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Find residue of sinz/z^4

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the residue of sinz/z^4

    2. Relevant equations
    The residue for a pole of order m at 0 is given my:
    lim z->0 [1/(m-1)!dm-1/dzm-1[(zmf(z)]]


    3. The attempt at a solution
    Clearly the pole has order 3:
    So we get:
    Re(0) = limz->0[1/2d2/dz2[sinz/z]]
    I get that the only term in the limit that doesn't go to zero is:
    Re(0) = limz->0[-1/2sinz/z] = -1/2
    But I should get -1/3!
    Apparantly something goes wrong somewhere as I should get a tree from differentiation. Can anyone see where that is? :S
     
  2. jcsd
  3. May 23, 2012 #2

    Office_Shredder

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    You didn't take the second derivative of sin(z)/z correctly
     
  4. May 23, 2012 #3
    hmm I thought I might have done something wrong there. Let's go over it:

    (sinz/z)'' = (cosz/z-sinz/z2)' = -sinz/z - 2cosz/z2+3sinz/z3

    Now as we let z->0 the only term that survives is sinz/z which ->-1. Where is my mistake?
     
  5. May 23, 2012 #4
    Your last term should be ##\frac{2 \sin{(z)}}{z^3}##. Go through that derivative again to find your mistake.

    Also, the other terms do approach values, they don't "disappear." Use l'Hospital's Rule.
     
  6. May 23, 2012 #5
    okay yes I see what I did wrong now. I accidently took the terms involving z of a higher order than one to go to zero when they really diverge.
     
  7. May 23, 2012 #6
    Well, no, they don't diverge. They converge to specific points (like I said, use l'Hospital's Rule). Also, you didn't take the second derivative properly which would yield bad results as well.
     
  8. May 23, 2012 #7

    HallsofIvy

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    Are you required to use that formula? Often the simplest way to find the residue of a function, at a given value of z, is to use the fact that the residue is the coefficient of [itex]z^{-1}[/itex] in the Laurent series for f(z) about the given value.

    Here, we can start with the Taylor's series for sin(z): [itex]z- (1/6)z^3+ \cdot\cdot\cdot[/itex] so that
    [tex]\frac{sin(z)}{z^4}= \frac{1}{z^3}- \frac{1}{6z}+ \cdot\cdot\cdot[/tex]
    so that the residue is, as you say, -1/6.
     
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