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Thread: A Fourier series problem from Rudin

  1. #1
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    A Fourier series problem from Rudin

    Hello,

    I am trying to prove some problem out of Rudin, chapter 8, problem 12, part d). In particular,

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

    You need part c) before the question itself makes sense,

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

    The proof of c) is not so bad.

    Parseval's theorem, specifically line (85) on page 191 says that: $$\frac{1}{2\pi}\int_{-\delta}^{\delta}|f(x)|^2 dx=\sum_{-\infty}^{\infty}|c_m|^2$$ We can rewrite this using the real form of the Fourier series, i.e., if $f(x)\sim \frac{a_0}{2}+\sum_{m=1}^\infty(a_m\cos(mx)+b_m\si n(mx))$, then $$\frac{1}{\pi}\int_{-\delta}^{\delta}f^2(x)dx=\frac{a_0^2}{2}+\sum_{m=1 }^\infty(a_m^2+b_m^2)$$ Here, since the $b_m$s are zero we have, $$\frac{1}{\pi}\int_{-\delta}^{\delta}f^2(x)dx=\frac{a_0^2}{2}+\sum_{m=1 }^\infty(a_m^2)$$ Hence, $$\frac{2\delta}{\pi}=\frac{1}{\pi}\int_{-\delta}^{\delta}f^2(x)dx=\frac{a_0^2}{2}+\sum_{m=1 }^\infty (a_m^2)=\frac{4\delta^2}{2\pi^2}+\sum_{m=1}^\infty \Big(\frac{2\sin(m\delta)}{\pi m}\Big)^2=\frac{\delta^2}{2\pi^2}+\sum_{m=1}^{\inf ty} \frac{2^2\sin^2(m\delta)}{\pi^2 m^2}$$ Which we can rewrite as, \begin{equation*}\begin{split}\frac{2\delta}{\pi}& =\frac{4\delta^2}{2\pi^2}+\sum_{m=1}^\infty \frac{4\sin^2(m\delta)}{\pi^2m^2}\\ \frac{\pi^2}{4\delta}\frac{2\delta}{\pi}&=\frac{4\ delta^2}{2\pi^2}\frac{\pi^2}{4\delta}+\sum_{m=1}^\ infty \frac{4\sin^2(m\delta)}{\pi^2m^2}\frac{\pi^2}{4\de lta}\\ \frac{\pi}{2}&=\frac{\delta}{2}+\sum_{m=1}^\infty \frac{\sin^2(m\delta)}{\delta m^2}\\ \frac{\pi-\delta}{2}&=\sum_{m=1}^\infty \frac{\sin^2(m\delta)}{\delta m^2}\end{split}\end{equation*}

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

    The idea is supposed to be to let the partial sums of this integral be the LHS of the equality proved in part c), then hopefully blah blah blah my way to replacing something with $\frac{\pi}{2}$. I'm having trouble justifying this or working at all with this stuff. I should have to evaluate

    $$\lim_{t \to \infty} \int_0^t \Big(\frac{\sin x}{x}\Big)^2 dx$$

    which has Riemann sums, $\sum_{i=1}^n \Big(\frac{\sin(nt/n)}{\sqrt{nt/n}}\Big)^2\Delta(t)$?

    So I'm not really sure how to justify this, and even then what to do next.
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  2. #2
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    Re: A Fourier series problem from Rudin

    $\begin{align*}
    &\sum \limits_{m=1}^\infty ~\dfrac{\sin^2(m \delta)}{m^2 \delta} = \\

    &\sum \limits_{m=1}^\infty ~\dfrac{\sin^2(m \delta)}{m^2 \delta^2}~\delta = \\

    &\sum \limits_{m=1}^\infty ~\left(\dfrac{\sin(m \delta)}{m \delta} \right)^2~\delta

    \end{align*}$

    and this is recognized as the Riemann sum of

    $\displaystyle \int_0^\infty~\left(\dfrac{\sin(x)}{x}\right)^2~dx $
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  3. #3
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    Re: A Fourier series problem from Rudin

    So then showing d) means to show that for every fixed $\varepsilon>0$ we can pick some $n^*$ such that for all $n\geq n^*$

    $$\Bigg|\int_0^t \Big(\frac{\sin(x)}{x}\Big)^2 dx- \sum_{m=1}^n \Big(\frac{\sin(m\delta)}{m\delta}\Big)^2\delta \Bigg|<\varepsilon$$
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  4. #4
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    Re: A Fourier series problem from Rudin

    No sorry this means that for every $\varepsilon>0$ we can FIND an $n^*$ such that for all $n\geq n^*$ we have $$\Bigg|\int_0^t \Big(\frac{\sin(x)}{x}\Big)^2 dx- \sum_{m=1}^n \Big(\frac{\sin(m\delta)}{m\delta}\Big)^2\delta \Bigg|<\varepsilon$$
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