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Math Help - The zeroes of a tricky continuous function

  1. #1
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    The zeroes of a tricky continuous function

    Here's a tricky one:
    I need to disprove that the function [f(x)=sum( sin(2*Pi*(x+1)/n) / (n*sin(Pi*x/n)*sin((x+2)*Pi/n), n=2..trunc(squareroot(x))
    =0] in every open interval (k, k+1) for large enough k and forever after. How?
    Last edited by AlexBot; November 10th 2010 at 06:16 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by AlexBot View Post
    Here's a tricky one:
    I need to disprove that the function [f(x)=sum(sin(2*Pi*(x+1)/n)/(n*sin(Pi*x/n)*sin((x+2)*Pi/n), n=2..trunc(sqrt(x))
    =0] in every open interval (k, k+1) for large enough k and forever after. How?
    \displaystyle f(x)=\sum_{n=2}^{\text{trunc}(\sqrt{x})}\sin\left(  \frac{2\pi(x+1)}{n}\right)\sin\left(\frac{\pi(x+2)  }{n}\right)?

    What is the truncation function? The floor function?
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  3. #3
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    no, there's a division sign in there and another term. I don't really know how to type using TEX. Sorry. ANy ideas though?
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  4. #4
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    yes, the truncated function is the floor function.
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