Exercise 8.14

Question

Exercise 14: 
    If $f(x)=\big(\pi-|x|\big)^2$ on $[-\pi,\pi]$, prove that $$f(x)=\frac{\pi^2}{3}+\sum_{n=1}^\infty\frac{4}{n^2}\cos nx$$ and deduce that $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6},
    \quad\quad\sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}.$$

analambanomenos · Accepted Answer

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The maximum local rate of change of $f$ will occur at the multiples of $2\pi$, where $\lim|f^{\prime}|=2\pi$. Hence $f$ satisfies the condition of Theorem 8.14 with constant $M=2\pi$, so
    we can conclude that the Fourier series of $f$ converges to $f(x)$ for all $x$.

Here are the Fourier coefficients of $f$ (where $n
e 0$, and using $e^{in\pi}=e^{-in\pi}$ for all integers $n$):
    \begin{align*}
        c_0 &= \frac{1}{2\pi}\int_{-\pi}^0(\pi+x)^2\,dx+\frac{1}{2\pi}\int_0^{\pi}(\pi-x)^2\,dx \
        &= \frac{1}{2\pi}\bigg(\frac{\pi^3}{3}+\frac{\pi^3}{3}\bigg) \
        &= \frac{\pi^2}{3} \
        c_n &= \frac{1}{2\pi}\int_{-\pi}^0(\pi^2+2\pi x+x^2)e^{-inx}\,dx+\frac{1}{2\pi}\int_0^{\pi}(\pi^2-2\pi x+x^2)e^{-inx}\,dx \
        &= \frac{1}{2\pi}\int_{-\pi}^{\pi}(\pi^2+x^2)e^{-inx}\,dx+\int_{-\pi}^0xe^{-inx}\,dx-\int_0^{\pi} xe^{-inx}\,dx \
    \end{align*}
    \begin{align*}
        \frac{1}{2\pi}\int_{-\pi}^{\pi}(\pi^2+x^2)e^{-inx}\,dx &= \frac{1}{2\pi}\cdot\frac{i}{n}(2\pi^2e^{-in\pi}-2\pi^2e^{in\pi})-\frac{1}{2\pi}\cdot\frac{2i}{n}\int_{-\pi}^{\pi} xe^{-inx}\,dx \
        &= 0-\frac{i}{\pi n}\cdot\frac{i}{n}(\pi e^{-in\pi}+\pi e^{in\pi})+\frac{i}{\pi n}\cdot\frac{i}{n}\int_{-\pi}^{\pi} e^{-inx}\,dx \
        &= \frac{1}{n^2}(e^{-in\pi}+e^{in\pi})-\frac{i}{\pi n^2}(e^{-in\pi}-e^{in\pi}) \
        &= \frac{1}{n^2}(e^{-in\pi}+e^{in\pi}) \
    \end{align*}
    \begin{align*}
        \int_{-\pi}^0 xe^{-inx}\,dx &= \frac{i}{n}(0e^0+\pi e^{in\pi})-\frac{i}{n}\int_{-\pi}^0 e^{-inx}\,dx \
        &= \frac{i\pi}{n}e^{in\pi}-\frac{i}{n}\cdot\frac{i}{n}(e^0-e^{in\pi}) \
        &= \frac{i\pi}{n}e^{in\pi}+\frac{1}{n^2}(1-e^{in\pi}) \
    \end{align*}
    \begin{align*}
        \int_0^{\pi} xe^{-inx}\,dx &= \frac{i}{n}(\pi e^{-in\pi}-0e^0)-\frac{i}{n}\int_0^{\pi} e^{-inx}\,dx \
        &= \frac{i\pi}{n}e^{-in\pi}-\frac{i}{n}\cdot\frac{i}{n}(e^{-in\pi}-e^0) \
        &= \frac{i\pi}{n}e^{-in\pi}+\frac{1}{n^2}(e^{-in\pi}-1) \
    \end{align*}
    $$c_n=\frac{1}{n^2}(e^{-in\pi}+e^{in\pi})+\frac{i\pi}{n}e^{in\pi}+\frac{1}{n^2}(1-e^{in\pi})-\frac{i\pi}{n}e^{-in\pi}-\frac{1}{n^2}(e^{-in\pi}-1)=\frac{2}{n^2}$$
    Hence
    \begin{align*}
        f(x) &= \frac{\pi^2}{3}+\sum_{|n|
e 0}\frac{2}{n^2}e^{inx} \
        &= \frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac{2}{n^2}(e^{inx}+e^{-inx}) \
        &= \frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac{4}{n^2}\cos nx
    \end{align*}

Letting $n=0$, we get 
    \begin{align*}
        \frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^2} &= \pi^2 \
        \sum_{n=1}^{\infty}\frac{1}{n^2} &= \frac{\pi^2}{6}
    \end{align*}

And by Parseval's theorem, we get
    \begin{align*}
        \frac{\pi^4}{9}+2\sum_{n=1}^{\infty}\frac{4}{n^4} &= \frac{1}{2\pi}\int_{-\pi}^0(\pi+x)^4\,dx+\frac{1}{2\pi}\int_0^{\pi}(\pi-x)^4\,dx \
        \frac{\pi^4}{9}+8\sum_{n=1}^{\infty}\frac{1}{n^4} &= \frac{1}{2\pi}\cdot\frac{1}{5}(\pi^5+\pi^5) \
        \sum_{n=1}^{\infty}\frac{1}{n^4} &= \frac{1}{8}\bigg(\frac{\pi^4}{5}-\frac{\pi^4}{9}\bigg) \
        \sum_{n=1}^{\infty}\frac{1}{n^4} &= \frac{\pi^4}{90}
    \end{align*}

Exercise 8.14

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Exercise 8.14

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