From part (a) the coefficients of the Fourier series were found to be:
![{\displaystyle {\begin{aligned}c_{0}&={\frac {1}{2}}\\c_{n}&={\begin{cases}0&{\text{if }}n{\text{ is even, }}n\not =0\\{\frac {-i}{\pi n}}&{\text{if }}n{\text{ is odd}}\end{cases}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6f27abc0deb6c4532ac48bb4fec23d59b2956a5f)
This yields:
Parseval’s Theorem states:
![{\displaystyle \int _{a}^{b}\left|f(x)\right|^{2}=\sum _{n=-\infty }^{\infty }\left|c_{n}\right|^{2}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6f1ae2bc4c0eefc7aa21cb3f818370f62ecc6a9e)
Computing the left side:
![{\displaystyle \mathrm {LHS} =\int _{a}^{b}\left|f(x)\right|^{2}=\int _{0}^{1/2}1^{2}={\frac {1}{2}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/97fbedf273116f47782edf2598c6124009905d97)
Computing the right side:
![{\displaystyle \mathrm {RHS} =\sum _{n=-\infty }^{\infty }\left|c_{n}\right|^{2}=({\frac {1}{2}})^{2}+\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}n^{2}}}+\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}(-n)^{2}}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/12edc014fae6f99c9a5714781589f33fe047c706)
Here the sum of odds from
to
was split into two sums from 1 to
and -1 to
. It can be observed that the second sum is the same as the first where
and since
, these two sums are equivalent.
![{\displaystyle \mathrm {RHS} =({\frac {1}{2}})^{2}+2(\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}n^{2}}})}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/996d844a5b9b96542928da402b99e0dc53451fa4)
To get the expression that we want we can make a substitution of variables to remove the odd restriction in our sum. Let
. It can be seen that for k=0,1,2,3,..., n will always be odd.
Note: Making this substitution will change summation range.
At
. So the new summation range will be
to
.
![{\displaystyle \mathrm {RHS} ={\frac {1}{4}}+2(\sum _{k=0}^{\infty }{\frac {1}{\pi ^{2}(2k+1)^{2}}})}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/dca7de1d21b3b704c6897e900d7488ade7bee18b)
Using the left side that was computed above we get:
![{\displaystyle {\frac {1}{2}}={\frac {1}{4}}+{\frac {2}{\pi ^{2}}}(\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}})}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ad8842bd76333681d02b3dcd2d035d8353285407)
Rearranging this to get the final answer:
![{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0c80092abe56dcc60b9703266661ad4ca09f3147)