- Thread starter
- #1

- Thread starter pantboio
- Start date

- Thread starter
- #1

- Admin
- #2

- Jan 26, 2012

- 4,197

I think it's not too difficult to show that the $z_{k}$'s are simple poles. [EDIT] This is incorrect. $z_{0}$ is simple, but the others are double.

Last edited:

- Admin
- #3

- Jan 26, 2012

- 4,197

Whoops. You're right. So you're going to need the next order formula, assuming that works. You could try$z_k$ are zeros of order two of denominator, but $z_0=0$ is also a zero of numerator. So $z_0=0$ is a simple pole, but all the others $z_k=2k+1$ with $k\neq 0$ are 2-poles.

$$ \text{Res}(f,z_{k})= \lim_{z \to z_{k}} \frac{d}{dz} \left[(z-z_{k})^{2}f(z) \right]$$

for the second-order poles. That would come out to

$$ \text{Res}(f,2k+1)= \lim_{z \to 2k+1} \frac{d}{dz} \left[(z-2k-1)^{2} \, \frac{z-1}{1+ \cos( \pi z)} \right]$$

$$= \lim_{z \to 2k+1} \frac{d}{dz} \left[ \frac{(z-1)(z-2k-1)^{2}}{1+ \cos( \pi z)} \right]$$

$$= \lim_{z \to 2k+1} \frac{d}{dz} \left[ \frac{z^{3}-(4k+3)z^{2} +(4k^{2}+8k+3) z-(2k+1)^{2}}{1+ \cos( \pi z)} \right].$$

Looks like a fair amount of algebra. Hmm. You might be better off going for the series expansion.

- Jan 29, 2012

- 661

$\dfrac{z-1}{1+\cos \pi z}=\dfrac{2k+[z-(2k+1)]}{\dfrac{\pi^2}{2}[z-(2k+1)]^2+\ldots}=\ldots +\dfrac{A_{-2}}{[z-(2k+1)]^2}+\dfrac{2/\pi^2}{z-(2k+1)}+A_0+\ldots$

So, $\mbox{Res }(f,2k+1)=\mbox{coef }\left(\dfrac{1}{z-(2k+1)}\right)=\dfrac{2}{\pi^2}$

- Thread starter
- #5

so the residue is the same for all poles, maybe because of $\cos$ periodicity

$\dfrac{z-1}{1+\cos \pi z}=\dfrac{2k+[z-(2k+1)]}{\dfrac{\pi^2}{2}[z-(2k+1)]^2+\ldots}=\ldots +\dfrac{A_{-2}}{[z-(2k+1)]^2}+\dfrac{2/\pi^2}{z-(2k+1)}+A_0+\ldots$

So, $\mbox{Res }(f,2k+1)=\mbox{coef }\left(\dfrac{1}{z-(2k+1)}\right)=\dfrac{2}{\pi^2}$

- Jan 29, 2012

- 661

More general: the coefficient $A_{-1}$ in the Laurent series expansion of $f$ does not depend on $k$.so the residue is the same for all poles, maybe because of $\cos$ periodicity