Science:Math Exam Resources/Courses/MATH307/April 2013/Question Section 202 06 (b)/Solution 1

From part (a) the coefficients of the Fourier series were found to be:

${\begin{aligned}c_{0}&={\frac {1}{2}}\\c_{n}&={\begin{cases}0&{\text{if }}n{\text{ is even, }}n\not =0\\{\frac {-i}{\pi n}}&{\text{if }}n{\text{ is odd}}\end{cases}}\end{aligned}}$

This yields: $f(x)=c_{0}+\sum _{n{\text{ odd, }}n=-\infty }^{\infty }{\frac {-i}{\pi n}}e^{2\pi inx}$

Parseval’s Theorem states:

$\int _{a}^{b}\left|f(x)\right|^{2}=\sum _{n=-\infty }^{\infty }\left|c_{n}\right|^{2}$

Computing the left side:

$\mathrm {LHS} =\int _{a}^{b}\left|f(x)\right|^{2}=\int _{0}^{1/2}1^{2}={\frac {1}{2}}$

Computing the right side:

$\mathrm {RHS} =\sum _{n=-\infty }^{\infty }\left|c_{n}\right|^{2}=({\frac {1}{2}})^{2}+\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}n^{2}}}+\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}(-n)^{2}}}$

Here the sum of odds from $-\infty$ to $\infty$ was split into two sums from 1 to $\infty$ and -1 to $-\infty$. It can be observed that the second sum is the same as the first where $n=-n$ and since $(-n)^{2}=n^{2}$, these two sums are equivalent.

$\mathrm {RHS} =({\frac {1}{2}})^{2}+2(\sum _{n{\text{ odd, }}n=1}^{\infty }{\frac {1}{\pi ^{2}n^{2}}})$

To get the expression that we want we can make a substitution of variables to remove the odd restriction in our sum. Let $n=2k+1$. 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 $n=1,1=2k+1\Rightarrow k=0$. So the new summation range will be $k=0$ to $\infty$.

$\mathrm {RHS} ={\frac {1}{4}}+2(\sum _{k=0}^{\infty }{\frac {1}{\pi ^{2}(2k+1)^{2}}})$

Using the left side that was computed above we get:

${\frac {1}{2}}={\frac {1}{4}}+{\frac {2}{\pi ^{2}}}(\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}})$

Rearranging this to get the final answer:

$\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}$