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Math Help - How to find the order of the pole

  1. #1
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    How to find the order of the pole

    Find the order of the pole f(z) = 1/(2cosz-2+z^2)^2 at z= 0
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  2. #2
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    Quote Originally Posted by essedra View Post
    Find the order of the pole f(z) = 1/(2cosz-2+z^2)^2 at z= 0
    Lets analyze the denominator without the square

    2\cos(z)-2+z^2=-2+z^2+2\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}=2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}

    z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!} \right)=z^4g(z)

    Where g(0) \ne 0

    So now your fuction looks like
    \frac{1}{[z^4g(z)]^2}=\frac{1}{z^8}\cdot \frac{1}{[g(z)]^2}

    where the 2nd factor is non zero at zero.
    Last edited by TheEmptySet; June 5th 2011 at 04:10 PM. Reason: error
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  3. #3
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    Re: How to find the order of the pole

    Hi.

    Why have you gone from  2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}

    to  z^4\left(2\sum_{n=2}^{\infty}\frac{(-1)^nz^{2n-4}}{(2n)!}\right) = z^4g(z)

    and why can't g(0) be 0. Thanks
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: How to find the order of the pole

    Here is z=0 as pole of order 8

    Attached Thumbnails Attached Thumbnails How to find the order of the pole-poles.png  
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