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Finding the Residue of this function

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data

    The original problem is to find the integral from 0 to 2[itex]\pi[/itex] of [itex]\frac{1}{2\pi}e^{cosθ}cos(2θ)[/itex]

    2. Relevant equations

    I have it down to two equations which I believe I have to find the residues of and add the results, multiply by 2[itex]\pi[/itex]i and I will have my answer.

    3. The attempt at a solution

    Have 1/2(∫e[itex]^{1/2(z+1/z)}[/itex]*z + ∫e[itex]^{1/2(z+1/z)}[/itex]*(1/z[itex]^{3}[/itex])

    so I believe I have to find Res(e[itex]^{1/2(z+1/z)}[/itex]*z) and Res(e[itex]^{1/2(z+1/z)}[/itex]*(1/z[itex]^{3}[/itex])

    If this is correct I really am unsure of how to find these residues..
     
  2. jcsd
  3. Mar 17, 2012 #2
    You can find the residue of [itex] z e^{\frac{z^2+1}{2z}}[/itex] by using the Cauchy product:

    [tex]ze^{\frac{z^2+1}{2z}}=z\sum_{n=0}^{\infty} \frac{z^n}{2^n n!} \sum_{n=0}^{\infty} \frac{1}{2^n n! z^n}=z\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{2^k k!} \frac{1}{2^{n-k} (n-k)! z^{n-k}}[/tex]

    Now that double sum, get all the z's together on one side. Then that's [itex]z^{1+k-(n-k)}[/itex] right? When does that exponent equal -1? When ever 2k+2=n right? Now we only want the coefficients of 1/z for that to sum them up (the resisue is an infinite sum). So we need to sum all the k-values when n=2k+2. So form the inner sum using only the coefficients when n=2k+2 and compute:

    [tex]\sum_{k=0}^{\infty} ? [/tex]
     
  4. Mar 17, 2012 #3
    If I do all of that I get [itex]\sum^{∞}_{k=0}\frac{1}{(k+2)!(2k+2)2^{3k+4}}[/itex] as the coefficient of [itex]\frac{1}{z}[/itex]

    so this would be the residue right? Then I just have to do the same thing for the other residue and add them and multiply by 2[itex]\pi[/itex]i according to the residue theorem...
     
  5. Mar 18, 2012 #4
    Try to check them in Mathematica. It's not hard:

    Code (Text):

    In[2]:=
    NIntegrate[z*Exp[(z^2 + 1)/(2*z)]*I*Exp[I*t] /.
       z -> Exp[I*t], {t, 0, 2*Pi}]

    Out[2]=
    5.551115123125783*^-16 + 0.8529277641641219*I

    In[3]:=
    Sum[1/(2^k*k!*(k + 2)!*2^(k + 2)), {k, 0, Infinity}]

    Out[3]=
    BesselI[2, 1]

    In[4]:=
    N[2*Pi*I*BesselI[2, 1]]

    Out[4]=
    0. + 0.8529277641641214*I
     
     
  6. Mar 18, 2012 #5
    i'm sorry i'm kind of lost with the mathematica stuff (don't have the program).. but why did you integrate z*e^((z^2+1)/2z)*i*e^(it) ? I don't understand where the i*e^(it) came from at the end..
     
  7. Mar 18, 2012 #6
    I understand everything you said about using the Cauchy product.. and went through all of that and then created a single sum from 0 to infinity substituting n=2k+2 to create the coefficient of the 1/z term in the sum, which I thought was the residue... where am I going wrong here?

    Thanks for the help
     
  8. Mar 18, 2012 #7
    The residue of [itex]\displaystyle z e^{\frac{z^2+1}{2z}}[/itex] is [itex]\displaystyle\sum_{k=0}^{\infty} \frac{1}{2^k k! (k+2)! 2^{k+2}}[/itex]

    That's what you should get when you make the substitution n=2k+2. So we can check that by numerically integrating:

    [tex]\mathop\oint\limits_{|z|=1} z e^{\frac{z^2+1}{2z}}dz[/tex]

    Now let [itex]z=e^{it}[/itex] so that [itex]dz=ie^{it}[/itex]

    and substitute that into the integral above to get:

    [tex]\int_0^{2\pi} z e^{\frac{z^2+1}{2z}}ie^{it}dt,\quad z=e^{it}[/tex]

    That's what I coded in Mathematica.
     
  9. Mar 18, 2012 #8
    ahh ok I see now.. so I did the same for the other residue (with 1/z^3 instead of z) and found that the sum is 1/z when n=2k-2 so substituted and got [itex]\sum^{∞}_{k=0}\frac{1}{k!(k-2)!2^{k-1}}[/itex]

    then added the two residues and multiplied by 1/2 to get [itex]\sum^{∞}_{k=0}\frac{1}{k!(k-2)!2^{k}}[/itex]

    (multiplied by [itex]\frac{1}{2}[/itex] because the original integral had [itex]\frac{1}{4*\pi*i}[/itex] in front before finding residues so [itex]\frac{1}{4*\pi*i}[/itex] * 2[itex]\pi[/itex]*i = [itex]\frac{1}{2}[/itex]

    I think this is correct.. Thanks again for your help
     
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