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Math Help - confusing limit

  1. #1
    Junior Member
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    confusing limit

    What is
    <br />
\lim_{n\to \infty} \frac{\pi}{2^n} \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})

    I'm getting 0 but I think i'm going wrong somewhere. Can anyone give me an idea?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by nahduma View Post
    What is
    <br />
\lim_{n\to \infty} \frac{\pi}{2^n} \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})

    I'm getting 0 but I think i'm going wrong somewhere. Can anyone give me an idea?
    What about this: \sin\frac{j\pi}{2^n} is the imaginary part of \left(\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}\right  )^j. Thus, if you make a detour through complex numbers, you can explicitly evaluate \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n}), because you can then apply the formula for a geometric series.
    Last edited by Failure; April 29th 2010 at 10:51 AM.
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  3. #3
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    Answer please

    Thanks. Now I'm getting 2 as the answer.
    Can anyone confirm that?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by nahduma View Post
    Thanks. Now I'm getting zero (which I think is the answer).
    Can someone please confirm that?
    No, I'm sorry but I get the answer 2. Quite possibly I am mistaken, therefore let me give you some of the my steps:

    \frac{\pi}{2^n}\sum_{k=1}^{2^n}\mathrm{e}^{\mathrm  {i}\frac{k\pi}{2^n}}=\frac{\pi}{2^n}\left[\frac{\mathrm{e}^{\mathrm{i}\left(\pi+\frac{\pi}{2  ^n}\right)}-1}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}-1\right]
    =\frac{\pi}{2^n}\left[\frac{-\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}-1\right]=-\frac{\pi}{2^n}\frac{2\mathrm{e}^{\mathrm{i}\frac{  \pi}{2^n}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}
    =-\frac{\pi}{2^n}\frac{2\mathrm{e}^{\mathrm{i}\frac{  \pi}{2^n}}}{\mathrm{i}\frac{\pi}{2^n}+o\left(\math  rm{i}\frac{\pi}{2^n}\right)}\rightarrow 2\mathrm{i}

    where I have used Landau's little-o. Now this last expression seems to have the limit 2i, for n\rightarrow\infty, and the imaginary part of this would be 2.

    P.S: I have confirmed that a numeric approximation quite rapidly converges to this solution.
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  5. #5
    MHF Contributor
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    Another way of looking at this problem is that it consists of a Riemann sum converging to the integral \int_0^\pi\!\!\!\sin x\,dx = 2.
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