Science:Math Exam Resources/Courses/MATH307/December 2010/Question 06 (c)/Solution 1

According to Parseval's identity it holds that

$\sum _{n=-\infty }^{\infty }|c_{n}|^{2}=\int _{0}^{1}|f(x)|^{2}dx$

where c_n are the Fourier coefficients of ƒ(x). In part (b) we calculated the right hand side and found that the above equals to 1/3.

In part (a) we found that the Fourier coefficients are $c_{0}={\frac {1}{2}}$ and $c_{n}=-{\frac {1}{2\pi in}}$. Hence

${\begin{aligned}{\frac {1}{3}}&=\sum _{n=-\infty }^{\infty }|c_{n}|^{2}\\&=|c_{0}|^{2}+2\sum _{n=1}^{\infty }\left|c_{n}\right|^{2}\\&=\left|{\frac {1}{2}}\right|+2\sum _{n=1}^{\infty }\left|-{\frac {1}{2\pi in}}\right|^{2}\\&={\frac {1}{4}}+{\frac {1}{2\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}\end{aligned}}$

where we have noted that $\displaystyle {}|c_{-n}|^{2}=|c_{n}|^{2}$ in splitting up the sum. Solving for the remaining series yields the final answer

$\sum _{n=1}^{\infty }n^{-2}={\frac {\pi ^{2}}{6}}$