Find the Residue for sin(z)/z^4

[zezima1]

Homework Statement

Calculate the residue of sinz/z^4

Homework Equations

The residue for a pole of order m at 0 is given my:
lim z->0 [1/(m-1)!d^m-1/dz^m-1[(z^mf(z)]]

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

 

[Office_Shredder]

You didn't take the second derivative of sin(z)/z correctly

 

[zezima1]

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?

 

[scurty]

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?

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.

 

[aaaa202]

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.

 

[scurty]

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.

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.

 

[HallsofIvy]

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 $z^{-1}$ in the Laurent series for f(z) about the given value.

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