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Thread: confusing limit

  1. #1
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    confusing limit

    What is
    $\displaystyle
    \lim_{n\to \infty} \frac{\pi}{2^n} \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})$

    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
    $\displaystyle
    \lim_{n\to \infty} \frac{\pi}{2^n} \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})$

    I'm getting 0 but I think i'm going wrong somewhere. Can anyone give me an idea?
    What about this: $\displaystyle \sin\frac{j\pi}{2^n}$ is the imaginary part of $\displaystyle \left(\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}\right )^j$. Thus, if you make a detour through complex numbers, you can explicitly evaluate $\displaystyle \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})$, because you can then apply the formula for a geometric series.
    Last edited by Failure; Apr 29th 2010 at 09:51 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:

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

    where I have used Landau's little-o. Now this last expression seems to have the limit 2i, for $\displaystyle n\rightarrow\infty$, 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
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    Another way of looking at this problem is that it consists of a Riemann sum converging to the integral $\displaystyle \int_0^\pi\!\!\!\sin x\,dx = 2$.
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