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Math Help - Finding the residue of a given function

  1. #1
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    Finding the residue of a given function

    Hi

    I was wondering whether someone would be able to have a look at this question and my answer to see if I am correct. If I am not, or the solution requires more information, I would greatly appreciate it if you could show me how solve this problem.


    QUESTION:

    Find Res(g,0) for g(z) = z^-2 . cosh(z) (I mean z squared multiplied by cosh(z))




    MY SOLUTION

    = cosh(z) / z^2

    cosh(z) = 1 + z^2/2! + z^4/4! + z^6/6!

    cosh(z)/z^2 = 1/z^2 + 1/2! + z^2/4! + z^4/6! + .....

    This has a pole of order 2 at z = 0

    Res(g,0) = 1/2


    Thank you
    Doug
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  2. #2
    Super Member wingless's Avatar
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    No.. Do you remember what residue is?
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  3. #3
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    Quote Originally Posted by dougwm View Post
    Hi

    I was wondering whether someone would be able to have a look at this question and my answer to see if I am correct. If I am not, or the solution requires more information, I would greatly appreciate it if you could show me how solve this problem.


    QUESTION:

    Find Res(g,0) for g(z) = z^-2 . cosh(z) (I mean z squared multiplied by cosh(z))
    \cosh z = 1 + \frac{z^2}{2!}+... so \frac{\cosh z}{z^2} = \frac{1}{z^2}+\frac{1}{2!}+....
    The coefficient for z^{-1} is 0.
    Thus, what does this mean ?
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  4. #4
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    That it is a removable singularity at z = 0? Therefore the residue of this is 0?
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  5. #5
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    Quote Originally Posted by dougwm View Post
    That it is a removable singularity at z = 0? Therefore the residue of this is 0?
    It is not a removable singularity. It is a singularity it just happens to have the residue 0.
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  6. #6
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    How come the residue is 0? Is it directly from the definition?

    I dont understand how you would work out the residue for this equation
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  7. #7
    Super Member wingless's Avatar
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    See this post to see a derivation of Residue Theorem and why a_{-1} is called residue.

    It's called residue because when you integrate the Laurent series of f(z), only the a_{-1}\frac{1}{z-z_0} term is left.

    Definition: Residue is the negative first terms coefficient.

    Can you see any term in the form a_{-1}\frac{1}{z-z_0} in your expansion?
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  8. #8
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    no, there is only: 1/z^2 and then 1/2! therefore thers is no 1/z with coefficient...

    so the residue is zero, but it is not a removable singularity as there is 1/z^2 (therefore not all principle parts are 0) so it is just a singularity.
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  9. #9
    Super Member wingless's Avatar
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    Exactly. Such singularities are called poles. See here.
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  10. #10
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    How about this, since \lim_{z\to 0}\left(z^2\frac{\cosh(z)}{z^2}\right)=1, then the function has a pole of order 2 at the origin. Thus:

    \mathop\text{Res}_{z=0}\left\{\frac{\cosh(z)}{z^2}  \right\}=\frac{d}{dz} cosh(z)\Bigg|_0=0
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