Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble with a specific problem actually. I have a function $$f(z) = \frac{1}{z\cdot \sin{z}}$$

Now I want to find the residues of this. The Laurent series expanded about $0$ shows that $0$ is a pole of order $2$. The expansion looks something like this $$\frac{1}{z^2}+\frac{1}{6} + \frac{7z^2}{360} + \cdots $$

so since the first coefficient of $z$ is just zero, the residue of this function is $0$.

BUT I want to know why zero is the ONLY pole. Clearly $2\pi$ is a singularity point. Then when you expand about $2\pi$ you get the following expansion $$\frac{1}{2\pi (z - 2\pi)} - \frac{1}{4\pi^2} + \frac{(3+2\pi^2)(z-2\pi)}{24\pi^3} + \cdots $$

Again, it looks to me that the first negative power of $z$ has the coefficient $\frac{1}{2\pi}$.

So why is it that when I type in "poles of function 1/(z*sin(z))" wolfram only identifies 0 as the pole. If I type in "poles of function 1/(sin(z))" then it identifies the poles as $n\pi$. Furthermore if you type in "residues of 1/(z*sin(z))" it only identifies 0 as a residue when we just saw above that $\frac{1}{2\pi}$ is also a residue. Whats even more weird is that if you type in "residues of 1/(z*sin(z)) at 2pi" it does give the right residue. Weird.

share|improve this question
6  
Basically, computers are stupid. –  anon Apr 18 '12 at 0:09
    
so I am right? The residue at $z = 0$ is $0$, the residue at $z = 1$ is $\frac{-1}{\pi}$ and so on? –  Tyler Hilton Apr 18 '12 at 0:12
    
wolframalpha.com/input/… –  Aryabhata Apr 18 '12 at 0:14
    
Hi Aryabhata. I added that into my post. I saw that too. This is exactly my question. We know other poles and residues exist so why is that wolfram just identifies 0 as a pole. I mean if it can identify $n\pi$ as poles for $\frac{1}{\sin{z}}$ then it should be able to do that if you multiply the denominator by z. –  Tyler Hilton Apr 18 '12 at 0:19
    
FWIW, Maple 16 says: > singular(1/(z*sin(z))); {z = 0}, {z = Pi*_Z1} –  Robert Israel Apr 18 '12 at 1:14
show 2 more comments

2 Answers

up vote 0 down vote accepted

Rather than expanding the function $f(z)$ around the point $z=2\pi$, let's rearrange $f(z)$ instead by looking at the Taylor series expansion of $\sin(z)$

$$ f(z) = \frac{1}{(z)(z-\frac{z^3}{3!}+\frac{z^5}{5}+\cdots)}$$ $$= \frac{1}{(z^2)(1-\frac{z^2}{3!}+\frac{z^4}{5}+\cdots)}$$ $$ f(z) = \frac{1}{(z^2)}\left(1+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)^2+\cdots\right)$$ $$ f(z) = \left(\frac{1}{z^2} + \frac{1}{z}\right) \left(1+\frac{z^2}{3!}+\cdot\right) = \frac{1}{z^2}+\frac{1}{z}+\cdots $$

Rather than expanding at (what seems) an undefined point, a quick series alteration gives us a different "picture" of this function.

share|improve this answer
    
What am I looking at though? This is just expansion around z=0 no? –  Tyler Hilton Apr 18 '12 at 4:11
    
You are correct on everything you've said except that zero is not the only pole, n* pi for n integer (and not zero) gives a pole of order 1, you can verify this (as you have) with a Laurent Series expansion or by famous results/theorems on poles/singularities. –  Arbias Hashani Apr 19 '12 at 13:35
2  
The last displayed identity for $f(z)$ is wrong. –  Did May 21 '12 at 11:54
add comment

Multiply the function by z upstairs and downstairs:

f(z)= z / z^2⋅sinz = (1/over z^2) { z / sinz}

Now, the second factor is a regular function with no poles. It remains just the factor in front which is:

(1/over z^2)

A function with a single pole of order two, as expected.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.