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Math Help - [SOLVED] Find a residue (checking my result and clearing doubts)

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Find a residue (checking my result and clearing doubts)

    Find the residue in z=0 of f(z)=exp\left ( z+ \frac{1}{z}  \right ).

    My attempt : I notice that f is analytic except in 0. (well I think so! I guess I should show that f is analytic. Is that sufficient that f satisfy C-R equations? Because as you'll see, I need the analyticity of f for the coming part)
    I believe that \int _{\gamma} f(z)dz = 2\pi i= Res f(0).
    I forgot something very important : why does \int _{\gamma} f(z)dz = 2\pi i? I mean, I think there's a theorem that justify it (it seems it's not Cauchy-Goursat theorem) but I forgot which one.

    Am I right?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    Find the residue in z=0 of f(z)=exp\left ( z+ \frac{1}{z} \right ).

    My attempt : I notice that f is analytic except in 0. (well I think so! I guess I should show that f is analytic. Is that sufficient that f satisfy C-R equations? Because as you'll see, I need the analyticity of f for the coming part)
    I believe that \int _{\gamma} f(z)dz = 2\pi i= Res f(0).
    I forgot something very important : why does \int _{\gamma} f(z)dz = 2\pi i? I mean, I think there's a theorem that justify it (it seems it's not Cauchy-Goursat theorem) but I forgot which one.

    Am I right?
    I think (I haven't checked) that z = 0 is an essential singularity of f(z) = exp\left ( z+ \frac{1}{z} \right ). To get the residue you'll need to calculate the coefficient of \frac{1}{z} in the Laurent series about z = 0.
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    That's a particular example of what happens when a Taylor series is multiplied by a singular series: the term on 1/z turns out to be an infinite sum sometimes. You have:

    e^{z} e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\sum_{n=0}^{  \infty}\frac{1}{z^n n!}=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{k!}\frac{1}{z^{n-k}(n-k)!}

    But we can figure out which terms have 1/z in them. It's whenever k-(n-k)=-1 right? Can you then see:

    \mathop\text{Res}\limits_{z=0} e^{z+1/z}=\sum_{k=0}^{\infty}\frac{1}{k!(k+1)!}
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  4. #4
    MHF Contributor arbolis's Avatar
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    Ah ok so I was all wrong.
    Next time I'll think about writing exp(a(z)+b(z)) as exp(a(z)) \cdot exp (b(z)).
    shawsend, I follow you until you stated
    Can you then see:

    . No, I don't see this.

    Ans thanks mr fantastic, I'll try.
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  5. #5
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    The powers of z in the double sum have the form \frac{z^k}{z^{n-k}}=z^{k-(n-k)}. We want all the 1/z terms or when k-(n-k)=-1 or when n=2k+1 and when n is that value, the coefficient on 1/z is \frac{1}{k!(k+1)!}. I'm just replacing the (n-k) expression in the double sum with (2k+1-k). But since k is going from 0 to n and n is going to infinity, that's the same as k going from 0 to infinity (eventually) and that's why I use the summation expression.

    Try and work through the same process for the following:

    \oint_{|z|=1} \cos(z)e^{1/z}dz
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  6. #6
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by shawsend View Post
    The powers of z in the double sum have the form \frac{z^k}{z^{n-k}}=z^{k-(n-k)}. We want all the 1/z terms or when k-(n-k)=-1 or when n=2k+1 and when n is that value, the coefficient on 1/z is \frac{1}{k!(k+1)!}. I'm just replacing the (n-k) expression in the double sum with (2k+1-k). But since k is going from 0 to n and n is going to infinity, that's the same as k going from 0 to infinity (eventually) and that's why I use the summation expression.

    Try and work through the same process for the following:

    \oint_{|z|=1} \cos(z)e^{1/z}dz
    Thanks for the clarification. I've tried your integral, which gave me \mathop\text{Res} \cos (z) e ^{\frac{1}{z}} in z=0 is worth \sum _{k=0}^{\infty} \frac{(-1)^k}{(2k)!(2k+1)!}.
    I hope I'm right.
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  7. #7
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    Very good Arbolis. I think you have it now. Here's the numeric check in Mathematica which expresses the sum in terms of some obscure KevinBel function.

    Code:
    In[208]:=
    Sum[(-1)^k/((2*k)!*(2*k + 1)!), 
      {k, 0, Infinity}]
    
    NIntegrate[Cos[z]*Exp[1/z]*I*Exp[I*t] /. 
       z -> Exp[I*t], {t, 0, 2*Pi}]
    N[2*Pi*I*((KelvinBei[1, 2] - 
         KelvinBer[1, 2])/Sqrt[2])]
    
    Out[209]=
    (KelvinBei[1, 2] - KelvinBer[1, 2])/
      Sqrt[2]
    
    Out[210]=
    -4.440892098500626*^-16 + 
      5.761766462098244*I
    
    Out[211]=
    0. + 5.7617664620982385*I
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  8. #8
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    Quote Originally Posted by shawsend View Post
    That's a particular example of what happens when a Taylor series is multiplied by a singular series: the term on 1/z turns out to be an infinite sum sometimes. You have:

    e^{z} e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\sum_{n=0}^{  \infty}\frac{1}{z^n n!}=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{k!}\frac{1}{z^{n-k}(n-k)!}

    But we can figure out which terms have 1/z in them. It's whenever k-(n-k)=-1 right? Can you then see:

    \mathop\text{Res}\limits_{z=0} e^{z+1/z}=\sum_{k=0}^{\infty}\frac{1}{k!(k+1)!}
    Of related interest:

    Wolfram|Alpha 1

    Wolfram|Alpha 2
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