equating $\sqrt{14}$ to the function, but that is not getting me anywhere.
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$
| The locus of points whose distance from $(-1,2)$ is $\sqrt{14}$ is the circle
$(x+1)^2+(y-2)^2=14$
that is
$\color{red}{x^2+2 x}+y^2-4 y-9=0$
The parabola equation can be written as $\color{red}{x^2+2 x}=\color{blue}{3-y}\quad(*)$
Therefore the intersection points, the points of the parabola which are $\sqrt{14}$ from the given point, can be found solving
$\color{blue}{3-y} +y^2-4 y-9=0$
which gives $y_1=-1;\;y_2=6$
plugging back $y_1$ in $(*)$
$x^2+2x=4$ gives $x=-1\pm\sqrt 5$
while the other gives no real solutions
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right triangle with sides $1$, $7$ and $5\sqrt{2},$ I get that $$\begin{array}{lcl}
\sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\
\cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\
\end{array}$$
But this will not get me standard angles for $A.$
| Hint
$$\arctan x +\arctan y
=\arctan\left(\frac{x+y}{1-xy}\right) $$
See [Additivity of $\arctan(\frac{x+y}{1-xy})$]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate the integral $\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$ I'm looking for the value of this integral:
$$\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$$
I try to integrate it:
\begin{align}
I&=\int_0^1\arctan x\left(\frac{1}{x}-\frac{2x^2}{1+x^3}\right)dx\\
&=\int_0^1\frac{\arctan x}{x}dx-2\int_0^1\frac{x^2\arctan x}{1+x^3}dx\\
&=G-\frac{2}{3}\left(\arctan x\ln (1+x^3)|_0^1-\int_0^1\frac{\ln (1+x^3)}{1+x^2}dx\right)\\
&=G-\frac{\pi}{6}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (1+x)+\ln (x^2-x+1)}{1+x^2}dx\\
&=G-\frac{\pi}{6}\ln 2+\frac{2}{3}\frac{\pi}{8}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (x^2-x+1)}{1+x^2}dx\\
&=G-\frac{\pi}{12}\ln 2+\frac{2}{3}\int_0^1\frac{\ln (x^2-x+1)}{1+x^2}dx\\
\end{align}
Where G is the Catalan's constant,but I don't know how to do it next .
| $$\color{blue}{I = \frac{2}{9}\pi \ln (2 + \sqrt 3 ) - \frac{{\pi \ln 2}}{{12}} - \frac{G}{9}}$$
Let $$I_1 = \int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx \quad \quad I_2 = \int_1^\infty \frac{\ln(1+x^3)}{1+x^2}dx$$
Then subsituting $x\to 1/x$ in $I_1$ gives $$\tag{1} I_1 - I_2 = -3\int_1^\infty \frac{\ln x}{1+x^2} dx = -3G$$
Next we calculate $$I_1+I_2 = \int_0^\infty \frac{\ln(1+x^3)}{1+x^2} dx $$
Denote $$F(a) = \int_0^{ + \infty } {\frac{{\ln (a + {x^3})}}{{1 + {x^2}}}dx} $$
then, after some works, we obtain $$F'(a) = \int_0^{ + \infty } {\frac{1}{{(a + {x^3})(1 + {x^2})}}dx} = \frac{\pi }{2}\frac{a}{{1 + {a^2}}} + \frac{1}{6}\frac{{\ln a}}{{1 + {a^2}}} + \frac{{2\sqrt 3 \pi }}{9}\frac{{{a^{2/3}} - {a^{ - 2/3}}}}{{1 + {a^2}}}$$
Hence, after some calculations, $$\tag{2} I_1 + I_2 = F(0) + \int_0^1 {F'(a)da} = - \frac{G}{3} + \frac{{\pi \ln 2}}{4} + \frac{{2\pi }}{3}\ln (2 + \sqrt 3 ) $$
Combining (1) and (2) gives the value of $I_1$, and hence by what you have done
$$I = G-\frac{\pi}{6}\ln2 + \frac{2}{3} I_1 $$ we can find the value of $I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.
| Let $x = \tan\alpha$ and $y = \tan\beta$ for some $\alpha,\beta\in(-\pi/2,\pi/2)$. Then $\sqrt{x^2+1} = \sec\alpha$ and $\sqrt{y^2+1} = \sec\beta$, so the condition can be rewritten as
$$(\tan\alpha+\sec\alpha)(\tan\beta+\sec\beta) = 1. $$
Multiplying by $\cos\alpha\cos\beta$ yields
$$(\sin\alpha+1)(\sin\beta+1) = \cos\alpha\cos\beta. $$
Multiplying by $(1-\sin\alpha)(1-\sin\beta)$ yields
\begin{align} (1-\sin^2\alpha)(1-\sin^2\beta) &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies\cos^2\alpha\cos^2\beta &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta).
\end{align}
Thus
\begin{align} (1+\sin\alpha)(1+\sin\beta) = \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta) \\
\implies 1+\sin\alpha+\sin\beta+\sin\alpha\sin\beta &= 1-\sin\alpha-\sin\beta+\sin\alpha\sin\beta \\
\implies 2(\sin\alpha+\sin\beta) &= 0.
\end{align}
This implies that $\alpha = -\beta$, and hence $x = \tan\alpha = \tan(-\beta) = -\tan\beta = -y$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} $$
This limit will go to zero (I had tried using calculator manually).
I have tried to open the trigonometric identities and the fact that $ \lim\limits_{x \rightarrow 0}\frac{\sin (ax)}{x} = a $ gives :
$$ \lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{\cos (3x)( \cos (3x) - 1)}{6 x^{2}} $$
$$ \cos (3x) = \cos (2x) \cos(x) - \sin (2x) \sin(x) = \cos^{3}(x) - 3\sin^{2}(x) \cos(x) $$
$$ \lim_{x \rightarrow 0 } \frac{ \cos (3x) }{x^{2}} = \lim_{x \rightarrow 0} \frac{\cos^{3}(x)}{x^{2}} - 3 $$
Any view on this and more efficient way to solve this? Thanks before.
| \begin{align}
\lim_{x\to0} \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)}
&= \lim_{x\to0} \frac{ -2\sin^2 (\dfrac{3x}{2}) \cos (3x) }{ \sin (2x) \sin (3x)} \\
&= -2\lim_{x\to0} \dfrac{\sin^2 (\dfrac{3x}{2}) }{(\dfrac{3x}{2})^2} \dfrac{(2x)}{\sin (2x)} \frac{ (3x)}{ ( \sin (3x)}.\dfrac{(\dfrac{3x}{2})^2}{(2x)(3x)}\cos (3x) \\
&=\color{blue}{-\dfrac{3}{4}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find Taylor series of: $x^x$ Find the Taylor seriespolynomial form of:
$$x^x$$
My attempt:
I started calculating derivatives of $x^x$ for the series,
$$f(x)=x^x$$
$$f'(x)=x^x(\ln x+1)$$
$$f''(x)=x^{x-1}+x^x(\ln x+1)$$
$$f'''(x)=\cdots$$
$$\vdots$$
But i could get only till certain terms and got frustrated,
Attempt no. 2:
$$f(x)=x^x$$
$$f(x)=e^{x\ln x}$$
$$f(x)=1+x\ln x+\frac{x^2\ln^2 x}{2!}+\cdots$$
$$f(x)=1+x({x-1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots)+\frac{x^2}{2}({x-1}-\frac{(x-1)^2}{2}+\frac{(x-2)^3}{3}+\cdots)^2+\cdots$$
Now let's try to get $x^n$'s coefficient,
$$C(x^0)=1$$
In the bracket multiplied to x,
$$1+x(x-1-x^2+2x-1+x^3-3x+3x-1+\cdots)+\cdots$$
Following the lead
$$1+x(g(x)-1-1-1-1\cdots)+\cdots$$
$$C(x^1)=-1-1-1-\cdots$$
So,
$$C(x^1)=-\infty\cdots?$$
How do we write this equation?
Attempt 3:
$$f(x)=1+(x+1)\ln (x+1)+\frac{(x+1)^2}{2!}\ln^2(x+1)+\cdots$$
Now,
$$f(x)=1+(x+1)(x+\frac{x^2}{2}+\frac{x^3}{3}\cdots)+\cdots$$
Now,
$$C(x^0)=1$$
Considering $(1+x)^2,(1+x)^3\cdots$ contribution to$C(x^1)$
$$C(x^1)=1+1+1+1+\cdots$$
$$C(x^1)=\infty\cdots ?$$
Still.....stuck
| Hint: We can find an elaboration of the $n$-th derivative of $x^x$ in an example (p. 139) of Advanced Combinatorics by L. Comtet. The idea is based upon a clever Taylor series expansion. Using the differential operator $D_x^j:=\frac{d^j}{dx^j}$ the following holds:
The $n$-th derivative of $x^x$ is
\begin{align*}
\color{blue}{D_x^n x^x=x^x\sum_{i=0}^n\binom{n}{i}(\ln(x))^i\sum_{j=0}^{n-i}b_{n-i,n-i-j}x^{-j}}\tag{1}
\end{align*}
with $\color{blue}{b_{n,j}}$ the Lehmer-Comtet numbers.
These numbers follow the recurrence relation
\begin{align*}
b_{n+1,j}=(j-n)b_{n,j}+b_{n,j-1}+nb_{n-1,j-1}\qquad\qquad n,j\geq 1
\end{align*}
and the first values, together with initial values are listed below.
\begin{array}{c|cccccc}
n\setminus k&1&2&3&4&5&6\\
\hline
1&1\\
2&1&1\\
3&-1&3&1\\
4&2&-1&6&1\\
5&-6&0&5&10&1\\
6&24&4&-15&25&15&1\\
\end{array}
The values can be found in OEIS as A008296. They are called Lehmer-Comtet numbers and were stored in the archive by N.J.A.Sloane by referring precisely to the example we can see here.
Example: $n=2$
Let's look at a small example. Letting $n=2$ we obtain from (1) and the table with $b_{n,j}$:
\begin{align*}
D_x^2x^x&=x^x\sum_{i=0}^2\binom{2}{i}(\ln(x))^i\sum_{j=0}^{2-i}b_{2-i,2-i-j}x^{-j}\\
&=x^x\left(\binom{2}{0}\sum_{j=0}^2b_{2,2-j}x^{-j}+\binom{2}{1}\ln(x)\sum_{j=0}^1b_{1,1-j}x^{-j}\right.\\
&\qquad\qquad\left.+\binom{2}{2}\left(\ln(x)\right)^2\sum_{j=0}^0b_{0,0-j}x^{-j}\right)\\
&=x^x\left(\left(b_{2,2}+b_{2,1}\frac{1}{x}+b_{2,0}\frac{1}{x^2}\right)+2\ln(x)\left(b_{1,1}+b_{1,0}\frac{1}{x}\right)
+(\ln(x))^2b_{0,0}\right)\\
&=x^x\left(1+\frac{1}{x}+2\ln(x)+\left(\ln(x)\right)^2\right)
\end{align*}
in accordance with the result of Wolfram Alpha.
Note: A detailed answer is provided in this MSE post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Multivariable $\epsilon-\delta$ proof verification Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$
Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this case I have $x$ and $y$ approaching something else.
$(x,y) \rightarrow (1,1) \Leftrightarrow \lvert\lvert (x,y)-(1,1)\lvert\lvert \rightarrow 0$ which can be written as $0 < \sqrt {(x-1)^2+(y-1)^2} < \delta$ for some arbitrarily small $\delta >0$.
Goal: show that $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that
$0 < \sqrt {(x-1)^2+(y-1)^2}<\delta\Rightarrow0<|xy-1|<\epsilon$
Proof:
If $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$, then
$|xy-1|=|xy-x-y+1+x+y-2|=|(x-1)(y-1)+(x-1)+(y-1)|$
$\le|(x-1)(y-1)|+|x-1| +|y-1|=|x-1||y-1|+|x-1|+|y-1|$
$=(\sqrt{(x-1)^2})(\sqrt{(y-1)^2})+\sqrt{(x-1)^2}+\sqrt{(y-1)^2}$
$\le(\sqrt{(x-1)^2+(y-1)^2})^2+2\sqrt{(x-1)^2+(y-1)^2}<\delta^2+2\delta$
If $\delta=$$\epsilon \over4$, then $\delta^2+2\delta=\frac{\epsilon^2}{16}+\frac{8\epsilon}{16}=\frac{\epsilon^2+8\epsilon}{16}$
Now, $\frac{\epsilon^2+8\epsilon}{16}<\epsilon \Leftrightarrow \epsilon^2+8\epsilon<16\epsilon\Leftrightarrow\epsilon(\epsilon-8)<0$
Since $\epsilon>0$, we have $\epsilon(\epsilon-8)<0$ if and only if $\epsilon<8$
So if we have $\epsilon<8$, then we can pick $\delta=\frac{\epsilon}{4}$ which gives us $\delta^2+2\delta<\epsilon$. If $\epsilon \ge8$, then we can pick $\delta=2$ which gives us $\delta^2+2\delta\le\epsilon$.
Therefore, if we pick $\delta =$ min {${\frac{\epsilon}{4}, 2}$}, then $0 < \sqrt {(x-1)^2+(y-1)^2}<\delta$ implies $0<|xy-1|<\epsilon$
Thus, $\lim\limits_{(x,y) \to (1,1)} xy=1$
| A shorthand you might find useful:
Since
$$(x-y)^2 \gt 0 \Rightarrow xy \lt \frac{x^2+y^2}{2} $$
Also, note that
$$\lVert (x-1, y-1) \rVert \lt \delta \Rightarrow |x-1| \lt \delta \;\land\; |y-1| \lt \delta \Rightarrow x \lt \delta+1 \;\land\; y \lt \delta+1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $, $\forall$ $x$ $\in$ $[0,n]$
Prove that:
$$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $$ $\forall$ $x$ $\in$ $[0,n].$
I think it's easy, but I'm very stuck in this question, I need to prove that
$$\left(1- \frac{x}{n} \right)^n \leq \left(1- \frac{x}{n+1} \right)^{n+1}, \hspace{0.1cm} \forall \hspace{0.1cm} x \in [0,n]. $$
Any help?
| $$\left(1-\frac{x}{n+1}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n}-\frac{x}{n+1}\right)^{n+1}=$$
$$=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^n\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)\geq$$
$$\geq\left(\left(1-\frac{x}{n}\right)^n+n\left(1-\frac{x}{n}\right)^{n-1}\cdot\frac{x}{n(n+1)}\right)\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right).$$
Thus, it remains to prove that
$$\left(\left(1-\frac{x}{n}\right)^n+\left(1-\frac{x}{n}\right)^{n-1}\cdot\frac{x}{n+1}\right)\left(1-\frac{x}{n+1}\right)\geq\left(1-\frac{x}{n}\right)^n,$$
which is
$$\frac{x^2}{n(n+1)^2}\left(1-\frac{x}{n}\right)^{n-1}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$ Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I split into partial fractions hoping that some of them will cancel out and the sum might telescope.
I hope theres a simpler method to this.
Edit
As a side-question, can we extend this to $k$ factors in denominator, something like
$$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)...(r+(k-1)d)}$$
Using N.S answer, we can extend it easily:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$
|
And solving them takes much time.
There is a trick so solving partial fractions quickly.
$$\frac{1}{(r)(r + d)(r + 2d)(r + 3d)}
=
\frac{A}{r} + \frac{B}{r + d} + \frac{C}{r + 2d} + \frac{D}{r + 3d}$$
$$1
=
A(r + d)(r + 2d)(r + 3d) +
B(r)(r + 2d)(r + 3d) +
C(r)(r + d)(r + 3d) +
D(r)(r + d)(r + 2d)$$
The above is true for all $r$. So plug in the roots of the original equation. $r = 0$ gives:
$$1 = A(d)(2d)(3)$$
$$A = \frac{1}{6d^3}$$
Then $r = -d$ :
$$1 = B(-d)(d)(2d)$$
$$B = \frac{-1}{2d^3}$$
Then $r = -2d$:
$$1 = C(-2d)(-d)(d)$$
$$C = \frac{1}{2d^3}$$
Last $r = -3d$:
$$1 = D(-3d)(-2d)(-d)$$
$$D = \frac{-1}{6d^3}$$
So you get:
$$S = \frac{1}{6d^3}\left( \frac{1}{r} - \frac{3}{r + d} + \frac{3}{r + 2d} - \frac{1}{r + 3d}\right)$$
After you do it a few times and get the pattern, you can almost do it in your head. I suggest looking at it as 2 telescoping series:
$$\text{Sum} = \left(\frac{1}{6d^3} \sum_{r = 1}^n \frac{1}{r} - \frac{1}{r + 3d}\right) +
\left(\frac{-1}{2d^3} \sum_{r = 1}^n \frac{1}{r + d} - \frac{1}{r + 2d}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(t) = rt + s$, then for a general circle $\mathrm C(x, y) = (x - \alpha)^2 + (y - \beta)^2 - \gamma^2$ we can substitute the the parameters of the line and get
$$\mathrm C(x(t), y(t)) = a t^2 + bt + c = 0,$$
where $a,b,c \in \Bbb R$ and they depend on $p$,$q$.
Now if this quadratic has $2$ roots then the circle intersects the line at $2$ points and same for $1$ and $0$ roots.
If we are given that $\mathrm Z$ is tangent then we can using completing the square and get $(t - d(p, q) ) ^2 + e(p, q) = 0$.
So our condition for the line tangent to the circle would be $e(p, q) = 0$.
Parameterising the three equations,
$\mathbf L^\prime (l) = (2,0) + l (0,1)$, $\mathbf M^\prime(m) = (0, 5) + m(-1, 0)$ and $\mathbf N^\prime(n) = (2 , -1) + n(4,3)$.
Substituting these in the general equation of a circle, $\mathrm C(x, y) = (x - h)^2 + (y - k)^2 - r^2$.
$$\left\{ \begin{align} \mathrm{C(L_x, L_y)} &= (2-h)^2 +(l-k)^2 -r^2 \\ \mathrm{C(M_x, M_y)} &= (m+h)^2 +(5-k)^2 -r^2 \\ \mathrm{C(N_x, N_y)} &= (2+4n-h)^2 +(3n-1-k)^2 -r^2 \end{align} \right.$$
As per the above method,
From first equation we get $r^2 = (2 - h) \iff r = |2 - h|$ and from second equation, $r = |5 - k|$.
Completing the square on the third equation we get,
$$\mathrm{C(N_x, N_y)} = \left(5n + \dfrac{5 - 4h - 3k}{5}\right) + 5 - 4h + 2k - r^2 + h^2 + k^2 - \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
Therefore,
$$ 2 + 2k + h^2 + k^2 = r^2 + 4h + \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$
So here we got three equations:
\begin{align} 2 +2k +h^2 +k^2 &= r^2 +4h +\left[1 - \dfrac{4h-3k}{5}\right]^2 \\ r &= |5-k| \\ r &= |2-h|. \end{align}
I don't mind solving the third equation after substituting for $r$ from first in it, it is just a quadratic. What I don't understand is whether I should take $r = + (5 - k)$ or $r = - (5 - k)$ and $r = +(2 - h)$ or $r = - (2 - h)$?
Which two out of those should I take? And why?
| converting the equation in the Hessian Normalform we have
$$\frac{|3x-4y-10|}{5}=R$$ where $$(x,y)$$ denotes the middle Point of our searched circle; and it must be $$x=2+R,y=5-R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve $\arcsin{(2x^2-1)}+2\arcsin{x} = -\frac{\pi}{2}$ First of, we have to restrict the domain of the equation by looking at the argument of the first term. The domain for $\arcsin$ is $[-1,1]$, so the inequality $ -1\leq 2x^2-1 \leq1$ has to hold. So $x\geq0$ and $-1\leq x\leq 1$. It follow that the domain is $[-1,1]\cap[0,\infty)=[0,1].$ So the real domain of the equation remains $[0,1]$ and if there exists $x\in\mathbb{R}$ that solves the equation it should belong to that domain.
Lets solve it by rewriting $$\arcsin{(2x^2-1)}=-\frac{\pi}{2}-2\arcsin{x}=-\left(\frac{\pi}{2}+2\arcsin{x}\right).$$
Sine of both sides gives $$2x^2-1=\sin\left(\frac{\pi}{2}+2\arcsin{x}\right)=\cos{(2\arcsin{x})}=\cos^2{(\arcsin{x})}-\sin^2{(\arcsin{x})}.$$
Using $\cos^2{x}=1-\sin^2{x}$ we get
$2x^2-1=1-2\sin^2(\arcsin{x})=1-2x^2 \Longleftrightarrow2x^2-1\Longleftrightarrow x_{1,2}=\pm\frac{1}{\sqrt{2}}.$
Only $x=\frac{1}{\sqrt{2}}\in{[0,1]},$ which is the only root.
According to wolfram alpha, the only real solutions are $x=0$ and $x=-1.$ Where did I go wrong?
| Hint :
Let $\arcsin x=u\implies-\dfrac\pi2\le u\le\dfrac\pi2\iff-\pi\le2u\le\pi$ and $x=\sin u$
$\arcsin(2x^2-1)=\arcsin(-\cos2u)=-\arcsin(\cos2u)=\arccos(\cos2u)-\dfrac\pi2$
Now $\arccos(\cos2u)=\begin{cases}2u &\mbox{if } 0\le2u\le\pi \\-2u& \mbox{if } 0\le-2u\le\pi\iff-\pi\le 2u\le0&\\2\pi-2u& \mbox{if } 0\le2\pi-2u\le\pi\iff\pi\le 2u\le2\pi\\2\pi+2u& \mbox{if } 0\le2\pi+2u\le\pi\iff-2\pi\le2u\le-\pi\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $y(1)$ if $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt$ Consider the integral equation $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt, x\in[0,\pi].$ Then the value of y(1) is
$1. \;\;19/20 \\
2. \;\;1\\
3. \;\;17/20\\
4. \;\;21/20 $
My Attempt: $y(x)=x^3+\int_0^x \sin(x-t) \ y(t)dt \\ \Rightarrow y(x)=x^3+\int_0^x(\sin x \cos t-\cos x \sin t)\ y(t)dt \\ \Rightarrow y(x)=x^3+\sin x\int_0^x \cos t\ y(t)dt -\cos x\int_0^x \sin t\ y(t)dt................(1)\\
y'(x)=3x^2+ \sin x \cos x\ y(x)+\cos x\int_0^x \cos t\ y(t)dt-\cos x\sin x \ y(x)+\sin x\int_0^x \sin t\ y(t)dt \\ y'(x)=3x^2+\cos x\int_0^x \cos t \ y(t)dt+\sin x\int_0^x \sin t\ y(t)dt\\ y''(x)=6x+\cos x(\cos x\ y(x))-\sin x\int_0^x \cos t\ y(t)dt+\sin x(\sin x\ y(x))+\cos x\int_0^x \sin t\ y(t)dt \\ y''(x)=6x+(\cos^2x+\sin^2x)\ y(x)-(\sin x\int_0^x \cos t \ y(t)dt-\cos x\int_0^x \sin t \ y(t)dt) \\ \\ y''(x)=6x+y(x)-(y(x)-3x^2)\\ y''(x)=6x+3x^2\\$
now integrating twice we get
$$y(x)=x^3+ \frac{x^4}{4}+c_1x+c_2 $$
Since $y(0)=0$, we get $c_2=0$.
I don't know how to solve further to find $y(1)$.
| In the next-to-last line, you should have $y(x) - x^3$, so that $y(x) = x^3 + \frac{x^5}{20} + c_1x + c_2$. Then since line 5 implies $y'(0) = 0$, both $c_1$ and $c_2$ are zero, so that $y(x) = x^3 + \frac{x^5}{20}$, and $y(1) = \frac{21}{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$(xy+yz+zx)(1+6xyz) \geq 11xyz$ I've came across this inequality:
Let $x>$, $y>0$ and $z > 0$ with $x+y+z=1$.
Prove that
$$(xy+yz+zx)(1+6xyz) \geq 11xyz.$$
I don't know where to take it from, I've tried means inequality, but it didn't help me.
Some hints would be great!
| It's $$6xyz(xy+xz+yz)+(xy+xz+yz)(x+y+z)^3\geq11xyz(x+y+z)^2$$ or
$$\sum_{sym}(x^4y+3x^3y^2-2x^3yz-2x^2y^2z)\geq0,$$
which is true by Muirhead.
Also, we can prove the last inequality by AM-GM.
Indeed, $$\sum_{cyc}x^4y=\frac{1}{13}\sum_{cyc}(9x^4y+y^4z+3z^4x)\geq$$
$$\geq\frac{1}{13}\cdot13\sum_{cyc}\sqrt[13]{\left(x^4y\right)^9\left(y^4z\right)\left(z^4x\right)^3}=\sum_{cyc}x^3yz;$$
$$\sum_{cyc}x^4z=\frac{1}{13}\sum_{cyc}(9x^4z+y^4x+3z^4y)\geq$$
$$\geq\frac{1}{13}\cdot13\sum_{cyc}\sqrt[13]{\left(x^4y\right)^9\left(y^4z\right)\left(z^4x\right)^3}=\sum_{cyc}x^3yz;$$
$$\sum_{cyc}(x^3y^2+x^3z^2)\geq2\sum_{cyc}\sqrt{x^3y^2x^3z^2}=2\sum_{cyc}x^3yz$$
and
$$\sum_{cyc}x^3yz=\frac{1}{2}xyz\sum_{cyc}(x^2+y^2)\geq\frac{1}{2}xyz\cdot2\sum_{cyc}\sqrt{x^2y^2}=xyz\sum_{cyc}xy=\sum_{cyc}x^2y^2z.$$
Also, there are easy way by SOS:
$$\sum_{sym}(x^4y+3x^3y^2-2x^3yz-2x^2y^2z)=$$
$$=\sum_{cyc}(x^4y+x^4z+3x^3y^2+3x^3z^2-4x^3yz-4x^2y^2z)=$$
$$=\sum_{cyc}(x^4y+xy^4-x^3y^2-x^2y^3)+$$
$$+2\sum_{cyc}(z^3x^2+z^3y^2-2z^3xy+x^3z^2+y^3z^2-x^2z^2y-y^2z^2x)=$$
$$=\sum_{cyc}(x-y)^2(xy(x+y)+2z^3+2z^2(x+y))\geq0.$$
There are also easy proofs by uvw and by Rearrangement, but Muirhead it's the best, I think.
| {
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"url": "https://math.stackexchange.com/questions/2445128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find generating function for the sequence 0, 0, 0, 0, 3, 4, 5, 6, ...
1.Derive the generating function for the sequence $$0, 0, 0, 0, 3, 4, 5, 6, . . .$$
2.Derive the generating function for the sequence $$0, 0, −12, 36, −108, 324, .. .$$
So the first function looks like $3x^4 + 4x^5+5x^6...$ = $x^4(3+4x+5x^2...)$ and that looks like the generating function for sequence $0,1,2,3,...$
I assume the answer for the first one is something like $\frac{x^4()}{1-x}$
, what am I missing?
For the second one if I factor $12$ out, ($-x^2+3x^3-9x^4...)$, is it something like $\frac{12x^2}{1+3x}$
| For $|x|<1$ we obtain:
$$3x^4+4x^5+...=x^2(3x^2+4x^3+...)=x^2(x^3+x^4+...)'=$$
$$=x^2\cdot\left(\frac{x^3}{1-x}\right)'=\frac{x^4(3-2x)}{(1-x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a $2 \times 2$ real matrix $A$ such that $A^2=-4I$? Does there exist a $2 \times 2$ matrix $A$ with real entries such that $A^2=-4I$ where $I$ is the identity matrix?
Some initial thoughts related to this question:
*
*The problem would be easy for complex matrices, we could simply take identity matrix multiplies by $2i$.
*There is another question on this site showing that this has no solution for $3\times3$ matrices, since the LHS has determinant $\det(A^2)=\det^2(A)$ which is a square of real number, but determinant of $-4I_3$ is negative. But the same argument does not work for $2\times2$ matrices, since $\det(-4I_2)=4$ is positive.
| A 2-by-2 matrix is not too bad to work out the algebraic equation for its coefficients:
Let
$
A=\begin{bmatrix} a & b \\c & d\end{bmatrix}
$. We end up with the matrix equation:
$$
A^2 = \begin{bmatrix} a^2+bc & b(a+d) \\c(a+d) & d^2 + bc\end{bmatrix}
= -4 \, I = \begin{bmatrix} -4 & 0 \\0 & -4\end{bmatrix}.
$$
So we have 4 equations
$$
\begin{equation}
a^2+bc = -4 \quad (1)\\
b(a+d) = 0 \quad (2) \\
c(a+d) = 0 \,\quad (3) \\
d^2+bc = -4 \quad (4).
\end{equation}
$$
It is easy to see that $a + d = 0$. Since if otherwise $a+d \not=0$, equations (2) and (3) it would imply that: $b=c=0$, then from equation (1): $a^2 = -4$ which can't happen.
Since $a+d=0$, we have $a^2 = d^2$. Thus equations (1) and (4) are identical. Therefore, the conclusion is that there are infinitely many solutions, as long as $a,b,c,d$ satisfy that
$$
\begin{cases}
a+d=0 \quad &(5)\\
a^2 = -(4+bc) \quad &(6).
\end{cases}
$$
Since there are 4 variables with 2 equation, one can choose 2 free variables and derive the others from the free ones. The simplest is from suggestion of user "zwim":
$$
A=\begin{bmatrix} a & -\frac{a^2+4}{c} \\c & -a\end{bmatrix},
$$
where $a$ and $c$ are any real numbers.
Example: choose a = 1, c = -5 we have:
$
A=\begin{bmatrix} 1 & 1 \\-5 & -1\end{bmatrix}.
$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Multivariate random vector with normal distribution Let $X=(X_1, X_2)'\in N(\mu, \Lambda),$ where
$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_1=3).$
Don't know even where to start.
| Currently your question says
$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_3=3).$
I will assume you just forgot $X_3.$ You have
$$
\begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} \sim N\left( \begin{bmatrix} 1 \\ 1 \\ \mu_3 \end{bmatrix} , \begin{bmatrix} 3 & 1 & \sigma_{1,3} \\ 1 & 2 & \sigma_{2,3} \\ \sigma_{1,3} & \sigma_{2,3} & \sigma_{3,3} \end{bmatrix} \right).
$$
Let $Y = X_2+3X_3.$ Then
$$
\begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix},
$$
so that
$$
\operatorname{E} \begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \mu_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 + 3\mu_3 \end{bmatrix}
$$
and
$$
\operatorname{var} \begin{bmatrix} X_1 \\ Y \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 & \sigma_{1,3} \\ 1 & 2 & \sigma_{2,3} \\ \sigma_{1,3} & \sigma_{2,3} & \sigma_{3,3} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 1+3\sigma_{1,3} \\ 1+3\sigma_{1,3} & 2+6\sigma_{2,3} + 9\sigma_{3,3} \end{bmatrix}.
$$
Find that bivariate normal density and plug in $3$ in place of the second argument, for $Y=3.$ Don't worry about the normalizing constant; you can read of the expected value and the variance from other aspects of the form of the density function.
Or alternatively, use one of the usual formulas for the conditional distribution of one component of a bivariate normal random variable given the value of the other component.
| {
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Solve the inequality for $x$: $\log_4 (x^2 − 2x + 1) < \log_2 3$ Solve the inequality for $x$: $$\log_4 (x^2 − 2x + 1) < \log_2 3$$
I got two answers and I'm not sure if I did it correctly.
1st ans: $(-2,1)\cup (1,4)$
2nd ans: $x \neq -2,4$
| By the change of base formula, $\log_4{x^2-2x+1}$ is equal to $\frac{\log_2{x^2-2x+1}}{\log_2{4}}$, or $\frac{1}{2} \log_2{x^2-2x+1}$. Since both sides are in the same base, we can directly solve the inequality:
$$\frac{1}{2} \log_2{x^2-2x+1} < \log_2{3}$$
$$\log_2{x^2-2x+1} < 2 \log_2{3}$$
$$x^2-2x+1 < (2^{\log_2{3}})^2$$
$$x^2-2x+1<9$$
$$x^2-2x-8<0$$
Now factorise the quadratic equation to get its roots, then set up the inequality. For the second question, when does $\log_4{x^2+2x+1}$ equal $0$? How can you apply this constraint to your inequality?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Other idea to find range of $y=x+\frac{1}{x-4}$ $$y=x+\frac{1}{x-4}$$ I tried to find range of this function as below
$$y(x-4)=x(x-4)+1\\yx-4y=x^2-4x+1\\x^2-x(4+y)+(4y+1)=0\\\Delta \geq 0 \\(4+y)^2-4(4y+1)\geq 0 \\(y-4)^2-4\geq 0\\|y-4|\geq 2\\y\geq 6 \cup y\leq2$$ this usuall way. but I am interested in finall answer... is the other idea to find function range ?
| use this fact : $|a+\frac 1a|\geq 2$
$$\quad{y=x+\frac{1}{x-4}\\y-4=x+\frac{1}{x-4}-4\\y-4=(x-4)+\frac{1}{x-4}=a+\frac 1a\\|y-4|\geq 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorize polynomials function The task was:
Find the set of zeros (roots) of the following function $$f(x)=x^4-6x^2-8x+24$$
What I did:
I found the possibles roots $$\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24 $$
and I found that X=2 but I cannot get the other 3 roots
| You found that $x = 2$ is a root of $f(x) = x^4 - 6x^2 - 8x + 24$. This means $x - 2$ is a factor. Dividing $f(x)$ by $x - 2$ yields
\begin{align*}
f(x) & = x^4 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^3 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) + 4x^2 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) - 2x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 4x - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 12x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) - 2x(x - 2) - 12(x - 2)\\
& = (x - 2)(x^3 + 2x^2 - 2x - 12)
\end{align*}
Now apply the Rational Roots Theorem and Factor Theorem to
$$g(x) = x^3 + 2x^2 - 2x - 12$$
to find another root, which may be the same as the first root.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$, define $\tan x$ and angle $x$. If: $$\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$$ $(0^\circ < x < 90^\circ$), define $\tan x$ and angle $x$.
Because of this in the parenthesis I know that I have to use some relation between complementary angles ex. $3\cos x = 3\sin(90^\circ-x)$ but when I include that in my problem I don't see any progress? I also don't really know what to do with the $2$ on the RHS. Any hints?
| \begin{align}
& \frac {9\sin x-3\cos x}{2\sin x + \cos x}=2 \\[10pt]
\text{Therefore } & \frac{9\tan x - 3}{2\tan x + 1} = 2 \\[10pt]
& \text{and so on.}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2454622",
"timestamp": "2023-03-29T00:00:00",
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} |
How did x become -sqrt(x^2)?
In computing the limit as $x \to -\infty$, we must remember that for $x<0$, we have $\sqrt{x^2} = |x| = -x$. So when we divide the numerator by $x<0$, we get
$$
\frac{\sqrt{2x^2+1}}{x}
= \frac{\sqrt{2x^2+1}}{-\sqrt{x^2}}
= - \sqrt{\frac{2x^2+1}{x^2}}
= - \sqrt{2 + \frac{1}{x^2}}
$$
How did the denominator change from $x$ to $-\sqrt{x^2}$ ? I can't seem to understand.
I'm mainly confused by where the negative came from.
| For $x<0$ $$\sqrt{x^2}=|x|=-x,$$
which says $x=-\sqrt{x^2}.$
| {
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"url": "https://math.stackexchange.com/questions/2454812",
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How can I show $z^2 = 6x^2 + 2y^2$ has no non-trivial integer solutions? EDIT: I missed a part of the question as there was a typo in my notes (this is part of my working on a proof using the Bruck–Ryser–Chowla theorem which was presented incorrectly in my version of the lecture notes). I have updated the question and shown some new working.
I was thinking of supposing that $x^2$, $y^2$ & $z^2$ have no common factors (because if they did, just divide out by that factor), and then considering possible solutions to $z^2 - 2y^2 = 6x^2$ modulo 6. See, we know $z^2 \equiv 2y^2 \mod 6$ which gives us six options.
*
*$z^2 \equiv 2y^2 \equiv 0 \mod 6$
*$z^2 \equiv 2y^2 \equiv 1 \mod 6$
*$z^2 \equiv 2y^2 \equiv 2 \mod 6$
*$z^2 \equiv 2y^2 \equiv 3 \mod 6$
*$z^2 \equiv 2y^2 \equiv 4 \mod 6$
*$z^2 \equiv 2y^2 \equiv 5 \mod 6$
In each case, I was thinking I could arrive to a contradiction, but I'm having a little trouble with the last five cases.
*
*Let $z^2 = 6 \alpha$ and $2y^2 = 6 \beta$. Then $x^2$, $y^2$ & $z^2$ all have a factor of 3 which is impossible.
*Let $z^2 = 6 \alpha + 1$ and $2y^2 = 6 \beta + 1$ ...
*Let $z^2 = 6 \alpha + 2$ and $2y^2 = 6 \beta + 2$ ...
*Let $z^2 = 6 \alpha + 3$ and $2y^2 = 6 \beta + 3$ ...
*Let $z^2 = 6 \alpha + 4$ and $2y^2 = 6 \beta + 4$ ...
*Let $z^2 = 6 \alpha + 5$ and $2y^2 = 6 \beta + 5$ ...
Any comments on the general proof structure or help with the last two cases would be greatly appreciated! Cheers.
| Assume $(x,y,z)$ is a solution to $z^2 = 6x^2 + 2y^2 $ with smallest positive $|z|$, then $\gcd(x,y) = 1$. Putting $z=2z_1$ gives
$$2z_1^ 2 = 3x^2 + y^2 $$ Since $x,y$ are both odd, RHS is divisible by $4$, so $z_1$ is even, that is $8\mid (2z_1^2)$. Hence $3x^2 + y^2 \equiv 0 \pmod{8}$. But the square of an odd number modolo $8$ must be $1$, a contradiction.
Hence $(0,0,0)$ is the only solution.
| {
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Prove that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ without calculus? I'm stuck with this sample RMO question I came across:
Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...
In the solution to this problem, I found the solver making the assumption,
$\sqrt[n]{n}>\sqrt[n+1]{n+1}$ for $n \geq 3$ How would you prove this?
Any help would be greatly appreciated.
EDIT: In this competition, you aren't allowed to use calculus. Non-calculus methods would be appreciated.
| Assume $n\geq 3.$ Start with this calculation:
$$\frac{(n+1)^n}{n^n} = \left(1+\frac{1}{n}\right)^n = 1+\binom{n}{1}\frac{1}{n} +\binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} +\cdots + \frac{1}{n^n}.$$
In the $k$th term, the numerator of the binomial coefficient is $n(n-1)(n-2)\cdots (n-k+1)$ which is less than $n\cdot n\cdots n=n^k.$ So the binomial expansion above is less than
$$1+1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \frac{1}{n!},$$
which, in turn is less than
$$1+1+ \frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n} < 3\leq n. $$
So we have $$n > \frac{(n+1)^n}{n^n}$$
$$n^{n+1} > (n+1)^n$$
$$n^{(n+1)/n} > n+1$$
$$n^{1/n} > (n+1)^{1/(n+1)}.$$
| {
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Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$
is this method correct?
| Setting $$y=tx$$ then we get
$$\frac{x^3+t^3x^3}{x^2+t^2x^2}=x\frac{1+t^3}{1+t^2}$$ this tends to Zero if $x$ tends to zero
| {
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Need hint for solving the following problem
If $f(x^{500}-1)=5x^{1015}+3x^{244}+7x+10$. Find the sum of coefficients of $f(x^5+1)$.
Let $x^{500}-1=y$, then $f(y)=5(y+1)^\frac{1015}{500}+3(y+1)^\frac{244}{500}+7(y+1)^\frac{1}{500}+10$.
But, I don't think that it is a right approach as each term of $f(y)$ gets transformed into an infinite series via binomial expansion, in that case it is not possible to sum the coefficients.
Another approach(This may be wrong!!,Please check)
Let,$x^{500}-1=x^5+1\implies x^{500}-x^5-1=1$.So,$f(x^5+1)=5(x^{500}-x^5-1)x^{1015}+3(x^{500}-x^5-1)x^{244}+7(x^{500}-x^5-1)x+10(x^{500}-x^5-1)$
So,the sum of coeffecients$=5-5-5+3-3-3+7-7-7+10-10-10=-25$
Please give some hints.
| I think a good hint is to take a look at factor theorem!
| {
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Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{align}
(x-a)^3 = (\sqrt[3]{2}(b+c\sqrt[3]{2}))^3 = 2 (b+c\sqrt[3]{2})^3
\end{align}
but this does not seems good.
From the argument of
\begin{align}
x^3-2 = (x-\sqrt[3]{2})(x-\sqrt[3]{2}w)(x-\sqrt[3]{2}w^2)
\end{align}
where $w^2+w+1=0$,
I came up with some idea, and do the computation via mathematica. And i figure out the minimal polynomial for that is
\begin{align}
&(x-(a+b\sqrt[3]{2}+c\sqrt[3]{4}))(x-(a+bw\sqrt[3]{2}+cw^2\sqrt[3]{4}))(x-(a+bw^2\sqrt[3]{2}+cw\sqrt[3]{4})) \\
&=x^3 +3 a^2 x-3 a x^2-6 b c x-2 b^3-4 c^3 +6 a b c-a^3
\end{align}
The question is how to obtain this polynomial from the starting point.
| I don't get it, where is a problem, a solution is direct. Lift the starting equation
$$x-a=b\sqrt[3]{2}+c\sqrt[3]{4}$$
on $3$, so have:
$$(x-a)^3=(b\sqrt[3]{2}+c\sqrt[3]{4})^3$$
so
$$ (x-a)^3=2b^3+4c^3+6bc(\underbrace{b\sqrt[3]{2}+c\sqrt[3]{4}}_{x-a})$$
and we are done:
$$ p(x)=(x-a)^3-6bc(x-a)-2b^3-4c^3$$
So we don't need any fancy identity's like $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$ or some linear algebra stuff.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Third order convergence of an iteration scheme Consider the iteration scheme
$x_{n+1}=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a}{x_n^2})$
For third order convergence to $\sqrt 2$, the values of $\alpha$ and $\beta$ are ......
I tried it by plugging $x_{n+1}=x_n=\sqrt a$ as $n\rightarrow\infty$ and got $\alpha +\beta =1$.
After this I tried to put it in the form
$Lim_{n\rightarrow\infty}$$\frac{|x_{n+1}-\sqrt a|}{|x_n-\sqrt a|^3}=M$ where $M\ne 0$
but couldn't succeed. Any suggestion and hints please!
The given answer is $\alpha=1/8$,$\beta=3/8$.
| We start by considering two sequences -
$$ A \rightarrow x_{n+1} = \frac{x_{n}}{2}(1 + \frac{a}{x_{n}^2})$$
and
$$ B \rightarrow x_{n+1} = \frac{x_{n}}{2}(3 - \frac{x_{n}^2}{a}) $$
We can observe that by letting $\lim_{n \rightarrow \infty} x_{n} = \xi$ and $\lim_{n \rightarrow \infty} x_{n+1} = \xi$ , $\xi^2 = a$ where $\xi$ is the exact solution.
Next we substitute $x_{n} = \xi + \epsilon_{n}$ and $x_{n+1} = \xi + \epsilon_{n+1}$ and $a = \xi^2$ to both $A$ and $B$ and we obtain the following equations respectively.
In case of
$$A : \epsilon_{n+1} = \frac{\epsilon_{n}^2}{2 \xi} + O(\epsilon^3)$$
which implies that iti s 2nd oreder convergent with error constant $\frac{1}{2\xi}$.
for the case
$$B : \epsilon_{n+1} = \frac{-3}{2\xi} (\epsilon^{n})^2 + O(\epsilon_{n}^2)$$
We next do this operation - adding three times equation $A$ to equation $B$,in order to get third order convergence as after this operation $\epsilon_{n+1} = O(\epsilon_{n}^3)$.
so after the above operation we get the following equation
$$x_{n+1} = \frac{1}{8}x_{n}(6 + 3\frac{a}{x_{n}^2} - \frac{x_{n}^2}{a})$$
and comparing to your given formula
we get the coefficients $\alpha = \frac{1}{8} , \beta = \frac{3}{8}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$ Let $x,y,z$ be real numbers all greater than $1$, then prove that
$$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$
My Attempt:
I am trying to use $A.M>GM$ but am not able to do it
| Let $z=\min\{x,y,z\}$, $x-1=a$, $y-1=b$ and $z-1=c$.
Thus, $c=\min\{a,b,c\}>0$ and we need to prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+2}{b+2}+\frac{b+2}{c+2}+\frac{c+2}{a+2}$$ or
$$\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{b+2}{c+2}+\frac{c+2}{a+2}-\frac{b+2}{a+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(b+2)}+\frac{(c-a)(c-b)}{(a+2)(c+2)},$$
which is obvious.
Another way.
We need to prove that:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+2}{b+2}+\frac{b+2}{c+2}+\frac{c+2}{a+2}$$ or
$$\sum_{cyc}(a^3b^2-a^2b^2c+2a^3b+2a^2b^2-4a^2bc+4a^2c-4abc)\geq0,$$
which is true by AM-GM and Rearrangement.
Indeed,
$$\sum_{cyc}a^3b^2=a^2b^2c^2\sum_{cyc}\frac{a}{c^2}\geq a^2b^2c^2\sum_{cyc}\frac{a}{a^2}=\sum_{cyc}a^2b^2c;$$
$$\sum_{cyc}a^3c=abc\sum_{cyc}\frac{a^2}{b}\geq abc\sum_{cyc}\frac{a^2}{a}=\sum_{cyc}a^2bc;$$
$$\sum_{cyc}(2a^2b^2-2a^2bc)=\sum_{cyc}(a^2c^2-2a^2bc+a^2b^2)=\sum_{cyc}a^2(b-c)^2\geq0$$ and
$$\sum_{cyc}a^2c\geq3\sqrt[3]{a^3b^3c^3}=\sum_{cyc}abc.$$
Done again!
Another way.
Let $k>0$, $\frac{2}{k}=t$ and $x-1=ka$, $y-1=kb$ and $z-1=kc$.
Thus, we need to prove that
$$\sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{ka+2}{kb+2}$$ or
$$\sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{a+t}{b+t}.$$
Now, let $f(t)=\sum\limits_{cyc}\frac{a+t}{b+t}$, where $t>0$.
Since $\lim\limits_{t\rightarrow0^+}f(t)=\sum\limits_{cyc}\frac{a}{b}$, we need to prove that $f$ decreases for $t\rightarrow0^+$.
Indeed,
$$\lim_{t\rightarrow0^+}f'(t)=\lim_{t\rightarrow0^+}\sum_{cyc}\frac{b-a}{(b+t)^2}=\sum_{cyc}\frac{b-a}{b^2}=$$
$$=\sum_{cyc}\left(\frac{b-a}{b^2}+\frac{1}{b}-\frac{1}{a}\right)=-\sum_{cyc}\frac{(a-b)^2}{b^2a}\leq0.$$
Done!
| {
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Solving the equation $\tan{x} + \tan \frac{x}{4}=2$ I am trying to solve the equation $\tan{x} + \tan \frac{x}{4}=2$.
1st attempt: I use the following identity $\tan2x=\frac{2\tan x}{1-\tan^{2}x}$. Then get the following equation
$$
u^5 -2u^4-8u^3+12u^2+3u+2=0,
$$
where $u=\tan \frac{x}{4}$. But i can't find any root for this equation.
2nd attempt: $\tan{4x} + \tan x=2\Rightarrow \sin{4x}\cos{x}+\sin{x}\cos{4x}=2\cos{4x}\cos{x}\Rightarrow \sin{5x}=2\cos{4x}\cos{x}$
$\sin{5x}=2\cos{4x}\cos{x}\Rightarrow \frac{1}{2}\sin{5x}=\cos{4x}\cos{x}\Rightarrow \cos({2k\pi\pm\frac{\pi}{3}})\sin{5x}=\cos{4x}\cos{x}$.
I can't continue from here.
| you can use $$\tan(x)=\frac{2\tan(x/2)}{1-\tan^2(x/2)}$$ and
$$\tan(x/2)=\frac{2\tan(x/4)}{1-\tan^2(x/4)}$$
putting all things together and factorizing we get
$$\left( {u}^{2}-4\,u+1 \right) \left( {u}^{3}+2\,{u}^{2}-3\,u-2
\right)
=0$$ where $$u=\tan(x/4)$$
| {
"language": "en",
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$\sin(\frac{\pi}{3})+\sin(\frac{2\pi}{3})+...+\sin(\frac{n\pi}{3})=2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$ I need to prove that
$\forall n\in\mathbb N:$
$$\sum_{i=1}^n \sin(\frac{i\pi}{3}) = 2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$$
Using induction only leads me to proving quite complicated trig identities. I guess it also can be solved using reminders because sins on the LHS are equal to:
$$\frac{\sqrt3}{2},\frac{\sqrt3}{2},0,-\frac{\sqrt3}{2},-\frac{\sqrt3}{2},0$$
so per every 6th term they cancel out.
Could You give me any hint?
| Induction way:
Assume that we have $$\sum_{k=0}^n \sin \frac{k\pi}{3} = 2\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6}.$$
We prove that
$$\sum_{k=0}^{n+1} \sin \frac{k\pi}{3} = 2\sin\frac{(n+1)\pi}{6}\sin\frac{(n+2)\pi}{6}.$$
Indeed, one has $$LHS = \sum_{k=0}^n \sin \frac{k\pi}{3} + \sin\frac{(n+1)\pi}{3} = 2\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6} + 2\cos\frac{(n+1)\pi}{6}\sin\frac{(n+1)\pi}{6}$$
$$=2\sin \frac{(n+1)\pi}{6}\left(\sin\frac{n\pi}{6} + \cos\frac{(n+1)\pi}{6}\right) = 2\sin \frac{(n+1)\pi}{6}\left(\sin\frac{n\pi}{6} + \sin(\frac{\pi}{2}-\frac{(n+1)\pi}{6})\right)$$
$$=2\sin \frac{(n+1)\pi}{6}\left(2 \sin (\frac{\pi}{6})\cos(\frac{n\pi}{6}-\frac{\pi}{6})\right) = 2\sin \frac{(n+1)\pi}{6}\sin \frac{(n+2)\pi}{6}.$$
Shorter way
One has $$\sin \frac{\pi}{3}\sum_{k=0}^n \sin \frac{k\pi}{3} = \sum_{k=0}^n \sin \frac{k\pi}{3}\sin \frac{\pi}{3} =\sum_{k=0}^n\frac{1}{2}\left( \cos \frac{(k-1)\pi}{3} - \cos \frac{(k+1)\pi}{3}\right)$$
$$ = \frac{1}{2}\left(\cos\frac{-\pi}{3}+1-\cos\frac{n\pi}{3}-\cos\frac{(n+1)\pi}{3}\right) = \frac{1}{2}\left(2\cos^2\frac{\pi}{6}-2\cos\frac{(2n+1)\pi}{6}\cos\frac{\pi}{6}\right)$$
$$=\cos\frac{\pi}{6}\left(\cos\frac{\pi}{6}-\cos\frac{(2n+1)\pi}{6}\right)=2\sin\frac{\pi}{3}\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6}.$$
| {
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$|\sin z|^2+|\cos z|^2=1$ I have a task here:
Show that $|\cos z|^{2}+|\sin z|^{2} = 1$ if and only if $z\in \mathbb R$.
I've tried to different ways.
1:
Since $z\in \mathbb R$ $z^2\geq0$
$-1\leq \cos z\leq1\leftrightarrow |\cos z|\leq1$
$(\cos z)^2\leq(1)^2\leftrightarrow |\cos z|^2\leq1^2$
$(\cos z)^2\leq1\leftrightarrow |\cos z|^2\leq1$ Which also means that $|\cos z|^2=(\cos z)^2$
$-1\leq \sin z\leq1\leftrightarrow |\sin z|\leq1$
$(\sin z)^2\leq(1)^2\leftrightarrow |\sin z|^2\leq1^2$
$(\sin z)^2\leq1\leftrightarrow |\sin z|^2\leq1$ Which also means that $|\sin z|^2=(\sin z)^2$
$|\cos z|^{2}+|\sin z|^{2} = (\cos z)^2+(\sin z)^2=1 \blacksquare$
2:
$|\cos z|^{2}+|\sin z|^{2} = |\frac{e^{iz}+e^{-iz}}{2}|^2+|\frac{e^{iz}-e^{-iz}}{2i}|^2=||\frac{e^{iz}}{2}|+|\frac{e^{-iz}}{2}||^2+||\frac{e^{iz}}{2i}|+|\frac{-e^{-iz}}{2i}||^2$
$|\frac{e^{iz}}{2}|^2 + 2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+|\frac{e^{-iz}}{2}|^2+|\frac{e^{iz}}{2i}|^2 + 2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|+|\frac{e^{-iz}}{2i}|^2$
$=|\frac{e^{iz}}{2}|^2 -|\frac{e^{iz}}{2}|^2+2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|+|\frac{e^{-iz}}{2}|^2-|\frac{-e^{-iz}}{2}|^2=2|\frac{e^{iz}}{2}||\frac{e^{-iz}}{2}|+2|\frac{e^{iz}}{2i}||\frac{-e^{-iz}}{2i}|$
$=2|\frac{e^0}{4}|+2|\frac{-e^{0}}{-4}|=\frac{1}{2}+\frac{1}{2}=1 \blacksquare$
Is these two proofs sufficent?
| Imagine a right angle triangle with sides a, b and c, with c as hypotenuse, whereby $a^2 + b^2 = c^2$ and
$sin z = \frac {a}{c}$
$cos z = \frac {b}{c}$
Thus, $(sinz)^2 + (cosz)^2)$
$=(\frac {a}{c})^2 + {\frac {b}{c}})^2$
$= \frac {a^2+b^2}{c^2}$
$=\frac {c^2}{c^2}$
$=1$
| {
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Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe I need to differentiate or integrate? But I'm not sure how that might work.
Any ideas? Thanks.
| This is $$\sum_{k=0}^{n}\frac{(-1)^{k}}{k+1}\binom{n}{k}.$$ Then multiplying each term by $\frac{n+1}{n+1},$ we get $$\sum_{k=0}^{n}\frac{(-1)^{k}}{n+1}\binom{n+1}{k+1}=\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k-1}\binom{n+1}{k}.$$ Adding and subtracting $1$ and applying the Binomial Theorem gives $$\frac{1}{n+1}-\frac{1}{n+1}\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}1^{n+1-k}=\frac{1}{n+1}(1-(1-1)^{n+1})=\frac{1}{n+1}.$$
| {
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Solving $n$th order determinant I have a determinant of nth order that I am not able to convert into a triangular shape. I believe that this determinant is quite easy, but I can't find a way to fully convert one of the corners into zeros. My other idea was to use the Laplace principle, but that didn't work as well.
$$
\begin{vmatrix}
4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
1 & 4 & 4 & 0 & \cdots & 0 & 0 \\
0 & 1 & 4 & 4 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 4 & 4 \\
0 & 0 & 0 & 0 & \cdots & 1 & 4 \\
\end{vmatrix}
$$
If someone could present a detailed way of converting this determinant into a triangular shape, it would be much appreciated.
In addition to that, maybe someone could give some tips for solving nth order determinant by converting it into a triangular shape, using the Laplace principle or any other more or less basic methods.
| Alternatively: doing row operations to get a diagonal matrix:
$$ R1\cdot (-\frac14)+R2\to R2; \ R2\cdot (-\frac13)+R3\to R3; \ etc$$
$$\begin{vmatrix}
4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
1 & 4 & 4 & 0 & \cdots & 0 & 0 \\
0 & 1 & 4 & 4 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 4 & 4 \\
0 & 0 & 0 & 0 & \cdots & 1 & 4 \\
\end{vmatrix}=\begin{vmatrix}
4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
0 & \frac{12}{4} & 4 & 0 & \cdots & 0 & 0 \\
0 & 0 & \frac{32}{12} & 4 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & \frac{n\cdot 2^{n-1}}{(n-1)\cdot 2^{n-2}} & 4 \\
0 & 0 & 0 & 0 & \cdots & 0 & \frac{(n+1)\cdot 2^n}{n\cdot 2^{n-1}} \\
\end{vmatrix}= \\ \prod_{k=1}^n \frac{(k+1)\cdot 2^k}{k\cdot 2^{k-1}}=(n+1)\cdot 2^n.$$
| {
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"url": "https://math.stackexchange.com/questions/2467629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| $$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6$$
But we know that $$(X-Y)^2\ge 0\Longleftrightarrow X^2+Y^2 \ge 2XY$$
taking $X= a^3$ and $Y=b^3$ we get
$$2a^3b^3 \le a^2b^4+b^2a^4 $$
so
$$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6 \le a^6 + \color{red}{a^2b^4+b^2a^4 } + b^6 = (a^2+b^2)(a^4+b^4) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dimension of image and kernel? If W is a subspace of $\mathbb R^4$ given by $W=\{(x,y,z,w):x+z+w=0,y+z+w=0\}$ . Then what will be the dimension of image and kernel of W.
I just learned that dimension of $W=\text{number of independent variables}-\text{number of constraints}=4-2=2$. I amcurious to know abou the dimension of kernel and image.
| You need a linear function for a kernel and an image. A set doesn't have one. If you consider $A=\begin{pmatrix}1 & 0 & 1& 1\\0&1&1&1\end{pmatrix}$ then it induces a linear map $L:\mathbb R^4\to\mathbb R^2$ defined by $$
L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=A\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}x+z+w\\y+z+w\end{pmatrix}.
$$
For this linear map we have a kernel
\begin{align}
ker(L)&=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in \mathbb R^4~:~L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\right\}=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in\mathbb R^4~:~\begin{pmatrix}x+z+w\\y+z+w\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\right\}\\
=&=\left\{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in\mathbb R^4~:~x+z+w=0,y+z+w=0\right\}
\end{align}
You see, this is your set $W$. Now consider $L$. You might can use the following formula
$$
\dim(W)=\dim(ker(L))=\dim(domain(L))-\text{rank}(L)=\dim(\mathbb R^4)-\text{rank}(A)=4-2=2.
$$
The rank of $A$ is the maximal linear amount of linear independent rows.
And the image of $L$ is given by
$$
image(L)=\left\{\begin{pmatrix}a\\b\end{pmatrix}\in\mathbb R^2~:~\text{there exists }\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\text{ such that }L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}\right\}.
$$
The dimension of $image(L)$ is such the rank of $A$ and therefore it can't be greater than the dimenion of the codomain. Here it is $2$ and since $\dim(\mathbb R^2)=2$ you get $image(L)=\mathbb R^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a number is irrational From the fact that $\frac{1}{5}(3+4i)$ has infinite order in $(\mathbb{C},\cdot)$, I'm supposed to infer that $\frac{1}{\pi}\arctan{\frac{4}{3}}$ is irrational. Irrationality of $\arctan\frac{4}{3}$ follows immediately but I can't see why the irrationality of the product does. Any hints would appreciated!
| Let $\alpha$ be your complex number, then $$\alpha^n = \frac{a_n}{5^{n+1}} + \frac{b_n}{5^{n+1}}i$$
and since we have $$\alpha^n = \frac{1}{5}(3+4i)\left(\frac{a_{n-1}}{5^n} + i\frac{b_{n-1}}{5^n}\right) = \frac{1}{5^{n+1}}\left(3a_{n-1} - 4b_{n-1} + i(a_{n-1} + 3b_{n-1})\right)$$ it is clear that $a_n =3a_{n-1} - 4b_{n-1}$ and $b_n = a_{n-1} + 3b_{n-1}$, with $a_1 = 15$ and $b_1 = 20$.
Examining everything $\pmod{7}$ we find
$(a_1, b_1) = (1, 6), (a_2, b_2) = (0, 5), (a_3,b_3) = (6,1)$ and $(a_4,b_4)=(0,2), (a_5, b_5) = (1,6)$ so the cycle continues with no $b_n = 0$.
Hence $\alpha^n$ is never real, but if $\arctan(4/3) = \frac{p}{q}\pi$ we would have $\exp\left(\frac{p}{q}\pi i \cdot{2q}\right) = 1 = \alpha^{2q}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find out the following integral? I want to solve the integral $$\int_{0}^{\infty} \frac{dx}{(x+b^2)^{3/2}(x+a^2)^{1/2}}$$
I tried it via partial fraction but calculations goes out of my reach. So is there any easier way to do it?
| For $x\geq 0$, if $a^2\not=b^2$, the integrand
\begin{align*}\frac{1}{(x+b^2)^{3/2}(x+a^2)^{1/2}}
&=\frac{1}{2}\left(\frac{x+a^2}{x+b^2}\right)^{-1/2}\cdot\frac{2}{(x+b^2)^2}\\
&=\frac{1}{2}\left(\frac{x+a^2}{x+b^2}\right)^{-1/2}\cdot\frac{(x+b^2)-(x+a^2)}{(x+b^2)^2}\cdot\frac{2}{b^2-a^2}
\end{align*}
is the derivative of
$$\frac{2}{b^2-a^2}\cdot\left(\frac{x+a^2}{x+b^2}\right)^{1/2}.$$
Hence for $|b|>0$,
$$\int_{0}^{\infty} \frac{dx}{(x+b^2)^{3/2}(x+a^2)^{1/2}}=\frac{2}{b^2-a^2}\left[\left(\frac{x+a^2}{x+b^2}\right)^{1/2}\right]_0^{\infty}=\frac{2}{b^2-a^2}\left(1-\frac{|a|}{|b|}\right)=\frac{2}{|b|(|a|+|b|)}.$$
If $a^2=b^2>0$ then
$$\int_{0}^{\infty} \frac{dx}{(x+a^2)^{2}}=\left[-\frac{1}{x+a^2}\right]_0^{\infty}=\frac{1}{a^2}.$$
So the formula $\dfrac{2}{|b|(|a|+|b|)}$ holds also in this case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Epsilon-delta limit does not exist I am having difficult proving that the limit $$\lim_{x\to 1} \frac{x-2}{x^4-1}$$ does not exist using the epsilon-delta definition.
Clearly this must be true since the function is unbounded near 1, but I'm having difficult formalizing this.
Any help to point me in the right direction would be appreciated.
| $\lim_{x\to 1} \frac{x-2}{x^4-1}
$
Let $x = 1+y$.
then
$\begin{array}\\
\lim_{x\to 1} \frac{x-2}{x^4-1}
&=\lim_{y\to 0} \frac{(1+y)-2}{(1+y)^4-1}\\
&=\lim_{y\to 0} \frac{y-1}{1+4y+6y^2+4y^3+y^4-1}\\
&=\lim_{y\to 0} \frac{y-1}{4y+6y^2+4y^3+y^4}\\
&=\lim_{y\to 0} \frac{y-1}{y(4+6y+4y^2+y^3)}\\
&=\lim_{y\to 0} \frac{-1}{4y}\\
\end{array}
$
and this limit does not exist
(and has different signs
for $y > 0$ abd $y < 0$).
Note that
if the numerator was
$x-1$
instead of
$x-2$,
then the limit would be
$\dfrac14$
since the quotient would be
$\frac{y}{y(4+6y+4y^2+y^3)}
=\frac{y}{4y}
=\frac14
$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Foot of perpendicular proof. I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.
My book says:
Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$
My proof:
Slope of line $\perp$ to $ax+by+c=0$ is $\frac{b}{a}\implies\tan\theta=\frac{b}{a}\implies\begin{cases}\cos\theta &=\pm\frac{a}{\sqrt{a^2+b^2}}\\ \sin\theta &=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$
Let $L$ passes through point $(x_ 1,y_1)$ perpendicular to $ax+by+c=0$.
Let $r$ be the algebraic distance of $(x_ 1,y_1)$ from $ax+by+c=0$ $\implies r=\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$.
Now co-ordinate of any point on $L$ distance $r$ from point $(x_1,y_1)$ can be given as:$$\bigg(x_1+r\cos \theta,\ y_1+r\sin\theta\bigg)$$, where $\theta$ is angle $L$ makes with positive direction of $x$-axis.
Substituting the values:$$\bigg(x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}},\ y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}\bigg)$$ Now I didn't wrote $\pm$ with $\cos$, whose sign can be calculated as of $\tan$.
Now camparing gives :
$x=x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$, $\ \ \ y=y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$
Note, I didn't mixed the denominators $\bigg(\sqrt{a^2+b^2}\bigg)$ to $\bigg(a^2+b^2\bigg)$, as I've calculated $\frac{a}{\sqrt{a^2+b^2}}$ from $\tan\theta$ with sign.
When I applied this on some problems gives me the correct answers but is not in the bookish form please help me to do this.
| There are many way to find the foot of the perpendicular. The bookish formula is the same as yours! Just a little bit of reordering:
I pick up from where you stopped.
$$x=x_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$$
$$ y=y_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$$
I don't see a reason why you can't mix the denominators.
$$x=x_1-(ax_1+by_1+c)\cdot\frac{a}{{a^2+b^2}}$$
$$\frac{x-x_1}a=-\frac{ax_1+by_1+c}{a^2+b^2}$$
Similarly,
$$y=y_1-(ax_1+by_1+c)\cdot\frac{b}{{a^2+b^2}}$$
$$\frac{y-y_1}b=-\frac{ax_1+by_1+c}{a^2+b^2}$$
Or
$$\frac{x-x_1}a=\frac{y-y_1}b=-\frac{ax_1+by_1+c}{a^2+b^2}$$
Note:
The negative sign is due to our assumption that the foot of the perpendicular is below the given point. You probably may have assumed the foot of perpendicular is above the point.
Excuse me if I am wrong somewhere.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z = \tan(x/2)$, what is $\sin(x)$ and $\cos(x)$? While reading mathematical gazette, I noticed an interesting "theorem". If $z = \tan(x/2)$, then $\sin(x) = \frac{2z}{1+z^2}$ and $\cos(x) = \frac{1-z^2}{1+z^2}$.
How can I derive these so I don't have to remember them?
| $$\frac{2\tan(x/2)}{1+ \tan^2(x/2)} = \frac{2\tan(x/2)}{\sec^2(x/2)} = 2\tan(x/2) \cdot \cos^2(x/2) = 2\sin(x/2)\cos(x/2) = \sin(2\cdot x/2) = \sin(x)$$
$$\frac{1- \tan^2(x/2)}{1+ \tan^2(x/2)} = \frac{1- \tan^2(x/2)}{\sec^2(x/2)} = 1- \tan^2(x/2) \cdot \cos^2(x/2) = \cos^2(x/2) - \sin^2(x/2) = \cos(2 \cdot x/2) = \cos(x)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Last digit of $3^ {29}+11^{12}+15$ I found this exercise in Beachy and Blair: Abstract Algebra:
Find the units digit of
$$
3^{29}+11^{12}+15
$$
by choosing an appropriate modulus and reducing the sum.
I find $12$ being an appropriate one since
$$
3^{29}=3^2\cdot ((3^3)^3)^3
$$
and
$$
3^3\equiv3\ \ (mod \ 12)
$$
so
$$
3^{29}\equiv 3 \ (mod \ 12)
$$
furthermore
$$
11^{12}=(11^2)^{6}\equiv 1 \ (mod \ 12)
$$
finally
$$
15\equiv 3 \ (mod \ 12).
$$
So the last digit should be
$$
3+1+3=7.
$$
Is this right? And could I have chosen the modulus "more appropriately"?
| $(10+r)^n=\underbrace{10^n+C_n^1\,10^{n-1}r+C_n^2\,10^{n-2}r^2+...}_\text{all this is divisible by 10}+r^n$ by binomial formula.
So we have the identity : $(10+r)^n\equiv r^n\pmod{10}$
*
*$15\equiv 5\pmod{10}$
*$11^{12}=(10+1)^{12}\equiv 1^{12}\equiv 1\pmod{10}$
*$3^{28}=9^{14}=(10-1)^{14}\equiv(-1)^{14}\equiv 1\pmod{10}$
Finally $3^{29}+11^{12}+15\equiv 3\times 1+1+5\equiv 9\pmod{10}$
So the last digit is a $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $3a+b+3c+4d$ if $\left(\begin{smallmatrix}-4&-15\\2&7\end{smallmatrix}\right)^{100}=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Let
$\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^{100} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Find $3a + b + 3c + 4d$.
I've gotten this answer for the matrix to the 100th power(via calculator) , but it's HUGE. I need a simpler method to find it. Can I get some tips please?
| Let $A=\begin{pmatrix}-4&-15\\2&7\end{pmatrix}$. By Cayley-Hamilton theorem, we get $A^2-3A+2I=O$. Thus $A^n$ can be represented as the sum of $A$ and $I$; that is, there are $a_n$ and $b_n$ such that $A^n = a_n A + b_n I$. Then
\begin{align}
A^{n+1}&=a_n A^2+b_n A\\
&=a_n(3A-2I)+b_n A\\
&=(3a_n +b_n)A-2a_n I.
\end{align}
We get
$$\begin{cases}
a_{n+1}=3a_n+b_n\\
b_{n+1} = -2a_n.
\end{cases}$$
Now we can construct the linear recurrence relation of second order for $a_n$:
$$
a_{n+2}=3a_{n+1}-2a_n.
$$
Thus, $a_n = c_1 2^n + c_2$ for some $c_1,c_2$. Since $a_1=1$ and $a_2=3$, $a_n = 2^n -1$, and $b_n = -2^n+2$. Also,
\begin{align}
A^{100}&= a_{100}A+b_{100}I\\
&=\begin{pmatrix}
-4a_{100}+b_{100} & -15a_{100}\\
2a_{100} & 7a_{100} + b_{100}
\end{pmatrix}.
\end{align}
Therefore,
\begin{align}
3a+b+3c+4d&=3(-4a_{100}+b_{100})-15a_{100}+3\cdot 2a_{100} +4(7a_{100}+b_{100})\\
&=7a_{100}+7b_{100}\\
&=7(2^{100}-1)+7(-2^{100}+2)\\
&=7.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods other than function.
| $$x^2-4x+y^2-6x= (x-2)^2 + (y-3)^2-13$$
now this means you have to maximize distance from $(2,3)$
now lets analyze $|x-y| + |x+y|$
in quadrant 1: $x>y$
$$x-y+x+y=4 \rightarrow x=2$$
similiarly you have for $y>x$
$$y=2$$
now if you take negative of both $x$ and $y$ you will get same result, there will be symmetry around origin and figure will recreate as $x=-2$ and $y=-2$ in 3rd quadrant for $x<y$ and $y<x$ respectively
for 2nd quadrant we analyze $|x|>y$
$$ y-x -x-y=4 \rightarrow x=-2$$
after doing similar analysis for $y$ and using symmetry arguments you observe that you get a square of side $4$ centered around origin. Now you just need to pick a point from this square which gives you the maximum distance from $(2,3)$
| {
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Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| With spherical coordinate system, $r=\sqrt{6}$ and
$$z=-x-y=-\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi$$
so your parametric equation will be
$$
\begin{aligned}x&=\sqrt{6}\,\sin \theta \,\cos \varphi \\y&=\sqrt{6}\,\sin \theta \,\sin \varphi \\z&=\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi \end{aligned}$$
where $ θ ∈ [0, π], φ ∈ [0, 2π)$.
| {
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Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.
On the other end of spectrum is the reasonably easy to understand series
$$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$
But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.
A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.
Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.
| Equation (2) in the question is the result of plugging $x=1$ in the expansion for the arctangent, using
$$\frac{\pi}{4}=\tan^{-1}(1)$$
Instead, from
$$\frac{\pi}{6}=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$$
a series with faster convergence is obtained. Taking six terms,
$$\pi>2\sqrt{3}\left(1-\frac{1}{3·3}+\frac{1}{5·3^2}-\frac{1}{7·3^3}+\frac{1}{9·3^4}-\frac{1}{11·3^5}\right)=\frac{509024\sqrt{3}}{280665}>3.141$$
Four terms from
$$\frac{\pi}{8}=\tan^{-1}\left(\sqrt{2}-1\right),$$
give
$$\pi> 8\left(\sqrt{2}-1-\frac{(\sqrt{2}-1)^3}{3}+\frac{(\sqrt{2}-1)^5}{5}-\frac{(\sqrt{2}-1)^7}{7}\right)=\frac{32}{105}\left(716-499\sqrt{2}\right)>3.141,$$
while two terms from
$$ \frac{\pi}{12}=\sin^{-1}\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)$$
lead to
$$\pi > 12\left( \frac{\sqrt{3}-1}{2\sqrt{2}} + \frac{1}{6} \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^3 \right) = \frac{27\sqrt{3}-29}{4\sqrt{2}}>3.14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
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"answer_id": 6
} |
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$?
I would greatly appreciate it if people could please explain how one would go about this.
| Using difference of squares ...
$$z^2-2i= z^2 - 2e^{i \frac \pi 2} = z^2 - (\sqrt 2 e^{i \frac \pi 4})^2
\\ = (z+\sqrt 2 e^{i \frac \pi 4})(z-\sqrt 2 e^{i \frac \pi 4})$$
$$z^2+2i= z^2 - 2e^{i \frac {3\pi} 2} = z^2 - (\sqrt 2 e^{i \frac {3\pi} 4})^2
\\ = (z+\sqrt 2 e^{i \frac {3\pi} 4})(z-\sqrt 2 e^{i \frac {3\pi}4})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Largest rectangle that can fit isnide of equilateral triangle I have an equilateral triangle with a fixed side-length $x$.
What is the largest area of a rectangle that can be put inside the triangle?
| Consider the situation as in the image..
Then the side $b$ depends on $a$ by the formula $b = \tan(60°) \frac{x-a}{2} = \sqrt{3}\frac{x-a}{2}$, so the area of the rectangle is $S = ab = a\sqrt{3}\frac{x-a}{2}$.
Now look for the maximum of $S$ as a function of $a$ (calculate the first derivative and find the zero point or notice it is a parabola, so its maximum is in the middle of zero-points). In both cases you get that the maximum is in the point $a = \frac{x}{2}$, hence the maximum area is $\frac{x}{2}\sqrt{3}\frac{x-\frac{x}{2}}{2}=\frac{x^2\sqrt{3}}{8}$.
Note that the area of the entire triangle $ \frac{x^2\sqrt{3}}{4}$, so the maximum rectangle area is $1/2$ the triangle area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the solution for $x^{\frac {10x}{3} }=\frac {243} {32}$? I understand that $243$ is $3^5$ and $32$ is $2^5$, but I don't know where to go from there after converting it to logarithm.
| As you said, $\frac{243}{32}=(\frac{3}{2})^5$. So why not try $x=\frac{3}{2}$? Indeed, we have $\frac{10\cdot \frac{3}{2}}{3}=5$ which makes $x=\frac{3}{2}$ a solution. We will now Show that there are no other solutions.
Obviously, we cannot have $0<x<1$ because that would imply $x^{\frac{10x}{3}}<1$. Furthermore, if we had $x<-1$, it would follow that $x^{\frac{10x}{3}}<1$ as well (if the term even was defined). And then, $x$ can't be between $0$ and $-1$ because that would make the exponent no whole number - but powers with this kind of exponent are only defined for positive reals as the base. So we have $x>1$. Thus, we consider the function $x^{\frac{10x}{3}}$ on the interval $(1, \infty)$. The derivative of this function is $$(x^{\frac{10x}{3}})'=(e^{ln(x)\cdot \frac{10x}{3}})'=e^{ln(x)\cdot \frac{10x}{3}}\cdot (ln(x)\cdot \frac{10}{3})\cdot (\frac{1}{x}\cdot \frac{10x}{3})=\frac{100}{9}\cdot e^{ln(x)\cdot \frac{10x}{3}}\cdot ln(x)>0$$
for all $x\in (1, \infty)$. So our function is strictly monotonically increasing on the interval $(1, \infty)$ and thus there can only be this one solution we discovered above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| By C-S and C-S we obtain:
$$\sum_{cyc}\frac{x^3}{(y+2z)^2}=\frac{1}{3}\sum_{cyc}x\sum_{cyc}\frac{x^3}{(y+2z)^2}\geq\frac{1}{3}\left(\sum_{cyc}\frac{x^2}{y+2z}\right)^2\geq$$
$$\geq\frac{1}{3}\left(\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+2z)}\right)^2=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)$
$$\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)\tag1$$
$n\ge 1.$
How do we show that $$F(n)=(-1)^{n-1}\ln\left({\pi\over 2}\prod_{k=1}^{n-1}\left({k+1\over k}\right)^{(-1)^k}\right)?$$
$x=\tan(u)$ then $\mathrm dx=\sec^2(u)\mathrm du$
$\cos(2x)={1-\tan^2(x)\over 1+\tan^2(x)}$
$$-\int_{0}^{\pi\over 4}\cos(2u){\tan^{2n}(u)\over \ln(\tan(u))}{\mathrm du\over \cos^2(u)\tan(u)}\tag2$$
$\tan(u)\cos^2(u)={2\over \sin(2u)}$
$$-{1\over 2}\int_{0}^{\pi\over 4}\cot(2u){\tan^{2n}(u)}{\mathrm du\over \ln\tan(u)}\tag3$$
$${\cos(2x)\over \sin(2x)}{\sin^{2n}(x)\over cos^{2n}(x)}=\cos(2x){\sin^{2n-1}(x)\over \cos^{2n+1}(x)}=\tan^{2n-1}(x)-\tan^{2n+1}(x)$$
$$-{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n-1}(x)\over \ln\tan(x)}\mathrm du+{1\over 4}\int_{0}^{\pi\over 4}{\tan^{2n+1}(x)\over \ln\tan(x)}\mathrm du\tag4$$
| For any $s > 0$, we have
\begin{align*}
F(s)
&\stackrel{u=x^2}{=} \int_{0}^{1} \frac{u-1}{u+1} \frac{u^{s-1}}{\log u} \, du
= \int_{0}^{1} \frac{u^{s-1}}{u+1} \left( \int_{0}^{1} u^x \, dx \right) du \\
&= \int_{0}^{1} \int_{0}^{1} \frac{u^{x+s-1}}{u+1} \, dudx
= \int_{0}^{1} \left( \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1} u^{k+x+s-1} \, du \right) \, dx \\
&= \int_{0}^{1} \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k+x+s} \right) \, dx
= \sum_{k=0}^{\infty} (-1)^k \log\left( \frac{k+s+1}{k+s} \right).
\end{align*}
Now for positive integers $n$, we get
\begin{align*}
F(n)
&= (-1)^n \sum_{k=n}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \\
&= (-1)^{n-1} \left( \sum_{k=1}^{n-1} (-1)^k \log\left( \frac{k+1}{k} \right) -
\sum_{k=1}^{\infty} (-1)^k \log\left( \frac{k+1}{k} \right) \right).
\end{align*}
Finally the last infinite sum can be computed by Wallis product, hence the identity follows.
| {
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"url": "https://math.stackexchange.com/questions/2497786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Simplify: $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$ I am doing a pretty hard problem:
$$\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
So it is a pretty long and complicated problem. I got stuck though. My idea was to turn $\sqrt[4]{27} =\sqrt[4]{3}\sqrt{3}$ and since I cant make the second part easier with Langranges formula (it doesn't apply to this) I made it $\sqrt[4]{3}-1$.
I seemed happy that I was getting somewhere and I thought that I had it but later on I just got stuck primarily by the 1's that I don't know what to do with.
| \begin{align}
&=\frac{\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\bigg)\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber \\
&=\frac{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}-\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber\\
&=\frac{2\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber
\end{align}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Bernoulli Differential Equation with an $xy^3$ Term Below is my solution to a differential equation. The answer that I got was different than the book's answer so I am assuming mine is wrong. I am hoping somebody can tell me where I went wrong.
Problem:
Solve the following differential equation.
\begin{eqnarray*}
\frac{dy}{dx} + y &=& xy^3 \\
\end{eqnarray*}
Answer:
Observe that this is a Bernoulli equation.
\begin{eqnarray*}
y^{-3} \frac{dy}{dx} + y^{-2} &=& x \\
z &=& y^{-2} \\
\frac{dz}{dz} &=& -2y^{-3 } \frac{dy}{dx} \\
\frac{1}{2} \frac{dz}{dx} + z &=& x \\
\frac{dz}{dx} + 2z &=& 2x \\
\end{eqnarray*}
Now we have a linear differential equation so we look to find an integrating factor, $I$.
\begin{eqnarray*}
I &=& e ^{ \int 2 dx } = e^ {2x} \\
e^ {2x} \frac{dz}{dx} + 2e^ {2x}z &=& 2xe^{2x} \\
D( e{2x}z ) &=& 2xe^{2x} \\
e^{2x}z &=& \int 2xe^{2x} \\
\end{eqnarray*}
Now to evaluate this integral, we use integration by parts with $u = x$, $dv = 2e^{2x}dx$ and $v = e^{2x}$.
\begin{eqnarray*}
\int 2xe^{2x} &=& xe^{2x} - \int e^{2x} \,\, dx = xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\
e^{2x}z &=& xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\
\frac{e^{2x}}{y^2} &=& xe^{2x} - \frac{ e^{2x} }{2} + C_1 \\
\frac{2}{y^2} &=& 2x - 1 + C e^{-2x} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
\frac{2}{y^2} &=& 2x - 1 + C e^{2x} \\
\end{eqnarray*}
| Akiva Weinberger answered the question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $x^{x+y}=y^{y-x}$ in positive integers.
Let $x,y$ are positive integers. Solve the equation
$$x^{x+y}=y^{y-x}$$
smartplot(x^(x+y)=y^(y-x));
I used http://www.wolframalpha.com
| Let $x,y$ be integers, then $x^{x+y}$ is integer, and so must be $y^{y-x}$. Therefore, $y \geq x$.
Now from fundamental theorem of arithemtic it's evident that $x = n^a$ and $y = n^b$ for some positive integers $n, a, b$, where $b > a$. If $n = 1$ we get solution $x=y=1$. Otherwise, rewrite original equation in terms of $n,a,b$ and take $\log_n$ of both sides. You'll get
$$
a(n^a + n^b) = b(n^b - n^a) \quad\iff\quad
a+b = (a-b)n^{a-b}.
$$
Let $c = b-a$, then $a+b = c + 2a$ and
$$
2a = c(n^c - 1).
$$
So you can choose arbitrary positive integers $c$ and $n$, where $c(n-1)$ is even, and get a solution
$$
x = n^{c(n^c-1)/2}, \quad y = n^{c(n^c-1)/2 + c}.
$$
UPD: As user90369 mentioned in comments, if you denote $n^c = m$, the solution will take simpler form
$$
x = m^{(m-1)/2}, \quad y = m^{(m+1)/2},
$$
where $m = 2k+1$ or $(2k)^2$ for some integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to find $x$ given $\log_{9}\left(\frac{1}{\sqrt3}\right) =x$ without a calculator? I was asked to find $x$ when: $$\log_{9}\left(\frac{1}{\sqrt3}\right) =x$$
Step two may resemble:
$${3}^{2x}=\frac{1}{\sqrt3}$$
I was not allowed a calculator and was told that it was possible. I put it into my calculator and found out that $x$=-0.25 but how do you get that?
| There are already some great answers to this question, but just in case you need the basics of logs...
One nice way to think about logs is that it asks for the how many of a particular factor is in a number.
$$log_5(125)=3$$
Because there are three factors of 5 in the number 125.
$$125=5 \cdot 5 \cdot 5=5^3$$
Similarly,
$$log_3(81)=4$$
Because
$$81=9 \cdot 9=3 \cdot 3 \cdot 3 \cdot 3=3^4$$
Finding simple logs is just a matter of counting a particular factor.
If you have the log of a fraction like
$$log_3 \left(\frac{1}{81}\right)$$
Then we are dealing with a negative number of factors.
$$log_3 \left(\frac{1}{81}\right)=-4$$
Because
$$\frac{1}{81}=3^{-4}$$
Similarly
$$log_2 \frac{1}{8}=-log_2(8)=-3$$
We have one more concept to look at and then we'll be ready.
$$log_{25} (5)$$
We can think of 5 as one half of a factor of 25.
$$log_{25} (5)=\frac{1}{2}$$
Similarly
$$log_{25} (125)=\frac{3}{2}$$
Because there are 3 half factors of 25 in 125.
We can also look at
$$log_{81}(3)=\frac{1}{4}$$
Since
$$81=3 \cdot 3 \cdot 3 \cdot 3$$
We need four factors of 3 to make 81, so 3 is 1 out of the four factor of 3 we need to make an 81.
$$log_5{\sqrt 5}=\frac{1}{2}$$
Because $\sqrt 5$ is half a factor of 5.
$$log_{25}{\sqrt 5}=\frac{1}{4}$$
Because it takes four factors of $\sqrt{5}$ to make 25.
$$\sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5=25$$
So $\sqrt{5}$ is one fourth of a factor of 25.
So looking at your problem by building up,
$$log_9(3)=\frac{1}{2}$$
$$log_9(\sqrt{3})=\frac{1}{4}$$
$$log_9\left(\frac{1}{\sqrt{3}}\right)=-\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I prove $\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$ How can I prove(using sequence convergence definition):
$$\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$$
I need to cancel the n in the numerator, any hint will be appreciated.
| For a given $\epsilon > 0$, we have: $|a_n-1|=\left|\dfrac{n}{\sqrt{n^2+n+1}}-1\right|= \left|\dfrac{n-\sqrt{n^2+n+1}}{\sqrt{n^2+n+1}}\right|= \dfrac{n+1}{\sqrt{n^2+n+1}\left(\sqrt{n^2+n+1}+n\right)}< \dfrac{n+1}{n(n+n)}< \dfrac{2n}{2n^2} = \dfrac{1}{n}< \epsilon$, when $n > 1/\epsilon$. Thus you simply choose $N_0 = 1+\lfloor 1/\epsilon\rfloor$, then if $n \ge N_0$ then: $|a_n - 1| < \epsilon$ which means $a_n \to 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$ Find $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$$
I have tried the following:
$$(2n-1)!!=\frac{(2n)!}{2^{2n}n!}$$
$$(2n)!!=2^nn!$$
$$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}=\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}$$
Now using Stirling approximation,
$$n!=\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}$$
$$(2n)!=\sqrt {2\pi 2n}\left(\frac{2n}{e}\right)^{2n}$$
$$\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}=\lim_{n\rightarrow \infty}\frac{\sqrt{2\pi}\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{2\pi})^2(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$
$$=\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$
How to proceed with solving this limit?
| We have
$$ \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\frac{1}{2}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right) \tag{A}$$
and by squaring both sides
$$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=1}^{n-1}\left(1+\frac{1}{4k(k+1)}\right)\tag{B} $$
from which:
$$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\leq \frac{1}{4n}\prod_{k=1}^{n-1}\exp\left(\frac{1}{4k}-\frac{1}{4(k+1)}\right)\leq \frac{e^{1/4}}{4n}\tag{C} $$
implying that the wanted limit is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating expection of number of draws necessary to capture $m$ marked animals I'm working on the following problem, but I'm stuck and I hope anyone can provide some help:
A population of $N$ animals has had a certain number $a$ of its members captured, marked and then released. Show that the probability $p_n$ that it is necessary to capture $n$ animals in order to obtain $m$ which have been marked is $$p_n= \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{n-m}\left/\binom{N-1}{n-1}\right.,$$ where $m \leq n \leq N-a+m$. Hence, show that $$\frac{a}{N}\binom{a-1}{m-1}\frac{(N-a)!}{(N-1)!}\sum_{n=m}^{N-a+m}\frac{(n-1)!(N-n)!}{(n-m)!(N-a+m-n)!}=1,$$ and that the expectation of $n$ is $\frac{N+1}{a+1}m$.
Well, I was able to do every step, but I'm stuck at what to do to show that $\mathbb{E}(n)=\frac{N+1}{a+1}m$. Any help would be appreciated!
| Let $X$ be a random variable representing the number of animals it is necessary to capture in order to obtain $m$ marked animals, from a total population of $N$ animals, $a$ of which are marked. Then,
$$p_n = \mathbb{P}(X=n) = \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{n-m}\left/\binom{N-1}{n-1}\right.$$
for $n = m,m+1,\ldots, N-a+m$. The task is to find $\mathbb{E}(X)$.
By definition, $$\mathbb{E}(X) = \sum_{n=m}^{N-a+m} np_n$$
Let $X'=X-m$, $x = n-m$ and $p_x = \mathbb{P}(X'=x)$ for $x=0,1,\ldots,N-a$.
Then $p_x = p_n$ and
\begin{align}
\mathbb{E}(X) &= \sum_{x=0}^{N-a} (x+m)p_x \\
&= m + \sum_{x=0}^{N-a} xp_x \\
&= m + \sum_{x=1}^{N-a} x \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{x}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{a}{N}\binom{a-1}{m-1} \sum_{x=1}^{N-a} x \binom{N-a}{x}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m}{N}\binom{a}{m} \sum_{x=1}^{N-a} (N-a)\binom{N-a-1}{x-1}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m(N-a)}{a+1} \sum_{x=1}^{N-a} \frac{a+1}{N} \binom{a}{m}\binom{N-a-1}{x-1}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m(N-a)}{a+1} \sum_{n=m+1}^{N-(a+1)+(m+1)} \frac{a+1}{N} \binom{(a+1)-1}{(m+1)-1}\binom{N-(a+1)}{n-(m+1)}\left/\binom{N-1}{n-1}\right.
\end{align}
Let $Y$ be a random variable representing the number of animals it is necessary to capture in order to obtain $m+1$ marked animals, from a total population of $N$ animals, $a+1$ of which are marked. Then,
$$\mathbb{P}(Y=n) = \frac{a+1}{N} \binom{(a+1)-1}{(m+1)-1}\binom{N-(a+1)}{n-(m+1)}\left/\binom{N-1}{n-1}\right.$$
for $n = m+1,m+2,\ldots,N-(a+1)+(m+1)$. Therefore,
\begin{align}
\mathbb{E}(X) &= m + \frac{m(N-a)}{a+1} \sum_{n=m+1}^{N-(a+1)+(m+1)} \mathbb{P}(Y=n) \\
&= m + \frac{m(N-a)}{a+1} \\
&= \frac{N+1}{a+1}m
\end{align}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504353",
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"source": "stackexchange",
"question_score": "2",
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$ \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$ Evaluate:
$$ L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$$
My approach:
Each term can be written as
$$ \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{n^2+k}$$
$$ \therefore \lim_{n \to \infty}\frac k{n^2+k}=\lim_{n \to \infty}\left(1-\frac {n^2}{n^2+k}\right)=0$$
hence,
$$ L=0$$
Problem:
The correct answer is 1/2, please indicate the flaw in my approach or post a new solution.
Thank You
| HINT:
Using $\sum_{k=1}^n k=\frac{n(n+1)}{2}$ along with the estimates $n^2+1\le n^2+k\le n^2+n$ reveals
$$\frac{n(n+1)}{2(n^2+n)}\le\sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}$$
| {
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System of simultaneous equations $x^n+y^n+z^n=3$ ($n$=1,2,5) Does someone know the solution of the following system of simultaneous equations ($x$, $y$, $z$ are real numbers):
$$
x+y+z=3
$$
$$
x^2+y^2+z^2=3
$$
$$
x^5+y^5+z^5=3
$$
Presented at one of math competitions, don't remember when and where.
| Here is the algebraic solution. Consider z as parameter, so we have
$$
x+y=3-z
$$
$$
x^2+y^2=3-z^2
$$
Subtracting second equation from first squared gives:
$$
2xy=6-6z+2z^2
$$
or
$$
xy=z^2-3z+3
$$
Roots of the system
$$
x+y=3-z
$$
$$
xy=z^2-3z+3
$$
are the roots of polynomial
$$
t^2-(3-z)t+(z^2-3z+3)
$$
which has real roots if
$$
(3-z)^2-4(z^2-3z+3) \ge 0
$$
Simplifying the LHS expression gives:
$$
-3(z-1)^2 \ge 0
$$
or
$$
(z-1)^2 \le 0
$$
which may happen only if $z=1$.
Substituting $z=1$ gives $x=y=1$
| {
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$x\in \mathbb{R}\setminus\{0\}$ and $x + \frac{1}{x}$ is an integer. Prove $\forall n\in \mathbb{N}$ $x^n + \frac{1}{x^n}$ is also an integer.
Let $x$ be a non-zero real number such that $x + \frac{1}{x}$ is an integer. Prove that $\forall n\in \mathbb{N}$ the number $x + \frac{1}{x}$ is also an integer.
Attempt at solution using induction:
base case: $n = 1$ then $x^n + \frac{1}{x^n} = x + \frac{1}{x}$ which is an integer
Inductive assumption: Assume that for some $k\in \mathbb{N} : x^k + \frac{1}{x^k}$
We have to show that $ x^{k+1} + \frac{1}{x^{k+1}}$ is an integer.
$$ x^{k+1} + \frac{1}{x^{k+1}} = x^k\cdot x + \frac{1}{x^k\cdot x} = \frac{(x^k\cdot x)\cdot (x^k\cdot x)+1}{x^k\cdot x}$$
I can't find a way to seperate $x$ and $\frac{1}{x}$ from the term so I can use the inductive assumption.
Edit: I don't think that this question should count as a duplicate since the linked question is asking to specifically solve another problem and one of the answers of that question utilize the proof from this question in their answer but that question itself is different and not related to this question since that question can be solved without using this proof as other answers of that question don't include it.
| Both, $x$ and $\dfrac 1x$ are roots of the same equation $X^2-aX+1=0$
where $a$ is an integer. It follows that any equation deduced from it is also an equation of both $x$ and $\dfrac 1x$.
We have
$$X^2=aX-1$$ $$X^3=aX^2-X=a(aX-1)-X=(a^2-1)X-a$$
$$X^4=(a^2-1)X^2-aX=(a^2-1)(aX-1)-aX=(a^3-2a)X-(a^2-1)$$
For $X^n$ one has by iteration$$X^n=f_n(a)X+g_n(a)$$ where $f_n(a)$ and $g_n(a)$ are integers.
Since also $$\left(\frac{1}{X}\right)^n=f_n(a)\left(\frac{1}{X}\right)+g_n(a)$$ we conclude that
$$x^n+\frac{1}{x^n}=f_n(a)(x+\frac1x)+2g_n(a)=af_n(a)+2g_n(a)\in\mathbb Z$$
(Note that this mode allows us to calculate the integer values of $x^n+\frac{1}{x^n}$).
| {
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Generators of hyperbola
A variable generator meets two generators of the system through extremities $B$ & $B'$ of the minor axis of the principal elliptic section of the hyperboloid $$\frac{x^2} {a^2} +\frac{y^2}{b^2} -z^2c^2=1$$ in $P$ & $P'$. Prove that $BP$. $B'P'=a^2+c^2$
My attempt :
The point of intersection of p-sytem of generator with q-system of generator is
$$x=\frac{a(1+pq)}{p+q},\ y=\frac{b(p-q)}{p+q},\ z=\frac{c(1-pq)}{p+q}$$ for the hyperbola $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2} {c^2} =1$. So, for the given hyperbola, $z=\frac{1-pq}{c(p+q)}$.
At $B(0,b,0)$ & $B'(0, - b, 0)$, $x=0$, $z=0$. So, $1+pq=0$, $1-pq=0$.
Not able to get $2$ values of $p$ or $q$.
| We have the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2} {c^2} =1$
Generating lines of hyperboloid can be written in the standard form:
$$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}{-bcos\theta} = \frac{z}{c} \longrightarrow (1)$$ and
$$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}{-bcos\theta} = \frac{z}{-c} \longrightarrow (2)$$
From (1), we can obtain the equations of two generators passing through minor axis by putting $\theta=90^\circ$ and $\theta=-90^\circ$:
$$\frac{x}{a} = \frac{y - b}{0} = \frac{z}{c} \longrightarrow (3)$$
and
$$\frac{x}{-a} = \frac{y + b}{0} = \frac{z}{c} \longrightarrow (4)$$
According to the question $B=(0,b,0)$ and $B'=(0,-b,0)$. Further P is the intersection of (2) and (3):
$$P = (\frac{acos\theta}{1+sin\theta},b,\frac{cos\theta}{c(1+sin\theta)})$$
and P' is intersection of (2) and (4):
$$P' = (\frac{-acos\theta}{sin\theta-1},-b,\frac{cos\theta}{c(sin\theta-1)})$$
Now we just apply distance formula $BP.B'P'$ and we get this as $a^2+c^2$. Hence proved.
| {
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Injectivity and Surjectivity of a Function We have the set $S$ that is the set of odd positive integers. A function $F:\mathbb{N} \rightarrow S$ is defined by $F(n)=k$ for each $n\in \mathbb{N}$, where $k$ is an odd positive integer for which $3n+1=2^mk$ for some non-negative integer $m$. Prove or disprove the following:
(1) $F$ is injective. (2) $F$ is surjective.
I have thought of ways I could start to answer this question, but my attempts have been fruitless. Help would very much be appreciated for this question.
Million thanks in advance.
| For any integer $n$, calculate $3n+1$. Once you have $3n+1$ write that as a product by taking out as many factors of $2$ as possible. The remaining odd factor is the value of $F(n)$.
If $n=1$, then $3n+1 = 4$, and $4=2^2 \times 1$, so $m=2$ and $k=1$. Hence $F(1)=1$
If $n=2$, then $3n+1 = 7$, and $7 = 2^0 \times 7$, so $m=0$ and $k=7$. Hence $F(2)=7$
If $n=3$, then $3n+1 = 10$, and $10 = 2^1 \times 5$, so $m=1$ and $k=5$. Hence $F(3)=5$
If $n=4$, then $3n+1 = 13$, and $13 = 2^0 \times 13$, so $m=0$ and $k=13$. Hence $F(4) = 13$
If $n=5$, then $3n+1 = 16$, and $16 = 2^4 \times 1$, so $m=4$ and $k=1$. Hence $F(5) = 1$
| {
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How to expand $(x^{n-1}+\cdots+x+1)^2$ (nicely) sorry if this is a basic question but I am trying to show the following expansion holds over $\mathbb{Z}$:
$(x^{n-1}+\cdots+x+1)^2=x^{2n-2}+2x^{2n-3}+\cdots+(n-1)x^n+nx^{n-1}+(n-1)x^{n-2}+\cdots+2x+1$.
Now I can show this in by sheer brute force, but it wasn't nice and certainly wasn't pretty. So I am just wondering if there are any snazzy ways to show this? If it helps, I am assuming $x^m=1$ for some $m>n-1$.
| A kind of graphical proof: Consider the case of $(1+x+x^2+x^3+x^4+x^5)^2$ expanded in a square array in this way:
$$\begin{array}{|l||l|l|l|l|l|}
\hline
&\color{red}{1}&\color{red}{x}&\color{red}{x^2}&\color{red}{x^3}&\color{red}{x^4}&\color{red}{x^5}\\
\hline
\color{red}{x^5}&x^5&x^6&x^7&x^8&x^9&x^{10}\\
\hline
\color{red}{x^4}&x^4&x^5&x^6&x^7&x^8&x^{9}\\
\hline
\color{red}{x^3}&x^3&x^4&x^5&x^6&x^7&x^{8}\\
\hline
\color{red}{x^2}&x^2&x^3&x^4&x^5&x^6&x^{7}\\
\hline
\color{red}{x}&x&x^2&x^3&x^4&x^5&x^{6}\\
\hline
\color{red}{1}&1&x&x^2&x^3&x^4&x^{5}\\
\hline
\end{array}$$
Terms $x^k$ with the same exponent $k$ are situated "in a natural way" on a same diagonal and the "population" of these diagonals linearly increase:
$$1, \ 2 x, \ 3 x^2, \ 4 x^3, \ 5 x^4, \ 6 x^5, \ 5 x^6, \cdots 2 x^9, \ 1x^{10},$$
with a maximum along the main diagonal, then linearly decrease...
Remark: one mimicks here the (discrete) convolution of a uniform distribution with itself giving a "tent" function, as it is called in signal processing, with an evident application : the law of the sum of two dies (here with faces numbered $0$ to $5$) with a maximal probability for result 5.
| {
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Use recurrence relation to find the probability
Ram and Shyam have agreed to bet one dollar on each flip of a fair coin and to continue
playing until one of the them wins all of other’s money. Use a recurrence relation to find
the probability that Ram will win all of Shyam’s money if Ram starts with $a$ dollars and
Shyam starts with $b$ dollars.
My attempt:
Probability of Ram winning at $b$-th try = $\frac{1}{2} \times \frac{1}{2} \times \dots
= (\frac{1}{2})^b$
Probability of Ram winning at $(b + 1)$-th try = $(\frac{1}{2})^{b+1}$
.
.
.
Summing up all probabilities till infinite tries, I get probability to be $\frac{\frac{1}{2}^b}{1-\frac{1}{2}}$ = $\frac{1}{2}^{b-1}$
Now I am pretty sure, my approach is incorrect. Can someone tell me how to solve this?
| Hint: the following equation is the solution:
$$p(a,b)=0.5\times\left(p(a-1,b+1)+p(a+1,b-1)\right)$$
$$p(1,1)=0.5$$
$$p(x,0) = 1, x\geq 1$$
$$p(0,y) = 0, y\geq 1$$
You can expand the equation:
$$p(a,b) = 0.5\times(p(a,b) + 0.5(p(a-2,b+2) + p(a+2,b-2)) = 0.5\times p(a,b) + \frac{1}{2^2}(p(a-2,b+2) + p(a+2,b-2)) \Rightarrow$$
$$(1-\frac{1}{2})p(a,b) = \frac{1}{2^2}(p(a-2,b+2) + p(a+2,b-2))$$
Expand another time:
$$(1-\frac{1}{2})p(a,b) = \frac{1}{2^2}(0.5(p(a-1,b+1) + p(a+1,b-1)) + 0.5(p(a-3,b+3) + p(a+3,b-3))) = \frac{1}{2^2}(p(a,b) + 0.5(p(a-3,b+3) + p(a+3,b-3)))\Rightarrow$$
$$(1-\frac{1}{2}-\frac{1}{2^2})p(a,b) = \frac{1}{2^3}(p(a-3,b+3) + p(a+3,b-3))$$
| {
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Series and integrals for Apery's constant Two integrals for Apery's constant $\zeta(3)$ are
$$\zeta(3)=\frac{16}{3} \int_0^1 \frac{x\log^2\left(x\right)}{1+x^2}dx$$
and
$$\zeta(3)=\frac{32}{7} \int_0^1 \frac{x\log^2\left(x\right)}{1-x^4}dx$$
How can series expressions be obtained from them?
Related questions:
A series to prove $\frac{22}{7}-\pi>0$
Series for $\zeta(3)-\frac{6}{5}$
| A variant of Marco's answer.
$\displaystyle J=\int_0^1 \frac{x\ln^2 x}{1+x^2}\,dx$
Perform the change of variable $y=x^2$,
$\begin{align} J=\frac{1}{8}\int_0^1 \frac{\ln^2 y}{1+y}\,dy\end{align}$
Let,
$\displaystyle K=\int_0^1 \frac{\ln^2 x}{1+x}\,dx$
$\displaystyle L=\int_0^1 \frac{\ln^2 x}{1-x}\,dx$
$\begin{align}
L-K=\int_0^1 \frac{2x\ln^2 x}{1-x^2}\,dx
\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}
L-K&=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{4}L
\end{align}$
Therefore,
$\displaystyle K=\frac{3}{4}L$
Therefore,
$\displaystyle J=\frac{3}{32}L$
Let,
$\begin{align}M=\int_0^1 \frac{x\ln^2 x}{1-x^4}\,dx\\
\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}M&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x^2}\,dx\\
&=\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1+x}\,dx\\
&=\frac{1}{16}L+\frac{1}{16}K\\
&=\frac{7}{64}L
\end{align}$
$\begin{align}L&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\int_0^1\left(\sum_{n=0}^{\infty} x^n\ln^2 x\right)\,dx\\
&=\sum_{n=0}^{\infty} \left(\int_0^1 x^n\ln^2 x\,dx\right)\\
&=2\sum_{n=0}^{\infty} \frac{1}{(n+1)^3}\\
&=2\zeta(3)
\end{align}$
(the exchange of the integral and the series is justified by Fubini-Tonelli theorem)
Therefore,
$\displaystyle J=\frac{3}{16}\zeta(3)$
$\displaystyle M=\frac{7}{32}\zeta(3)$
| {
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How do I solve Cauchy problem for $y''-2y'-3y = e^{4x}$? I have one:
$$y''-2y'-3y = e^{4x} \quad y(0) = 1 \quad y'(0) = 0$$
I've found the solution as a sum of general solution and particular one:
$$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$
Applying the first condition I got:
$$C_1+C_2+ \frac{1}{5} = 1$$
but I do not know how to proceed, techically.
| Using
$$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$
then differentiate to obtain
$$y'(x) = - C_1 e^{-x} + 3 C_2 e^{3x} + \frac{4}{5} e^{4x}.$$
Now set $x=0$ in these equations to obtain
\begin{align}
C_{1} + C_{2} + \frac{1}{5} &= 1 \\
- C_{1} + 3 C_{2} + \frac{4}{5} &= 0.
\end{align}
Solving this set for $C_{1}$ and $C_{2}$ yields $C_{1} = 4/5$ and $C_{2} = 0$ and leads to
$$y(x) = \frac{1}{5} \, [ 4 \, e^{-x} + e^{4 x} ].$$
| {
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A box contains a penny, two nickels, and a dime. If two coins are selected randomly from the box, without replacement, and if X is the sum... A box contains a penny (1¢), two nickels (5¢), and a dime (10¢). If two coins are selected randomly from the box, without replacement, and if $X$ is the sum of the values of the two coins,
*
*What is the probability distribution table of $X$?
$$\begin{array}{|c|c|c|c|c|}\hline X & 6¢ & 10¢ & 11¢ & 15¢ \\ \hline f(x) & 2/6 & 1/6 & 1/6 & 2/6\\\hline\end{array}$$
*What is the cumulative distribution function $F(x)$ of $X$?
The cumulative distribution function, $F(x)$ of $X$ is defined as: $F(x) = P(X ≤ x)$
So would that mean I just write:
$P(X ≤ 6) = 2/6$
$P(X ≤ 10) = 1/6$
$P(X ≤ 11) = 1/6$
$P(X ≤ 15) = 2/6$
\begin{align*}
P(X \leq 6) & = P(X = 6)=2/6\\
P(X \leq 10) & = P(X = 6) + P(X = 10)=2/6+1/6=1/2\\
P(X \leq 11) & = P(X = 6) + P(X = 10) + P(X = 11)=1/2+1/6=2/3\\
P(X \leq 15) & = P(X = 6) + P(X = 10) + P(X = 11) + P(X = 15)=2/3+2/6=1
\end{align*}
| Your answer to the first question is correct.
For the second question, the cumulative distribution function (CDF) of a random variable $X$ at $x$ is found by finding the probability that $X \leq x$. Since there are only four possible values for $X$,
\begin{align*}
P(X \leq 6) & = P(X = 6)\\
P(X \leq 10) & = P(X = 6) + P(X = 10)\\
P(X \leq 11) & = P(X = 6) + P(X = 10) + P(X = 11)\\
P(X \leq 15) & = P(X = 6) + P(X = 10) + P(X = 11) + P(X = 15)
\end{align*}
As a sanity check, note that $P(X \leq 15) = 1$ since all the possible values for $X$ are less than or equal to $15$.
| {
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Solve the equation $x^{2/3}-6x^{1/3} + 8 = 0$ I'm going around in circles on this one...
Solve the equation $$x^{2/3}-6x^{1/3} + 8 = 0$$
According to the book, the answer is 8, 64, but that makes no sense to me either...
Thanks
Gary
| This is just a quadratic in disguise. Don't you see that $x^{2/3}$ can be actually rewritten like this: $\left(x^{\frac{1}{3}}\right)^2$? There are typically two strategies that are used to approach solving this type of equation. You either assign your original expression under the square to a new variable and solve the equation with respect to that new variable and when done substitute back what you've got and finally solve it using the original expression. Or if you anything like me, you just don't care and treat $x^{1/3}$ as a single quantity.
$$
x^\frac{2}{3}-6x^\frac{1}{3} + 8 = 0\implies\\
\left(x^{\frac{1}{3}}\right)^2-6x^\frac{1}{3} + 8 = 0\implies\\
\left(x^{\frac{1}{3}}\right)^2-2\cdot 3\cdot x^\frac{1}{3} +3^2 -3^2 + 8 = 0\implies\\
\left(x^{\frac{1}{3}} - 3\right)^2=9-8\implies\\
\left(x^{\frac{1}{3}} - 3\right)^2=1\implies\\
x^{\frac{1}{3}} - 3=\pm 1\implies\\
x^{\frac{1}{3}} =3\pm 1\implies\\
x^{\frac{1}{3}} =3+ 1 \;\;\;\;\;or \;\;\;\;\;x^{\frac{1}{3}} =3- 1\implies\\
x^{\frac{1}{3}} =4 \;\;\;\;\;or \;\;\;\;\;x^{\frac{1}{3}} =2\implies\\
x =4^3 \;\;\;\;\;or\;\;\;\;\; x =2^3\implies\\
x =64 \;\;\;\;\;or \;\;\;\;\;x =8
$$
| {
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Probability of deck of cards such that each person receives one ace
Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.
The answer to this is given as$$\frac{13^4}{\binom {52}4}$$
My doubt is the following:
*
*The book justifies ${\binom {52}{4}}$ as number of possible different combinations of the four positions in the deck occupied by 4 aces. That sounds like a case of arrangements to me, so shouldn't we think about permutations and not combinations if we are concerned about how the aces are to be arranged in the deck ?.
*Shouldn't the denominator be ${\binom {52}{13}}$ since you are choosing 13 cards for 4 people.
| There are
$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$
ways to distribute $13$ cards to each of four people.
There are $4!$ ways to distribute the aces so that each person receives one and
$$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$
ways to distribute the remaining cards so that each person receives twelve of them. Hence, the desired probability is
\begin{align*}
\frac{4!\dbinom{48}{12}\dbinom{36}{12}\dbinom{24}{12}\dbinom{12}{12}}{\dbinom{52}{13}\dbinom{39}{13}\dbinom{26}{13}\dbinom{13}{13}} & = \frac{4! \cdot \dfrac{48!}{12!36!} \cdot \dfrac{36!}{12!24!} \cdot \dfrac{24!}{12!12!} \cdot \dfrac{12!}{12!0!}}{\dfrac{52!}{13!39!} \cdot \dfrac{39!}{13!26!} \cdot \dfrac{26!}{13!13!} \cdot \dfrac{13!}{13!0!}}\\[2mm]
& = \frac{4! \cdot \dfrac{48!}{12!12!12!12!}}{\dfrac{52!}{13!13!13!13!}}\\[2mm]
& = \frac{4!48!}{12!12!12!12!} \cdot \frac{13!13!13!13!}{52!}\\[2mm]
& = \frac{4!48!13^4}{52!}\\[2mm]
& = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm]
& = \frac{13^4}{\dbinom{52}{4}}
\end{align*}
Let's compare this solution with the approach of your author. As you stated, there are $\binom{52}{4}$ ways to choose the four positions occupied by the aces in the deck. Since each person receives $13$ cards, there are $13$ possible places for the position of the ace in each person's hand. Hence, the desired probability is
$$\frac{13^4}{\dbinom{52}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simplify to terms of generalized power mean I have the following expression${}^1$ I'd like to simplify
$$r \equiv \left[ \frac1{x\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{y\sqrt{y}} \left( 1 - \frac{k}{y} \right) \right]\left[ \frac1{\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{\sqrt{y}} \left( 1 - \frac{k}{y} \right)\right]^{-1}$$
where $x > y > k\geq 1$.
Some attempts led me to think that $r$ can be expressed nicely in terms of generalized means of $x,y$ like the harmonic ($p=-1$), geometric ($p \to 0$), and other powers:${}^2$
$$\begin{align}
H &\equiv \left( \frac1x + \frac1y\right)^{-1} & G &\equiv \sqrt{xy} & M \equiv \left( x^p + y^p \right)^{1/p} \quad \text{with perhaps} \quad p=\frac{-1}2
\end{align}$$
To be specific, my question is this:
Is there a way to write it in this form $$r \overset{?}{=} \frac1x + \frac1y + \frac1{\sqrt{xy}} + {}\color{red}{??} =\frac1H + \frac1G + {}\color{red}{??} $$ such that the $\color{red}{??}$ part is "nice" in terms of $k$ and maybe $H,G$ or $M$ defined above?
I also wonder if there's some standard techniques like examining the asymptotic form to guess the coefficients of the proposed terms. I tried a bit but didn't get very far.
Directly pulling out $\frac1H$ and $\frac1G$ just change $r$ into something equally inviting but not more compact.
Any suggestions will be appreciated. Honestly, one main reason for me to think that "oh there must be some nice and more compact forms" is the glaring symmetry. I will be okay if there turn out to be none. Thanks.
Footnote 1:
$\quad$For the record, $r$ is part of the expression of the ratio between $\frac{\partial^2 f(u)}{ \partial u^2}$ and $\frac{\partial f(u) }{ \partial u}$, evaluated at $u=0$, where $$f(u) = \left( \frac1{\sqrt{u+y}} - \frac1{\sqrt{u+x}} \right)\sqrt{u+k}$$
Footnote 2:
$\quad$This is a slight abuse of notations, as the generalized means should carry the averaging factors $\frac12$ or $\frac1{\sqrt{2}}$. My shorthands can also match the p-norms, which has the benefit of the norms ordered in powers have correspondingly decreasing magnitudes. However, p-norms doesn't cover $\sqrt{x y}$ like generalized mean.
Update (Nov.28th, 2017)
With all the inverses lying around, it turns out that the best way is to express $r$ in terms of the ... you guessed it ... inverses: $v \equiv 1/x$ and $w \equiv 1/y$. I'll spare the readers of the actual algebra. Suffice to say that this is also the natural course to take to adopt the notions of p-norms; one just have to give up on unifying $1 / \sqrt{xy} = 1/G = \sqrt{vw}$ formally.
| We can write
$$r=\frac 1H+\frac 1G+\color{red}{\frac{1}{M-G+\frac{GM}{k}}}$$
where
$$M=(x^{-1/2}+y^{-1/2})^{-2}$$
If we write
$$r=\frac{\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}-k\left(\frac{1}{x^2\sqrt x}-\frac{1}{y^2\sqrt y}\right)}{\frac{1}{\sqrt x}-\frac{1}{\sqrt y}-k\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)}=\frac 1x+\frac 1y+\frac{1}{\sqrt{xy}}+A$$ then we have
$$A=\frac{k\left(\frac 1x-\frac 1y\right)\left(\frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}\right)}{\frac{1}{\sqrt x}-\frac{1}{\sqrt y}-k\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)}$$
Using
$$x+y=\frac{G^2}{H},\quad xy=G^2,\quad \frac{1}{M}=\frac 1H+\frac 2G$$
we have
$$\left(\frac 1x-\frac 1y\right)^2=\left(\frac 1H+\frac 2G\right)\left(\frac 1H-\frac 2G\right)\implies \frac 1x-\frac 1y=-\sqrt{\frac 1M\left(\frac 1H-\frac 2G\right)}$$
$$\left(\frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}\right)^2=\frac{1}{G^2}\left(\frac 1H+\frac 2G\right)\implies \frac{1}{x\sqrt y}+\frac{1}{y\sqrt x}=\frac 1G\sqrt{\frac 1M}$$
$$\left(\frac{1}{\sqrt x}-\frac{1}{\sqrt y}\right)^2=\frac 1H-\frac{2}{G}\implies \frac{1}{\sqrt x}-\frac{1}{\sqrt y}=-\sqrt{\frac 1H-\frac{2}{G}}$$
$$\small\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)^2=\left(\frac 1H+\frac 1G\right)^2\left(\frac 1H-\frac 2G\right)\implies \frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}=-\left(\frac 1M-\frac 1G\right)\sqrt{\frac 1H-\frac 2G}$$
So, we get
$$A=\frac{k\left(-\sqrt{\frac 1M\left(\frac 1H-\frac 2G\right)}\right)\frac 1G\sqrt{\frac 1M}}{-\sqrt{\frac 1H-\frac{2}{G}}-k\left(-\left(\frac 1M-\frac 1G\right)\sqrt{\frac 1H-\frac 2G}\right)}=\frac{1}{M-G+\frac{GM}{k}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3? How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3?
If I find this by method of complementation then
Total number of arrangements without any restrictions =$\frac{7!}{4!×2!}$.
Then I should subtract total no of arrangements where 2 is beside either three
That means arrangements with 233,323,332 should be subtracted. if we take 233,323,332 as 3 possibilities then how should I proceed further
Please correct me if I am wrong here.
| As you observed, there are
$$\binom{7}{4}\binom{3}{2}\binom{1}{1} = \frac{7!}{4!2!1!} = \frac{7!}{4!2!}$$
distinguishable arrangements of four $1$s, one $2$, and two $3$s. From these, we must exclude those arrangements in which a $3$ is adjacent to the $2$.
A $3$ is adjacent to the $2$: We have six objects to arrange, four $1$s, a $3$, and a block containing a $2$ and a $3$. We choose four of the six positions for the $1$s, one of the remaining two positions for the $3$, and the final position for the block. The $2$ and $3$ can be arranged within the block in $2!$ ways. Hence, the number of such arrangements is
$$\binom{6}{4}\binom{2}{1}\binom{1}{1} \cdot 2! = \frac{6!}{4!1!1!} \cdot 2! = \frac{6!2!}{4!}$$
However, in counting arrangements in which a $3$ is adjacent to the $2$, we have counted arrangements in which both $3$s are adjacent to the $2$ twice, once when we designated $32$ as our block and once when we designated $23$ as our block. We only want to subtract such arrangements once. Therefore, we must add them back.
Both $3$s are adjacent to the $2$: This can only occur if the arrangement includes the block $323$. Thus, we have five objects to arrange, the four $1$s and the block. We choose four of those five positions for the $1$s. The block must be placed in the remaining position. Hence, there are
$$\binom{5}{4} = \frac{5!}{4!}$$
such arrangements.
By the Inclusion-Exclusion Principle, there are
$$\binom{7}{4}\binom{3}{2}\binom{1}{1} - \binom{6}{4}\binom{2}{1}\binom{1}{1}2! + \binom{5}{4}\binom{1}{1} = \frac{7!}{4!2!} - \frac{6!2!}{4!} + \frac{5!}{4!} = 50$$
permissible arrangements.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does this question on conditional probability have an incorrect solution? Going through some exercises on probability, and I chanced upon this problem.
An urn contains 6 red balls and 3 blue balls. One ball is selected at
random and is replaced by a ball of the other color. A second ball is
then chosen. What is the conditional probability that the first ball
selected is red, given that the second ball was red?
My solution seems correct. I do not know where I went wrong.
Let R1 be the event that a first red ball is selected.
Let R2 be the event that a second red ball is selected.
Then,
$P(R_2|R_1) = \dfrac{5}{9}\times\dfrac{6}{9}=\dfrac{30}{81}\Rightarrow P(R_1\cap R_2) = P(R_2|R_1)\times P(R_1)=\dfrac{30}{81}\times \dfrac{6}{9}=\dfrac{180}{729}$
$P(R_2|R_1^C) = \dfrac{7}{9}\times\dfrac{3}{9}=\dfrac{21}{81}\Rightarrow P(R_1^C\cap R_2) = P(R_2|R_1^C)\times P(R_1^C)=\dfrac{21}{81}\times \dfrac{3}{9}=\dfrac{63}{729}$
and thus, $P(R_2) = P(R_2 \cap R_1) + P(R_2 \cap R_1^C) = \dfrac{243}{729}$
$\therefore P(R_1|R_2)=\dfrac{P(R_1 \cap R_2)}{P(R_2)}=\dfrac{\dfrac{180}{729}}{\dfrac{243}{729}}=\dfrac{20}{27}$
The answer, however, is $\dfrac{10}{17}$, which seems like a typo to me, but I'm not sure. Thanks!
| With problems like this I like to turn $P(R_1 | R_2)$ into $P(R_2 | R_1)$ if it is an easier calculation. We have,
$$\begin{align*}
P(R_1 | R_2)
&= \frac{P(R_1 \cap R_2)}{P(R_2)}\\\\
&= \frac{P(R_2 | R_1)P(R_1)}{P(R_2)}\\\\
&= \frac{\frac{5}{9}\cdot\frac{6}{9}}{\frac{6}{9}\cdot\frac{5}{9}+\frac{3}{9}\cdot\frac{7}{9}}\\\\
&\approx .558 \\\\
&= \frac{10}{17}
\end{align*}$$
Just some explanations for these values:
$P(R_2 | R_1) = \frac{5}{9}$ since if we select red first, then we only have $5$ reds of the $9$ the second time around.
$P(R_2)=\frac{6}{9}\cdot\frac{5}{9}+\frac{3}{9}\cdot\frac{7}{9}$ since we can either select red and then red again or select blue and then red.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is to be expressed in terms of $x$ and $n$. I tried setting a formula for the $n$-th term and setting it into a difference but ran into a dead end.
| A supplement to the nice answer from @labbhattacharjee. We see an example of telescoping based upon
\begin{align*}
\color{blue}{\frac{2^k}{1-x^{2^k}}+\frac{2^k}{1+x^{2^k}}}
=\frac{2^k\left(1+x^{2^k}\right)+2^k\left(1-x^{2^k}\right)}{\left(1-x^{2^k}\right)\left(1+x^{2^k}\right)}
&\color{blue}{=\frac{2^{k+1}}{1-x^{2^{k+1}}}}\tag{1}
\end{align*}
We obtain according to (1)
\begin{align*}
\color{blue}{\sum_{k=0}^n\frac{2^k}{1+x^{2^k}}}&=\sum_{k=0}^n\left(\frac{2^{k+1}}{1-x^{2^{k+1}}}-\frac{2^k}{1-x^{2^k}}\right)\\
&=\sum_{k=0}^n\frac{2^{k+1}}{1-x^{2^{k+1}}}-\sum_{k=0}^n\frac{2^k}{1-x^{2^k}}\\
&=\sum_{k=1}^{n+1}\frac{2^{k}}{1-x^{2^{k}}}-\sum_{k=0}^n\frac{2^k}{1-x^{2^k}}\tag{2}\\
&\color{blue}{=\frac{2^{n+1}}{1-x^{2^{n+1}}}-\frac{1}{1-x}}\tag{3}
\end{align*}
Comment:
*
*In (2) we shift the index of the left-hand sum to start with $k=1$.
*In (3) we do the telescoping.
| {
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"source": "stackexchange",
"question_score": "9",
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What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and... Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.
| I suggest to go for $X=\frac 1x$, $Y=\frac 1y$ and $Z=\frac 1z$ because it has the advantage to simplify the denominators.
The equations become
$\dfrac{XY+YZ+ZX}{Y+Z}=\dfrac 12\quad;\quad\dfrac{XY+YZ+ZX}{X+Z}=\dfrac 13\quad;\quad\dfrac{XY+YZ+ZX}{X+Y}=\dfrac 14$
And it gives a simple system to solve
$\begin{cases}
Y+Z=2k\\
X+Z=3k\\
X+Y=4k\\
XY+YZ+ZX=k\end{cases}$
For instance $(2)-(1)$ gives $X-Y=k$ and reporting in $(3)$ gives $2Y=3k$.
This is all similar easy calculation and we end up with $X=\frac 52 k,Y=\frac 32k,Z=\frac 12k$ and $\dfrac{23k^2}4=k$
Finally $x=\frac 1X=\frac 25\times\frac{23}{4}=\frac {23}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Transformation to $\log\left(\frac a2\cos x\right)$
Question: How does$$T=-\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\frac {\log\left(\tfrac {a^2}4\cos^2x\right)}{\sqrt{\tfrac {a^2}4\cos^2x}}\left(\frac a2\cos x\right)=4\int\limits_0^{\tfrac {\pi}2}dx\,\log\left(\frac a2\cos x\right)\tag1$$
I was evaluating an integral, namely$$I=\int\limits_0^adx\,\frac {\log x}{\sqrt{ax-x^2}}$$And I'm having trouble seeing how to get from the left-hand side to the right-hand side. Doesn't the denominator and numerator cancel to leave you with$$T=-\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\log\left(\frac {a^2}4\cos^2x\right)=-2\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\log\left(\frac a2\cos x\right)$$But I don't see how that can be substituted to get the second expression of $(1)$. What kind of substitution should be made to make the transformation from where I left off to $(1)$?
| First, $\cos(x)\ge 0$ and is even for $x\in[-\pi/2,\pi/2]$. Hence, $\sqrt{\cos^2(x)}=\cos(x)$. If $a>0$, then $\sqrt{a^2}=a$. Therefore,
$$\frac{\log\left(\frac{a^2}{4}\cos^2(x)\right)\left(\frac a2\cos(x)\right)}{\sqrt{\frac{a^2}{4}\cos^2(x)}}=\frac{\log\left(\left(\frac{a\cos(x)}{2}\right)^2\right)\left(\frac a2\cos(x)\right)}{\frac a2\cos(x)}=2\log\left(\frac{a\cos(x)}{2}\right)$$
Finally, exploiting the evenness of the cosine function, we find
$$\begin{align}
\int_{-\pi/2}^{\pi/2}\frac{\log\left(\frac{a^2}{4}\cos^2(x)\right)\left(\frac a2\cos(x)\right)}{\sqrt{\frac{a^2}{4}\cos^2(x)}}\,dx&=\int_{-\pi/2}^{\pi/2}2\log\left(\frac{a\cos(x)}{2}\right)\,dx\\\\
&=4\int_0^{\pi/2}\log\left(\frac{a\cos(x)}{2}\right)\,dx
\end{align}$$
as expected!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...$
But, But I'm stuck here.
| We can rewrite the product as a telescoping one. The end result is:
$$\prod_{k=0}^6 \left((1+x)^{2^k} + x^{2^k}\right) =
\prod_{k=0}^6 \frac{(1+x)^{2^{k+1}} - x^{2^{k+1}}}{(1+x)^{2^k} - x^{2^k}}
= \frac{(1+x)^{2^7} - x^{2^7}}{(1+x)^{2^0} - x^{2^0}} = (x+1)^{128} - x^{128}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^4}})}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}\\&
=\lim_{n\rightarrow\infty}\dfrac{n+1-n\sqrt{1+\dfrac{1}{n^4}}}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}
\end{align}$
Given that $\dfrac{1}{n}$ tends to $0$ (so denominator is 2), can I reduce $n$ and $-n\sqrt{1+\dfrac{1}{n^4}}$ and say that the limit is $\dfrac{1}{2}$?
I mean $\dfrac{1}{n^4}$ tends to $0$, so $\sqrt{1+\dfrac{1}{n^4}}$ tends to $1$ and in this case $n-n\sqrt{1+\dfrac{1}{n^4}}$ can be simplified to $n-n$.
Solution given in my book uses conjugate multiplication twice to get rid of all the roots in nominator, but I am curious if my answer is correct or my teacher will tell me that simplifying the way I did it is incorrect.
| Let $t = \frac1n$
Then use the derivative formula at $t = 0$ ,
$$ \begin{align}\lim_{n\to\infty}\sqrt{n^2+n}-\sqrt[4]{n^4+1} &= \lim_{n\to\infty} n\left(\sqrt{\frac{1}{n}+1}-\sqrt[4]{\frac{1}{n^4}+1}\right) \\&= \lim_{t\to0}\frac{1}t(\sqrt{t+1}-\sqrt[4]{t^4+1})\\&= \lim_{t\to 0}\frac{\sqrt{t+1}-1}{t} - \frac{\sqrt[4]{t^4+1}-1}{t} \\&=\left(\sqrt{t+1}\right)'\bigg|_{t =0 } -\left(\sqrt[4]{t^4+1}\right)'\bigg|_{t =0 }= \frac{1}{2}\end{align} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I tried rewriting the expression as $$\sum_{i=1}^n \frac{\bar{x_i}-1}{|x_i-1|^2}$$
I then used the fact that $$x^{n+1}-1=(x-1)(x^n+x^{n-1}+...+x+1=0$$ so $x_i$ are the complex nth roots of unity.Using cosine formula I found that $$|x_i-1|^2=2-2\cos(\frac{2i\pi}{n+1})=4(\sin \frac{\pi}{n+1})^2$$
After substituting this I couldn't simplify the resulting expression.Any ideas?Thanks.
| For the first part :
Write $$x^n+x^{n-1}+...+x+1 =(x-x_1)(x-x_2)\ldots (x-x_{n})$$
Take log both the sides
$$\log (x^n+x^{n-1}+...+x+1)= \log(x-x_1) +\log(x-x_2)+\ldots +\log(x-x_n)$$
Differentiating w.r.t. $x$, we get
$$\frac{ nx^{n-1}+(n-1)x^{n-2}+\ldots+ 1}{x^n+x^{n-1}+...+x+1}= \frac{1}{x-x_1}+\frac{1}{x-x_2}+\ldots +\frac{1}{x-x_n}$$
Now put $x=1$
$$\frac{(n)+(n-1)+\ldots + 1}{1+1+\ldots +1}=-\left(\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}\right)$$
$$\color{red}{\boxed{ \frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}=- \frac{n}{2} }}$$
| {
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How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| Compute the minimal polynomial of $\alpha=-27-3\cdot3^{1/3}+9\cdot3^{2/3}$:
$$
\begin{align}
\alpha&=-27-3\cdot3^{1/3}+9\cdot3^{2/3}\tag1\\
\alpha^2&=567+405\cdot3^{1/3}-477\cdot3^{2/3}\tag2\\
\alpha^3&=-81-25515\cdot3^{1/3}+16767\cdot3^{2/3}\tag3
\end{align}
$$
Since $(-3,405,-25515)\times(9,-477,16767)=-2214(1,81,2430)$, we get that
$$
\alpha^3+81\alpha^2+2430\alpha+19764=0\tag4
$$
Thus, dividing $(4)$ by $19764\alpha$, we get
$$
\begin{align}
\frac1\alpha
&=-\frac{\alpha^2+81\alpha+2430}{19764}\\
&=-\frac{45+9\cdot3^{1/3}+14\cdot3^{2/3}}{1098}\tag5
\end{align}
$$
| {
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"answer_id": 2
} |
Doubts on obtaining orthonormal basis In finding the matrix $P$ that orthogonally diagonalizes $A$ and to determine $P^TAP$, where $$A =
\begin{pmatrix}
1 & -1 & 1 & -1\\
-1 & 1 & -1 & 1\\
1 & -1 & 3 & 1\\
-1 & 1 & 1 & 3\\
\end{pmatrix}
$$
I've found the eigenvalues to be $0$ and $4$.
In order to find the orthonormal basis, i've considered two scenarios where $\lambda = 0$ and $\lambda = 4$:
For $\lambda = 0$, on solving the system, we have $ \begin{pmatrix}
w\\
x\\
y\\
z\\
\end{pmatrix} $ = $s \begin{pmatrix}
1\\
1\\
0\\
0\\
\end{pmatrix} $ + $t \begin{pmatrix}
2\\
0\\
-1\\
1\
\end{pmatrix} $
For $\lambda = 0$, on solving the system, we have $ \begin{pmatrix}
w\\
x\\
y\\
z\\
\end{pmatrix} $ = $s \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
1\\
0\\
\end{pmatrix} $ + $t \begin{pmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
0\\
1\\
\end{pmatrix} $
My Doubts:
*
*As for $E_{4}$ as seen above, the answer considers the set of $ \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
1\\
0\\
\end{pmatrix} $, $ \begin{pmatrix}
-1\\
1\\
0\\
2\\
\end{pmatrix} $ to be the basis for $E_{4}$. However, why is there a need to scale the basis by two into its integer basis of $ \begin{pmatrix}
-1\\
1\\
0\\
2\\
\end{pmatrix} $ from $ \begin{pmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
0\\
1\\
\end{pmatrix} $. Isn't it the same?
*The answers present the orthonormal basis set of $E_{0}$ as $ \begin{pmatrix}
\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}\\
0\\
0\\
\end{pmatrix} $, $ \begin{pmatrix}
\frac{1}{2}\\
-\frac{1}{2}\\
-\frac{1}{2}\\
\frac{1}{2}\\
\end{pmatrix} $ and the orthonormal basis set for $E_{4}$ as $ \begin{pmatrix}
\frac{1}{\sqrt6}\\
-\frac{1}{\sqrt6}\\
\frac{2}{\sqrt6}\\
0\\
\end{pmatrix} $, $ \begin{pmatrix}
-\frac{1}{\sqrt12}\\
\frac{1}{\sqrt12}\\
\frac{1}{\sqrt12}\\
\frac{3}{\sqrt12}\\
\end{pmatrix} $
They stated that the orthonormal basis can be found by using the Gram-Schmidt Process. However, i'm not really sure how it can be computed. As for $E_{0}$, isn't it true that we divide each integer by $\sqrt (1^2 + 1^2)$ to get $ \begin{pmatrix}
\frac{1}{\sqrt2}\\
\frac{1}{\sqrt2}\\
0\\
0\\
\end{pmatrix} $ and $\sqrt (2^2+-1^2+1^2)$ to get $ \begin{pmatrix}
\frac{2}{\sqrt6}\\
0\\
-\frac{1}{\sqrt6}\\
\frac{1}{\sqrt6}\\
\end{pmatrix} $ but it was apparently not the case as seen above. Why is that so?
Thanks.
| If you have two vectors $u_1$ and $u_2$, the Gram-Schmidt algorithm produces two vectors $v_1$ and $v_2$ defined by
$$
v_1 = \frac{u_1}{\|u_1\|}
\quad
v_2 = \frac{u_2 - (u_2|v_1)v_1}{\|u_2 - (u_2|v_1) v_1\|}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some help me how to solve it , thanks
| I do not understand the phrase "A detailed canonical answer is required to address all the concerns" since I am unaware of any concerns about the answer given some time ago. However, here are the actual equations and how to solve them.
Let $V = \pmatrix{f(x)\\f(\frac{1}{x})\\f(\frac{1}{1-x})\\f(\frac{x}{x-1})\\f(\frac{x-1}{x})\\f(1-x)},$ $W = \pmatrix{x\\\frac{1}{x}\\1-x\\\frac{x-1}{x}\\\frac{x}{x-1}\\\frac{1}{1-x}\\}$
Then $$ \pmatrix{1&2&0&3&0&0\\2&1&3&0&0&0\\0&0&2&0&3&1\\0&0&0&2&1&3\\3&0&0&1&2&0\\0&3&1&0&0&2}V=W$$
Multiplying on the left by $ \pmatrix{-3&3&-5&1&7&1}$ gives $$24f(x)=\pmatrix{-3&3&-5&1&7&1}W$$
Thus giving the answer noted above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following is what I have done so far:
To show by induction: $a_{k+1} < a_k \forall k \in \mathbb N_0$
Induction start: $n=1$
$a_2= \frac{3^2}{5^2+1}=\frac{9}{25+1}=\frac{9}{26}=\frac{18}{52} < \frac{26}{52}=\frac{1}{2}=\frac{3}{6}=\frac{3^1}{5^1+1}=a_1$
Induction step:
$$\begin{align}
a_{k+1}&<a_k \\
\equiv \frac{3^{k+1}}{5^{k+1}+1} &< \frac{3^k}{5^k+1} \\
\equiv \frac{3^{k+1}}{5^{k+1}} &< \frac{3^k}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+1}} &< \frac{3^{k+1}}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+2}} &< \frac{3^{k+1}}{5^{k+1}} \\
\equiv \frac{3^{k+2}}{5^{k+2}+1} &< \frac{3^k+1}{5^{k+1}+1} \\
\equiv a_{k+2} &< a_{k+1} \\
\end{align}$$
To Show: $|a_k|= a_k$
$$
\begin{align}
|a_k| &= |\frac{3^k}{5^k+1}| \\
&= \frac{|3^k|}{|5^k+1|} \\
&= \frac{3^{|k|}}{5^{|k|}+|1|} \\
&= \frac{3^k}{5^k+1} \\
&= a_k
\end{align}$$
Because of the induction I can conclude that the sequence $\lim_{k \to \infty}a_k$ becomes smaller and smaller.
And because of $|a_k|=a_k$ are all values $\forall k \in \mathbb N$ positive.
$\Rightarrow $ The sequence $a_k$ is monotically decreasing.
$\Rightarrow $ The series $\sum_{k=1}^{\infty}a_k$ is monotically increasing.
$$
\begin{align}
a_k = \frac{3^k}{5^k+1} &< \frac{3^k}{5^k} \\
&< \frac{3^k}{3^k} \\
&= 1 \\
\end{align}$$
$$\lim_{k \to \infty} 1 = 1$$
And thus: $\exists N \in \mathbb N$, such that
$$|a_k| \le 1, \forall \quad k \ge N$$
Using the direct comparison test there can be concluded that $\sum_{k=1}^{\infty}{a_k}$ converges.
Question: Is my proof correct?
| you have $\frac{3^k}{5^k+1} < \frac{3^k}{5^k}= \left(\frac{3}{5}\right)^k$ than it follows $$ \sum_{k=1}^{\infty}a_k = \sum_{k=1}^{\infty} \frac{3^k}{5^k+1} \leq \sum_{k=1}^{\infty} \frac{3^k}{5^k} = \sum_{k=1}^{\infty} \left(\frac{3}{5}\right)^k = \sum_{k=0}^{\infty} \left(\frac{3}{5}\right)^{k+1}=\frac{3}{5} \left( \frac{1}{1-\frac{3}{5} } \right) = \frac{3}{2}$$
So you showed that a bigger series converges and after the comparison test you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\dfrac{x}{1-x^2}))+(-\dfrac{x^2}2)}{x^4}$$
$$\lim_{x \to 0} \frac{(\dfrac{(1-\cos (\dfrac{x}{1-x^2}))}{(\dfrac{x}{1-x^2})^2})(\dfrac{x}{1-x^2})^2+(-\dfrac{x^2}2)}{x^4}$$
now what ?
| Use the famous formula $1-\cos 2t=2\sin^{2}t$ to transform the expression into $$2\cdot\frac{\sin^{2}t-(x/2)^{2}}{x^{4}}$$ where $t=x/(2(1-x^{2}))$. Next the numerator needs to be split by adding and subtracting $t^2$. It is easy to check that $(t^2-(x/2)^{2})/x^4\to 1/2$ and hence the desired limit is equal to $$1+2\lim_{x\to 0}\frac{\sin^{2}t-t^2}{t^{4}}\cdot(t/x)^{4}$$ Since $t/x\to 1/2$ we can see that he desired limit is equal to $$1+\frac{1}{8}\lim_{t\to 0}\frac{\sin t +t} {t} \cdot\frac{\sin t-t} {t^3}$$ And then the answer is easily seen to be $1+(1/8)(2)(-1/6)=23/24$. Note that the last limit (which evaluates to $-1/6$) necessitates the use of tools like Taylor series or L'Hospital's Rule.
The limit can not be solved just using $\lim\limits_{x\to 0}\dfrac{1-\cos x} {x^{2}}=\dfrac{1}{2}$ and algebraic manipulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| Using $x^n - y^n = (x - y)\sum_{k=0}^n x^ky^{n-k}$, we have
$$
\sqrt{x+2}-\sqrt[3]{x+20} = \frac{(x+2)^3-(x+20)^2}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}} \\
\sqrt[4]{x+9}-2 = \frac{x-7}{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}
$$
Notice that $(x+2)^3-(x+20)^2 = (x - 7)(x^2+12x+56)$, so
$$
\begin{aligned}
\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = \frac{(x - 7)(x^2+12x+56)/\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}}{(x-7)/\sum_{k=0}^4(x+9)^{k/4}2^{4-k}} \\
& = (x^2+12x+56)\frac{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}}
\end{aligned}
$$
so
$$
\begin{aligned}
\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} & = (7^2+12\cdot7+56)\frac{\sum_{k=0}^416^{k/4}2^{4-k}}{\sum_{k=0}^69^{k/2}27^{2-k/3}} \\
& = 189\cdot\frac{2^4}{27^2} = \frac{112}{27}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Quadrilateral, two circumcircles, relation to prove In a convex quadrilateral $ABCD$ diagonals $AC$ and $BD$ intersect at point $S$. Denote with $P$ the center of circumcircle of triangle $ABS$. Denote with $Q$ the center of circumcircle of triangle $CDS$. Prove that:
$$4 PQ \geq AB + CD.$$
I was able to calculate the right side based on angle between diagonals and circle radii. I was also able to make some estimate of the distance between circle centers. But I could not make the match between two formulas.
| A graph is drawn below, together with the explanation of the notations. To simplify the explanation, we only restrict to the most non-trivial case: $\angle B \geq \angle A$, $\angle C \geq \angle D$.
It is not hard to see that $\angle PSA = \frac{\theta + \angle A - \angle B}{2}$ and $\angle QSD = \frac{\theta + \angle D - \angle C}{2}$. Therefore
$$
\angle PSQ = \frac{2\theta + \angle A - \angle B + \angle D - \angle C}{2} + 180^\circ - \theta \geq 180^\circ - \frac{\angle B + \angle C}{2} \geq \theta.
$$
By law of sines, $\overline{AB} + \overline{CD} = 2(r_1 + r_2)\sin\theta$.
By law of cosines,
$$
\begin{aligned}
\overline{PQ} &\geq \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos\theta} = \sqrt{(r_1+r_2)^2 - 2(1+\cos\theta)r_1r_2}\\
&\geq\sqrt{(r_1+r_2)^2 - \frac{1}{2}(1+\cos\theta)(r_1+r_2)^2}\\
&= (r_1+r_2)\sqrt{\frac{1-\cos\theta}{2}} = (r_1+r_2)\sin\frac{\theta}{2} \geq \frac{1}{2}(r_1+r_2)\sin\theta.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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