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Residues 2

  1. Feb 10, 2009 #1
    residues 2 :)

    1. The problem statement, all variables and given/known data

    Ok, so I have this function and I need to find the residues for it.
    I developed the function into Laurent series and my question is, how can I find a(-1) which is the residues?

    Thanks in advance.


    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 10, 2009 #2

    HallsofIvy

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    Re: residues 2 :)

    The function is
    [tex]f(z)= cos(\frac{1}{z-2})[/tex]
    and you correctly used the McLaurin series for cos(u), with u= 1/(z-2), to write the Laurent series
    [tex]\sum_{n= 0}^\infty \frac{(-1)^n}{(2n)!} (z-2)^{-2n}[/tex]
    Since the power of z- 2 is always even, there is no (z-2)-1 term: the coefficient is 0.

    HOWEVER, because that series has an infinite number of negative powers with non-zero coefficient, z= 2 is an essential singularity, not a pole. There is NO residue.
     
  4. Feb 10, 2009 #3
    Re: residues 2 :)

    Oh, so an essential singularity has no residue?

    Thanks.
     
  5. Feb 10, 2009 #4
    Re: residues 2 :)

    How about this one?

    Thanks again.
     

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  6. Feb 10, 2009 #5

    HallsofIvy

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    Re: residues 2 :)

    I am informed that, according to Wikipedia, I am wrong about this. That is, that the residue is still the coefficient of the -1 power in the Laurent series. The difference is just that the formula for the residue, taking the nth derivative where n is the order of the pole, cannot be used but you still can find it by finding the Laurent series itself. In the case of your first problem, the residue is 0 and for your new problem, it is 1.
     
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