Exercise 8.12

Question

Exercise 12: 
    Suppose $0<\delta<\pi$, $f(x)=1$ if $|x|\le\delta$, $f(x)=0$ if $\delta<|x|\le\pi$, and $f(x+2\pi)=f(x)$ for all $x$.

(a) Compute the Fourier coefficients of $f$.

(b) Conclude that $$\sum_{n=1}^\infty \frac{\sin(n\delta)}{n}=\frac{\pi-\delta}{2}.$$

(c) Deduce from Parseval's theorem that $$\sum_{n=1}^\infty \frac{\sin^2(n\delta)}{n^2\delta}=\frac{\pi-\delta}{2}.$$

(d) Let $\delta\rightarrow 0$ and prove that $$\int_0^\infty\bigg(\frac{\sin x}{x}\bigg)^2\,dx=\frac{\pi}{2}.$$

(e) Put $\delta=\pi/2$ in (c). What do you get?

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

(a) For $n
e 0$
    \begin{align*}
        c_n &= \frac{1}{2\pi}\int_{-\delta}^\delta e^{-inx}\,dx \
        &= \frac{1}{2\pi}\cdot\frac{1}{in}(e^{in\delta}-e^{-in\delta}) \
        &= \frac{\sin n\delta}{\pi n} \
        c_0 &= \frac{1}{2\pi}\int_{-\delta}^\delta 1\,dx \
        &= \frac{\delta}{\pi}
    \end{align*}

(b) By Theorem 8.14, the Fourier series for $f(x)$ converges for $x=0$ to $f(0)=1$, hence (noting that $(\sin x)/x$ is an even function)
    \begin{align*}
        1 &= \frac{\delta}{\pi}+\sum_{|n|
e 0}\frac{\sin n\delta}{\pi n} \
        1-\frac{\delta}{\pi} &= \frac{2}{\pi}\sum_{n=1}^\infty\frac{\sin n\delta}{n} \
        \frac{\pi-\delta}{2} & = \sum_{n=1}^\infty\frac{\sin n\delta}{n}
    \end{align*}
    
    (c) We have by Parseval's theorem, using $c_{-n}=c_{n}$ for $n
e 0$,
    \begin{align*}
        \frac{1}{2\pi}\int_{-\delta}^\delta\big|f(x)\big|^2\,dx &= \sum_{-\infty}^\infty|c_n|^2 \
        \frac{1}{2\pi}\int_{-\delta}^\delta 1\,dx &= \frac{\delta^2}{\pi^2}+2\sum_{n=1}^\infty\frac{\sin^2n\delta}{\pi^2n^2} \
        \frac{\delta}{\pi}-\frac{\delta^2}{\pi^2} &= \frac{2\delta}{\pi^2}\sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta} \
        \frac{\pi^2}{2\delta}\cdot\frac{\delta(\pi-\delta)}{\pi^2} &= \sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta} \
        \frac{\pi-\delta}{2} &= \sum_{n=1}^\infty\frac{\sin^2n\delta}{n^2\delta}
    \end{align*}

(d) First note that by L'Hospital's Rule, $$\lim_{xightarrow 0}\frac{\sin x}{x}=\lim_{xightarrow 0}\frac{\cos x}{x}=1,$$ so that the integral $$\int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx$$
    is well-defined. Also, for $\delta>0$,
    \begin{align*}
        \lim_{Aightarrow\infty}\int_\delta^A\bigg|\frac{\sin x}{x}\bigg|^2\,dx &\le \lim_{Aightarrow\infty}\int_\delta^A\frac{1}{x^2}\,dx \
        &= \lim_{Aightarrow\infty}\bigg(\frac{1}{\delta}-\frac{1}{A}\bigg) \
        &= \frac{1}{\delta}
    \end{align*}
    so that the improper integral converges. That is, if $\varepsilon>0$, there is an $A>0$ large enough so that $$\Bigg|\int_0^\infty\bigg(\frac{\sin x}{x}\bigg)^2\,dx-
    \int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx\Bigg|\le\frac{\varepsilon}{4}.$$

Let $\delta=A/M$ for some large integer $M$, and let $P=\{0,\delta,\ldots,M\delta=A\}$ be a partition of $[0,A]$. Then $$\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta^2}\cdot\delta=
    \sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}$$ is a Riemann sum of the finite integral. Hence there is an $M$ large enough so that $$\Bigg|\int_0^A\bigg(\frac{\sin x}{x}\bigg)^2\,dx-
    \sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}\Bigg|\le\frac{\varepsilon}{4}.$$

From part (c), we can make $M$ large enough so that $$\Bigg|\sum_{n=1}^M\frac{\sin^2(n\delta)}{n^2\delta}-\frac{\pi-\delta}{2}\Bigg|\le\frac{\varepsilon}{4}$$ and we can make $M$ large enough
    so that $\delta/2\le\varepsilon/4$, so that $$\bigg|\frac{\pi-\delta}{2}-\frac{\pi}{2}\bigg|\le\frac{\varepsilon}{4}.$$

Putting together these four inequalities, we get that, for all $\varepsilon>0$, $$\Bigg|\int_0^\infty\bigg(\frac{\sin x}{x}\bigg)^2\,dx-\frac{\pi}{2}\Bigg|\le\varepsilon,$$ so that
    the equality follows.

(e) Since $\sin^2(n\pi/2)=1$ if $n$ is odd, and 0 otherwise, we get from part (c) that $$\sum_{n=1}^\infty\frac{1}{(2n-1)^2}=\frac{\pi}{2}\cdot\frac{\pi-(\pi/2)}{2}=\frac{\pi^2}{8}.$$

Exercise 8.12

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Exercise 8.12

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