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

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MATH307 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  Q6 (e)  •  Q6 (f)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  

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Question 06 (c)
Explain how the orthogonality of the functions $e^{2\pi inx}$ allows you to relate the inner product in part (b) to the sum $\sum _{n=-\infty }^{\infty }|c_{n}|^{2}$ Use your answer to to calculate the infinite sum $\sum _{n=1}^{\infty }n^{-2}$

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Solution

According to Parseval's identity it holds that $\sum _{n=-\infty }^{\infty }|c_{n}|^{2}=\int _{0}^{1}|f(x)|^{2}dx$ where cn 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 $\sum _{n=1}^{\infty }n^{-2}={\frac {\pi ^{2}}{6}}$

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