Prove $sum_{n=1}^inftyfrac{1}{(2n-1)^4}=frac{pi^4}{96}$ [closed]
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Prove using Parseval identity applied to the functions: $x,,|x|, x^2$ the convergence of the sum:
$$sum_{n=1}^inftyfrac{1}{(2n-1)^4}=frac{pi^4}{96}tag1$$
My attempt:
The identity of Parseval is:
$$2a_0^2+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)$$
where $a_0,a_n,b_n $ are coefficients of fourier series.
Here i'm a little stuck. Can someone help me?
pde fourier-series parsevals-identity
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
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closed as off-topic by Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove using Parseval identity applied to the functions: $x,,|x|, x^2$ the convergence of the sum:
$$sum_{n=1}^inftyfrac{1}{(2n-1)^4}=frac{pi^4}{96}tag1$$
My attempt:
The identity of Parseval is:
$$2a_0^2+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)$$
where $a_0,a_n,b_n $ are coefficients of fourier series.
Here i'm a little stuck. Can someone help me?
pde fourier-series parsevals-identity
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
$endgroup$
closed as off-topic by Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove using Parseval identity applied to the functions: $x,,|x|, x^2$ the convergence of the sum:
$$sum_{n=1}^inftyfrac{1}{(2n-1)^4}=frac{pi^4}{96}tag1$$
My attempt:
The identity of Parseval is:
$$2a_0^2+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)$$
where $a_0,a_n,b_n $ are coefficients of fourier series.
Here i'm a little stuck. Can someone help me?
pde fourier-series parsevals-identity
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
$endgroup$
Prove using Parseval identity applied to the functions: $x,,|x|, x^2$ the convergence of the sum:
$$sum_{n=1}^inftyfrac{1}{(2n-1)^4}=frac{pi^4}{96}tag1$$
My attempt:
The identity of Parseval is:
$$2a_0^2+sum_{n=1}^infty (a_n^2+b_n^2)=frac{1}{L}int_{-L}^{L}f^2(x)$$
where $a_0,a_n,b_n $ are coefficients of fourier series.
Here i'm a little stuck. Can someone help me?
pde fourier-series parsevals-identity
pde fourier-series parsevals-identity
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
edited Dec 31 '18 at 0:49
Martín Vacas Vignolo
3,796623
3,796623
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
asked Dec 31 '18 at 0:36
Bvss12Bvss12
1,821719
1,821719
closed as off-topic by Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz Dec 31 '18 at 8:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, B. Mehta, RRL, ancientmathematician, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
add a comment |
$begingroup$
let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
add a comment |
$begingroup$
let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
let $f=x^2$ and evaluate all the fourier coefficients and the integral in parsevals identity. move things around and you will get the sum from $n=1$ to infinity of $1/n^4$, or zeta(4). Notice that your sum is zeta(4) with only odd indices. You can find this given zeta(4) if you realize that the even indices in zeta(4) are equal to $1/16*zeta(4)$. So your sum is $(1-1/16)*zeta(4)=15/16*zeta(4)=pi^4 /96$
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
edited Dec 31 '18 at 0:50
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
answered Dec 31 '18 at 0:45
Abhishek VangipuramAbhishek Vangipuram
694
694
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
add a comment |
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
$begingroup$
$LaTeX text{ Tip}:$ Use zeta, pi and frac {a}{b} to obtain $zeta$, $pi$ and $frac ab$
$endgroup$
– Mohammad Zuhair Khan
Dec 31 '18 at 6:12
add a comment |
