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

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

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