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

MATH307 April 2013

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Question Section 202 06 (b)
Find $\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{2}}}$ using Parseval`s formula.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}$