Science:Math Exam Resources/Courses/MATH307/December 2012/Question 04 (d)

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MATH307 December 2012

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

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Question 04 (d)
Consider the Fourier series $t^{2}-t=\sum _{n=-\infty }^{\infty }c_{n}e^{2\pi int}$ for $0\leq t\leq 1$. (d) Given that $c_{n}={\frac {1}{2\pi ^{2}n^{2}}}$ for $n\not =0$, use Parseval's formula to find the value of the infinite sum $\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}$. Calculate a numerical expression - you do not need to simplify your answer.

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Hint
Parseval's formula says that $\int _{0}^{1}|f(t)^{2}|dt=\left\langle f,f\right\rangle =\sum _{n=-\infty }^{\infty }|c_{n}|^{2}$ Calculate the integral on the left hand side and plug in the given and calculated values for cn on the right hand side. Then, isolate for $\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}$

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Solution

Consider the Fourier series $t^{2}-t=\sum _{n=-\infty }^{\infty }c_{n}e^{2\pi int}$ for $0\leq t\leq 1$. Parseval’s formula says $\int _{0}^{1}|f(t)^{2}|dt=\left\langle f,f\right\rangle =\sum _{n=-\infty }^{\infty }|c_{n}|^{2}$ Plugging $\displaystyle f(t)=t^{2}-t$ in the left hand side we calculate $\left\langle f,f\right\rangle =\int _{0}^{1}(t^{4}-2t^{3}+t^{2})dt={\frac {1}{5}}-{\frac {1}{2}}+{\frac {1}{3}}={\frac {1}{30}}$ Plugging in the given and calculated values for cn in the right hand side we compute ${\begin{aligned}\sum _{n=-\infty }^{\infty }|c_{n}|^{2}&=\sum _{n=-\infty }^{-1}|c_{n}|^{2}+|c_{0}|^{2}+\sum _{n=1}^{\infty }|c_{n}|^{2}\\&=\sum _{n=-\infty }^{-1}{\frac {1}{4\pi ^{4}n^{4}}}+\left|-{\frac {1}{6}}\right|+\sum _{n=1}^{\infty }{\frac {1}{4\pi ^{4}n^{4}}}\\&={\frac {1}{36}}+2\sum _{n=1}^{\infty }{\frac {1}{4\pi ^{4}n^{4}}}\\&={\frac {1}{36}}+{\frac {1}{2\pi ^{4}}}\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}\end{aligned}}$ Combining the left hand side and the right hand side of Parseval's formula we can find ${\frac {1}{30}}-{\frac {1}{36}}={\frac {1}{180}}={\frac {1}{2\pi ^{4}}}\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}$ Therefore, the infinite sum is $\sum _{n=1}^{\infty }{\frac {1}{n^{4}}}={\frac {\pi ^{4}}{90}}$

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