Exercise 8.13

Question

Exercise 13: 
    Put $f(x)=x$ if $0\le x<2\pi$, and apply Parseval's theorem to conclude that $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}.$$

analambanomenos · Accepted Answer

The Fourier coefficients for $f$ are (letting $n
e 0$ and using $e^{i2\pi n}=1$)
    \begin{align*}
        c_0 &= \frac{1}{2\pi}\int_0^{2\pi}x\,dx \
        &= \frac{1}{2\pi}\bigg(\frac{(2\pi)^2-0}{2}\bigg) \
        &= \pi \
        c_n &= \frac{1}{2\pi}\int_0^{2\pi}xe^{-inx}\,dx \
        &= \frac{1}{2\pi}\bigg(\frac{2\pi}{-in}-0\bigg)-\frac{1}{2\pi}\cdot\frac{1}{-in}\int_0^{2\pi}e^{-inx}\,dx \
        &= -\frac{1}{in}+\frac{1}{2\pi in}\cdot\frac{1-1}{-in} \
        &= \frac{i}{n}.
    \end{align*}
    Hence by Parseval's theorem
    \begin{align*}
        \sum_{-\infty}^\infty|c_n|^2 &= \frac{1}{2\pi}\int_0^{2\pi}x^2\,dx \
        \pi^2 + 2\sum_1^\infty\frac{1}{n^2} &= \frac{4\pi^2}{3} \
        \sum_1^\infty\frac{1}{n^2} &= \frac{1}{2}\bigg(\frac{4\pi^2}{3}-\pi^2\bigg) = \frac{\pi^2}{6}
    \end{align*}

Amit Awasthi · Answer

\begin{proof}
      First find the Fourier coefficients $c_n$ for $f$. For $n 
e 0$, using integration by parts,
      \begin{align*}
        c_n &= \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-inx} dx\
        &= \frac{1}{2\pi} \left( 2\pi \frac{e^{-2\pi in}}{-in} - \int_0^{2\pi} \frac{e^{-inx}}{-in} dx ight) = \frac{i}{n}.
      \end{align*}
      Then, for $n = 0$,
      \begin{equation*}
        c_0 = \frac{1}{2\pi} \int_0^{2\pi} f(x) dx = \pi.
      \end{equation*}
      Now apply Parseval's theorem. This gives us the equation:
      \begin{align*}
        \frac{1}{2\pi} \int_0^{2\pi} |f(x)| dx &= \sum_{-\infty}^\infty |c_n|^2\
        \frac{4\pi^2}{3} &= \sum_{|n| 
e 0} \frac{1}{n^2} + \pi^2\
        \frac{\pi^2}{6} &= \sum_{n = 1}^\infty \frac{1}{n^2}.\qedhere
      \end{align*} 
\end{proof}

Exercise 8.13

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Exercise 8.13

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