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  1. #1
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    series

    summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2


    can neone prove this??
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  2. #2
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    Quote Originally Posted by durham2 View Post
    summation n=0 to infinity r^ncosn(theta)=rcos(theta)-r^2/1-2rcos(theta)+r^2


    can neone prove this??

    This is true iff |r|< 1, so that then we have the sum of a geometric sequence with quotient less than 1 in absolute value:  r^ne^{n i\theta}=r^n(\cos n\theta + i\sin n\theta), \mbox{using DeMoivre's theorem, so}

    \sum\limits_{n=0}^\infty(re^{i\theta})^n=\frac{1}{  1-re^{i\theta}}=\frac{1-re^{-i\theta}}{r^2+1-2r\cos \theta} <--correction done here

    Now just separate real and imaginary parts above.

    Tonio
    Last edited by tonio; October 27th 2009 at 01:58 PM.
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