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Power of poles

  1. Jul 13, 2010 #1
    [tex]f=\frac{z}{1-\cos z}[/tex]

    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=[tex]\frac{1-\cos z}{z}[/tex]
    [tex]g'=\frac{(-\sin z)z-(1-\cos z)}{z^2}[/tex]
    [tex]g'(0)=0/0[/tex]
    [tex]g''=\frac{-\sin z z^2 -(cos z -1)2z}{z^4}[/tex]
    [tex]g''(0)=0/0[/tex]

    the book says that its a first order pole

    it should differ zero in order to be pole
     
    Last edited: Jul 13, 2010
  2. jcsd
  3. Jul 13, 2010 #2

    lanedance

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    once again, try expanding the cosine as a taylor series about zero, should help you see what is happening
     
  4. Jul 13, 2010 #3
    but i want to solve it this way
    where did i go wrong in this way

    i want to solve it by the derivative way
    not by developing into a series
     
    Last edited: Jul 13, 2010
  5. Jul 13, 2010 #4

    vela

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    0/0 is an indeterminate form. You need to use the hospital rule to get an actual value for the limit.
     
  6. Jul 13, 2010 #5
    but its not a limit
    its a derivative
     
  7. Jul 13, 2010 #6

    vela

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    You can't just plug in 0 to evaluate g'(0) because you get an indeterminate form. You have to find the limit of g'(z) as zā†’0.
     
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