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Math Help - residue

  1. #1
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    residue

    Find the residue of the following.

    (a) \text{Res}(f, z_0=0), where f(z)=\frac{e^z}{\sin(2z^2)}


    (b) \text{Res}(f, z_0=0), where f(z) = \frac{\cos(\frac{1}{z})}{e^z}<br />
    We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by pascal4542 View Post
    Find the residue of the following.

    (a) \text{Res}(f, z_0=0), where f(z)=\frac{e^z}{\sin(2z^2)}


    (b) \text{Res}(f, z_0=0), where f(z) = \frac{\cos(\frac{1}{z})}{e^z}<br />
    We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
    If z_0 is a pole of order m, then \text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right].

    In the case its a simple pole, we have \text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right).

    Are you allowed to use these?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    If z_0 is a pole of order m, then \text{Res}\!\left(f;z_0\right)=\frac{1}{m-1}\lim_{z\to z_0}\frac{\,d^{m-1}}{\,dz^{m-1}}\left[(z-z_0)^mf\!\left(z\right)\right].

    In the case its a simple pole, we have \text{Res}\!\left(f;z_0\right)=\lim_{z\to z_0}(z-z_0)f\!\left(z\right).

    Are you allowed to use these?
    Yes I am. I will try that. Thank you.
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  4. #4
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    Quote Originally Posted by pascal4542 View Post
    Yes I am. I will try that. Thank you.

    I was wondering if you can check my work.

    For (a) \text{Res}(f, z_0=0), where f(z)=\frac{e^z}{\sin(2z^2)} we have

    \frac{1+z+z^2/2!+z^3/3!+\cdots}{2z^2 -(2z^2)^3/3!+(2z^2)^5/5!- \cdots}

     = \frac{1}{2z^2}+\frac{1}{2z}+1/4+z/12


    So the residue is 1/2.


    For (b) \text{Res}(f, z_0=0), where f(z) = \frac{\cos(\frac{1}{z})}{e^z} we have


    \cos(\frac{1}{z}) \cdot (1-z+z^2/2-z^3/6+z^4/24+\cdots).


    I am stuck. I don't know how to get the residue. I know that z_0 is an essential singularity. I need help on part (b).
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  5. #5
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    You have a Taylor series multiplied by a non-terminating singular series so the residue is an infinite sum due to the term-wise multiplication of z^n by z^{-n} so then:

    \begin{aligned}\frac{\cos(1/z)}{e^z}&=\sum_{n=0}^{\infty}\frac{(-1)^n}{1}\frac{1}{(2n)! z^{2n}} \sum_{n=0}^{\infty}\frac{(-1)^n z^n}{n!}\\<br />
&=\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{(-1)^k}{(2k)!z^{2k}}\frac{(-1)^{n-k}z^{n-k}}{(n-k)!}<br />
\end{aligned}<br />

    and note the coefficients on the 1/z terms are when the exponent of z is n=3k-1. Now, can you extract out all those terms and come up with the expression:

    \mathop\text{Res}\limits_{z=0} \frac{\cos(1/z)}{e^z}=\sum_{k=0}^{\infty} \frac{(-1)^{3k-1}}{(2k)!(2k-1)!}
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