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Find all $n \in \mathbb{N}$ such that ${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}$
Find all $n \in \mathbb{N}$ such that $${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}.$$
I know that if $2n+1$ were prime number, then
$${{2n}\choose{n}} = \frac{(2n) (2n-1) \cdots (n+1)}{n!} \equiv \frac{(-1)(-2)\cdots(-n)}{n!} = (-1)^n \pmod{2n+1}.$$
However, I'm not sure whether $\{ n \,|\, 2n+1 \text{ are prime}\}$ are the only possible solutions.
| Consider $N=(2n+1)$, an odd number. If it is a prime number, then with the argument in the OP we have the congruence $\binom {2n}n=(-1)^n$ modulo $N$.
In case $N$ is not a prime we have to divide first in the above "scheme". I will give an example. For $n=10$, $N=21=3\cdot 7$, we have to compute modulo $21$
$$
\binom{2n}n=
\binom{20}{10}=
\frac{11}{10}\cdot
\frac{12}{ 9}\cdot
\frac{13}{ 8}\cdot
\frac{14}{ 7}\cdot
\frac{15}{ 6}\cdot
\frac{16}{ 5}\cdot
\frac{17}{ 4}\cdot
\frac{18}{ 3}\cdot
\frac{19}{ 2}\cdot
\frac{20}{ 1}\ .
$$
Now we have to simplify first with all the divisors of $21$, these are $3,7$, and instead of $\frac{14}7$ we get $\frac 21$. This contribution is not $(-1)$ "as in the pattern" for a prime number. There are also more complicated effects like $\frac{12}9=\frac 43$ where we cannot fully simplify inside of these fraction. We need the contribution of $18$ to get rid of the "second factor" $3$. Note that performing the calculus in $R=\Bbb Z/21$ we can replace the fractions which belong to $R^\times$ with the corresponding $(-1)$ factors modulo $21$, i.e.
$$
\begin{aligned}
\binom{20}{10}
&=
\frac{11}{10}\cdot
\frac{12}{ 9}\cdot
\frac{13}{ 8}\cdot
\frac{14}{ 7}\cdot
\frac{15}{ 6}\cdot
\frac{16}{ 5}\cdot
\frac{17}{ 4}\cdot
\frac{18}{ 3}\cdot
\frac{19}{ 2}\cdot
\frac{20}{ 1}
\\
&=
(-1)\cdot
\frac{12}{ 9}\cdot
(-1)\cdot
\frac{14}{ 7}\cdot
\frac{15}{ 6}\cdot
(-1)\cdot
(-1)\cdot
\frac{18}{ 3}\cdot
(-1)\cdot
(-1)\ .
\end{aligned}
$$
But the remained not $(-1)$ factors contribute in a chaotic way to the final result. With a probability $\sim 1/N$ (or $\sim 1/\varphi(N)$) we may get by fortune the $(-1)$. Here is an example, obtained with minimal sage code:
sage: for n in [3..10**6]:
....: N = 2*n+1
....: if N.is_prime(): continue
....: R = Zmod(N)
....: if R(binomial(2*n,n)) == R((-1)^n):
....: print n, N
....:
2953 5907
and i better stop here, there is already a better answer.
Pari/gp code finding it:
? for(n=1, 10^6, N=2*n+1; if( (1-isprime(N)) & binomial(2*n,n) % N == (-1)^n % N, print("n = ", n, " N = ", N, " with factors ", factor(N))))
n = 2953 N = 5907 with factors [3, 1; 11, 1; 179, 1]
(and it still runs right now...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Let $W_{1},W_{2}$ be sub-spaces of $\mathbb{R}^{4}$, find a subspace $W_{3}$ s.t $W_3\subset W_{2}$ and $W_{1}\oplus W_{3}=W_{1}+W_{2}$ Let $W_{1},W_{2}$ be linear sub-spaces of $\mathbb{R}^{4}$.
$W_{1}=\text{sp}\{(1,2,3,4),(3,4,5,6),(7,8,9,10)\}$
$W_{2}=\text{sp }\{(x,y,z,w)| \ x+y=0\}$
Find a linear subspace of $ \ \mathbb{R}^{4} \ ; W_{3}$ , such that: $W_{3}\subset W_{2} \ \text {*and*} \ W_{1}\oplus W_{3}=W_{1}+W_{2}$
My attempt:
I applied Gaussian elimination on the vectors of $W_1$, such that:
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
0\\
-2\\
-1
\end{array}\right)+b\left(\begin{array}{c}
0\\
1\\
-2\\
-1
\end{array}\right)|a,b\in\mathbb{R}\right\} $$
At that point I got stuck. I'm not sure how to continue.
| HINT
We should write as a span (check again you derivation by RREF)
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=s\left(\begin{array}{c}
1\\
1\\
1\\
1
\end{array}\right)+t\left(\begin{array}{c}
0\\
1\\
2\\
3
\end{array}\right)\right\}$$
and we can also easily find that
$$W_{2}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=r\left(\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right)+s\left(\begin{array}{c}
0\\
0\\
1\\
0
\end{array}\right)+t\left(\begin{array}{c}
0\\
0\\
0\\
1
\end{array}\right)\right\}$$
then check by RREF on the $5$ basis vectors that $W_{1}+W_{2}$ has dimension $4$ and finally select 2 basis vectors for $W_3$ from the basis vectors of $W_2$ such that $W_{1}\oplus W_{3}=\mathbb{R^4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determining the null space of the matrix
Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$
My try:
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\rightarrow 5R_3-2R_2}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 0 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{5}}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
From this I got $$x-y=0\implies x=y\\y=0$$
$$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$
So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?
| Right Nullspace
If vector $\alpha = [x,y]$ is in the right nullspace of A then
$$\begin{bmatrix}
1 & -1 \\
2 & 3 \\
1 & 1 \\
\end{bmatrix}[x,y]^T = \begin{bmatrix} 0 \\ 0 \\ 0
\end{bmatrix}$$
This gives
$$x - y = 0$$
$$2x + 3y = 0$$
$$x + y = 0$$
First and third equation tells us that $x = y = -x$, i.e. the only solution is $[0,0]$. So, the right nullspace has dimension zero
Left Nullspace
If vector $\alpha = [x,y,z]$ is in the left nullspace of A then
$$\begin{bmatrix}
x & y & z
\end{bmatrix}
\begin{bmatrix}
1 & -1 \\
2 & 3 \\
1 & 1 \\
\end{bmatrix} = \begin{bmatrix}
0 & 0
\end{bmatrix}$$
So we get
$$x + 2y + z = 0$$
and
$$-x + 3y + z = 0$$
First equation tells us $$x = -(2y+z)$$
and second one says
$$x = 3y+z$$
So by equating $x=x$, we get
$$-2y-z = 3y+z$$
that is
$$5y = -2z$$
or $$y = - \frac{2}{5}z$$. Plugging this back into one of the two x equations, we get
$$x= 3y + z = 3(- \frac{2}{5}z) + z = \frac{-6+5}{5}z = -\frac{1}{5}z$$
So, the Null space takes the form
$$\begin{bmatrix}
-\frac{1}{5} \\
- \frac{2}{5} \\
1
\end{bmatrix}
z$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$$
So, maximum value $m$ is $\frac{7}{a+3b}+{\sqrt{3ab}}$ so $m=\frac{7}{2{\sqrt{3ab}}}+{\sqrt{3ab}}$ is $2{\sqrt{\frac{7}{2}}}$ so $2m^2=28$.
But it didn't satisfy equal condition so how do I get it?
| Your answer is right!
Indeed, we'll prove that $\sqrt{14}$ is a maximal value.
Let $a=x^2$ and $b=3y^2$, where $x$ and $y$ are positives.
Thus, $2a^2+7ab+3b^2=2x^4+21x^2y^2+27y^4$ and we need to prove that
$$x^2+3xy+6y^2\leq\sqrt{14}$$ or
$$(x^2+3xy+6y^2)^2\leq2(2x^4+21x^2y^2+27y^4)$$ or
$$(x-y)^2(x^2+6y^2)\geq0,$$ which is obvious.
The equality occurs for $b=3a$, which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d) None of these
Attempt:
I have expanded and checked but couldn't spot any exact differentials.
Secondly, it's not a homogeneous equation, so couldn't use $y = vx$.
How do I go about solving this problem?
| hint: $x+y = (x+2) + (y-2)$ , and use $(A+B)^2 = A^2 + 2AB +B^2, A = x+2,B = y - 2$ to expand the numerator and simplify
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansion... Also got no more clues.
| $x(1-x)^{-\frac 12}$
Generalized binomial theorem.
$x(1-x)^{-\frac 12} = x(1^{-\frac {1}{2}} + \frac {-1}{ 2}1^{-\frac {3}{2}}(-x) + \frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}1^{-\frac {5}{2}}(-x)^2+\frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}\frac {-5}{2}\frac {1}{3}1^{-\frac {5}{2}}(-x)^3\cdots$
$x(1-x)^{-\frac 12} = x + \frac {1}{ 2}x^2 + \frac {3}{8}x^3+\frac {5}{16}x^4\cdots$
$a_1 = 1\\
a_{n+1} = a_n{\frac {2n-1}{2n}}\\
a_n= \frac {(2n-1)!}{2^{2n}((n-1)!)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Counting problem - fault in my reasoning. The problem is as follows:
The dean of science wants to select a committee consisting of mathematicians and physicists to discuss a new curriculum. There are $15$ mathematicians and $20$ physicists at the faculty; how many possible committees of $8$ members are there, if there must be more mathematicians than physicists (but at least one physicist) on the committee?
The given solution splits up the choices into the cases where there is $1$ physicist, $2$ physicists, and $3$ physicists. Then it's simply a matter of summing up $\binom{15}{5}\cdot\binom{20}{3}$ and $\binom{15}{6}\cdot\binom{20}{2}$ and $\binom{15}{7}\cdot\binom{20}{1}$, giving $4503070$.
This is what I tried to do:
first I pick a physicist, making sure I have at least one on the team. I have $20$ choices for this. Then I pick $5$ mathematicians, making sure there are more mathematicians than physicists on the team. I have $15\cdot14\cdot13\cdot12\cdot11$ choices for this. Then I pick two people from the $29$ people that are left because it doesn't matter whether they are mathematicians or physicists. I have $29\cdot28$ choices for this. After I multiply all those numbers, I need to divide by the number of ways I can permutate them, because order doesn't matter. I get $$\dfrac{20\cdot15\cdot14\cdot13\cdot12\cdot11\cdot29\cdot28}{8!} = 145145,$$ which is obviously not the right solution. Where did I go wrong?
Thanks in advance.
| You made a calculation error.
$$\frac{20 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 29 \cdot 28}{8!} = \color{red}{1161160}$$
Let's examine your numerator:
$$20 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 29 \cdot 28 = 20 \cdot \frac{15!}{10!} \cdot \frac{29!}{27!} = \binom{20}{1} \cdot \binom{15}{5} \cdot 5! \cdot \binom{29}{2} \cdot 2!$$
Since the order in which we choose the five designated mathematicians does not matter, we should divide your numerator by $5!$. Since the order in which the two additional committee members does not matter, we should also divide your numerator by $2!$. Doing so yields
$$\binom{20}{1}\binom{15}{5}\binom{29}{2} = 24384360 = \frac{8!}{5!2!} \cdot 1161160$$
which is too large.
Let's see why.
A committee of eight people drawn from $20$ physicists and $15$ mathematicians that contains a majority of mathematicians but at least one physicist must contain either five mathematicians and three physicists or six physicists and two physicists or seven mathematicians and one physicist. The number of ways of selecting exactly $k$ mathematicians and $8 - k$ physicists from the $15$ mathematicians and $20$ physicists is
$$\binom{15}{k}\binom{20}{8 - k}$$
Hence, the number of admissible committees is
$$\binom{15}{5}\binom{20}{3} + \binom{15}{6}\binom{20}{2} + \binom{15}{7}\binom{20}{1} = 450370$$
By designating a particular physicist as the physicist on the committee, you count each committee with $p$ physicists $p$ times, once for each way you could designated one of them as the designated physicist. By designating five of the mathematicians on the committee as the five designated mathematicians on the committee, you count each committee with $m$ mathematicians $\binom{m}{5}$ times, once for each way you could designate five of the $m$ mathematicians as the designated mathematicians. Observe that
$$\binom{20}{1}\binom{15}{5}\binom{29}{2} = \color{red}{\binom{5}{5}\binom{3}{1}}\binom{15}{5}\binom{20}{3} + \color{red}{\binom{6}{5}\binom{2}{1}}\binom{15}{6}\binom{20}{2} + \color{red}{\binom{7}{5}\binom{1}{1}}\binom{15}{7}\binom{20}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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multidimensional chain rule, example Given $f:\mathbb{R}^3\to\mathbb{R}$, $f(u,v,w)=uv+vw-uw$ and
$g:\mathbb{R}^2\to\mathbb{R}^3$, $g(x,y)=(x+y, x+y^2, x^2+y)$, I have to give $D(f\circ g)(x,y)$
I calculated like this:
$Df(u,v,w)=(v-w, u+w, v-u)$ and
$Dg(x,y)=\begin{pmatrix} 1&1&2x\\ 1&2y& 1\\ 0&0&0\end{pmatrix}$
Furthermore it is $Df(g(x,y))=(x+y^2-x^2-y, x+y+x+y^2, x+y^2-x-y)$
Now with the multidimensional chain rule, we have:
$D(f\circ g)(x,y)=Df(g(x,y))\cdot Dg(x,y)$
$(y^2-x^2+x+y, y^2+y+2x, y^2-y)\cdot\begin{pmatrix} 1&1&2x\\ 1&2y& 1\\ 0&0&0\end{pmatrix}$
$$=(\color{red}{y^2-x^2+x-y+y^2+y+2x}, \color{blue}{y^2-x^2+x-y+2y^3+2y^2+2xy}, \color{orange}{2xy^2-2x^3+2x^2-2xy+y^2+y+2x})$$
$$=(\color{red}{2y^2+3x-x^2}, \color{blue}{2y^3+3y^2-y-x^2+x+4xy}, \color{orange}{y^2+y-2x^3+2x^2+2x+2xy^2-2xy})$$
Which is wrong accoring to:
http://www.wolframalpha.com/input/?dataset=&i=d%2Fdx+ReplaceAll%5Buv%2Bvw-uw,%7Bu-%3Ex%2By,v-%3Ex%2By%5E2,w-%3Ex%5E2%2By%7D%5D
[You might check Multidimensional chain rule, online calculator for further reference.]
But I do not see my mistake myself, unfortunatly.
Also I have a question, because I got a little bit confused by the notation.
If you write $g:\mathbb{R}^2\to\mathbb{R}^3$, $g(x,y)=(x+y, x+y^2, x^2+y)$, dont you have to be more clear and note $(x+y, x+y^2, x^2+y)^t$, since a vector in $\mathbb{R}^3$ should be normally noted like $\begin{pmatrix}x\\y\\z\end{pmatrix}$ this [where I just use $x,y,z$ because of simplicity], or do I mess something up here?
Thanks in advance.
| $Dg(x,y)$ should be a matrix of $3\times 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derive the following identity $1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
Count the elements of the following set
$$A=\{(x,y,z): 1\leq x,y,z \leq n+1, z>\max\{x,y\}\}.
$$
From this derive the following identity:
$$1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
In the same manner find the formula for $1^k + 2^k + \ldots + n^k$ for $k=3,4$.
It is easy to see that $|A| = 1^2 + 2^2 + \ldots + n^2$, since from the sum rule we have $$|A| = \sum_{i=0}^{n+1} |\{(x,y,i): 1\leq x,y,z \leq n+1, i> \max\{x,y\}\}| = \sum_{i=0}^{n+1} i^2$$ (as we can choose $x$ and $y$ in $i \times i$ ways for each $i$).
However I can't see why is $|A|$ equals $\dfrac{n(n+1)(2n+1)}{6}$.
| One method is this: When n= 0, the sum is 0. When n= 1, the sum is 1. When n= 2, the sum is 1+ 4= 5. When n= 3, the sum is 1+ 4+ 9= 14. When n= 4 the sum is 1+ 4+ 9+ 16= 30. When n= 5 the sum is 1+ 4+ 9+ 16+ 25= 55, etc.
Obviously, at each step we are just adding the next square so the "first differences" are: n= 0, 1- 0= 1; n= 1, 5- 1= 4; n= 2, 14- 5= 9; n= 3, 30- 14= 16; n= 4, 55- 30= 25, etc.
The "second differences" are: n= 0, 4- 1= 3; n= 1, 9- 4= 5; n= 2, 16- 9= 7; n= 3, 25- 16= 9.
The "third differences" are: n= 0, 5- 3= 2; n= 1, 7- 5= 2; n= 2, 9- 7= 2.
The third differences are the constant 2 so all succeeding differences are 0. By "Newton's divided difference formula" the sequence is given by $a_n= 0+ 1(n)+ 3(n)(n-1)/2!+ 2(n)(n-1)(n-2)/3!= \frac{1}{6}\left(6n+ 9n^2- 9n+ 2n^3- 6n^2+ 4n\right)= \frac{1}{6}\left(2n^3+ 3n^2+ n\right)= \frac{1}{6}n(n+1)(2n+1)$.
As a check, when n= 1, that is $\frac{1}{6}(2+ 3+ 1)= 1$; when n= 2, that is $\frac{1}{6}(16+ 12+ 2)= 5$; when n= 3, that is $\frac{1}{6}(54+ 27+ 3)= 14$, etc.
| {
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"url": "https://math.stackexchange.com/questions/2888652",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate, $f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$
If $$f(x)=\frac{4^x}{4^x+2}$$
Calculate,
$$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots
f\bigg(\frac{1996}{1997}\bigg)$$
My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$f\bigg(\frac{1}{2}\bigg)=\frac{1}{2}$$
$$f\bigg(\frac{1}{3}\bigg)+f\bigg(\frac{2}{3}\bigg)=1$$
$$f\bigg(\frac{1}{4}\bigg)+f\bigg(\frac{2}{4}\bigg)+f\bigg(\frac{3}{4}\bigg)=\frac{3}{2}$$
I could see that,
$$f\bigg(\frac{1}{n}\bigg)+f\bigg(\frac{2}{n}\bigg)+f\bigg(\frac{3}{n}\bigg)\ldots
f\bigg(\frac{n-1}{n}\bigg)=\frac{n-1}{2}$$
So, $$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots
f\bigg(\frac{1996}{1997}\bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?
| Given $f(x)=\dfrac{4^x}{4^x+2}$
From that we get $f(1-x)=\dfrac{2}{4^x+2}$
First let us take the last term $f\left(\dfrac{1}{1997}\right)$
Notice that $f\left(\dfrac{1}{1997}\right)=f\left(1-\dfrac{1}{1997}\right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{1997}\right)+........+f\left(1-\dfrac{2}{1997}+f\left(1-\dfrac{1}{1997}\right)\right)$
all of them makes pairs.
So, the total pairs $=\dfrac{1996}{2}=998$
So, the sum $=1+1+1+......998$ times $=998$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Distribution function of $X_1 + X_2$ given probability distributions of $X_1$ and $X_2$ Suppose I have a probability distribution for random variable $X_1$ given by:
$$
P(x) =
\begin{cases}
1/a, & \text{if} \;\; 0 \leq x \leq a\\
0, & \text{otherwise}
\end{cases}
$$
and $X_2$ with probability distribution:
$$
P(x) =
\begin{cases}
1/b, & \text{if} \;\; 0 \leq x \leq b\\
0, & \text{otherwise}
\end{cases}
$$
What is the probability distribution function of $X_1 + X_2$.
| These are both uniform random variables if they independent we can write it like this. First $X_{1} \sim U(0,a)$ $X_{2} \sim U(0,b)$ . Also note the proof assumes that $ a \leq b$ The pdf of a uniform random variable $X \sim U(a,b)$ is given by
$$ f_{X}(x) =\begin{align}\begin{cases} \frac{1}{b-a} & \textrm{ for } x \in [a,b] \\ \\ 0 & \textrm{ for everywhere else } \end{cases} \end{align} \tag{1}$$
then the pdf of $X_{1}$ is given by
$$f_{X_{1}}(x_{1}) =\begin{align}\begin{cases} \frac{1}{a} & \textrm{ for } x \in [0,a] \\ \\ 0 & \textrm{ for everywhere else } \end{cases} \end{align}\tag{2}$$
so the pdf of $X_{2}$ is
$$f_{X_{2}}(x_{2}) =\begin{align}\begin{cases} \frac{1}{b} & \textrm{ for } x \in [0,b] \\ \\ 0 & \textrm{ for everywhere else } \end{cases} \end{align} \tag{3}$$
Let the density $S = X_{1} + X_{2}$ be given by
$$ f_{S}(s) = \int_{\mathbb{R}} f_{X_{1}}(s-x_{2})f_{X_{2}}(x_{2}) dx_{2} \tag{4}$$
$$f_{X_{1}}(s-x_{2}) =\begin{align}\begin{cases} \frac{1}{a} & 0 \leq s-x_{2} \leq a \\ \\ 0 & \textrm{ for everywhere else } \end{cases} \end{align} \tag{5}$$
$$f_{X_{2}}(x_{2}) =\begin{align}\begin{cases} \frac{1}{b} & 0 \leq x_{2} \leq b \\ \\ 0 & \textrm{ for everywhere else } \end{cases} \end{align} \tag{6}$$
when $ 0 \leq x_{2} \leq a$
$$ f_{S}(s) = \int_{0}^{s} \frac{1}{ab} dx_{2} = \frac{x_{2}}{ab}\Big|_{0}^{s} = \frac{s}{ab} \tag{7}$$
when $ a \leq x_{2} < b $
$$ f_{S}(s) =\int_{s-a}^{s} \frac{1}{ab} dx_{2} = \frac{1}{ab}\Big|_{s-a}^{s} = \frac{s}{ab} - \frac{s-a}{ab}= \frac{1}{ab}(s-(s-a)) = \frac{a}{ab} =\frac{1}{b}
\tag{8} $$
when $ b \leq x_{2} \leq a+b $
$$ f_{S}(s) = \int_{s-a}^{b} \frac{1}{ab} dx_{2} =\frac{1}{ab}\Big|_{s-a}^{b} = \frac{b}{ab} - \frac{s-a}{ab} = \frac{1}{a} - \frac{s-a}{ab} = \frac{1}{a} - \frac{s}{ab}+\frac{1}{b} \tag{9}$$
$$f_{S}(s) =\begin{align}\begin{cases} \frac{1}{ab}s & 0 \leq s < a \\ \\ \frac{1}{b} & a \leq s < b \\ \frac{1}{a} -\frac{s}{ab}+\frac{1}{b} &b\leq s < a+b \\ 0 & \textrm{ otherwise} \end{cases} \end{align} \tag{10}$$
In order to visualize this here is some code
import numpy as np
import matplotlib.pyplot as plt
def sum_uni(a,b,n):
z1 = np.linspace(0,a,n)
z2 = np.ones(n)
z3 = np.linspace(b,a+b,n)
f1 = (1/(a*b))*z1
f2 = (1/b)*z2
f3 = (1/a)*z2-(1/(a*b))*z3 +(1/b)*z2
g1 = np.linspace(0,a,n)
g2 = np.linspace(a,b,n)
g3 = np.linspace(b,a+b,n)
my_plot = np.concatenate((g1,g2,g3),axis=0)
my_distribution = np.concatenate((f1,f2,f3),axis=0)
return my_plot, my_distribution
my_plot, my_dist = sum_uni(2,3,100)
plt.plot(my_plot,my_dist)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$
| In these cases, in order to avoid mistakes with the sign, we can let $y=-x\to \infty$ to obtain
$$\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}=\lim_{y\to \infty} \frac{-7y}{\sqrt {4y^2-7y}+2y}=-\frac 74$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3y±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.
| Hint:
$$2x^2-3xy-2y^2=2x^2-4xy+xy-2y^2=2x\underbrace{(x-2y)}+y\underbrace{(x-2y)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
(Modular Arithmetic) Congruences With Exponents I am trying to find the following:
The least positive integer $n$ for which
$3^n \equiv 1$ (mod $7$)
And hence $3^{100}$ (mod $7$).
The least positive integer $n$ for which
$5^n \equiv 1$ (mod $17$) or $5^n \equiv -1$ (mod $17$)
And hence evaluate $5^{243}$ (mod $17$).
Evaluate $2^4$ (mod $18$) and hence evaluate $2^{300}$ (mod $18$).
I've learned how to use the Euclidean algorithm, but I have no idea how to compute congruence when they have exponents, as above. These seem very daunting to me. I would greatly appreciate it if people could please take the time to explain how this is done.
I will then post my work for each of these for, if possible, feedback on the solutions, since I do not have any solutions for these.
Thank you.
| Regarding congruences with exponents - the cool thing about them is you can raise 'each side' to some power, or multiply them by a common factor and it remains true. It's often useful to show that $x^n\equiv \pm 1 \pmod k $. Regarding your first question, we know $3\equiv 3\pmod 7 $ so $3^2\equiv 2\pmod 7$ since $9\pmod 7=2$. So $3^3=3^2\times 3\equiv 2\times 3=6\equiv -1 \pmod 7$. So $3^3\equiv -1 \pmod 7$. Hence $(3^3)^2=3^6\equiv (-1)^2=1\pmod 7$. We know $3^{100}=3^{6\cdot 16+4}=3^{96}\cdot 3^3\cdot 3$, and since we know the value of all these modulo $7$, $3^{100}$ is congruent mod $7$ to the product of them.
For your 2nd question, we can apply these same principles - it just takes a bit longer $$\begin{align*}5&\equiv 5\pmod {17} \\ \implies 5^2&\equiv 8 \pmod {17} \\ \implies 5^3&\equiv 6\pmod {17} \qquad \text{since $40\pmod {17}=6$} \\ \implies 5^{4}&\equiv 13\pmod {17}\qquad \text{since $30\pmod{17}=13$} \\ &\vdots \\ \implies 5^8&\equiv -1\pmod{17}\end{align*}$$Note that we could have directly obtained the final result by squaring both sides of the fourth line. So now $5^{243}=(5^8)^{30}\cdot 5^3=\dots$
However, using Euler's theorem is usually faster - there's examples of solutions to similar problems within that link.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the solution.
| Thank you for your answers that helped me make a solution
$M_{red}=\dbinom{0}{3/2}$ , $r_{red}=\dfrac{3}{2}$ , $M_{blue}=\dbinom{0}{2}$ , $r_{blue}=2$, $M_{green}=\dbinom{2}{3/2}$ , $r_{green}=\dfrac{5}{2}$
$M_{black}=M=\dbinom{x}{y}$ , $r_{black}=r$
$length(v)=\dfrac{v}{\sqrt{v_{x}^{2}+v_{y}^{2}}}$ , $unitvec(v)=\dfrac{v}{length(v)}$
$d_{red}=length[M_{red}+unitvec(M_{red}-M)\times r_{red}-M]$
$d_{blue}=length[M_{blue}+unitvec(M_{blue}-M)\times r_{blue}-M]$
$d_{green}=length[M_{green}+unitvec(M_{green}-M)\times r_{green}-M]$
$\dbinom{0}{3/2}+\dfrac{\dbinom{0}{3/2}-\dbinom{x}{y}}{\sqrt{(0-x)^{2}+(3/2-y)^{2}}}\times \dfrac{3}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(-x-\dfrac{3x}{\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}+\left(3/2-y+\dfrac{9-6y}{2\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}}$
$\dbinom{2}{0}+\dfrac{\dbinom{2}{0}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(0-y)^{2}}}\times 2-M$ $\Rightarrow $ $d_{blue}=\sqrt{\left(2-x+\dfrac{4-2x}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}+\left(-y-\dfrac{2y}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}}$
$\dbinom{2}{3/2}+\dfrac{\dbinom{2}{3/2}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(3/2-y)^{2}}}\times \dfrac{5}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(2-x-\dfrac{5(-2+x)}{\sqrt{25+4x(-4+x)+4y(-3+y)}}\right)^{2}+\left(3/2-y+\dfrac{5(3/2-y)}{2\sqrt{(-2+x)^{2}+(-3/2+y)^{2}}}\right)^{2}}$
Setting these 3 equations to r and solving for x,y,r of the black circle results in:
$x=\dfrac{36}{23} , y=\dfrac{24}{23} , r=\dfrac{72}{23}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Last digit of sequence of numbers
We define the sequence of natural numbers
$$
a_1 = 3 \quad \text{and} \quad a_{n+1}=a_n^{a_n},
\quad
\text{ for $n \geq 1$}.
$$
I want to show that the last digit of the numbers of the sequence $a_n$ alternates between the numbers $3$ and $7$. Specifically, if we symbolize with $b_n$ the last digit of $a_n$, I want to show that
$$
b_n =
\begin{cases}
3, & \text{if $n$ is odd}, \\
7, & \text{if $n$ is even}.
\end{cases}
$$
There is a hint to prove that for each $n \in \mathbb{N}$, if $a_n \equiv 3 \pmod{5}$ then $a_{n+1} \equiv 2 \pmod{5}$ and if $a_n \equiv 2 \pmod{5}$ then $a_{n+1} \equiv 3 \pmod{5}$.
First of all, if we take $a_n \equiv 3 \pmod{5}$, then $a_{n+1}=3^3\pmod{5} \equiv 2 \pmod{5}$.
If $a_n \equiv 2 \pmod{5}$, then $a_{n+1}=2^2 \pmod{5}=4$. Or am I doing something wrong?
And also how does it follow, if we have shown the hint, that $b_n$ is $3$ when $n$ is odd, and $7$ if $n$ is even?
| First, you should prove that $a_n$ is odd for all $n$ : this follows from the fact that $a_1$ is odd, and if $a_n$ is odd, then $a_{n+1}=a_n^{a_n}$ is an odd number multiplied with itself some number of times, so it is odd.
Let us notice something more. First, $a_1 = 3 \equiv 3 \mod 4$. Next, if $a_n \equiv 3 \mod 4$, then $a_{n+1} \equiv 3^{a_n} \equiv 3^1 \mod 4$ (as $3^2 = 9 \equiv 1 \mod 4$, we may take remainder when $a_n$ is divided by $2$, and this is $1$ since $a_n$ is odd).
Thus, $a_n \equiv 3 \mod 4$ for all $n$.
To the main part.
If $a_n \equiv 3 \mod 5$, then $a_{n+1} = a_n^{a_n} \equiv 3^{a_n} \mod 5$ is all you can say using the fact that $a_n \equiv 3 \mod 5$. But note that $3^{4} \equiv 1 \mod 5$, so you can now say that $a_{n+1} \equiv 3^{(a_n \mod 4)} \mod 5$.
But the point is, that $a_n$ is odd, so $a_n \mod 4$ is either $1$ or $3$. Consequently, $a_{n+1} \equiv 3^1 = 3 \mod 5$ or $a_{n+1} \equiv 3^3 = 27 \mod 5$ must happen. We appear somewhat stuck here. Let me relieve you : we are doing fine.
With the above observation combined with the second part, we have $a_{n+1} \equiv 27 = 2 \mod 5$.
Now, if $a_n \equiv 2 \mod 5$, then $a_{n+1} \equiv 2^{a_n} \mod 5$, and again we equate this to $2^{a_n \mod 4} \mod 5 = 2^3 =8 \equiv 3 \mod 5$.
Thus, $a_{n+1} \equiv 3 \mod 5$.
So the remainders of $a_n$ and $a_{n+1}$ alternate modulo $5$. Since all $a_n$ are odd, this forces their last digits to alternate, and from knowing that the last digit of $a_1$ is $3$, the sequence of last digits must go $373737373...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $p(x) = x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of $f$ around $x=1$, then $1$ is a local minimum
Consider $f:\mathbb{R} \to \mathbb{R} \in C^4$. Show that $p(x) =
x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of order $4$ of $f$ around $x=1$, then
$1$ is a local minimum.
I'm not sure how to proceed. I know that $p(x) = \sum\limits_{k=0}^{4} \dfrac{f^{(n)}(1)(x-1)^n}{n!} = f(1)+f'(1)(x-1)+\dfrac{f''(1)(x-1)^2}{2} + \dfrac{f'''(1)(x-1)^{3}}{6} + \dfrac{f^{(4)}(1)(x-1)^4}{24}=x^4-4x^3+6x^2-4x+1$
and I can see that $p(1) = 0 \implies f(1)=0$. But how can I obtain information about $f'(1)$ and $f''(1)$?
| By the Remainder theorem of the Taylor polynomials we have $$ f(x)=P(x) + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$
$$ = (x-1)^4 + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$
$$ = (x-1)^4 \big[1+\frac {f^{(5)}(\xi)}{5!} (x-1)\big]$$
Let $x$ be close enough to $1$ such that $$ \big| \frac {f^{(5)}(\xi)}{5!} (x-1)\big|<1/2$$
Then $$ \frac {(x-1)^4}{2}<f(x)<\frac{3(x-1)^4}{2}$$
Thus $f(x)$ has a minimum at $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896654",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.
| In this case, since the equation is linear and given the "clue" that the coefficient of $x^2$ is non-zero, then consider $f(x) = a + b x + c x^2$. This yields the possible polynomial solutions, but not necessarily all possible solutions.
\begin{align}
x^2 &= (a + b x + c x^2) - 2 \, \left( a + \frac{b x}{2} + \frac{c x^2}{4} \right) + \left( a + \frac{b x}{4} + \frac{c x^2}{16} \right) \\
&= \frac{b x}{4} + \frac{ 9 \, c \, x^2}{16}.
\end{align}
This leads to $b=0$ and $c = 16/9$ and the general solution of the form
$$f(x) = a + \frac{16 \, x^2}{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit as $(x,y,z)\to (0,0,0)$ of $f(x,y,z) = \dfrac{xy+yz+xz}{\sqrt{x^2+y^2+z^2}}$ To find this limit, I converted to spherical coordinates and rewrote:
$$\lim_{r\to 0} \dfrac{r^2(\sin^2\theta \cos\phi \sin \phi + \sin\theta \cos \theta \sin \phi + \sin\theta \cos \theta \cos \phi)}{r} = 0$$
Is this method alright? Our teacher did using epsilon delta proof, so how can we use something similar to spherical coordinates if say we had four variable limit of kind:
$$\lim_{(w,x,y,z) \to (0,0,0,0)} \frac{xy+yz+xz+wx}{
\sqrt{x^2+y^2+z^2+w^2}}$$
| Notice that $x,y$ and $z$ are less and equal than $\sqrt{x^2+y^2+z^2}$ so that
$$ |xy+yz+xz| \leq 3\left(\sqrt{x^2+y^2+z^2}\right)^2.$$
Hence, by a straightforward application of sandwich theorem, there holds that the aforementioned limit equals zero, since
$$ 0\leq \lim_{(x,y,z)\rightarrow (0,0,0)}\left|\dfrac{xy+yz+xz}{\sqrt{x^2+y^2+z^2}}-0 \right|\leq \lim_{(x,y,z)\rightarrow (0,0,0)}3\sqrt{x^2+y^2+z^2}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Tough Divisibility Problem
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B \equiv 12 \pmod{19}$$
$$20000 + 1000A + 100 + 30 + B \equiv 12 \pmod{19}$$
$$ 5 + 12A + 5 + 11 + B \equiv 12 \pmod{19}$$
$$ 21+ 12A+ B \equiv 12 \pmod{19}$$
$$ 12A+ B + 9 \equiv 0 \pmod{19}$$
This is where I'm stuck.
| You have some errors, which I will fix:
$$\begin{align}\overline{2A13B}=20000 + 1000A + 100 + 30 + B &\equiv 12 \pmod{19} \Rightarrow \\
(19\cdot 1052+12)+(19\cdot 52+12)A+(19\cdot 5+5)+(19\cdot 1+11)+B&\equiv 12 \pmod{19} \Rightarrow \\
12+12A+5+11+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+19\cdot 1+9+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+9+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+B&\equiv 3\pmod{19}.\end{align}$$
Since $0\le A,B\le 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$\begin{align}\overline{3A21B}=30000 + 1000A + 200 + 10 + B &\equiv x \pmod{19} \Rightarrow \\
(19\cdot 1578+18)+(19\cdot 52+12)A+(19\cdot 10+10)+10+B&\equiv x \pmod{19} \Rightarrow \\
18+12A+10+10+B&\equiv x \pmod{19} \Rightarrow \\
12A+19\cdot 2+B&\equiv x \pmod{19} \Rightarrow \\
12A+B&\equiv x \pmod{19}.\end{align}$$
So, $x=3$.
For example, take $\overline{2A13B}=20133\equiv 12 \pmod{19}$ and $\overline{3A21B}=30213\equiv 3\pmod{19}$.
| {
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"url": "https://math.stackexchange.com/questions/2901824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding all complex roots of polynomial Given $\ i $ is a root for the polynom $\ p(x) = x^4 +2x^3 +3x^2 +2x +2 $ find all the roots of $\ p(x) $ in the $\ \mathbb C $ field.
$\ x^4 + 2x^3 + 3x^2 + 2 \ / \ x-i $
I get confused everytime I try to divide $\ p(x) $ by $\ x -i $. I end up with this residual: $\ 2+x^3i +2x^2i + 3xi +2i $
| As the coefficients are real, when $a$ is a root so $\bar{a}$ is a root too. Then
$$\dfrac{x^4 +2x^3 +3x^2 +2x +2}{x^2+1}=x^2+2x+2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902947",
"timestamp": "2023-03-29T00:00:00",
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Finding Jordan canonical form of a matrix given the characteristic polynomial I am trying to find the Jordan canonical form of a matrix $A$ given its characteristic polynomial. Suppose $A$ is a complex $5\times 5$ matrix with minimal polynomial $X^5-X^3$. The end goal of the problem is to find the characteristic polynomial of $A^2$ and the minimal polynomial of $A^2$.
I know that since the minimal polynomial of a matrix divides the characteristic polynomial of a matrix, then $A$ has the same minimal and characteristic polynomial, namely $X^5-X^3$. Now I am trying to find the JCF (Jordan Canonical Form) of $A$ to make it easier to compute $A^2$, since $A$ is conjugate to its JCF. So, since the characteristic polynomial of $A$ splits into $X^3(X+1)(X-1)$, then I know that the Jordan canonical form will have three Jordan blocks, 2 of size 1 corresponding to $1$ and $-1$ and one of size 3. Now, my problem is that I can't figure out the form of this third Jordan block. How do I know that it has the from $$\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
or the form
$$ \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}$$
or the form
$$\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
Thanks for all your help!
| The minimal polynomial is $m_A(x) = x^3(x-1)(x+1)$ and the characteristic polynomial is the same up to sign change $\chi_A(x) = -x^3(x-1)(x+1)$.
We can assume that $A$ is equal to its Jordan form, which is:
$$A = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}$$
We know that the $0$-block is of size $3 \times 3$ because $x^3$ is its power in $m_A$.
Squaring gives:
$$A^2 = \begin{bmatrix}
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}$$
Hence the characteristic polynomial is $\chi_{A^2}(x) = -x^3(x-2)^2$. The minimal polynomial can be seen to be $m_{A^2}(x) = x^2(x-1)$ because squaring the above matrix annihilates the $0$-block, and deducting $1$ annihilates the $1$-block.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$ QUESTION
Find Where The Curvature has an extremum ?
$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$$
MY APPROACH
$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}. . . . . (1)$$
$$\Rightarrow y^\frac{1}{2}=a^\frac{1}{2}-x^\frac{1}{2}$$
now differntiating both sides we will get:
$$\frac{1}{2}{y^\frac{-1}{2}}\frac{dy}{dx}=(-1)\frac{1}{2}x^\frac{-1}{2}$$
$$\Rightarrow \frac{dy}{dx}=-(\frac{y}{x})^\frac{1}{2}. . . . . .(2)$$
Now differentiating again with respect to x again:
$$\frac{d^2y}{dx^2}=-\bigg(\frac{1}{2}(\frac{y}{x})^\frac{-1}{2}.\frac{d}{dx}(\frac{y}{x})\bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{d}{dx}(\frac{1}{x})y-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{-y}{x^2}-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
now put the value of $\frac{dy}{dx}$ :
$$=\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{y}{x^2}-(\frac{y}{x})^\frac{1}{2}\frac{1}{x}\bigg)\Bigg)$$
After simplifying i got :
$$\frac{d^2y}{dx^2}=\frac{1}{2x}\bigg(\frac{y^\frac{1}{2}}{x^2}-1\bigg). . . . .(3)$$
but from simplification of equation (2) in terms of $a$ wil result :
$$\frac{dy}{dx}=1-(\frac{a}{x})^\frac{1}{2}$$
here i can easily simlify this to get $\frac{d^2y}{dx^2}$ i.e.
$$\Rightarrow\frac{d^2y}{dx^2}=\frac{\sqrt a}{2x\sqrt x}. . . . .(4)$$
I dont know why i am unable to reduce (3) to (4).May be there exists some calculation error,even thats not my question.
proceeding to find radius of curvature and curvature :
FROM FORMULA
$$\rho=\frac{\bigg(1+(\frac{dy}{dx})^2\bigg)^\frac{3}{2}}{\frac{d^2y}{dx^2}}$$
putting the value of (2) and (4):
$$\Rightarrow \rho=\frac{\bigg(1+\frac{y}{x}\bigg)^\frac{3}{2}2x\sqrt x}{\sqrt a}$$
$$\Rightarrow\rho=\frac{2(x+y)^\frac{3}{2}}{\sqrt a}$$
so curvature is
$$\frac{1}{\rho}=\kappa=\frac{\sqrt a}{2(2x+a-2\sqrt a\sqrt x)^3/2}$$
[notice that i have put y in terms of a nad x]
NOW BEGINS THE PROBLEM
for being extremum
$\frac{d\kappa}{dx} =0 $ and i have to check the sign of $\frac{d^2\kappa}{dx^2}$ :
as you can see in the image :
$$\frac{d\kappa}{dx}=\frac{3\sqrt a(2-\frac{\sqrt a}{\sqrt x})}{4(2x-2\sqrt x\sqrt a+a)^\frac{3}{2}}$$
Now letting this to zero we have :
$$2=\frac{\sqrt a}{\sqrt x}$$
thus i am getting $x=\frac{a}{4}$ as a critical point.
BUT the answer is given as $\frac{\sqrt 2}{a}$
you can even see by inspection $x=\frac{a}{4}$ is not a critical point.
please help and let me know where i have made mistake.
THIS IS MY HUMBLE REQUEST.
| Perhaps not a complete answer, but below you will find my solution to the problem, which leads to the same answer as yours, so we are either both wrong or the answer sheet is wrong.
Parameterize your curve as
$$ x(t) = a\cos^4 t, \quad y(t) = a\sin^4 t. $$
Then
\begin{align}
\dot x(t) &= -4a\cos^3 t \sin t,\\
\ddot x(t) &= -4a(-3\cos^2t\sin^2t + \cos^4 t)\\
&= -4a\cos^2 t(\cos^2t -3\sin^2 t),
\end{align}
and
\begin{align}
\dot y(t) &= 4a\sin^3 t \cos t,\\
\ddot y(t) &= 4a(3\sin^2t\cos^2t - \sin^4 t)\\
&= 4a\sin^2 t(3\cos^2t -\sin^2 t).
\end{align}
Then
\begin{align}
\dot x \ddot y - \ddot x \dot y
&= -16a^2\cos^3 t \sin^3 t (3\cos^2t -\sin^2 t) + 16a^2\sin^3 t \cos^3 t(\cos^2t -3\sin^2 t)\\
&= 16a^2\cos^3 t \sin^3 t (-4\cos^2 t - 4\sin^2t) = -64 a^2 \cos^3 t \sin^3 t,
\end{align}
and
\begin{align}
(\dot x^2 + \dot y^2 )^{3/2}
&= (16a^2\cos^6t\sin^2t+16a^2\sin^6t\cos^2t)^{3/2}\\
&= 64a^3\cos^3t\sin^3t (\cos^4t+\sin^4t)^{3/2}.
\end{align}
This gives
$$ \kappa(t) = \frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y^2 )^{3/2}} = - \frac{1}{a(\cos^4t+\sin^4t)^{3/2}}.$$
Now
\begin{align}
\dot\kappa(t) &= \frac{3}{2a}\frac{-4\cos^3t\sin t + 4\sin^3 t\cos t}{(\cos^4t+\sin^4t)^{5/2}}\\
&= \frac{3}{2a}\frac{4\cos t\sin t(\sin^2t- \cos^2 t}{(\cos^4t+\sin^4t)^{5/2}}\\
&= \frac{3}{2a}\frac{-2\sin(2t)\cos(2t)}{(\cos^4t+\sin^4t)^{5/2}}\\
&= -\frac{3\sin(4t)}{2a(\cos^4t+\sin^4t)^{5/2}}.
\end{align}
So $\dot \kappa = 0$ whenever $\sin(4t) = 0,$ which happens for
$$ 4t = \pi n \Leftrightarrow t = \frac{\pi}{4}n, \quad n\in \mathbb{Z}. $$
Since (I presume that) $x,y > 0$, we are considering $t \in (0, \pi/2)$, so the only valid option is $t = \pi/4$, and then
$$ x\left(\frac{\pi}{4}\right) = a\cos^4\left(\frac{\pi}{4}\right) = \frac{a}{\sqrt{2}^4} = \frac{a}{4}, $$
which confirms your answer.
Also, the graph does seem to indicate that the curvature has an extremum at this point. Note that we are not looking for an extremum of the curve, but rather the extremum of its curvature.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of a recursive complex sequence I am currently struggling with the following exercise:
Let $z_0 = x_0+iy_0 \ne 0$ be a complex number and let the sequence $(z_n)_n$ be recursively defined as
$$z_{n+1} = \frac{1}{2} \left( z_n+\frac{1}{z_n} \right)$$
for $n \ge 0$. Show that if $x_{0} > 0$ then $\lim_{n \to \infty} \ z_n = 1$.
$\\$
I do not know how to work with complex recursive sequences, but I suppose that, as in the real-valued case, a solution will require proving that the limit exists and then to solve the recurrency
$$l = \frac{1}{2} \left( l + \frac{1}{l} \right)$$
In the real valued case I would try to prove that $(z_n)_n$ is monotone and bounded, but since there is no order relation in $\mathbb{C}$ I can not use this apporoach here. Could you help me?
Edit: Thanks to Ahmad Bazzi's answer I now have an idea on how to prove that $(z_n)_n$ converges: If I can prove that both sequences $(x_n)_n$ and $(y_n)_n$ (these sequences are the real and imaginary parts of each $z_n$) converge I could deduce that $(z_n)_n$ converges.
To prove that both $(x_n)_n$ and $(y_n)_n$ converge I would show that these sequences are both bounded and monotone (via induction). But unfortunately I do not get far with this approach because the definitions of these sequences is interwined, i.e.:
\begin{align}
x_{n+1} &= \frac{1}{2}(x_n + \frac{x_n}{x_n^2 + y_n^2})\\
y_{n+1} &= \frac{1}{2}(y_n - \frac{y_n}{x_n^2 + y_n^2})
\end{align}
and I do not know how to go on from there. Any help would be appreciated.
| The way I see it is that you got two recursions going on
\begin{equation}
z_n = x_n + iy_n
\end{equation}
So
\begin{equation}
x_{n+1} + iy_{n+1}
=
\frac{1}{2}
(x_n + iy_n + \frac{1}{x_n + iy_n})
\end{equation}
The right hand side is
\begin{equation}
\frac{1}{2}
(x_n + iy_n + \frac{1}{x_n + iy_n})
=
\frac{1}{2}
(x_n + iy_n + \frac{x_n - iy_n}{x_n^2 + y_n^2})
\end{equation}
which is
\begin{equation}
\frac{1}{2}
(x_n + iy_n + \frac{1}{x_n + iy_n})
=
\frac{1}{2}\big( x_n + \frac{x_n}{x_n^2 + y_n^2} \big) + i\frac{1}{2}\big(y_n - \frac{y_n}{x_n^2 + y_n^2} \big)
\end{equation}
This means that
\begin{align}
x_{n+1} &= \frac{1}{2}(x_n + \frac{x_n}{x_n^2 + y_n^2})\\
y_{n+1} &= \frac{1}{2}(y_n - \frac{y_n}{x_n^2 + y_n^2})
\end{align}
Solve the system
\begin{align}
X &= \frac{1}{2}(X + \frac{X}{X^2 + Y^2})\\
Y &= \frac{1}{2}(Y - \frac{Y}{X^2 + Y^2})
\end{align}
where the limit $L$ is
\begin{equation}
L = X + iY
\end{equation}
First equation tells you that $X - \frac{1}{2}X = \frac{1}{2} \frac{X}{X^2 +Y^2}$, which means that
$$X^2 + Y^2 = 1$$Replace the above condition in the $Y$ equation, you will get $Y = \frac{1}{2}(Y-Y) = 0$, hence $X^2 = 1$ so $X = \pm 1$, i.e. $L = \pm 1$. So, the limit in your problem will depend on the initial value $z_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Express a matrix $A$ as the sum of a symmetric and a skew symmetric matrix I'm trying to express
$$A = \left[ \begin{array} { r r r } { 2 } & { - 2 } & { - 4 } \\ { - 1 } & { 3 } & { 4 } \\ { 1 } & { - 2 } & { - 3 } \end{array} \right]$$
as the sum of a symmetric and a skew symmetric matrix.
So far I have tried this:
converting into transpose
$A ^ { \prime } = \left[ \begin{array} { c c c } { 2 } & { - 1 } & { 1 } \\ { - 2 } & { 3 } & { 4 } \\ { - 4 } & { 4 } & { - 3 } \end{array} \right]$
well this dont seem to be symmteric matrix according to me. are there steps to be taken.
and for skew symmteric matrix i tried to do like this
transpose of equation
$A ^ { \prime } = \left[ \begin{array} { c c c } { 2 } & { - 1 } & { 1 } \\ { - 2 } & { 3 } & { 4 } \\ { - 4 } & { 4 } & { - 3 } \end{array} \right]$
took minus as common
$A ^ { \prime } = \left[ \begin{array} { c c c } { - 2 } & { 1 } & { - 1 } \\ { 2 } & { - 3 } & { 2 } \\ { 4 } & { - 4 } & { 3 } \end{array} \right]$
can you guide me how to proceed with them
| HINT
Recall that for any square matrix $A$
$$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$$
and
$$\left(\frac{A+A^T}{2}\right)^T=\frac{A^T+A}{2}=\frac{A+A^T}{2}$$
$$\left(\frac{A-A^T}{2}\right)^T=\frac{A^T-A}{2}=-\frac{A-A^T}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Verify Elementary Number Theory Proof (Romanian Olympiad) Can you check that there are no errors in this proof I wrote?
Here's the problem (from the 2002 Romanian Olympiad):
Let $p, q$ be two distinct primes. Prove that there are
positive integers $a, b$ so that the arithmetic mean of all the divisors of the number $n = p^a q^b$ is also an integer.
Here's my attempt:
Since $p$ and $q$ are distinct, either $p \neq 2$ or $q \neq 2$. Assume WLOG that $q \neq 2$.
Choose
$$a = (q+1)/2 - 1 $$
$$b=1.$$
Note that $a$ is a positive integer because $q$ is odd and $q \geq 3$.
It now follows that $p^a q^b = p^a q$ and has divisors
$$p^0, p^1, \dots, p^a,$$
$$p^0 q, p^1 q, \dots, p^a q.$$
There are $2(a+1)$ of them, so their arithmetic mean is
$$
\frac{1}{2(a+1)}\left[ \left( p^0 + p^1 + \dots + p^a \right) + \left( p^0 q + p^1 q + \dots + p^a q \right) \right]\\
= \frac{1}{2\left(\frac{q+1}{2}\right)}\left[ \left( p^0 + p^1 + \dots + p^a \right) + q \left( p^0 + p^1 + \dots + p^a \right) \right]\\
= \frac{1}{q+1}\left[ (q+1) \left( p^0 + p^1 + \dots + p^a \right) \right]\\
= p^0 + p^1 + \dots + p^a.
$$
This is an integer as desired.
| Your proof is fine,
but the magic values
of $a$ and $b$
bother me.
So,
I'm going to try
to find all solutions.
Spoiler:
I don't,
but do give some conditions.
The sum of the divisors
of $p^aq^b$ is
$\sigma(p^aq^b) =\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)}
$
and the number of divisors is
$d(p^aq^b) =(a+1)(b+1)$.
So the average is
$r(p^aq^b)
=\dfrac{\sigma(p^aq^b)}{d(p^aq^b)}
=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}
$.
In your case,
this is
$a=(q-1)/2, b=1$,
so
$a+1 = (q+1)/2$ and
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{(q+1)/2}-1)(q^{2}-1)}{(p-1)(q-1)((q+1)/2))2}\\
&=\dfrac{(p^{(q+1)/2}-1)(q^{2}-1)}{(p-1)(q-1)(q+1)}\\
&=\dfrac{(p^{(q+1)/2}-1)}{(p-1)}\\
\end{array}
$
and this is an integer.
If $b=1$,
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{a+1}-1)(q^{2}-1)}{(p-1)(q-1)(a+1)2}\\
&=\dfrac{(p^{a+1}-1)(q+1)}{2(a+1)(p-1)}\\
\end{array}
$
For this to be an integer
it is sufficient that
$2(a+1) | (q+1)$,
so
$2(a+1) = q+1$ works
if $q$ is odd.
In general,
if $q+1 = 2m(a+1)$,
$a
= \dfrac{q+1}{2m}-1
$.
If $q = 2^uv-1$
where $u \ge 1$ and $v$ is odd,
then
$a
= \dfrac{2^uv}{2m}-1
= \dfrac{2^{u-1}v}{m}-1
$.
In particular,
choosing $m=v$ gives
$a = 2^{u-1}-1$
and
$m = 1$ gives
OP's solution.
Another possibility is having
$a+1 = 2m$ so that
$\dfrac{(p^{a+1}-1)(q+1)}{2(a+1)(p-1)}
=\dfrac{(p^{2m}-1)(q+1)}{2(a+1)(p-1)}
=\dfrac{(p^{m}-1)(p^{m}+1)(q+1)}{4m(p-1)}
$
so that
$4m | (p^m+1)$
would work.
For $m=1$, this requires that
$p \equiv 3 \bmod 4$;
for $m=2$, this requires that
$p^2 \equiv 7 \bmod 8$
which can't happen since
$p^2 \equiv 1 \bmod 8$.
If $a+1=2m$ and $b+1 = 2n$ then
$\begin{array}\\
r(p^aq^b)
&=\dfrac{(p^{a+1}-1)(q^{b+1}-1)}{(p-1)(q-1)(a+1)(b+1)}\\
&=\dfrac{(p^{2m}-1)(q^{2n}-1)}{(p-1)(q-1)4mn}\\
&=\dfrac{(p^{m}-1)(q^{n}-1)(p^{m}+1)(q^{n}+1)}{(p-1)(q-1)4mn}\\
\end{array}
$
so it is enough if
$4mn | (p^{m}+1)(q^{n}+1)
$.
At this point,
I don't see any
easy solutions,
so I'll stop here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $
Why can we just add $ 2 - x ^ { 2 } $ in the numerator?
Step 1 to step 2, as well as step 3 to step 4 is clear to me.
| Because for $a > 0$ we know $\sqrt a = \frac a{\sqrt a}$ so $\sqrt{2 -x^2} = \frac {2-x^2}{\sqrt{2-x^2}}$ (assuming that $2 - x^2 > 0$ which must be the case as you have had $\sqrt{2 - x^2}$ in the denominator elsewhere in the equation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve $2a_{n-2} = a_n + a_{n-1}$ using generating function Need to solve:
$$2a_{n-2} = a_n + a_{n-1}$$
with: $a_0 = 0$ and $a_1=1$
I get:
$$f(x) = \frac{2x^3-x^2-x}{2x^2-x-1}$$ so I tried to scompose the denominator and I get:
$$f(x) = \frac{2x^3-x^2-x}{(x-1)(x+\frac{1}{2})}$$ now I think I have to use partial fraction but in this case I do not see any $x^3$ coming out so how should I procede?
$$\frac{A}{x-1}+\frac{B}{x+\frac{1}{2}}$$
| I think you committed a typo in calculations. From:
$$a_n=2a_{n-2}-a_{n-1}$$
we should have:
$$f(x)=\sum\limits_{n=0}\color{blue}{a_{n}}x^{n}=
x+\sum\limits_{n=2}a_nx^n=
x+\sum\limits_{n=2}\left(2a_{n-2}-a_{n-1}\right)x^n=\\
x+2x^2\left(\sum\limits_{n=2}a_{n-2}x^{n-2}\right)-x\left(\sum\limits_{n=2}a_{n-1}x^{n-1}\right)=\\
x+2x^2\left(\sum\limits_{n=0}a_{n}x^{n}\right)-x\left(\sum\limits_{n=1}a_{n}x^{n}\right)=x+2x^2f(x)-xf(x)$$
and
$$\color{red}{f(x)=\frac{x}{1+x-2x^2}}=\frac{1}{3(1-x)}-\frac{1}{3(1+2x)}=\\
\sum\limits_{n=0}\frac{1}{3}x^n-\sum\limits_{n=0}\frac{1}{3}(-2)^nx^n=
\sum\limits_{n=0}\color{blue}{\frac{1-(-2)^n}{3}}x^n$$
and (also agrees with wolfram)
$$a_n=\frac{1-(-2)^n}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show $\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx = 2\pi e^{2/3}$ Real Methods Taken from the post: The Integral that Stumped Feynman?
I want to know if the integral:
$$\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx = 2\pi e^{2/3}$$
can be evaluated using strictly real methods.
I've tried series of $e^x$ and $\cos x$ but to no avail. I tried differentiating under the integral, but nothing seemed to come out of it. Is there any wizardry that can conjure up this answer without complex analysis.
| Start with $\tan \left(\frac{x}{2} \right)$$=2\tan \left( \frac{t}{2} \right)$ $$I=2\int_0^{\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=8e^{5/8}\int_0^{\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)\frac{dt}{5-3\cos t} $$ Using $$ \sum_{n=1}^{\infty} a^{n} \cos(nx) = \frac12\left(\frac{1-a^{2}}{1-2 a \cos x + a^{2}}-1\right)$$ we can rewrite $$\frac{1}{5-3\cos x} =\frac14 +\frac12 \sum_{n=1}^\infty \frac{1}{3^n} \cos (nx) $$ thus we have $$I=2e^{5/8} \int_0^\pi \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) dt +4e^{5/8} \sum_{n=1}^\infty \frac{1}{3^n} \int_0^\pi \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) \cos(nt )dt $$
$$=2\pi e^{5/8}+4e^{5/8}\sum_{n=1}^\infty \frac{1}{3^n} I(n)$$ I dont know how to evaluate $I(n)$, but maybe someone can help.
$$I(0)=\pi,I(1)=\frac{\pi}{2^4}, I(2)=\frac{\pi}{2^8}, I(3)=\frac{\pi}{3\cdot 2^{11}}, I(4)=\frac{\pi}{3\cdot2^{16}},I(5)=\frac{\pi}{3\cdot 5 \cdot 2^{19}}$$$$ I(6)=\frac{\pi}{3^2\cdot5\cdot 2^{23}}, I(10)=\frac{\pi}{3^4\cdot5^2\cdot 7 \cdot 2^{39}}, I(20)=\frac{\pi}{3^8\cdot5^4 \cdot 7^2\cdot11\cdot 13\cdot 17\cdot 19\cdot 2^{79}}$$
Edit: As seen here: https://math.stackexchange.com/a/2913057/515527 $ I(n) =\frac{\pi} {2^{3n+1}n!}$, plugging this into the sum and using the series for $e^x$ will give the result immediately..
Another approach to evaluate $I(n)$ is to use that $$\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)=\sum_{n=0}^{\infty} \frac{\cos(nt)}{8^nn!}$$ Since $$\int_0^\pi \cos(nx) \cos(mx) dx=\begin{cases} \frac{\pi} {2} & n=m \\ 0 & n \neq m\end{cases}$$ We get that $I(n) =\frac{\pi} {2} \frac{1} {8^n n!} $ and the result follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Roll a die probability question
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this way:
Total $6P6 = 720$.
1,2 and 3 can be placed in $2$ locations $= 3^2$
4 and 5 can be placed in 1 location $= 2^1$
and 6 can be placed in two locations $= 2^2$
The probability is $= 72/720 = 0.1$
Is this correct?
| There are six possible outcomes for each of the five throws, so your denominator should be $6^5$.
For the favorable cases, we must have a sequence of five throws in which three of the outcomes are 1, 2, or 3, two of the outcomes are 4 or 5, and two of the outcomes are a six. There are $\binom{5}{2}$ ways to select which two positions in the sequence are occupied by a 1, 2, or 3 and $3$ ways to fill each of those positions. That leaves three positions to fill. There are $\binom{3}{1}$ ways to choose which of these positions will be occupied by a 4 or 5 and two ways to fill that position. The remaining two positions of the sequence must be filled with a 6. Therefore, the number of favorable cases is
$$\binom{5}{2}3^2 \cdot \binom{3}{1}2^1 \cdot \binom{2}{2}1^2 = \binom{5}{2}\binom{3}{1}\binom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2$$
Hence, the probability of obtaining a 1, 2, or 3 exactly twice and a 4 or 5 exactly once in five rolls of a fair die is
$$\frac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}$$
It is illustrative to compare this result with those provided by drhab and InterstellarProbe.
Observe that
$$\binom{5}{2}\binom{3}{1}\binom{2}{2} = \frac{5!}{2!3!} \cdot \frac{3!}{1!2!} \cdot \frac{2!}{2!0!} = \frac{5!}{2!1!2!} = \binom{5}{2, 1, 2}$$
where the multinomial coefficient
$$\binom{5}{2,1,2}$$
counts the number of ways of choosing a sequence of length five composed of two objects of one type (in this case, a roll that produces a 1, 2, or 3), one object of a second type (in this case, a roll that produces a 4 or 5), and two objects of a third type (in this case, a roll that produces a 6).
Therefore,
\begin{align*}
\frac{\dbinom{5}{2}\dbinom{3}{2}\dbinom{2}{2} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5} & = \frac{\dfrac{5!}{2!1!2!} \cdot 3^2 \cdot 2^1 \cdot 1^2}{6^5}\\
& = \frac{5!}{2!1!2!}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2\\
& = \binom{5}{2,1,2}\left(\frac{3}{6}\right)^2\left(\frac{2}{6}\right)^1\left(\frac{1}{6}\right)^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\pi/4} y(\sin y+\cos y)^n dy.
\end{equation*}
Although some terms may be canceled, the summation of the others seems terrible.
Another method is integration by parts. Here I found the recurrence formula for $\sin^n x$:
Integrating $\int \sin^n{x} \ dx$
So the first step is to have
\begin{equation*}
\left.\left(-\frac{1}{n}x\cos x\sin^{n-1} x+\frac{n-1}{n}x\int \sin^{n-2} xdx\right)\right|_0^{\pi/2}-\int_0^{\pi/2}\left(-\frac{1}{n}\cos x\sin^{n-1} x+\frac{n-1}{n}\int_0^x \sin^{n-2} t dt\right)dx
\end{equation*}
the evaluations at $0$ and $\pi/2$ and the integral of $\cos x\sin^{n-1}x$ seem OK, but what about the second term? Any better method for this problem?
| Assuming that $n$ is odd ($n=2m+1$), we may consider the triangle wave
$$ t(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\,\sin((2n+1)x)$$
and the fact that, by symmetry,
$$ \int_{0}^{\pi/2}x\sin^n(x)\,dx = \frac{1}{4}\int_{0}^{2\pi}t(x)\sin^n(x)\,dx. $$
On the other hand the Fourier sine series of $\sin^n(x)$ is straightforward to find:
$$ \sin^n(x) = \frac{1}{2^{n-1}}\sum_{k=0}^{m}\binom{n}{m-k} (-1)^k \sin((2k+1)x)$$
hence by the orthogonality relations:
$$\boxed{ \int_{0}^{\pi/2}x \sin^{2m+1}(x)\,dx = \frac{1}{2^{2m+1}}\sum_{k=0}^{m}\binom{2m+1}{m-k}\frac{1}{(2k+1)^2}}\tag{odd}$$
If $n$ is even we may similarly consider the Fourier cosine series of $\sin^n(x)$ and replace $x$ with the $\pi$-periodic continuation of $\frac{\pi}{2}-\left|\frac{\pi}{2}-x\right|$ over $[0,\pi]$, i.e. just a different triangle wave:
$$ T(x) = \frac{\pi}{4}-\frac{2}{\pi}\sum_{n\geq 0}\frac{\cos((4n+2)x)}{(2n+1)^2} = |t(x)|$$
leading to:
$$ \boxed{\int_{0}^{\pi/2}x \sin^{2m}(x)\,dx = \frac{\binom{2m}{m}\pi^2}{8\cdot 4^m}+\frac{1}{2^{2m}}\sum_{\substack{1\leq k\leq m\\k\text{ odd}}}\binom{2m}{m-k}\frac{1}{k^2}}\tag{even}$$
We may notice that the substitution $x\mapsto \arcsin x$, together with the Maclaurin series of $\arcsin^2$, provides a further hypergeometric representation for the given integral:
$$\begin{eqnarray*} \int_{0}^{\pi/2}x\sin^n(x)\,dx = \int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^2}}x^n\,dx &=&\sum_{m\geq 1}\frac{4^m}{2m(2m+n)\binom{2m}{m}}\\&=&\frac{1}{n+2}\cdot\phantom{}_3 F_2\left(1,1,1+\tfrac{n}{2};\tfrac{3}{2},2+\tfrac{n}{2};1\right).\end{eqnarray*}$$
The equality between the RHS of the last line and $(\text{odd})$ or $(\text{even})$ is probably a consequence of general transformation identities for hypergeometric $\phantom{}_3 F_2(\ldots;1)$ functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sum of series $\frac{ 4}{10}+\frac{4\cdot 7}{10\cdot 20}+\frac{4\cdot 7 \cdot 10}{10\cdot 20 \cdot 30}+\cdots \cdots$
Sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$
Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$
But i want to solve it using Beta ganmma function.
My Process follows
Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$
Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$
$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$
And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$
So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$
using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and
$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$
So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$
$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$
Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$
So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$
Could some Help me to solve it, Thanks
although it has solved Here
| For a complex number $z$ such that $|z|<1$, let
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\,z^n\,.$$
Then, the required sum $S:=\displaystyle\sum_{n=1}^\infty\,\prod_{k=1}^n\,\left(\frac{3k+1}{10k}\right)$ satisfies
$$S=f\left(\frac{3}{10}\right)-1\,.$$
We claim that $f(z)=(1-z)^{-\frac43}$ for all complex numbers $z$ such that $|z|<1$. As a consequence,
$$S=\left(\frac{7}{10}\right)^{-\frac43}-1=\left(\frac{10}{7}\right)^{\frac43}-1\approx0.608926\,.$$
Note that, for $n=0,1,2,3,\ldots$,
$$\begin{align}t_n&:=\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\\&=\frac{\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+1)}\\&=\frac{(n+1)\,\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+2)}\,,\end{align}$$
where $\Gamma$ is the usual gamma function.
That is,
$$t_n=\frac{n+1}{\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)}\,\text{B}\left(n+\frac43,\frac23\right)\,,$$
where $\text{B}$ is the usual beta function.
From the identity
$$\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi\,x)}\,,$$
we conclude that
$$\Gamma\left(\frac43\right)\,\Gamma\left(-\frac13\right)=\frac{\pi}{\sin\left(\frac{4\pi}{3}\right)}=-\frac{2\pi}{\sqrt{3}}\,.$$
Ergo,
$$\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)=\left(-\frac13\right)\,\left(-\frac{2\pi}{\sqrt{3}}\right)=\frac{2\pi}{3\sqrt{3}}\,.$$
In other words,
$$t_n=\frac{3\sqrt{3}}{2\pi}\,(n+1)\,\int_0^1\,u^{n+\frac13}\,(1-u)^{-\frac13}\,\text{d}u\,.$$
We now obtain
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\left(\sum_{n=0}^\infty\,(n+1)\,u^{n+\frac13}\,z^n\right)\,(1-u)^{-\frac13}\,\text{d}u\,.$$
This gives
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\frac{u^{\frac13}\,(1-u)^{-\frac13}}{(1-zu)^2}\,\text{d}u\,.$$
Let $v:=\dfrac{u}{1-u}$. Then,
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v\,.$$
The integral
$$J:=\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v$$
can be evaluated using the same keyhole contour as in this thread. It is not difficult to see that
$$\begin{align}J&=\left(\frac{2\pi\text{i}}{1-\exp\left(\frac{2\pi\text{i}}{3}\right)}\right)\,\text{Res}_{v=-\frac{1}{1-z}}\left(\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\right)
\\&=-\frac{\pi\,\exp\left(-\frac{\pi\text{i}}{3}\right)}{\sin\left(\frac{\pi}{3}\right)}\,\Biggl(-\frac{\exp\left(\frac{\pi\text{i}}{3}\right)}{3}\,(1-z)^{-\frac43}\Biggr)
\\&=\frac{2\pi}{3\sqrt{3}}\,(1-z)^{-\frac{4}{3}}\,.\end{align}$$
That is,
$$f(z)=(1-z)^{-\frac43}\text{ for all }z\in\mathbb{C}\text{ such that }|z|<1\,.$$
Well, the OP requested a long and combersome solution. The result can be easily proven via the binomial theorem:
$$(1-z)^{-\frac43}=\sum_{n=0}^\infty\,\binom{-\frac43}{n}\,(-1)^n\,z^n=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac13}{k}\right)\,z^n$$
for every complex number $z$ with $|z|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| $$7\times 2^k+1=n^2\iff7\times2^k=(n-1)(n+1)$$
$n$ must be odd so, $7\times 2^{k-2}=(\frac{n-1}{2})(\frac{n+1}{2})$ and $\gcd(\frac{n-1}{2},\frac{n+1}{2})=1$, since $\gcd(\frac{n-1}{2},\frac{n+1}{2})$ divides $\frac{n+1}{2}-\frac{n-1}{2}=1$.
This gives two possibilities:
*
*$\frac{n-1}{2}=7\Rightarrow n=225\Rightarrow 2^k=32$
*$\frac{n+1}{2}=7\Rightarrow n=169\Rightarrow 2^k=24$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the substitution of variables allowed in proofs? EDIT: I later realized no one saw that substituting $\sqrt{x} = x$ is not valid because they are not equal, you would have to assume $x$ as a completely different variable (i.e. $a$) and not the same one because they have no relation to each other. However, the solution to my proof problem is given.
So for example I have to prove:
$2y \leq \frac{y^2}{x^2} + x^2$ (for any real numbers $x$ and $y$ and $x$ $\neq$ $0$),
and I do some rough work at start with this and end up with the arithmetic-geometric mean inequality:
$\sqrt{xy} \le \frac{x+y}{2}$ (using substitution of variables).
Now to prove it, I start with the arithmetic-geometric mean inequality, then do the following steps:
$\sqrt{xy} \leq \frac{x+y}{2}$
$\sqrt{x}\sqrt{y} \leq \frac{x+y}{2}$
Now, substitute $\sqrt{x}$ with $x$ and so:
$xy \leq x^2+\frac{y^2}{2}$
Then throughout, I do one more substitution:
$\sqrt{x} = x$
to end up with the first (1) equation. Would this be allowed or is there another way of proving it? Such as squaring the right side of arithmetic-geometric mean inequality? I am not too sure. Thanks.
| $x^2 > 0$ we have
$2y \le \frac {y^2}{x^2} + x^2 \iff yx^2 \le \frac{x^2 + x^4}2$.
Now do the substition. Substitute $y = \sqrt a$ and $x^2 = \sqrt b$.
EXCEPT we don't know that $y$ is positive.
Let $y = \pm \sqrt a$ and $x^2 = \sqrt b$ so $y^2 = a \ge 0$ and $b = x^4 > 0$
And we know $\sqrt ab \le \frac {a+b}2$ for $a,b \ge 0$ so
$\sqrt {y^2} \sqrt {x^4} \le \frac {y^2 + x^4}2$ so
$|y|x^2 \le \frac {y^2 + x^4}2$ so
$2|y| \le \frac {y^2}{x^2} + x^2$ and finally
$2y \le 2|y| \le \frac {y^2}{x^2} + x^2$.
(remember $k \le |k|$ becuase if $k < 0$ then $k < 0 < |k|$. And if $k \ge 0$ then $k = |k|$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$ but cant figure how to apply those here since the centre is not at origin.
| given hyperbola is $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$ now differentiate the equation with respect to x to get the slope of tangent at that point $$\implies {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} = 1 \right)\implies \frac{2}{9}(x- 3)- \frac{1}{2}(y- 2){dy\over dx}= 0 \\ \implies{dy\over dx}= -\frac{5}{6} \ \ \ at\ (x_1,y_1)=(-2,{14\over3})$$ and the tangent line is $$y-y_1=m(x-x_1)\implies y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926446",
"timestamp": "2023-03-29T00:00:00",
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Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize denominator and then find factor of nominator in a way that these two cancel each other.
factorizing denominator gives:
$$ \lim_{x \rightarrow -2}\frac{3x^2+ax+a+3}{(x-1)(x+2)} $$
Now if it is possible to find solution to $3x^2+ax+a+3=0$ when $x=-2$
$$ 3(-2)^2+a(-2)+(-2)+3=0 $$
$$ 12-2a-2+3=0 $$
$$ 2a=13 \iff a = \frac{13}{2} $$
factorizing nominator gives:
$$ 3x^2+\frac{13x}{2}+\frac{19}{2}=0 $$
$$ x= \frac{-\frac{13}{2}\pm \sqrt{(\frac{13}{2})^2-4\cdot 3 \cdot (\frac{19}{2})}}{2\cdot 3} $$
Only problem is this equation is never zero with $x\in \mathbb{R}$ since there is negative value under square root. Now there is contradiction between $a=\frac{13}{2}$ and that equation $3x^2+ax+a+3=0$ when $a=\frac{13}{2}$ and $x=-2$
$$ 3(-2)^2+\frac{13}{2}\cdot (-2)+\frac{13}{2}+3 \neq 0 $$
There is clearly something wrong but i cannot see where this went wrong.
| Since the denominator at $x=-2$ is equal to $0$ it is necessary that also the numerator at $x=-2$ is equal to $0$ that is
$$3(-2)^2+a(-2)+a+3 =0 \implies a=15 $$
and therefore
$$\lim_{x \rightarrow -2}\frac{3x^2+15x+18}{x^2+x+2}=\lim_{x \rightarrow -2}\frac{3(x+2)(x+3))}{(x-1)(x+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The pdf of $\tan X$ when $X$ is Uniform Let $X$ be a random variable with pdf
$$ f(x) = \begin{cases} \frac{1}{\pi}, & \text{if } -\frac{\pi}{2} < x <\frac{\pi}{2} \\ 0, & \text{elsewhere} \end{cases} $$
Find the pdf of $Y = \tan(x)$.
My attempt:
$$F_Y(y) = P(Y\le y) = P(\tan(x)\le y) = P(x\le \tan^{-1}(y))=F_X(\tan^{-1}y) = \begin{cases} 0,&\text{if } x<-\frac{\pi}{2} \\ \frac{1}{\pi}{\tan^{-1}y}, & \text{if } -\frac{\pi}{2} \leq x <\frac{\pi}{2} \\ 1, & \text{if } x\geq \frac{\pi}{2} \end{cases}$$
$$ f_Y(y) = \begin{cases} \frac{1}{\pi} \frac{1}{1+x^2}, & \text{if } -\frac{\pi}{2} < x <\frac{\pi}{2} \\ 0, & \text{elsewhere} \end{cases} $$
Am I on the right path?
| \begin{align}F_Y(y) &= P(Y\le y) \\&= P(\tan(\color{red}X)\le y) \\&= P(X\le \tan^{-1}(y))\\&=F_X(\tan^{-1}y) \\&= \frac{1}{\pi}\left({\tan^{-1}y}+\frac{\pi}2 \right)\end{align}
$$ f_Y(y) = \frac{1}{\pi} \frac{1}{1+y^2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.
I know there is possible by derivate, but
$$f'(x)=n \left(\left(\cos ^2(x)+1\right) \sec ^2(x)\right)^{n-1} \left(2 \left(\cos ^2(x)+1\right)\tan (x) \sec ^2(x)-2 \tan (x)\right)+n \left(\left(\sin ^2(x)+1\right) \csc^2(x)\right)^{n-1} \left(2 \cot (x)-2 \left(\sin ^2(x)+1\right) \cot (x) \csc^2(x)\right).$$
I think this is not the best way.
| Let $A = \dfrac{1+\sin^2x}{\sin^2x}, B = \dfrac{1+\cos^2x}{\cos^2x}$. Then $A+B = 2+\dfrac{1^2}{\sin^2x}+\dfrac{1^2}{\cos^2x}\ge 2+\dfrac{(1+1)^2}{\sin^2x+\cos^2x}=6$. Thus: $A^n+B^n\ge \dfrac{(A+B)^n}{2^{n-1}}= \dfrac{6^n}{2^{n-1}}=2\cdot 3^n$. Equality occurs when $\sin x = \cos x$ or $x = \dfrac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Difficulty Finding $A^k$ Let $A=
\begin{bmatrix}
1& -1 & 1\\
0 & 1 & 1 \\
0 & 0 & 1\\
\end{bmatrix}$. Compute $A^k$.
My attempt
I'm trying to compute $A^k$ using this approach as follows:
$$
A=I+N=
\begin{bmatrix}
1& 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}+
\begin{bmatrix}
0& -1 & 1\\
0 & 0 & 1 \\
0 & 0 & 0\\
\end{bmatrix}
$$
with
$$
N^2=
\begin{bmatrix}
0& 0 & -1\\
0 & 0 & 0 \\
0 & 0 & 0\\
\end{bmatrix}, \, \text{and} \, \,
N^3=
\begin{bmatrix}
0& 0 & 0\\
0 & 0 & 0 \\
0 & 0 & 0\\
\end{bmatrix}
$$
Then,
$$
A^2=(I+N)^2=I+2N+N^2, \\
A^3=(I+N)^3=I+3N+3N^2, \\
A^4=(I+N)^4=I+4N+6N^2, \\
A^5=(I+N)^5=I+5N+10N^2, \\
A^6=(I+N)^5=I+6N+15N^2,
$$
By induction, we can see $A^k=(I+N)^k=I+kN+f[k]N^2$. But, I couldn't figure out what $f[k]$ is. Any help?
| You have written $A=I+N$, and you know that $N^3$ (and hence all higher powers) are zero. If $X$ and $Y$ are two matrices that commute with each other, then you can still use the bionomial theorem:
$$(X+Y)^n=\sum_{i+j=n}\binom{n}{i} X^i Y^j$$
Because $I$ commutes with $N$, and because all the higher powers of $N$ vanish, we can apply the formula to get
$$(I+N)^n=\sum_{i+j=n}\binom{n}{j} N^j=I+nN+\binom{n}{2}N^2$$
| {
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The minimum value of perimeter of the triangle Let there be $\triangle ABC$ having integral side lengths such that angle $$ \angle A=3 \angle B $$ Then find the minimum value of its perimeter.
I did this with trigonometry and got an equation as: $$ 2p = 4b[2\cos^3( \theta) + \cos^2(\theta) - \cos(\theta)] $$ where $b$ = side opposite to $\angle B$, $\theta$ is the $\angle B $ and $p$ is half of the perimeter. How should I proceed?
Also show any other appropriate method.
| My try:
$$\frac{a}{\sin3\beta}=\frac{b}{\sin\beta}$$
$$\sin3\beta=3\sin\beta-4\sin^3\beta=\frac{a}{b}\sin\beta$$
$$3-4\sin^2\beta=\frac{a}{b}$$
$$\sin^2\beta=\frac14(3-\frac ab)\tag{1}$$
$$\frac{b}{\sin\beta}=\frac{c}{\sin4\beta}$$
$$\frac{\sin4\beta}{\sin\beta}=\frac{2\sin2\beta \cos2\beta}{\sin\beta}=4\cos\beta (1-2\sin^2\beta)=\frac cb$$
$$16\cos^2\beta(1-2\sin^2\beta)^2=\frac{c^2}{b^2}$$
$$16(1-\sin^2\beta)(1-2\sin^2\beta)^2=\frac{c^2}{b^2}\tag{2}$$
Replace (1) into (2) and after some simplifications you get:
$$\frac{a^3}{b^3}-\frac{a^2}{b^2}-\frac{a}{b}+1=\frac{c^2}{b^2}$$
$$a^3-a^2 b-a b^2+b^3=c^2b$$
$$(a+b)(a-b)^2=c^2 b$$
However I was not able to proceed and solve this equation, except by guessing. Obviously 10, 8 and 3 fit the equation (I have started with $a=b=1$)
| {
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"source": "stackexchange",
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Discontinuity of $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ at $(x,y) = (0,0)$ Let $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ , when $(x,y) \neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x \rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks
| Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -\frac{x^2}{2}+x^n$ where $n >0$,then we have
\begin{align}\lim_{x \to 0}\frac{2(x^3+(-\frac{x^2}{2}+x^n)^3)}{2x^n} &= \lim_{x \to 0} \frac{(x^3+(-\frac{x^2}{2}+x^n)^3)}{x^n} \\
&= \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]\end{align}
If $n \in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, \begin{align} \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]=1\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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The tangent at $P$ to $y = x^2 - x^3$ meets the curve again at $Q$. Show that locus of midpoint of $PQ$ is $y=1-9x+28x^2-28x^3$
The tangent at a variable point $P$ of the curve $y = x^2 - x^3$ meets it again at $Q$. Show that the locus of the middle point of $PQ$ is $$y = 1 - 9x + 28x^2 - 28x^3$$
My approach
$$y^\prime=2x-3x^2$$
Equation of tangent: $$2\ x_1 -3\ x_1^2=\frac{y-\ x_1^2+\ x_2^3}{x-\ x_1}$$
From the equation of tangent:
$$(2\ x_1 -3\ x_1^2)(x-\ x_1)=y-\ x_1^2+\ x_2^3$$
Substituting the value of $y$ in the conic section we can find the value of $x$ but still it is getting complicated.
| Let $P(h,h^2-h^3),Q(k,k^2-k^3), h\ne k$
Now the gradient of $PQ$ is $$=\dfrac{k^2-k^3-(h^2-h^3)}{k-h}=k+h-(k^2+kh+h^2)$$
The gradient of the tangent at $P$ is $$=2h-3h^2$$
$\implies2h-3h^2=k+h-(k^2+kh+h^2)$
$\iff2h^2-(k+1)h+k-k^2=0$
$$\implies h=\dfrac{k+1\pm\sqrt{(k+1)^2+8(k^2-k)}}4=\dfrac{k+1\pm(3k-1)}4=k,\dfrac{1-k}2$$
As $h\ne k, h=\dfrac{1-k}2$
If $R(p,q)$ is the midpoint, $$2p=h+k=h+(1-2h)\iff h=?$$
$2q=h^2+k^2-(h^3+k^3)=h^2+(1-2h)^2-h^3-(1-2h)^3$
Replace the value of $h$ in terms of $p$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)$$
I have tried to factor out the $(1+\frac{1}{x})$ use the distributive property, and failed multiple times. Every online calculator I tried gives me different results.
| Eqn 1
$$(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$$
$$(-\frac{6}{x^2})+(-\frac{6}{x^3})+(\frac{6}{x^3})(1+\frac{1}{x^2}+\frac{2}{x})$$
$$(-\frac{6}{x^2})+ (-\frac{6}{x^3}) + \frac{6}{x^3}+\frac{6}{x^5}+\frac{12}{x^4}$$
$$\frac6{x^2}[\frac{1}{x^3}+\frac{2}{x^2}-1]$$
Eqn 2
$$\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$$
$$(\frac{6}{x^3}+\frac{6}{x^4})(1+\frac2{x})$$
$$\frac6{x^3}[1+\frac3{x} +\frac2{x^2}]$$
On comparision we know that both the equations are different and there is is no way to convert Eqn 1 to Eqn 2.
| {
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"timestamp": "2023-03-29T00:00:00",
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Condition for primeness of a degree 2 Solinas Prime Given $2^{2k+1}+2^{k+1}+1$ is prime, this necessarily imply $2k+1$ is a prime.
This result is assumed true in Primality tests for $2^p \pm 2^{(p+1)/2}+1$ using elliptic curves by YU TSUMURA in Proceedings of the American mathematical society vol. 139 num. 8 August 2011 Pages 2697-2703 and also considered easy to prove in that paper, but I didnt find anything in the web and the only fact I found that may help to prove this is $(2^{2k+1}+2^{k+1}+1)(2^{2k+1}-2^{k+1}+1)=2^{2(2k+1)}+1$.
| We begin with the following identity applied on $x=2^k$.
$$4x^4+1=(2x^2+2x+1)(2x^2-2x+1).$$
This gives that
$$\left(2^{2k+1}+2^{k+1}+1\right)\left(2^{2k+1}-2^{k+1}+1\right)=2^{4k+2}+1.$$
Assume for contradiction that $2k+1$ is not prime, but our first factor is prime. Let $p$ be a prime dividing $2k+1$. We have that, as $\frac{2k+1}{p}$ is odd,
$$x^p+1\big| x^{2k+1}+1,$$
which applied at $x=4$ gives
$$2^{2p}+1\big | 2^{4k+2}+1.$$
However, as $p$ is a proper divisor of $2k+1$, $2p\leq 2k+1$, so
$$2^{2p}+1\leq 2^{2k+1}+1 < 2^{2k+1}+2^{k+1}+1,$$
so $2^{2p}+1$ is not our first factor, and, as our first factor is prime, it cannot share any common factors with $2^{2p}+1$. This means that
$$2^{2p}+1 \big|2^{2k+1}-2^{k+1}+1.$$
As $2k+1=mp$ for some odd $m=2n+1$, we have that
$$2^{2k+1} = 2^{mp} \equiv 2^{2np+p}\equiv (-1)^n2^p\bmod \left(2^{2p}+1\right),$$
and similarly
$$2^{k+1} = 2^{\frac{(2n+1)p+1}{2}} = 2^{np+\left(\frac{p+1}{2}\right)}\equiv 2^{\frac{p+1}{2}}K\bmod \left(2^{2p}+1\right),$$
where $K$ is in the orbit of $2^p\bmod \left(2^{2p}+1\right)$; in particular, we may choose
$$K\in\left\{1,2^p,-1,-2^p\right\}.$$
Thus, we have
$$2^{2p}+1\bigg|(-1)^n 2^p - 2^{\frac{p+1}{2}}K+1.$$
As $p\geq 1$, both of the first two terms of the right side are clearly even, while $1$ is odd, which implies that the right side is odd and thus nonzero. As a result, its magnitude is $\geq 2^{2p}+1$, which implies that
$$2^{2p}+1\leq \left|(-1)^n 2^p - 2^{\frac{p+1}{2}}K+1\right|\leq 2^p\left(1+2^{\frac{p+1}{2}}\right)+1$$
$$2^p\leq 1+2^{\frac{p+1}{2}}.$$
$$x^2\leq 2+2x$$
for $x=2^{\frac{p+1}{2}}$. This is clearly false for $x\geq 3$, and as $p$ is odd, $2^{\frac{p+1}{2}}\geq 4$, and we have a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}}
\end{align}
| Continuing from yours:
$$I=\int\frac{1}{\sqrt{(x-a)(x-b)}}dx$$
now if we look at this:
$$(x-a)(x-b)=x^2-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^2-\frac{a^2+b^2}{2}$$
and so we can rewrite our integral as:
$$I=\int\frac{1}{\sqrt{(x-\frac{a+b}{2})^2-\frac{a^2+b^2}{2}}}dx$$
now let $u=x-\frac{a+b}{2}$ so $dx=du$
and the integral becomes:
$$I=\int\frac{1}{\sqrt{u^2-\frac{a^2+b^2}{2}}}du$$
for ease at the moment I am going to let $c^2=\frac{a^2+b^2}{2}$ making our integral:
$$I=\int\frac{1}{\sqrt{u^2-c^2}}du=\frac{1}{c}\int\frac{1}{\sqrt{\left(\frac{u}{c}\right)^2-1}}du$$
now if we let $u=c\cosh(v)$ we get $du=c\sinh(v)dv$ and our integral becomes:
$$I=\frac{1}{c}\int\frac{c\sinh(v)}{\sqrt{\cosh^2(v)-1}}dv=\int1dv$$
$$=v+C=\text{arcosh}\left(\frac{u}{c}\right)+C=\text{arcosh}\left(\frac{x-\frac{a+b}{2}}{c}\right)+C=\text{arcosh}\left(\frac{2x-(a+b)}{a^2+b^2}\right)+C$$
| {
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"source": "stackexchange",
"question_score": "3",
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Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
| A different appraoch:
Note that
\begin{align}
\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} &= \underbrace{\frac{n(n+1)(2n+1)}{6}}_{\displaystyle=\sum_{k=1}^n k^2 }-2\,\underbrace{\frac{n(n+1)}{2}}_{\displaystyle=\sum_{k=1}^{n}k}+n=\sum_{k=1}^n(k^2-2k+1)\in\Bbb Z
\end{align}
| {
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Find value of n such the function has local mimima at x=1. If $f$ is defined by
$$f(x)=(x^2-1)^n(x^2+x-1)$$
then $f$ has a local minimum at $x=1$, when
*
(i) $n=2$
(ii) $n=3$
(iii) $n=4$
(iv) $n=5$
Multiple options are correct.
The given answer is $n=2$ and $n=4$.
I tried putting derivative equal to zero and double derivative greater than zero, but both of them were independent of $n$.
| The second derivative test is inconclusive for $n > 2$.
As a fall back, you can use the first derivative test.
Let $n > 1$ be a positive integer.
Computing $f'(x)$, we get the factored form
$$f'(x)=(x^2-1)^{n-1}g(x)$$
where
$$g(x)= (2n+2)x^3+(2n+1)x^2-(2n+2)x-1$$
Noting that $g(1)=2n$, it follows that $g(x) > 0$ near $x=1$.
Since $n > 1$, we have $f'(1)=0$, so $x=1$ is a critical value of $f$.
Consider two cases . . .
Case $(1)$:$\;n\;$is odd.
Then $n-1$ is even, so $(x^2-1)^{n-1} > 0$ both to the left and to the right of $x=1$ near $x=1$.
Then since $g(x) > 0$ near $x=1$, it follows that $f'(x) > 0$ both to the left and to the right of $x=1$ near $x=1$, so by the first derivative test, the critical value $x=1$ does not correspond to a relative extremum.
Case $(2)$:$\;n\;$is even.
Then $n-1$ is odd, so $(x^2-1)^{n-1} < 0$ to the left of $x=1$ near $x=1$, and $(x^2-1)^{n-1} > 0$ to the right of $x=1$ near $x=1$.
Then since $g(x) > 0$ near $x=1$, it follows that $f'(x) < 0$ to the left of $x=1$ near $x=1$, and $f'(x) > 0$ to the right of $x=1$ near $x=1$, so by the first derivative test, the critical value $x=1$ corresponds to a relative minimum.
Therefore, for the stated question, the answers are $n=2$ and $n=4$.
As an alternative, much simpler solution, one can argue as follows . . .
Let $n$ be a positive integer.
First note that $f(1)=0$.
At $x=1$, we have $x^2+x-1 > 0$, hence, near $x=1$, the sign of $f(x)$ is the same as the sign of $(x^2-1)^n$.
If $n$ is even, $(x^2-1)^n > 0$, both to the left and to the right of $x=1$ near $x=1$, hence the same is true for $f(x)$. It follows that $f$ has a relative minimum at $x=1$.
If $n$ is odd, $(x^2-1)^n < 0$, to the left of $x=1$ near $x=1$, hence the same is true for $f(x)$. It follows that $f$ does not have a relative minimum at $x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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why using $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to factorize a polynomial degree 2 dose not always work i tried to factorize $5x^3-11x^2+2x$ so i took out $x$ and used $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the roots 2 and $\frac{1}{5}$ but to my surprise multiplying the roots like so $x(x-2)\cdot(x-\frac{1}{5})$ produces a fifth of the original polynomial what did i do wrong?
| It’s related to the derivation of the quadratic formula, which is done by completing the square. So, we know the formula.
$$ax^2+bx+c = 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
In reality, you aren’t solving for the original equation, which is $ax^2+bx+c = 0$. You’re solving for $x^2+\frac{b}{a}x+\frac{c}{a} = 0$, which gives the same solutions. You might want to check the proof for the quadratic formula (the first step is dividing both sides by $a$).
For example, looking at the following quadratic polynomials, it’s easy to figure out that their solutions are the same.
$0 = (x-2)(x-3) = x^2-5x+6$
$0 = 2(x-2)(x-3) = 2x^2-10x+12$
$0 = 3(x-2)(x-3) = 3x^2-15x+18$
$0 = a(x-2)(x-3) = ax^2-5ax+6a; a\neq 0$
For all the quadratic polynomials above, the solutions are $x = 2$ and $x = 3$, but when you use the quadratic function, you end up solving for the FIRST one, but it doesn’t matter, since they share the same solutions. The same applies to any $n^{th}$ degree polynomial, as long as $a$ is a non-zero value.
Another way of thinking about it is that in a polynomial equation $y = a(x-x_a)\cdot(x-x_b)\cdot(x-x_c)\cdot ... \cdot (x-x_n)$, $a$ affects stretch, not the $x$-intercepts, so the solutions remain the same for all values of $a$.
So, for your example, you found the solutions $0$, $\frac{1}{5}$, and $2$. You also know that $a = 5$, so rewrite it using $5$ at the beginning.
$$y = 5(x)(x-\frac{1}{5})(x-2)$$
Now, it’s the correct function!
| {
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Where does this derivation of the Fourier Series for csc(x) go wrong? In this post, the following derivation for the Fourier series of csc(x) is given:
\begin{align}
\csc x
&= \dfrac{1}{\sin x}\\
&= \dfrac{2i}{e^{ix}-e^{-ix}}\\
&= \dfrac{2ie^{-ix}}{1-e^{-2ix}}\\
&= 2ie^{-ix}\sum_{n\geq0}e^{-2inx}\\
&= 2i\sum_{n\geq0}e^{-(2n+1)ix}\\
&= 2\sum_{n\geq0}(i\cos(2n+1)x+\sin(2n+1)x)
\end{align}
now take the real part of both sides.
This seems to give the right answer, but has the very confusing property that somehow we end up, midway through the derivation, with a complex function on the right side, which we need to correct at the end by taking the real part. Since each step is supposed to be equal to the last, where does the complex part come from?
One obvious problem is in using the Taylor series at step 4, since $e^{-2ix}$ is right on the boundary of the radius of convergence of $\frac{1}{1-x}$. But how does this then yield the right answer so long as you just take the real part?
| The imaginary part actually seems to vanish — except at $x=0$ where it's infinite.
The Fourier cosine series of $\delta(x-\epsilon)$ is given by
$\frac{1}{\pi} + \sum_{n=1}^{\infty} \frac{2}{\pi} \cos n\epsilon \cos nx.$
Letting $\epsilon \to 0$ gives
$\delta(x) = \frac{1}{\pi} + \sum_{n=1}^{\infty} \frac{2}{\pi} \cos nx.$
Thus,
$$
\cos x + \cos 2x + \cos 3x + \cdots
= \sum_{n=1}^{\infty} \cos nx
= \frac{\pi}{2} \delta(x) - \frac12.
$$
We get the terms for even $n$ by replacing $x$ with $2x$:
$$
\cos 2x + \cos 4x + \cos 6x + \cdots
= \sum_{n=1}^{\infty} \cos 2nx
= \frac{\pi}{2} \delta(2x) - \frac12
= \frac{\pi}{4} \delta(x) - \frac12,
$$
where in the last step $\delta(ax) = \frac{1}{|a|} \delta(x)$ has been used.
The sum of terms with odd $n$ is therefore
$$
\sum_{n=0}^{\infty} \cos (2n+1)x
= \cos x + \cos 3x + \cos 5x + \cdots \\
= (\cos x + \cos 2x + \cos 3x + \cdots) - (\cos 2x + \cos 4x + \cos 6x + \cdots) \\
= \left( \frac{\pi}{2} \delta(x) - \frac12 \right) - \left( \frac{\pi}{4} \delta(x) - \frac12 \right)
= \frac{\pi}{4} \delta(x)
$$
Thus, for $x \neq 0$ we have $\sum_{n=0}^{\infty} \cos (2n+1)x = 0.$
| {
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Integrate $\int\frac{x^3+1}{x^4+x^3+x^2+x}\,dx$ How do I integrate $\displaystyle\int\dfrac{x^3+1}{x^4+x^3+x^2+x}\,\mathrm dx$?
I tried by splitting the equation in two parts like $\dfrac{x^3}{x^4+x^3+x^2+x}$ and $\dfrac1{x^4+x^3+x^2+x}$ and then cancelling out the $x$ terms from the first part and then trying to integrate further.
But this did nothing much to it, and also the other part became more difficult to solve.
How do I solve this integration?
| Hint:
$$\dfrac{x^3+1}{x^4+x^3+x^2+x}=\dfrac{1}{x}-\dfrac{1}{x^2+1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$.
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It is now necessary to prove that $x$-coordinates of extrema of $f(x)$ fall between $a, b$ and $c$. Or otherwise stated,
$a<x_1<b<x_2<c$, where $x_1$ and $x_2$ are $x$-coordinates of extrema.
The problem is, however, in the fact that this should be proven without the usage of Rolle's Theorem. I can prove this fact for quadratic, I get $x=(a+b)/2$, which suggests that extremum is halfway between the roots. With cubic, however, it is much more difficult. Any ideas?
| Translating and scaling, you can assume that $a=0$ and $b=1$. Then
$$f(x)=x^3-(c+1)x^2+cx$$
$$f'(x)=3x^2-2(c+1)x+c$$
The roots of $f'$ are
$$\frac{2c+2\pm\sqrt{4c^2-4c+4}}6=\frac{c+1\pm\sqrt{c^2-c+1}}3$$
Note that the discriminant is increasing for $c>1/2$. Then, since $c>b=1$, we have
$$\frac{c+1+\sqrt{c^2-c+1}}3>\frac{1+1+1}3=1$$
and
$$\frac{c+1-\sqrt{c^2-c+1}}3=\frac{c+1-\sqrt{(c-\frac12)^2+\frac34}}3<\frac{c+1-(c-\frac12)}3=\frac12<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951681",
"timestamp": "2023-03-29T00:00:00",
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3 digit numbers whose one digit is the average of the other two
Question: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
How can this be solved with an arithmetic sequence?
I can see that any two digits must be same parity to produce the even sum. Also any two of the digits must be divisible by $2$. if $\overline{abc}$ is the digit, $a+b=2c$ and doing that for all three digits gives three equations. Thats as far as I can get.
| Can be done by simple trial and error
(Note that plus sign is only used to distinguish the triplets of digits that can be used to form the desired number)
$0+2,4,6,8+1,2,3,4$
$1+3,5,7,9+2,3,4,5$
$2+4,6,8+3,4,5$
$3+5,7,9+4,5,6$
$4+6,8+5,6$
$5+7,9+6,7$
$6+8+7$
$7+9+8$
In total we get 20 possible triplets of digits that can be permuted in $3!=6$ different ways over 3 possible places. This gives us $20×6=120$
However, the numbers formed from the 4 triplets with $0$, ie, $0+2,4,6,8+1,2,3,4$, will have $2/6$ permutations in which the unit digit is the one $=0$. But this can't be since the numbers is a 3 digit number.
Hence we will have to subtract $4×2=8$ from $120$ to give our final answer of $112$ which matches the data provided in the other answer
| {
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$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ if and only if $a,b,c$ are the lengths of a right angled triangle Here is a problem from the book: Everything connected to Pithagoras
It is well known that in every right angled triangle $ABC$, $a^2+b^2=c^2$. Howevee, there are some more complicated equations as well. Here is one:
Prove that $$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$$ if and only if $a,b,c$ are the lengths of a right angled triangle.
I can’t prove it. I tried to write $a_4=x$ and so on, but only algebra didn’t help for me. Aldo it is interesting because in this problem it doesn’t matter which side is the hypotenuse...
Please help!
| Hint:
$$
2(a^4b^4+b^4c^4+c^4a^4)-(a^8+b^8+c^8)\\
=(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).
$$
Edit: I arrived at the above by guessing. My intuition is that: the given identity $2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ is symmetric in its info about $a$, $b$, and $c$, therefore, the conclusion has to be symmetric about $\angle A$, $\angle B$, and $\angle C$. So a product
$$
(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)
$$
came to mind. But it is 2 degrees lower than the degree of the given identity. So I needed a 2nd degree factor. Again, that factor has to be symmetric in $a$, $b$, and $c$. So $a^2+b^2+c^2$ was the immediate candidate (I would have tried $(a+b+c)^2$ next if $(a^2+b^2+c^2)$ hadn't worked.). Then, it remained for me to expand and simplify
$$
(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)
$$
using Mathematica.
| {
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How can I simplify the derivative further to match the correct answer? I've been stuck on trying to simplify my expression in order to match the correct answer but I can't seem to get the correct solution. Guidance towards the proper steps would be greatly appreciated!
The original question:
Differentiate and simplify completely
$$g(x)=\sqrt[3]{\frac{x^3-1}{x^3+1}}$$
My solution:
$$g^\prime(x) = \frac{2x^2}{\sqrt[3]{\left(x^3-1\right)^2}\sqrt[3]{\left(x^3+1\right)^4}}$$
I need help simplifying this to become this solution:
$$g^\prime(x) = \frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{(x+1)^2\left(x^2-x+1\right)^2(x-1)\left(x^2+x+1\right)} $$
Original image of solution.
| To remove some visual clutter, I'll define $a:=x^3-1$ and $b:=x^3+1$. So, your solution looks like:
$$\frac{2x^2}{\sqrt[3]{a^2}\;\sqrt[3]{b^4}} \tag{1}$$
Note that $\sqrt[3]{b^4}=\sqrt[3]{b^3 b} = b\sqrt[3]{b}$, so that $(1)$ becomes
$$\frac{2x^2}{b\;\sqrt[3]{a^2}\;\sqrt[3]{b}}\tag{2}$$
From here, we want to rationalize the denominator. We'll multiply that denominator by things that eliminate the cube roots, but then multiply the numerator by the same things for balance. Since
$$\sqrt[3]{w^2}\cdot\sqrt[3]{w}= \sqrt[3]{w^3} = w \tag{3}$$
multiplying by $\sqrt[3]{a}$ and $\sqrt[3]{b^2}$ will do what we want; thus,
$$\frac{2x^2}{b\;\sqrt[3]{a^2}\;\sqrt[3]{b}}\cdot\frac{\sqrt[3]{a}}{\sqrt[3]{a}}\cdot\frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{2x^2\;\sqrt[3]{a}\;\sqrt[3]{b^2}}{b\cdot ab} = \frac{2x^2\;\sqrt[3]{a}\;\sqrt[3]{b^2}}{ab^2}\tag{4}$$
That's essentially it, except for replacing $a$ and $b$ with their $x$-forms.
However, the target solution factors those $x$-forms (although, curiously, only in the denominator; go figure), using identities that you may want to commit to memory:
$$x^3-1=(x-1)(x^2+x+1) \qquad\qquad x^3+1=(x+1)(x^2-x+1) \tag{5}$$
This gives the target form, which I won't bother to type-out again. $\square$
| {
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Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake? Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:
Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.
My workings are:
Let $y=x-4$. Then when $x \to 4$, we have that $y \to 0$. Thus:
$$\lim_{y \to 0} \frac{(y+4)^4 - 4^{y+4}}{y} = \\ = \lim_{y \to 0}\frac{(y+4)^4}{y} - \lim_{y \to 0} \frac{4^{(y+4)}}{y} $$
And this step is not allowed from the get go, as I am deducting infinities, which is indeterminate. What I should have done though:
$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\
=4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$
| You can break up this limit under certain circumstances.
$\lim_\limits{x\to a} (f(x) + g(x)) = \lim_\limits{x\to a} f(x) + \lim_\limits{x\to a}g(x)$
You can do it if $\lim_\limits{x\to a} f(x), \lim_\limits{x\to a}g(x)$ both exist and are finite.
But if $\lim_\limits{x\to a} f(x) = \infty$ and $\lim_\limits{x\to a}g(x) = -\infty,$ then you have just given yourself and indeterminate form.
You could do this:
$\lim_\limits{y\to 0} \frac {(y+4)^4 - 4^{y+4}}{y} = \lim_\limits{y\to 0} \frac {y^4}{y}+ \lim_\limits{y\to 0} \frac {4y^3}{y}+\lim_\limits{y\to 0} \frac {6y^2}{y} +\lim_\limits{y\to 0} \frac {4y}{y} + \lim_\limits{y\to 0} \frac {4^4(1 - 4^y)}{y}$
Because each of those limits are defined and finite.
You could also use L'Hopital's rule.
| {
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Constructing the graph from parametric equations
$$x(\theta)=|\cos4\theta|\cos\theta$$
$$y(\theta)=|\cos 4\theta|\sin\theta$$
$$0\le \theta\le2\pi$$
I have to graph the equation.
Now, I have no idea as the parameter $\theta$ cannot be eliminated. At most I found that $x^2+y^2=\cos^24\theta$. Now, suppose I let $\theta=0$.So we get $x^2+y^2=1$. So this is a circle But of course the equation does not represent the whole circle. What to do?
| $\theta$ can be eliminated with $\tan\theta=\dfrac yx$, then
$$\cos\theta=\frac x{\sqrt{x^2+y^2}},
\\\sin\theta=\frac y{\sqrt{x^2+y^2}},
\\\cos4\theta=8\frac{x^4}{(x^2+y^2)^2}-8\frac{x^2}{x^2+y^2}+1.$$
Then
$$x^2+y^2=\left(8\frac{x^4}{(x^2+y^2)^2}-8\frac{x^2}{x^2+y^2}+1\right)^2,$$
or
$$(x^2+y^2)^5=(x^2+y^2)^2-8x^2y^2.$$
Admittedly, this is not helpful, the polar representation $\rho=|\cos4\theta|$ is much easier.
| {
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Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
Moreover,
$\theta=\tan^{-1}(1)=\frac{\pi}{4}$
Then, the polar form of $z$ is
$z=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Let $w\in\mathbb{C}$ such that $w^2=z$
Then
$w_k=\sqrt{\sqrt{2}}(\cos(\frac{\frac{\pi}{4}+2k\pi}{2})+i\sin(\frac{\frac{\pi}{4}+2k\pi}{2})$ for $k=0,1$.
Then the square roots, are:
$w_0=\sqrt[4]2(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8})=1,84+0,76i.$
$w_1=\sqrt[4]2(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8})=-1,84-0,76i.$
But, here i'm stuck because the $w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}\not = 1,84+0,76i$
| Well, you could cheat and just do
$$
w_0^2=\frac{1+\sqrt{2}}{2}-\frac{-1+\sqrt{2}}{2}+2i\sqrt{\frac{1+\sqrt{2}}{2}\cdot\frac{-1+\sqrt{2}}{2}}=1+i
$$
and note that $w_1=-w_0$.
Without cheating, you did quite well: $|1+i|=\sqrt{2}$, so
$$
1+i=\sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)=
\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)
$$
Hence we have
$$
w_0=\sqrt[4]{2}\left(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\right)
$$
and you can also compute $w_1$ (but it must be $w_1=-w_0$).
It remains to compute $\cos(\pi/8)$ and $\sin(\pi/8)$:
$$
\cos\frac{\pi}{8}=\sqrt{\frac{1+\cos(\pi/4)}{2}}=
\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}
$$
and
$$
\sqrt[4]{2}\sqrt{\frac{\sqrt{2}+1}{2\sqrt{2}}}=
\sqrt{\sqrt{2}\frac{\sqrt{2}+1}{2\sqrt{2}}}=
\sqrt{\frac{1+\sqrt{2}}{2}}
$$
You can do similarly, with
$$
\sin\frac{\pi}{8}=\sqrt{\frac{1-\cos(\pi/4)}{2}}
$$
| {
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Compute the following limit if they exist: $\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$ Compute the following limit if they exist:
$\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$
Attempt:
Given $\epsilon > 0$. Choose $\delta = \text{Not sure yet}$. Then $||(x, y, z) - (0,0,0)|| = \sqrt{x^2 + y^2 + z^2}$, and so $||(x, y, z) - (0,0,0)|| < \delta$
$\left| \frac{2x^2y c oz(z)}{x^2 + y^2} \right| \leq \frac{2x^2|y|}{x^2+y^2} \leq \frac{2 (x^2 + y^2)(\sqrt{x^2+y^2})}{x^2 + y^2} = 2 \sqrt{x^2 + y^2}$
Okay, I am stuck here. As you can see I need $\sqrt{x^2 + y^2 + z^2}$ in order for $\delta = \frac{\epsilon}{2}$.
Also what does it compute to?
| Well, we have $$2\sqrt{x^2+y^2} \le 2\sqrt{x^2+y^2+z^2} $$
and you can pick $\delta = \frac{\epsilon}2$.
| {
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Is the matrix $A = I - \frac{J}{n+1}$ idempotent? I am supposed to figure out if the following statement is false or true.
If $I$ is the $n \times n$ identity matrix, and $J$ is an $n \times n$ matrix consisting entirely of ones, then the matrix $$A = I - \frac{J}{n+1}$$ is idempotent (i.e., $A^{2} = A$).
I understand obviously what $I$ and $J$ are, my issue is with the $A = I - \frac{J}{n+1}$. I searched my textbook and found no reference to it. What does it mean?
| The expression $A = I - \frac{J}{n+1}$ is matrix addition, where $\frac{J}{n+1}$ represents scalar multiplication of the matrix $J$ by $\frac{1}{n+1}$. Since the matrix $J$ consists completely of $1$s, each entry of the scaled matrix $\frac{J}{n+1}$ is $\frac{1}{n+1}$. Here is the matrix addition:
$$ I - \frac{J}{n+1} =
\left[\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{array}\right]
-
\left[\begin{array}{cccc}
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{1}{n+1} & \frac{1}{n+1} & \cdots & \frac{1}{n+1} \\
\end{array}\right]$$
$$ \implies A = \left[\begin{array}{cccc}
\frac{n}{n+1} & \frac{-1}{n+1} & \cdots & \frac{-1}{n+1} \\
\frac{-1}{n+1} & \frac{n}{n+1} & \cdots & \frac{-1}{n+1} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{-1}{n+1} & \frac{-1}{n+1} & \cdots & \frac{n}{n+1} \\
\end{array}\right] $$
An interesting matrix. You could describe it by $A_{ij}=\frac{n}{n+1}$ if $i=j$ and $A_{ij}=\frac{-1}{n+1}$ if $i\neq j$.
| {
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How can I solve this inequality using convexity? Given $a, b, c\ge 0$ and $x, y, z> 0$ and $a + b + c = x + y + z$.
Show that $$a ^ 3 / x ^ 2 + b ^ 3 / y ^ 2 + c ^ 3 / z ^ 2 \ge a + b + c$$ prove inequality using convexity
| By Holder
$$\left(\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\right)(x+y+z)^2\geq(a+b+c)^3$$ and we are done!
Also, since $f(x)=x^3$ is a convex function on $[0,+\infty),$ by Jensen we obtain:
$$\sum_{cyc}\frac{a^3}{x^2}=(x+y+z)\sum_{cyc}\frac{x}{x+y+z}\left(\frac{a}{x}\right)^3\geq$$
$$\geq(x+y+z)\left(\sum_{cyc}\frac{x}{x+y+z}\frac{a}{x}\right)^3=a+b+c.$$
| {
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
| $$\sqrt[3]{n^3+1}-\sqrt{n^2-1}=\sqrt[3]{n^3+1}-n-\sqrt{n^2-1}+n
\\=\frac1{(n^3+1)^{2/3}+(n^3+1)^{1/3}n+n^2}+\frac1{\sqrt{n^2-1}+n}$$
and both terms are decreasing. The supremum is with $n=1$,
$$x_1=\sqrt[3]2.$$
| {
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Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ when $n \rightarrow \infty$? Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ ? How to solve such type of questions considering $n \rightarrow \infty$?
| Let $n = m^2$.
$a = \ln (\sqrt n)! = \ln 1 + \ln 2 + \cdots + \ln m$
$b = \ln \sqrt {n!} = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m + \ln (m+1) + \ln (m+2) + \cdots + \ln m^2\}$
$a - b = \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m\}
- \frac{1}{2} \{\ln (m+1) + \cdots + \ln (m+m) + \cdots + \ln m^2\}$
$a - b < \frac{1}{2}\{\ln 1 + \ln 2 + \cdots + \ln m\}
- \frac{1}{2} \{\ln 1 + \ln 2 + \cdots + \ln m\} = 0$
Hence $\sqrt n! < \sqrt {n!}$.
| {
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
| Denote $p(n):=(1+\frac{1}{1^3})(1+\frac{1}{2^3})(1+\frac{1}{3^3})...(1+\frac{1}{n^3})$.
Claim:
$$p(n)\leq3-\frac2{n^2},\,\forall n\geq2.$$
For $n=2$, we have $\frac94\leq3-\frac12$.
Then suppose $p(n)\leq3-\frac2{n^2}$ for some $n$. We see
$$\eqalign{
p(n+1)&=p(n)(1+\frac1{(n+1)^3})\cr
&\leq3-\frac{2}{n^2}+\frac3{(n+1)^3}-\frac{2}{n^2(n+1)^3}\cr
&=3+\frac{3n^2-2(n^3+3n^2+3n+1)-2}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+3n^2+6n+4}{n^2(n+1)^3}\cr
&=3-\frac{2n^3+2n^2+(n^2+6n+4)}{n^2(n+1)^3}\cr
&\leq3-\frac{2n^2(n+1)}{n^2(n+1)^3}\cr
&\leq3-\frac2{(n+1)^2}}.$$
As pointed out by @saulspatz, one can prove that $p(n)\leq3-\frac1n,\forall n\geq1$ by the same method.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
} |
is there are any short cut method to find the determinant of A? Find the determinant of A
$$A=\left(\begin{matrix}
x^1 & x^2 & x^3 \\
x^8 & x^9 & x^4 \\
x^7 & x^6 & x^5 \\
\end{matrix}\right)$$
My attempts : By doing a laplace expansion along the first column i can calculate,but it is a long process My question is that is there are any short cut method to find the determinant of A?
Thanks u
| Use Gaussian elimination, while keeping track of changes to the determinant.
$$ \begin{aligned}
\left\lvert \begin{matrix} x^1 & x^2 & x^3 \\ x^8 & x^9 & x^4 \\ x^7 & x^6 & x^5 \end{matrix} \right\rvert
&= x^6 \left\lvert \begin{matrix} 1 & 1 & 1 \\ x^7 & x^7 & x^1 \\ x^6 & x^4 & x^2 \end{matrix} \right\rvert \\
&= x^6\left\lvert \begin{matrix} 1 & 0 & 0 \\ x^7 & 0 & x^1 - x^7 \\ x^6 & x^4 - x^6 & x^2 - x^6 \end{matrix} \right\rvert \\
&= x^6\left\lvert \begin{matrix}0 & x^1 - x^7 \\ x^4 - x^6 & x^2 - x^6 \end{matrix} \right\rvert \\
&= -x^{11}(1 - x^6)(1 - x^2)
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of an integer $a$ such that $ a^2 +6a +1 $ is a perfect square. I was able to solve this but it required me using hit and trial at one step. I was wondering if i could find a more solid method to solve it.
p.s. this is the first time im asking a question here so sorry if i couldn't construct the question properly.
this is my solution -
$a^2 + 6a + 1 = k^2$
$(a+3)^2 = k^2 + 8$
$k^2 + 8 = m^2 $ where [$m=a+3$]
$(m-k)(m+k) = 8$ ... here by inspection $m= 3$ and $k =1$
hence, $a = 0, -6$.
| $$
\\(a+1)^2=a^2+2\cdot a+1\leq a^2+6\cdot a + 1 <a^2+6\cdot a+9=(a+3)^2
\\(a+1)^2\leq a^2+6\cdot a+1\leq(a+2)^2
\\=>
$$
if $\>a^2+6\cdot a + 1\>$ is a perfect square if and only if $\>a^2+6\cdot a + 1=(a+1)^2\>$ or $\>a^2+6\cdot a + 1=(a+2)^2\>=>\>a=0\>$ or $\>a=-6$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the matrix A in Ax=b I am watching the linear algebra course of MIT presented by Gilbert Strang and in Lecture 13 he solves the equation $Ax = b$. This is not the usual "find $x$", instead $x$ and $b$ are given and we need to find the matrix $A$ The link is at https://www.youtube.com/watch?v=l88D4r74gtM&t=841s at about minute 14:00. We know:
$Ax = b$
$b=\begin{bmatrix}2\\4\\2\end{bmatrix}$
$x=\begin{bmatrix}2\\0\\0\end{bmatrix}$ + $c\begin{bmatrix}1\\1\\0\end{bmatrix}$ + $d\begin{bmatrix}0\\0\\1\end{bmatrix}$
$\dim N(A)=2$
$A$ is a $3\times 3$ matrix.
But you can see all of that in the video, it looks much clearer on a blackboard.
I am having problems understanding what did professor Gilbert do to get the answer $A=\begin{bmatrix}1&&-1&&0\\2&&-2&&0\\1&&-1&&0\end{bmatrix}$
I just don't seem to understand what happened, at all. How did he come up with the first column? Then how about the other 2? How did he use the nullspace? Why did it work? How did he use $x$ and $b$ ?
| It's again a linear system, with unknowns living in a vector space, precisely the $3\times 1$ column vectors. Write $A=[a_1\ a_2\ a_3]$; then you know that
$$
\begin{cases}
2a_1 = b & (c=0,\;d=0) \\[4px]
3a_1+a_2=b & (c=1,\;d=0) \\[4px]
2a_1+a_3=b & (c=0,\;d=1)
\end{cases}
$$
This immediately entails that $a_3=0$, $a_1=\frac{1}{2}b$ and
$$
a_2=b-3a_1=-\frac{1}{2}b
$$
You could even do (formal) Gaussian elimination!
\begin{align}
\left[\begin{array}{ccc|c}
2 & 0 & 0 & b \\
3 & 1 & 0 & b \\
2 & 0 & 1 & b
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 0 & b/2 \\
3 & 1 & 0 & b \\
2 & 0 & 1 & b
\end{array}\right]
&& R_1\gets \tfrac{1}{2}R_1
\\[4px]&\to
\left[\begin{array}{ccc|c}
1 & 0 & 0 & b/2 \\
0 & 1 & 0 & -b/2 \\
0 & 0 & 1 & 0
\end{array}\right]
&& \begin{aligned} R_2&\gets R_2-3R_1\\R_3&\gets R_3-2R_1\end{aligned}
\end{align}
| {
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"answer_count": 2,
"answer_id": 0
} |
Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I inserted some $n$'s to get an idea on how the sequence is going to look like.
I got:
\begin{align}
a_0&=3\\
a_1&=1\\
a_2&=\frac{7}{9}\approx 0.\overline{7}\\
a_3&=\frac{3}{4}=0.75
\end{align}
Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$
Therefore, I show that
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I am having problems when trying to prove the inequalities above:
\begin{align}
& a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\
& = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\
& = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\
& = \cdots \text{ not sure what steps I could do now}
\end{align}
Boundedness:
The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$:
\begin{align}
a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\
&=3
\end{align}
The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally.
Convergence
Assumption (s.a) $\lim_{ n \to \infty} a_n =1$
Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:
\begin{align}
\mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\
&= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\
&= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\
&= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\
&= \left|\frac{2-2n}{(n+1)^2}\right|\\
&= \cdots \text{(how to go on?)}
\end{align}
| For monotonic behavior:
$a_{n+1}-a_n=\frac{(n+1)^2+3}{(n+2)^2}-\frac{n^2+3}{(n+1)^2}=\frac{(n+1)^4+3(n+1)^2-n^2(n+2)^2-3(n+2)^2}{(n^2+3n+2)^2}=\frac{2(n^2-n-4)}{(n^2+3n+2)^2}.$
Observe that for $n \geq 3$, the numerator is always positive, so for all $n \geq 3$, the sequence will be an increasing sequence (so ultimately an increasing sequence). The non-monotonic behavior only occurs in the first few terms $a_0 > a_1 > a_2>a_3<a_4<a_5<\dotsb<a_n<a_{n+1}<\dotsb$
For boundedness
\begin{align*}
a_n & = \frac{n^2+3}{(n+1)^2}\\
& \leq \frac{n^2+3}{n^2}\\
& \leq 1+\frac{3}{n^2}\\
& \leq 4 & (\forall n \geq 1)
\end{align*}
This is also satisfied by $a_0=3$.
Since it is ultimately monotonic and bounded, hence convergent.
For convergence:
You already have
\begin{align*}
|a_n-1| & = \left|\frac{2-2n}{(n+1)^2}\right|\\
& \leq \frac{2n}{(n+1)^2}\\
& \leq \frac{2n}{n^2}\\
& \leq \frac{2}{n}.
\end{align*}
For an $\epsilon >0$. Let $N$ be the smallest integer bigger than $\frac{2}{\epsilon}$. Then for all $n \geq N$, we have
$$|a_n-1| \leq \epsilon.$$
This shows that $\lim_{n \to \infty}a_n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973669",
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"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_1=F_2=1$, we suppose that $n \geq 3$.
$$\begin{align}
F_{n+1}&=F_n+F_{n-1} \\
F_{n+1}^2&=(F_n+F_{n-1})^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2
\end{align}$$
$$\begin{align}
F_{n-1} &=F_{n-2}+F_{n-3} \\
F_{n-1}^2&=(F_{n-2}+F_{n-3})^2=F_{n-2}^2+2F_{n-2}F_{n-3}+F_{n-3}^2
\end{align}$$
so that
$$
F_{n+1}^2-F_{n-1}^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2-F_{n-2}^2-2 F_{n-2} F_{n-3}-F_{n-3}^2
$$
How can we deduce that the latter is divisible by $4$?
Or do we show it somehow else, for example by induction?
| $$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})
=F_n(F_{n+1}+F_{n-1}).$$
Also $F_{n+1}=F_n+F_{n-1}\equiv F_{n-1}\pmod{F_n}$, so
$F_{n+1}+F_{n-1}\equiv 2F_{n-1}\pmod{F_n}$. If $F_n$ is even,
then so is $F_{n+1}+F_{n-1}$.
Similarly
$$F_{n+1}^3-F_{n-1}^3=F_n(F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2)$$
and
$$F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2\equiv 3F_{n-1}^2\pmod{F_n}$$
etc.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove $\lim_{x\to 4}\sqrt{x^3}=8$ $\lim_{x\to 4}\sqrt{x^3}=8$
My attempt:
Given $\epsilon>0$, choose $\delta=\min(1, \frac{\epsilon}{61})$
$|\sqrt{x^3}-8|=|\frac{(x-4)(x^2+4x+16)}{\sqrt{x^3}+8}|=|x-4||x^2+4x+16||\frac{1}{\sqrt{x^3}+8}|$
If $|x-4|<1$ then $-5<x<5$
So, $|x^2+4x+16|<61$
And $|\frac{1}{\sqrt{x^3}+8}|<1$
So, $|\sqrt{x^3}-8|<61|x-4|<61\delta=\epsilon$
Is this valid?
| The proof is good. You could avoid some steps by factoring
$$
\sqrt{x^3}-8=(\sqrt{x}-2)(x+2\sqrt{x}+4)
$$
and observe that, if $|x-4|<1$, then $|x+2\sqrt{x}+4|<5+2\sqrt{5}+4<15$ and you just need to take $|\sqrt{x}-2|<\frac{\varepsilon}{15}$; since
$$
|\sqrt{x}-2|=\left|\frac{x-4}{\sqrt{x}+2}\right|<|x-4|
$$
when $x>3$, you can take $\delta=\min(1,\varepsilon/15)$.
| {
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+b^2)\cos^2x-2ac\cos x+(c^2-b^2)=0\\
\implies\cos^2x-\frac{2ac}{a^2+b^2}\cos x+\frac{c^2-b^2}{a^2+b^2}=0
$$
$$
a\cos\alpha+b\sin\alpha=c\implies a\cos^2\alpha\cos\beta+b\sin\alpha\cos\alpha\cos\beta=c\cos\alpha\cos\beta\\
a\cos\beta+b\sin\beta=c\implies a\sin\alpha\sin\beta\cos\beta+b\sin\alpha\sin^2\beta=c\sin\alpha\sin\beta\\
c\cos(\alpha+\beta)=a\cos\beta+a\sin\alpha\cos\beta.(\sin\beta-\sin\alpha)+b\sin\alpha+b\sin\alpha\cos\beta(\cos\alpha-\cos\beta)\\
$$
I think its getting complicated to solve now. What is the simplest way to solve this kind of problems?
| Answer in the image. Hopefully it is legible
| {
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Real and imaginary part of $\tan(a+bi)$ When dealing with complex trigonometric functions, it is quite natural to ask how the real/imaginary part of $\tan(a+bi)$ can be expressed using $a$ and $b$.
Of course, since $\tan z$ and $\tanh z$ are tightly linked for complex variables, we could derive the real/imaginary part for hyperbolic tangent from the corresponding results for $\tanh(a+bi)$ and vice-versa. (We have $\tanh(iz)=i\tanh z$ and $\tan(iz)=i\tanh(z)$.)
I was not able to find this in a few basic sources I looked at. For example, I do not see it in the Wikipedia articles Trigonometric functions (current revision) and Hyperbolic function (current revision). (And List of trigonometric identities (current revision) does not mention much about complex trigonometric functions other than the relation to the exponential function.)
I have at least tried to find what are a results for some specific value of $a$ and $b$. I have tried a few values in WolframAlpha, for example, tangent of $2+i$, tangent of $1+2i$, tangent of $1+i$.
On this site I found this question: Calculate $\tan(1+i)$.
I have tried to calculate this myself, probably my approach is rather cumbersome - I post it below as an answer. I will be grateful for references, different derivations, different expressions for this formula. (And I will also be grateful if I receive some corrections to my approach - but do not treat this primarily as a solution-verification question, it is intended as a general question.)
| We want to express $\tan(a+bi)$ in the form
$$\tan(a+bi)=A(a,b)+B(a,b)i,$$
the two functions $A(a,b)$ and $B(a,b)$ are what we are looking for.
We have
\begin{align*}
\tan(a+bi)&=\frac{\sin(a+bi)}{\cos(a+bi)}\\
&\overset{(1)}=\frac{\sin a\cos(bi)+\cos(a)\sin(bi)}{\cos a\cos(bi)-\sin a\sin(bi)}\\
&\overset{(2)}=\frac{\sin a\cosh b+i\cos a\sinh b}{\cos a\cosh b-i\sin a\sinh b}\\
&=\frac{(\cos a\cosh b+i\sin a\sinh b)(\sin a\cosh b+i\cos a\sinh b)}{\cos^2a\cosh^2b+\sin^2a\sinh^2b}\\
&=\frac{\cos a\sin a(\cosh^2b-\sinh^2b)+i(\cos^2a+\sin^2a)\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b}\\
&=\frac{\cos a\sin a+i\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b}
\end{align*}
$(1)$: Using sum angle formulas.
$(2)$: Using $\cos iz=\cosh z$ and $\sinh i = i\sinh z$.
We would like to simplify this further, it seems that double argument formulae might help. In the numerator we can use that $\cos a \sin a = \frac{\sin 2a}2$ and $\cosh b \sinh b = \frac{\sinh2b}2$. For the denominator let us try that from the double angle formulas $\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x$ and $\cosh2x=\cosh^2x+\sinh^2x=2\cosh^2x-1=2\sinh^2x+1$ we can get that
\begin{align*}
\cos^2x&=\frac{1+\cos2x}2\\
\sin^2x&=\frac{1-\cos2x}2\\
\cosh^2x&=\frac{\cosh2x+1}2\\
\sinh^2x&=\frac{\cosh2x-1}2\\
\end{align*}
Now we have for the denominator $D$:
\begin{align*}
D&=\cos^2a\cosh^2b+\sin^2a\sinh^2b\\
&=\frac{1+\cos2a}2\cdot\frac{\cosh2b+1}2+\frac{1-\cos2a}2\cdot\frac{\cosh2b-1}2\\
&=\frac{1+\cos2a+\cosh2b+\cos2a\cosh2b}4+\frac{1+\cos2a-\cosh2b-\cos2a\cosh2b}4\\
&=\frac{\cos2a+\cosh2b}2
\end{align*}
So altogether we get that
$$\tan(a+bi)=\frac{\cos a\sin a+i\cosh b\sinh b}{\cos^2a\cosh^2b+\sin^2a\sinh^2b} = \frac{\sin2a+i\sinh2b}{\cos2a+\cosh2b}.$$
So for the real and imaginary part we get
\begin{align*}
A(a,b)&=\frac{\sin2a}{\cos2a+\cosh2b}\\
B(a,b)&=\frac{\sinh2b}{\cos2a+\cosh2b}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Probability that the circumference of circle is larger than $\pi$
The joint probability density function of the random variables $X$ and
$Y$ is given by $f(x,y)=x+y$ for $0 \leq x,y \leq 1$ and $f(x,y)=0$
otherwise. Consider the circle centered at the origin and passing
through the point $(X,Y)$. What is the probability that the
circumference of the circle is larger that $\pi$?
try
We know circumference of circle is $2 \pi r = 2 \pi \sqrt{X^2+Y^2}$. We want $P( 2 \pi \sqrt{X^2+Y^2} > \pi ) = P(X^2+Y^2 > 1/4) = 1 - P(X^2+Y^2 \leq 1/4).$
To calculate $P(X^2+Y^2 \leq 1/4)$, we want to integrate over the intersection of the unit square (first quadrant) and the circle of radius one half. This gives
$$ \int\limits_0^{1/2} \int\limits_0^{\sqrt{1/4-x^2}} (x+y) dy dx = \int_0^{1/2} (xy+y^2/2 )_0^{\sqrt{1/4-x^2}} dy = \int_0^{1/2} x \sqrt{1/4-x^2}+ \frac{1}{2} (1/4-x^2) dx = \dfrac{1}{12}$$
Is this a correct solution?
| \begin{align}
P(X^2 + Y^2 \le \frac14) &=\int_0^{\frac12} \int_0^{\frac{\pi}2}r(\cos(\theta)+\sin(\theta) \cdot r \, \, dr d\theta \\
&= \int_0^\frac12r^2 \, dr \int_0^\frac{\pi}2 \cos(\theta) + \sin(\theta) \, d\theta\\
&=\frac13\left( \frac12\right)^3 [ \sin(\theta) - \cos(\theta)]_0^\frac{\pi}2 \\
&=\frac13\left( \frac12\right)^3 (1-(0-1))\\
&= \frac1{12}
\end{align}
Hence the probability of having a circumference that is larger than $\pi$ is $\frac{11}{12}$.
| {
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For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$ For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$
Since the power of the right hand side is 4, $p(x)$ has to be 2.
So I assumed a solution of: $p(x)=ax^2+bx+c$ and then i put it in $p(p(x))+p(x)$ and got:
$a(ax^2+bc+c)^2+b(ax^2+bx+c)+c+ax^2+bx+c=x^4+3x^2+3$
To find the coefficients I tried $x=0$ and got
$ac^2+bc+2c=3$. But here I'm stuck, how do I go from here?
| The polynomials
$$Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$
and
$$ax^5+bx^4+cx^3+dx^2+ex+f$$
are equal for all $x$ if and only if
$$A=a\\B=b\\C=c\\D=d\\E=e\\F=f$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988298",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
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If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| Start off with the identity $\sin^2(x)+\cos^2(x)=1$. Rearrange this to find $\sin(x)=\pm\sqrt{1-\cos^2(x)}$. Substitute this into your equation and then rearrange to find $\sqrt{1-\cos^2(x)}$. You should be able to take it from there.
Note that it's not an identity that $\cos^2(x)=\sin(x)+\sin^3(x)$.
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
} |
Error in calculation of number of ways to put $20$ identical balls in $4$ labelled boxes if each box contains at most $18$ balls The number of ways to put $20$ identical balls in $4$ labelled boxes in such a way that each box contains at most $18$ balls is,
$$(a)~~\binom{24}{4}-16~~(b)~~\binom{24}{4}-10~~(c)~~\binom{23}{3}-16~~(d)~~\binom{23}{3}-12~~(e)~~\binom{24}{4}-12$$
My solution:
Answer - Stars bars All balls can go in each box - $4C1\cdot$ Stars bars with $2$ balls and $3$ boxes (Fill $1$ box with $18$ balls leaves $2$ balls and $3$ boxes. AND $4C1$ ways of selecting full box )
$$\begin{align}
&= (23 C 3) - (4 C 1) \cdot ( 4 C 2)\\
&= (23 C 3) - 4 \cdot 6\\
&= (23 C 3) - 24
\end{align}$$
Where did I go wrong?
| We need the coefficient of $t^{20}$ in the expansion $(1+t+\cdots +t^{18})^4$.
We have
\begin{align*}
(1+t+\cdots +t^{18})^4 &= \left(\dfrac{1-t^{19}}{1-t}\right)^4 \\
&= (1-t^{19})^4(1-t)^{-4} \\
&= (1-4t^{19} + \cdots )\sum_{n=0}^\infty \binom{n+3}{n}t^n
\end{align*}
and hence equals $\binom{23}{20} - 4\binom{4}{1} = \binom{23}{3} - 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
To prove: $\frac {\log (1+x)}{(1+x)} = \sum_{k=1}^∞ (-1)^{k+1}H_kx^k$ For the open interval $(-1,1)$ we have:
$$\log (1+x) = x - \frac {1}{2}.x^2 + \frac {1}{3}.x^3 - ……$$
$$(1+x)^{-1} = 1 - x + x^2 - x^3 + ……$$
Multiplying the two series and collecting together the coefficients of like powers of $x$ we have:
$$\frac {\log (1+x)}{(1+x)} = x - (1+\frac {1}{2})x^2 + (1+\frac {1}{2} + \frac {1}{3})x^3 - …… =\sum_{k=1}^∞ (-1)^{k+1}H_kx^k$$
Here $H_n$ is the $n^{th}$ harmonic number. How can this result be proved analytically i.e. by the direct use of successive differentiation or by some other sounder method?
| $$
\\\frac{\log{(1+x)}}{1+x}=\sum_{n=1}^{+\infty}{\frac{(-1)^{n+1}}{n}x^n}\sum_{m=0}^{+\infty}{(-1)^{m}x^m}=
\\=\sum_{n=1}^{+\infty}\sum_{m=0}^{+\infty}{\frac{(-1)^{n+m+1}}{n}x^{n+m}}=
\\=\sum_{n=1}^{+\infty}(-1)^{n+1}H_nx^n
\\k={n+m}=>n=1,\;2,\;3,\dots k=>
\\H_k=\sum_{n=1}^{k}{\frac{1}{n}}
$$
Q.E.D
| {
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"source": "stackexchange",
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The $1997$ IIT JEE problem
Let $S$ be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$
This problem appeared in IIT JEE $1997$ (re-exam).I really do not know know how to approach this problem properly.I have managed to prove $a^2 + b^2 + c^2 +d^2 \leq 4$
but not the initial inequality.
| $a^2 = p^2 + s^2$
$ b^2 = (1 – p)^2 + q^2$
$c^2 = (1 – q)^2 + (1 – r)^2$
$ d^2 = r^2 + (1 – s)^2$
$$\implies a^2 + b^2 + c^2 + d^2 = p^2 + (1 - p)^2 + q^2 + (1 – q)^2 + r^2 + (1 – r)^2 + s^2 + (1 – s)^2$$
Where $p$, $q$, $r$, $s$ all vary in the interval $[0, 1]$.
Now consider the function
$y^2 = x^2 + (1 – x)^2, 0 ≤ x ≤ 1,$
$$2y \frac {dy} {dx} = 2x – 2(1 – x) = 0$$
$$\implies x = \frac 12$$
which$$\frac {d^2y} {dx^2} = 4$$ i.e. positive
Hence y is minimum at $x =\frac 1 2$ and its minimum
Value is $\frac 1 4 + \frac 1 4 = \frac 1 2$
Clearly value is maximum at the end points which is $1$.
Minimum value of $a^2 + b^2 + c^2 + d^2 = \frac {1}{2} + \frac {1}{2} + \frac {1}{2} + \frac {1}{2} = 2$
And maximum value is $1 + 1 + 1 + 1 = 4$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
Limits - Calculating $\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule
Calculate $\displaystyle\lim\limits_{x\to 1} \frac{x^a -1}{x-1}$, where $a \gt 0$, without using L'Hospital's rule.
I'm messing around with this limit. I've tried using substitution for $x^a -1$, but it didn't work out for me.
I also know that $(x-1)$ is a factor of $x^a -1$, but I don't know where to go from here.
EDIT: Solved it, posting here for future generations :)
a) We can write $x^a$ as $e^{a\ln x}$ ending up with $\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{x-1}$
b) Multiplying by $\frac{a\ln x}{a\ln x}$ we end up with: $\lim\limits_{x\to 1} \frac{e^{a\ln x} -1}{a\ln x} \cdot \frac{\ln x}{x-1} \cdot a$
c) Now we just have to show that the first 2 limits are equal 1, and $\lim\limits_{x\to 1} a = a$
| You can use Bernoulli's ineq (which can be proved by am-gm and by continuity argument, so minimal calculus knowledge i think) and the squeeze theorem. You can assume wlog that $a\geq 1$, or otherwise, if $0<a<1$, take $X=x^a$, so
$$\lim_{x\to 1}\frac{x^a-1}{x-1}=\lim_{X\to 1}\frac{X-1}{X^{\frac1a}-1}=\frac{1}{\lim_{X\to 1}\frac{X^{\frac1a}-1}{X-1}}.$$
So, if $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $a\geq 1$ holds, we also have $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $0<a<1$ too.
Suppose that $a\geq 1$. Let $x=1+y$. Wlog, we may say that $y>-1$ is a small real number. For $a\geq 1$, Bernoulli's ineq gives $$x^a-1=(1+y)^a-1 \geq ay=a(x-1).$$ So, $\frac{x^a-1}{x-1}\geq a$ if $y>0$ (or $x>1$). (The ineq is reversed for $y<0$, or $x<1$.)
Now write $x=\frac{1}{1-z}$ for some small real number $z<\frac1a$ (that is, $z=y/(1+y)$). By Bernoulli's ineq, $(1-z)^a\geq 1-az$, so $$x^a-1=\frac{1}{(1-z)^a}-1\leq \frac{1}{1-az}-1=\frac{az}{1-az}.$$
That is, if $y>0$ (i.e. $x>1$), we get $z>0$ and
$$\frac{x^a-1}{x-1}=\frac{\frac1{(1-z)^a}-1}{\frac1{1-z}-1}\leq \frac{az}{1-az}\left(\frac{1-z}{z}\right)=\frac{a(1-z)}{1-az}.$$
The ineq is reversed if $y<0$ (i.e., $x<1$ and $z<0$).
Therefore,
$$a\leq \frac{x^a-1}{x-1}\leq \frac{a(1-z)}{1-az}=\frac{a}{x}\left(\frac{1}{1-a\left(1-\frac1x\right)}\right)=\frac{a}{x-a(x-1)}$$
if $x>1$.
Similarly, for $a\geq1$ and $0<x<1$, we have
$$\frac{a}{x-a(x-1)}\leq \frac{x^a-1}{x-1}\leq a.$$
Taking $x\to 1$ and using the squeeze thm, we have $\lim_{x\to 1}\frac{x^a-1}{x-1}=a$ for $a\geq 1$.
| {
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"url": "https://math.stackexchange.com/questions/2999527",
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"source": "stackexchange",
"question_score": "1",
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Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ I need some help calculating this limit:
$$\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$$
I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Hopital. Thanks a lot.
| HINT
Use that
$$ \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} =\left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)\frac{ \sqrt{x + \sqrt{x}}+ \sqrt{x - \sqrt{x}} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }=$$
$$=\frac{ x + \sqrt{x}- x + \sqrt{x} }{ \sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)\left(\frac{x}{5}\right)} \: dx = \frac{\pi}{2}$$
To begin with, I made a simple rearrangement
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{x^3} \: dx = \frac{\pi}{30}$$
From here I used the Sine/Cosine Identities
$$ \int_{0}^{\infty} \frac{\frac{1}{4}\left(-\sin\left(\frac{7}{15}x\right)+ \sin\left(\frac{13}{15}x\right) + \sin\left(\frac{17}{15}x\right) -\sin\left(\frac{23}{15}x\right) \right)}{x^3} \: dx = \frac{\pi}{30}$$
Which when expanded becomes
$$ -\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx +
\int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx -
\int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx
= \frac{2\pi}{15}$$
Using the property
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^3}\:dx = a^2 \int_{0}^{\infty}\frac{\sin(x)}{x^3}\:dx$$
We can reduce our expression to
$$\left[ -\left(\frac{7}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(\frac{17}{15}\right)^2 - \left(\frac{23}{15}\right)^2\right] \int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
Which simplifies to
$$ -\frac{120}{15^2}\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
And from which we arrive at
$$\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = -\frac{\pi}{4}$$
Is this correct? I'm not sure but when I plug into Wolframalpha it keeps timing out...
| \begin{multline}
\int_0^\infty \frac{\sin(x)}{x^3}dx = \int_0^1 \frac{\sin(x)}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx \\> \int_0^1 \frac{x/2}{x^3}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \frac{1}{2}\int_0^1 \frac{1}{x^2}dx +\int_1^\infty \frac{\sin(x)}{x^3}dx = \infty
\end{multline}
The integral diverges.
| {
"language": "en",
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Integer solutions of a variable coefficient polynomial I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 \leq a \leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?
My attempts:
Taking the equation $\pmod b$ we get:
$$- 2 a \equiv 0 \pmod b$$
Which implies a = b/2
Taking the equation $\pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b \equiv 0 \pmod a$$
$$ b^3 + 2b \equiv b^4 + 2b^2 \pmod a$$
$$ b^3 + 2b \equiv b(b^3 + 2b) \pmod a$$
$$ 1 \equiv b \pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$
Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, \frac{3b}{5}, \frac{4b-5}{5})$
Plugging $a_1 = \frac{3b}{5}, a_2 = \frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben
| Say $b \ne 1$. Write $$ a (2b^3 - b^2 + b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 - b^2 + b - 2 \mid b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2\mid b(b^2+2)$$
If $b$ is odd then $\gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2\mid b^2+2 \implies 2b^2+b+2\leq b^2+2 $$
which is impossible.
So $b$ is even $b=2c$ and now $$4c^2+c+1\mid c(4c^2+2)$$
Since $\gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1\mid 4c^2+2\implies4c^2+c+1\leq 4c^2+2$$ so $c\leq 1$ and thus $c=1$, $b=2$ and $a=1$.
| {
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"source": "stackexchange",
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Definition of $\binom{\frac{1}{2}}{1}$? How interpret the combination
$\binom{\frac{1}{2}}{n}$ when $n$ is a positive integer ?
| The definition is
$$
\binom{x}{n}=\frac{x(x-1)\dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
\begin{align}
\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\dotsm\left(\frac{1}{2}-n+1\right)
&=\frac{(-1)^{n-1}}{2^n}\bigl(1\cdot3\cdot\dots\cdot(2n-3)\bigr) \\
&=\frac{(-1)^{n-1}}{2^n}\bigl(1\cdot3\cdot\dots\cdot(2n-3)\bigr)
\frac{2\cdot4\cdot\dots\cdot(2n-2)}{2^{n-1}(1\cdot 2\cdot\dots\cdot(n-1))} \\
&=\frac{(-1)^{n-1}}{2^{2n-1}}\frac{(2n-2)!}{(n-1)!}\\
&=\frac{(-1)^{n-1}}{2^{2n-1}}\frac{(2n)!}{n!}\frac{n}{2n(2n-1)}\\
&=\frac{(-1)^{n-1}}{2^{2n}(2n-1)}\frac{(2n)!}{n!}
\end{align}
and therefore
$$
\binom{1/2}{n}=\frac{(-1)^{n-1}}{2^{2n}(2n-1)}\frac{(2n)!}{(n!)^2}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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If $n$ is an integer , find all the possible values for $(8n+6,6n+3)$ I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
| $(1)$ A euclidean sequence is $\ \overbrace{8n\!+\!6,\,6n\!+\!3,\,2n\!+\!3,\,{-}\color{#c00}6}^{\Large a_{k-1} -\, j\ a_k\ =\ a_{k+1}}\,$ so the gcd is
$$(2n\!+\!3,\,\color{#c00}{2\cdot 3}) = (2n\!+\!3,\color{#c00}2)(2n\!+\!3,\color{#c00}3) = (3,2)(2n,3) = (n,3)\qquad\qquad $$
$(2)$ A euclidean sequence is $\ 2n^2\!+\!3n\!+\!5,\!\!\!\!\underbrace{n^2\!+\!n\!+\!1,\, n\!+\!3,\, \color{#0a0}7}_{\large f(n)\ \equiv\ \color{#0a0}{f(-3)}\,\pmod{\!n+3}}\!\!\!\!$ so the gcd $= (n\!+\!3,7)$
| {
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"question_score": "2",
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Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$ How to prove
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$
where $G$ is the Catalan's constant.
Attempt
For the first sum,
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\left\{\sum_{k=1}^{\infty}\frac{i^k}{(k+1)^2}H_{k}\right\},$$
which can be evaluated by using the formula in this post:
$$\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x),$$
but we cannot apply the similar approach to the second sum
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_k.$$
Then, I tried to write the sum as
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}\int_0^1\frac{2x^{2k}+x^k-3}{x-1}~\mathrm dx$$
and it become more complicated.
Edit:
Are we able to evaluate the sum directly (avoid calculating integrals and polylogs as much as possible)? The integral given by @Jack D'Aurizio is a bit complicated (see this post).
| \begin{align}
S&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}\\
&=2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\
&=2\Im\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\frac{\pi^2}{16}\\
&=S_1+S_2-\frac{\pi^3}{16}\tag{1}
\end{align}
Using the generating function: $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
then
\begin{align}
S_1&=2\Im\left(\operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\ln(1-i)\operatorname{Li}_2(1-i)+\frac12\ln(i)\ln^2(1-i)\right)\\
&\boxed{S_1=-2\Im\operatorname{Li}_3(1-i)-G\ln2-\frac{\pi}{8}\ln^22}
\end{align}
$$S_2=\sum_{n=1}^\infty\frac{(-1)^nH_n}{(2n+1)^2}=\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx$$
The last integral is evaluated here
$$\boxed{\int_0^1\frac{\ln(1+x^2)\ln x}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-G\ln2+2\text{Im}\operatorname{Li_3}(1-i)=S_2}$$
Plugging $S_1$ and $S_2$ in $(1)$, we get $$\color{blue}{S=\frac{\pi^3}{32}-2G\ln2}$$
note that we used:
$$\ln(i)=\frac{\pi}{2}i$$
$$\ln(1-i)=\frac12\ln2-\frac{\pi}{4}i$$
$$\operatorname{Li_2}(1-i)=\frac{\pi^2}{16}-\left(\frac{\pi}{4}\ln2+G\right)i$$
which give us:
$$\ln(i)\ln^2(1-i)=\frac{\pi^2}{8}\ln2-\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$
$$\ln(1-i)\operatorname{Li_2}(1-i) =-\frac{\pi}{4}G-\frac{\pi^2}{32}\ln2-\left(\frac12\ln2G+\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22\right)i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Integral of $1/(x^2 + y^2)^2$ under the parabola $y = x^2 - x - 1$ I'm trying to show that
\begin{equation}
\int \limits_{y < x^2 - x - 1} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, dxdy = \pi,
\tag{1}
\end{equation}
and more generally that
\begin{equation}
\int\limits_{y < ax^2 + bx + c} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, dxdy
= \left(\frac{b^2 + 1 - 2ac}{4c^2}\right) \pi,
\tag{2}
\end{equation}
for $a > 0$ and $c < 0$. I believe these equations are true based on numerical computations.
It is pretty straightforward to integrate with respect to $y$ using the substitution $y = x\tan(v)$, but that turns the integral in equation (1) into
\begin{equation}
\frac{1}{2} \int \limits_{-\infty}^{\infty} \frac{1}{x^3} \left(\arctan(x - 1 - 1/x) + \mathrm{sgn}(x)\frac{\pi}{2} + \frac{x - 1 - 1/x}{1 + (x - 1 - 1/x)^2}\right)dx,
\end{equation}
which looks very difficult. I hope there is a more clever approach to the original integral.
| We can factor the parabola into the form $y=a(x-A)(x-B)$. If $A=B$ or $A$ and $B$ are not real numbers, out integral will diverge. If $AB<0$, then polar substitution converts the integral into the form of $\iint\frac{dr}{r^3}d\theta$. The limits of integration are from $r=0$ to $\frac{\sin(\theta)-b\cos(\theta)+\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$ and either from $\theta=0$ to $\theta=\pi$ or from $\theta=\pi$ to $\theta=2\pi$, depending on the sign of $a$. If $AB>0$, then the same substitution can be done, with the limits of integration from $r=\frac{\sin(\theta)-b\cos(\theta)-\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$ to $r=\frac{\sin(\theta)-b\cos(\theta)+\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$. The limits with respect to $\theta$ are based on the location of the vertex of the parabola, and either begin or end at $\theta=0$ or $\theta=\pi$.
Depending on the values of $a,b,c$, this integral might be easier to solve
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{HTHTHTHTH}$
Then we'll be left with 1 tail which can be placed at 6 places.
So number of cases = $6$
2) $4$ heads, $6$ tails
$\mathrm{HTHTHTH}$
Left with $3$ tails.
So all $3$ can be placed together at $5$ places.
Or $3 = 2+1$, in which case we have $5 \times 4 = 20$ cases.
Total number for #2 $= 20 + 5 = 25$
3) 3 heads, 7 tails.
$\mathrm{HTHTH}$
Left with $5$ tails.
All 5 can be placed together, so in that case we have $4$ cases.
Or $5 = 2+3= 1+4$
For $5 = 2+3 = 1+4$, we have $4\times 3$ cases each.
Therefore total number of cases for #3 $= 12+12+ 4 = 28$
4) 2 heads 8 tails.
$\mathrm{HTH}$
$7$ left
$7 = 7+ 0$ , $3$ cases.
$7 = 6+1= 5+2= 3+4$ , $6$ cases each
So number of cases for #4 = $6\times 3 + 3 = 21$
5) 1 head 9 tail, 10 cases
6) 0 heads, 1 case.
So total number of cases = $6+25+28+21+10 +1 = 91$
I tried thrice and got $91$ cases only.
So answer should be $P(E) = \dfrac{91}{2^{10}}$
But that's not the right answer.
Please tell me my mistake.
| Given that a fair coin is tossed $10$ times. So, the possible outcomes are $2^{10}$
Now to find the probability that no two heads appear consecutively, then the possible outcomes are
$$\begin{matrix} Heads & Tails & Total \\ 0 & 10 & \dbinom{11}{0} \\ 1 & 9 & \dbinom{10}{1} \\ 2 & 8 & \dbinom{9}{2} \\ 3 & 7 & \dbinom{8}{3} \\ 4 & 6 & \dbinom{7}{4} \\ 5 & 5 & \dbinom{6}{5} \\ \end{matrix}$$
We shouldn't consider the cases where heads are $>6$, otherwise there will be two heads consecutively. So, we neglect those cases.
Since, all the outcomes are mutually exclusive$$\left(1+\dbinom{10}{1}+\dbinom{9}{2}+\dbinom{8}{3}+\dbinom{7}{4}+\dbinom{6}{5}\right)\left(\dfrac{1}{2^{10}}\right)$$
$$(1+10+36+56+35+6)\left(\dfrac{1}{2^{10}}\right)=\dfrac{144}{2^{10}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $\lim_{x\to\pi/2}\left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)$ using algebra of limits Let $$
F(x) = \left(\frac{1-\sin x}{(\pi-2x)^4}\right)(\cos x)(8x^3 - \pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $\pi/2$.
How can we find the limit using algebra of limits?
The limit of $\dfrac{1-\sin x}{(\pi-2x)^4}$ as $x$ tends to $\pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-\frac{3\pi^2}{16}$.
How to apply algebra of limits here?
| Make life simpler using $x=y+\frac \pi 2$ which makes
$$\lim_{x\to\frac \pi2}\left(\frac{1-\sin (x)}{(\pi-2x)^4}\right)\cos (x)(8x^3 - \pi^3)=\lim_{y\to 0}\frac{\left(4 y^2+6 \pi y+3 \pi ^2\right) \sin (y) (\cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$\frac{\left(4 y^2+6 \pi y+3 \pi ^2\right) \sin (y) (\cos (y)-1)}{8 y^3}=-\frac{3 \pi ^2}{16}-\frac{3 \pi y}{8}+O\left(y^2\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$ $$\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$ should be calculated using complex numbers I think, the Wolfram answer is :
$ \frac{1}{3} (e^x + 2 e^{-x/2} \cos(\frac{\sqrt{3}x}{2})) $
How to approach this problem?
| We have that by $f(x)=\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$
$$f'(x)=\frac{d}{dx}\sum\limits_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}$$
$$f''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-1}}{(3n-1)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}$$
$$f'''(x)=\frac{d}{dx}\sum\limits_{n=1}^{\infty} \frac{x^{3n-2}}{(3n-2)!}=\sum\limits_{n=1}^{\infty} \frac{x^{3n-3}}{(3n-3)!}=f(x)$$
and $f'''(x)=f(x)$ has solution
$$f(x)=c_1e^x+c_2e^{-x/2}\cos\left(\frac{\sqrt 3 x}{2}\right)+c_3e^{-x/2}\sin\left(\frac{\sqrt 3 x}{2}\right)$$
with the initial conditions $f(0)=1$, $f'(0)=0$, $f''(0)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?
| Your method is also really good. Consider the two points $A(-1,2)$ and $B(2,5)$. We need to find the point $(x,x+2)$ on the line $y=x+2$ for which the sum $AD+BD$ is minimum (see the graph below).
$\hspace{3cm}$
You reflect the point $A$ over the line $y=x+2$ to find the point $C(0,1)$. The line through $C$ and $B$ is $y=2x+1$. The two lines intersect at $D(1,3)$, which is the solution of the given problem. Hence, when $x=1$, the sum will be minimum $\sqrt{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x^3-1)(x^2-x+1)\\
&=(x-1)(x^2+x+1)(x^2-x+1).
\end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
| The derivative has a single real root, namely $1$. The other two factors never vanish on $\mathbb{R}$ (and they are actually everywhere positive).
So we have $Q'(x)<0$ for $x<1$ and $Q'(x)>0$ for $x>1$. This implies $1$ is an absolute minimum for $Q(x)$. Note that the limits of $Q$ at $-\infty$ and $\infty$ are both $\infty$, so the absolute minimum must exist.
Since
$$
Q(1)=1-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}=\frac{30-20+15-12+10}{60}=\frac{23}{60}>0
$$
we can conclude that $Q(x)>0$ for every $x$.
The complex roots of $Q'$ play no role in this problem, which is about finding the (real) intervals in which $Q$ is monotonic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What are the complex roots of $x^3-1$? What are the complex roots of $x^3-1$?
Work I've done so far:
I've set $x = a + bi$. Since $x^3-1=0$, I set $x^3 = (a+bi)^3=1$.
This gives me the following:
(1) $(-ab^2 + a^3) + (2ab^2 + 2a^2b + a^2b - b^3)i$
Which means that I set $(-ab^2 + a^3) = a(a^2-b^2)= 1$ which is also equivalent to
(2) $a(a-b)(a+b)=(a-b)(a^2+ab)=1$.
I also set
(3) $(2ab^2 + 2a^2b + a^2b - b^3) = 0$.
I simplify (3) to
(4) $2b(a^2 + ab) + (a^2 -b^2)b = 0 $
which gives me
(5) $\frac{2b}{a-b} + \frac{b}{a} = 0$ using (2).
Then I get
(6) $\frac{2b}{a-b} = -\frac{b}{a}$. Then I get that $3a=b$. Plugging into (2) I get
(7) $a(a^2 - (3a)^2)=1 = a(a^2 -9a^2) = -8a^3$. So that $a= \frac{-1}{2}$. Now I get that $b= \frac{3}{2}$, which would give me $\frac{1}{2} + \frac{3}{2}i$. But on Wolfram, the imaginary component is close to $.9$. Where am I going wrong?
| You can just use difference of cubes, which gets the answer much more quickly:
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
Applying it here, you get
$$x^3-1 = (x-1)(x^2+x+1) = 0$$
The first factor obviously gives the real root of $x = 1$, so solve for the second factor. It should be pretty straightforward.
As for your error, you have expanded incorrectly in $(1)$. Recall that
$$(a+b)^3 = a^3+3a^2b+3ab^2+b^3$$
and you apply it to
$$(a+bi)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding magnitude of a complex number
$$z = \dfrac{2+2i}{4-2i}$$
$$|z| = ? $$
My attempt:
$$\dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = \dfrac{4+12i}{20} = \dfrac{4}{20}+\dfrac{12}{20}i = \dfrac{1}{5} + \dfrac{3}{5}i$$
Now taking its magnitude and we have that
$$|z| = \sqrt{\biggr (\dfrac 1 5 \biggr ) ^2 +\biggr (\dfrac 3 5 \biggr )^2} = \sqrt {\dfrac 2 5 }$$
Am I right?
| It is better to use $$\left|\dfrac ab\right|=\dfrac{|a|}{|b|}$$
$|2+2i|=\sqrt{2^2+2^2}=2\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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if such three condition find the $\sin{\angle OPA}$ Convex quadrilateral $ABCD$,and circumcenter is $O$,if Point $P$ lie on sides $AD$,and such
$$\dfrac{AP}{PD}=\dfrac{8}{5},~~PA+PB=3AB,~~PB+PC=2BC,~~PC+PD=\dfrac{3}{2}CD$$find $\sin{\angle OPA}$
I try let $AB=a,BC=b,CD=c,DA=d,,AP=x,PD=y,x+y=d$,and $OA=1$,then I get very ugly
use this wiki
$$1=R=\dfrac{1}{4}\sqrt{\dfrac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$
where $s$ is semiperimeter.
I want use this identity
$$\cos{(\angle APB+\angle CPB+\angle CPD)}=0$$
or $\angle APB=\angle 1,\angle CPB=\angle 2,\angle CPD=\angle 3$
$$\cos{\angle 1}\cos{\angle 2}\cos{\angle 3}=\sin{\angle 1}\sin{\angle 2}\cos{\angle 3}+\sin{\angle 1}\sin{\angle 3}\cos{\angle 2}+\sin{\angle 2}\sin{\angle 3}\cos{\angle 1}$$
| I have used coordinate geometry to attach this problem.
Use an unit circle at the original $x^2 + y^2 = 1$ and the line y = -a. In this way we have :
$A(-\sqrt{1 - a^2}, -a)$ and $D(\sqrt{1 - a^2}, -a)$
Using the ratio formula $P(\frac{3\sqrt{1 - a^2}}{13}, -a)$
Using distance formula we have
$PA = \frac{3}{13}\sqrt{1 - a^2} + \sqrt{1 - a^2} = \frac{16}{13}\sqrt{1 - a^2}$
$PD = \sqrt{1 - a^2} - \frac{3}{13}\sqrt{1 - a^2} = \frac{10}{13}\sqrt{1 - a^2}$
Also let B be $(x_1, \sqrt{1 - x_1^2})$ and C be $(x_2, \sqrt{1 - x_2^2})$
Hence $AB = \sqrt{(x_1 + \sqrt{1 - a^2})^2 + (\sqrt{1 - x_1^2} + a)^2}$
$PB = \sqrt{(x_1 - \frac{3}{13}\sqrt{1 - a^2})^2 + (\sqrt{1 - x_1^2} + a)^2}$
$BC = \sqrt{(x_1 - x_2)^2 - (\sqrt{1 - x_1^2} - \sqrt{1 - x_2^2})^2}$
$PC = \sqrt{(x_2 - \frac{3}{13}\sqrt{1 - a^2})^2 + (\sqrt{1 - x_2^2} + a)^2}$
$CD = \sqrt{(x_2 - \sqrt{1 - a^2})^2 + (\sqrt{1 - x_2^2} + a)^2}$
Put the above distances into PA + PB = 3AB theoretically we can express $x_1$ in terms of a.
Similarly for PC + PD = $\frac{3}{2}CD$ We also can express $x_2$ in terms of a.
By putting $x_1$ and $x_2$ into PB + PC = 2BC the equation can be solved for a.
Once $a$ is found we can find the answer :
$$\sin\angle OPA = \frac{13}{\sqrt{\frac{9}{a^2} + 160}}$$
The 3 equations are too difficult to solve and can only be resolved by tries and error.
My estimation gives the following results :
$a = 0.1$, $x_1 = -0.75$, $x_2 = 0.35$, $\sin \angle OPA = 0.399$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\tag {ii}$$
For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$
| (i)
$$(\sec^2x + \tan^2x)(\csc^2x + \cot^2x)$$
=>
$$(\sec^2x\csc^2x + \tan^2x\csc^2x + \sec^2x\cot^2x + \tan^2x\cot^2x)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x}{\cos^2x\sin^2x} + \frac{\cos^2x}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{1}{\cos^2x} + \frac{1}{\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{\sin^2x + \cos^2x}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \frac{1}{\cos^2x\sin^2x} + 1)$$
=>
$$(\sec^2x\csc^2x+ \sec^2x\csc^2x + 1)$$
=>
$$(1 + 2\sec^2x\csc^2x)$$
ii) $$\frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x}$$
=>
$$\frac{\cos x(1 - \cot x) + \sin x(1 - \tan x)}{(1 - \tan x)(1 - \cot x)}$$
=>
$$\frac{\frac{\cos^2x \sin x - \cos x\cos^2x + \sin^2 x\cos x - \sin x\sin^2x}{\sin x \cos x}}{(1 - \tan x)(1 - \cot x)}$$
=>
$$\frac{\cos^2 x(\sin x - \cos x) + \sin^2 x(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
$$\frac{(\sin^2 x - \cos^2 x)(\cos x - \sin x)}{\cos x\sin x(1 - \tan x)(1 - \cot x)}$$
=>
$$\frac{(\sin^2 x - \cos^2x)(\cos x - \sin x)}{(\cos x - \sin x)(\sin x - \cos x)}$$
=>
$$\frac{(\sin^2 x - cos^2 x)}{(\sin x - \cos x)}$$
=>
$$\frac{(\sin x + \cos x)(\sin x - \cos x)}{(\sin x - \cos x)}$$
=>
$$\sin x + \cos x$$
| {
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"source": "stackexchange",
"question_score": "2",
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find closed form for following double integral containing radicals question:
To prove :
$\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}=\dfrac{3\left(\Gamma{\dfrac{1}{3}}\right)^6}{2^{\dfrac{17}{3}}\pi^2}$
i tried attempting it by transforming into polar co-ordinates but it didn't
helped
i couldn't get rid of radicals and also there is no scope for changing order of
integration for it's evaluation.
please help . thank you
| This double integral is first transformed into a single integral of a complete elliptic integral:
\begin{align}
I&=\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}\\
&\stackrel{x=\cos t}{=}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt{y^2+\left( 1-\sin^2t \right)\left( 4-y^2 \right)}}\\
&=\frac{1}{2}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-\left( 1-y^2/4 \right)\sin^2t}}
\end{align}
We introduce the complete elliptic integral using its integral representation (DLMF):
\begin{align}
I&=\frac{1}{2}\int_0^1\frac{\mathbf{K}\left( \sqrt{1-y^2/4} \right)}{\sqrt{1-y^2}}dy\\
&\stackrel{y=\cos \phi}{=}\frac{1}{2}\int_0^{\pi/2}\mathbf{K}\left( \sqrt{1-\frac{1}{4}\cos^2\phi} \right)\,d\phi
\end{align}
(the argument of the elliptic function corresponds to its modulus). Similar integrals are calculated in these articles here and here where the following result is given:
\begin{equation}
\int_0^{\pi/2}\mathbf{K}\left( \sqrt{1-\sin^2\alpha\cos^2\phi} \right)\,d\phi=
\mathbf{K}\left(\sin \frac{\alpha}{2} \right)\mathbf{K}\left(\cos \frac{\alpha}{2} \right)
\end{equation}
Here we choose $\alpha=\pi/6$ and then
\begin{equation}
I=\frac{1}{2} \mathbf{K}\left(\sin \frac{\pi}{12} \right)\mathbf{K}\left(\cos \frac{\pi}{12} \right)
\end{equation}
But $\sin \frac{\pi}{12}=\frac{\sqrt{2}}{4}\left( \sqrt{3}-1 \right)=\lambda^*(3)$, where $\lambda^*(r)$ gives a singular value of the elliptic modulus $k_r$ for which $\mathbf{K}\left( \sqrt{1-k_r^2} \right)=\sqrt{r}\mathbf{K}(k_r)$ see here, here and here. Then
\begin{align}
\mathbf{K}\left(\sin \frac{\pi}{12} \right)&=\frac{3^{1/4}\Gamma^3\left( \frac{1}{3} \right)}{2^{7/3}\pi}\\
\mathbf{K}\left(\cos \frac{\pi}{12} \right)&=\frac{3^{3/4}\Gamma^3\left( \frac{1}{3} \right)}{2^{7/3}\pi}
\end{align}
and thus
\begin{equation}
I=\frac{3\Gamma^6\left( \frac{1}{3} \right)}{2^{17/3}\pi^2}
\end{equation}
as expected. This result is related to some Random Walk Integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$ The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $\rho$. Show that $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$.
My attempt:
$$E[\max(X^2,Y^2)]=
E\left(\frac{X^2+Y^2-|X^2-Y^2|}{2}\right)=
1-\frac{E|X^2-Y^2|}{2}$$
which is certainly lesser than the rhs. Is there anything wrong in my reasoning?
| Actually, the correct expression is
$$\max(X^2,Y^2)=\frac{1}{2}\left[X^2+Y^2\color{red}+|X^2-Y^2|\right]\tag{1}$$
Now simply note that
\begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\\&=E\left[\sqrt{(X+Y)^2(X-Y)^2}\right]
\\&\le \sqrt{E\left[(X+Y)^2(X-Y)^2\right]}\tag{2}
\\&\le\sqrt{E\,[(X+Y)^2]E\,[(X-Y)^2]}\tag{3}
\end{align}
In $(2)$, Jensen's inequality was used since $x\mapsto\sqrt x$ is a concave function for all $x\ge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Using Vieta's formula to evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$ Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,
$$x_1 + x_2 + x_3 = -\frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = \frac{c}{a},$$
$$x_1x_2x_3 = -\frac{d}{a},$$
How would one evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$?
Edit: I'm at this point currently:
$$\frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$
and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.
| Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= \frac{r}{s}+\frac{s}{t}+\frac{t}{r}$ and $q=\frac{s}{r}+\frac{t}{s}+\frac{r}{t}$. Then,
$$p+q=\frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
$$pq=3+\frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
That is,
$$p+q=\frac{BC-3D}{D}$$
and
$$pq=3+\frac{C^3+B^3D-6BCD+6D^2}{D^2}=\frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
Therefore, $p$ and $q$ are the roots of
$$x^2-\frac{BC-3D}{D}x+\frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
which is the same as
$$x^2+\frac{3ad-bc}{ad}x+\frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$
For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+\frac{575}{36}=\left(x-\frac{23}{6}\right)\left(x-\frac{25}{6}\right).$$
Indeed, $\{r,s,t\}=\{1,2,3\}$, so $$\{p,q\}=\left\{\frac12+\frac23+\frac31,\frac21+\frac32+\frac13\right\}=\left\{\frac{25}{6},\frac{23}{6}\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The limit of the sequence is $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
For our sequence:
$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{n!}-\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\cdot(1+\sqrt{n+1})-(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{(n-1)!}\cdot(\sqrt{n-1})}{\left((1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\right)\cdot(\sqrt{n}+1)}$
Which got me nowhere.
| We have that
$$\frac{\sqrt{(n-1)!}}{\prod_{k=1}^n\big(1+\sqrt{k}\big)}=\frac1{\sqrt n}\prod_{k=1}^n\frac{\sqrt{k}}{1+\sqrt{k}}\le \frac1{\sqrt n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
prove there exist postive integers $a,b$ such $p^2|a^2+ab+b^2$ Problem 1: Let prime $p\equiv 1\pmod 3$.show that:there exist postive integers $a\le b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<a\leq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.
The key claim is that if $p\equiv1\pmod3$, then $$p(x^2+xy+y^2)^2\;\mathrm{divides}\;(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $\color{red}{\underline{\color{black}{\text{Since it's known that}}}}$ one can select $a$ and $b$ such that $\color{red}{\underline{\color{black}{p^2\mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1\equiv0\pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2\;\mathrm{divides}\;F(x)\overset{\mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $\binom pk$ are clearly divisible by $p$. Let $\zeta$ be a third root of unity. Then $F(\zeta)=(1+\zeta)^p-\zeta^p-1=-\zeta^2-\zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(\zeta)=p-p=0$. Hence $\zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
Image that replaced the text
| Alternate solution via Thue's Lemma
Proposition. Let $p\equiv 1\pmod 3$ be a prime. Then there is a solution to
$$
x^2+x+1\equiv 0 \pmod{p^2}
$$
Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $\pmod p$, hence
$$
u^2 \equiv -3 \pmod p
$$
for some integer $u$. Setting $u\equiv 2c+1\pmod p$, this becomes
$$
c^2+c+1 \equiv 0 \pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1\equiv 0\pmod p$. Hensel's Lemma tells us that if
$$
f'(c)\not\equiv 0 \pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) \equiv 0 \pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 \equiv -3 \pmod p
$$
If $f'(c)\equiv 0 \pmod p$ then $0\equiv -3\pmod p$ which implies $p=3$. This contradicts $p\equiv 1\pmod 3$, therefore $f'(c)\not\equiv 0 \pmod p$.
Hence there is some $d$ such that
$$
d^2+d+1\equiv 0 \pmod{p^2}
$$
$$\tag*{$\square$}$$
We now solve our main problem.
Lemma. Let $p\equiv 1\pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
$$
a^2+ab+b^2\equiv 0 \pmod{p^2}
$$
Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 \equiv 0\pmod{p^2}
$$
Now by Thue's Lemma, since $\gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db \equiv a \pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 \equiv d^2b^2+db^2+b^2 \equiv b^2(d^2+d+1)\equiv 0 \pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-a\neq b$, otherwise $d\equiv -1\pmod{p^2}$ contradicts $d^2+d+1\equiv 0\pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.
$$\tag*{$\square$}$$
This completes the proof.
Extra. Let $0<a,b<p$ and
$$a^2+ab+b^2\equiv 0 \pmod{p^2}$$
for some odd $p$. Then
$$a^2+ab+b^2 = p^2$$
Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2\equiv 0 \pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 \equiv 0 \pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2\equiv 0 \pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$\tag*{$\square$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is it possibile to obtain the sum of the following series without using hypergeometric functions? I know that the following series:
$$
\sum_{n=1}^{+\infty}\frac{ (n!)^2}{(2n)!}
$$
converges. If I plug it in Wolphram Alpha, I can see that its sum is
$$
\frac{1}{27} \left(9 + 2 \sqrt{3} \pi\right).
$$
Is it possibile to obtain it without the use of the hypergeometric functions?
| I started from the fact that
$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt{\pi}$
}
Divide both side by $n!$ and express $\frac{(n!)^2}{(2n)!}$
We get
$\frac{(n!)^2}{(2n)!}=\frac{n!}{4^n \Gamma(n+\frac{1}{2})}\sqrt\pi=\frac{\Gamma(n+1)}{4^n \Gamma(n+\frac{1}{2})}\Gamma(\frac{1}{2})$
Introduce $\beta$ function and take the sum of both sides from $0$ to $\infty$
$S=\sum\limits_{n=1}^\infty\frac{(n!)^2}{(2n)!}=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\beta(n+1,\frac{1}{2})-1$
Based on the definition of $\beta$ function can be written
$S=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\int\limits_0^1\frac{t^n}{ \sqrt{1-t}} dt-1=\int\limits_0^1\sum\limits_{n=0}^\infty\frac{2n+1}{2^{2n+1}}\frac{t^n}{ \sqrt{1-t}}dt-1$
The integral can be devided into two parts:
$\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty n(\frac{t}{4})^n dt+\frac{1}{2}\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty (\frac{t}{4})^n dt-1 $
Take $\frac{t}{4}\sum\limits_{n=0}^\infty n(\frac{t}{4})^{n-1}=t\frac{d}{dt}\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4t}{(4-t)^2}$ and
$\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4}{4-t}$ into account we get:
$S=\int\limits_0^1\frac{2(4+t)}{\sqrt{1-t}(4-t)^2}dt-1$
Let $x=\sqrt{1-t}$ then $S=4\int\limits_0^1 \frac{(5-x^2)}{(3+x^2)}dx-1$
Forming the integral in the following way:
$S=\frac{8}{9}\int\limits_0^1\frac{1}{(1+(\frac{x}{\sqrt3})^2)^2}dx+\frac{4}{3}\int\limits_0^1\frac{1-(\frac{x}{\sqrt3})^2}{(1+(\frac{x}{\sqrt3})^2)^2}dx-1$
Applying the following substitution:$\frac{x}{\sqrt3}=\tan \theta$ we receie:
$S=\frac{8\sqrt3}{9}\int\limits_0^{\frac{\pi}{6}}\cos^2 \theta d\theta+\frac{4\sqrt3}{3}\int\limits_0^{\frac{\pi}{6}}\cos2\theta d\theta-1=\frac{2\sqrt3\pi}{27}+\frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
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