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Math Help - power of poles

  1. #1
    MHF Contributor
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    power of poles

    f=\frac{z}{1-\cos z}
    the singular points are z=2pik
    and poles because there limit is infinity
    now i want to determine te power of the pole
    g=1/f= \frac{1-\cos z}{z}
    g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}
    g'(2\pi k)=0
    g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}
    g''(2\pi k)=0

    the book says that its a second order pole
    which is not true
    because g''(2\pi k)=0
    where it should differ zero in order to be pole
    Last edited by transgalactic; July 13th 2010 at 01:04 AM.
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  2. #2
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    i have solved the 2pi k part

    for zero i have a problem

    f=\frac{z}{1-\cos z}

    the singular points are z=2pik and zero
    i solved for z=2pik
    and poles because there limit is infinity
    now i want to determine te power of the pole
    g=1/f= \frac{1-\cos z}{z}
    g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}
    g'(0)=0/0
    g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}
    g''(0)=0/0

    the book says that its a first order pole

    it should differ zero in order to be pole
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    f=\frac{z}{1-\cos z}
    the singular points are z=2pik
    and poles because there limit is infinity
    now i want to determine te power of the pole
    g=1/f= \frac{1-\cos z}{z}
    g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}
    g'(2\pi k)=0
    g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}
    g''(2\pi k)=0

    the book says that its a second order pole
    which is not true
    because g''(2\pi k)=0
    where it should differ zero in order to be pole
    It's always good to have different methods for doing the same thing, that way you can check if your answers coincide. I'm going to give you how I would do it, and then you figure out where your argument went wrong.

    Take g(z)=1-\cos(z) expand in Taylor series and you get that the first term is a_1z^2 form which it follows that at 0 f has a first order pole. Since \cos is periodic and the numerator doesn't vanish at any point other than 0, we get that the poles for 2\pi k with k\neq 0 are of order 2
    Last edited by Jose27; July 13th 2010 at 08:38 AM.
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  4. #4
    A Plied Mathematician
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    Reply to Jose27: I think you meant, "... and the numerator doesn't vanish...", right?
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