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Math Help - Compute a sum

  1. #1
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    Compute a sum

    Hi! I need to show that
    \frac{8}{\pi^2} \sum_{k=0}^{\infty}\frac{\sin(\frac{1}{2}(2k+1)\pi t) \sin(\frac{1}{2}(2k+1)\pi s)}{(2k+1)^2}= \min(t,s) for t,s \in (0,1) .
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  2. #2
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    Quote Originally Posted by DavidEriksson View Post
    Hi! I need to show that
    \frac{8}{\pi^2} \sum_{k=0}^{\infty}\frac{\sin(\frac{1}{2}(2k+1)\pi t) \sin(\frac{1}{2}(2k+1)\pi s)}{(2k+1)^2}= \min(t,s) for t,s \in (0,1) .
    This looks like the sort of sum that arises in Fourier theory, so I looked for a function that might have this as its Fourier series.

    Let 0<T<\pi/2, and define a function by f(x) = \begin{cases}x&(0\leqslant x\leqslant T),\\ T&(T\leqslant x\leqslant\pi-T),\\ \pi-x&(\pi-T\leqslant x\leqslant\pi).\end{cases}

    Make f into an odd function on the interval [-\pi,\pi] by defining f(-x)=-f(x). Notice that f(x) = \min\{x,T\} when x\in [0,\pi/2].

    Since f is odd, its Fourier cosine coefficients will all be 0. The sine coefficients will be given by

    b_n = \displaystyle\frac2\pi\Bigl(\int_0^T\!\!x\sin nx\,dx + \int_T^{\pi-T}\!\!\!T\sin nx\,dx + \int_{\pi-T}^\pi(\pi-x)\!\sin nx\,dx\Bigr).

    Grind out thtose integrals, using integration by parts as necessary. You will find that b_n = \tfrac{4\sin nT}{\pi n^2} if n is odd, and b_n=0 if n is even. The function f is continuous and piecewise smooth, so it is the sum of its Fourier series. Therefore f(x) = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k  +1)T\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)x\bigr).

    Now put x = \tfrac12\pi s and T = \tfrac12\pi t to see that \min\{\tfrac12\pi s,\tfrac12\pi t\} = \displaystyle\sum_{k=0}^\infty\frac{4\sin\bigl((2k  +1)\tfrac12\pi t\bigr)}{\pi(2k+1)^2}\sin\bigl((2k+1)\tfrac12\pi s\bigr). That's close enough to the answer that I needn't continue.
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