1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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):

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

    5.551115123125783*^-16 + 0.8529277641641219*I

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

    BesselI[2, 1]

    N[2*Pi*I*BesselI[2, 1]]

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook