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:

$\begin{array}{rl}{c}_0& =\frac{1}{2}\\ c_n& =\{\begin{array}{ll}0& \text{if }n\text{ is even, }n=0\\ \frac{-i}{\pi n}& \text{if }n\text{ is odd}\end{array}\end{array}$

This yields: $f(x)=c_0+\sum _{n\text{ odd, }n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{-i}{\pi n}{e}^{2\pi inx}$

Parseval’s Theorem states:

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

Computing the left side:

$\mathrm{L}\mathrm{H}\mathrm{S}=\underset{a}{\overset{b}{\int }}{|f(x)|}^2=\underset{0}{\overset{1/2}{\int }}{1}^2=\frac{1}{2}$

Computing the right side:

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

Here the sum of odds from $-\mathrm{\infty }$ to $\mathrm{\infty }$ was split into two sums from 1 to $\mathrm{\infty }$ and -1 to $-\mathrm{\infty }$. 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.

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

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 $\mathrm{\infty }$.

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

Using the left side that was computed above we get:

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

Rearranging this to get the final answer:

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