Singularities

**pols89** · January 4th 2009, 04:02 AM

Hey guys
Im having a panic as i should be revising, but instead im stuck on a question from my coursework, can anyone help

Find and classify the singularities of
(pi*cot(pi*x))/x^2

thanks

**Moo** · January 4th 2009, 04:19 AM

Hello,

Originally Posted by pols89

Hey guys
Im having a panic as i should be revising, but instead im stuck on a question from my coursework, can anyone help

Find and classify the singularities of
(pi*cot(pi*x))/x^2

thanks

cot=cos/sin

$\frac{\pi \cot(\pi x)}{x^2}=\frac{\pi \cos(\pi x)}{x^2 \sin(\pi x)}$

For any $x \in \mathbb{Z}$ , there is a singularity, because thereafter, $\sin(\pi x)=0$
Now I believe there is a triple pole for 0, and simple pole (not too sure) for any integer in $\mathbb{Z}^\times$
you can check it by finding the Laurent series near these points.

**mr fantastic** · January 4th 2009, 02:44 PM

Originally Posted by Moo

Hello,

cot=cos/sin

$\frac{\pi \cot(\pi x)}{x^2}=\frac{\pi \cos(\pi x)}{x^2 \sin(\pi x)}$

For any $x \in \mathbb{Z}$ , there is a singularity, because thereafter, $\sin(\pi x)=0$
Now I believe there is a triple pole for 0, and simple pole (not too sure) for any integer in $\mathbb{Z}^\times$
you can check it by finding the Laurent series near these points.

The smallest value of $m$ such that $\lim_{x \rightarrow 0} x^m f(x) \neq \infty$ is $m = 3$ so $x = 0$ is a pole of order 3.

The smallest value of $m$ such that $\lim_{x \rightarrow n} (x-n)^m f(x) \neq \infty$ is $m = 1$ so $x = n$ is a simple pole (the more PC expression is pole of order 1).

**Moo** · January 4th 2009, 11:05 PM

Originally Posted by mr fantastic

The smallest value of $m$ such that $\lim_{x \rightarrow 0} x^m f(x) \neq \infty$ is $m = 3$ so $x = 0$ is a pole of order 3.

The smallest value of $m$ such that $\lim_{x \rightarrow n} x^m f(x) \neq \infty$ is $n = 1$ so $x = n$ is a simple pole (the more PC expression is pole of order 1).

And the limit has to be $\neq 0$ , isn't it ?

Actually, I couldn't be sure for the second poles, because I didn't study the limit

And I thought it was $\lim_{x \to n} (x-n)^m f(x)$ ?

**mr fantastic** · January 4th 2009, 11:20 PM

Originally Posted by Moo

And the limit has to be $\neq 0$ , isn't it ? Mr F says: No. Think about the Laurent series of f(x) around ${\color{red}x = x_0}$ and what you need so that ${\color{red}(x - x_0) f(x)}$ is analytic ....

Actually, I couldn't be sure for the second poles, because I didn't study the limit

And I thought it was $\lim_{x \to n} (x-n)^m f(x)$ ? Mr F says: Yes, it is.

Whoops. Actually there were a couple of typos from my rushed cut and paste. Fixed now. Thanks.

**Moo** · January 4th 2009, 11:29 PM

Woops sorry. Actually, it's just that $f(n)$ has to be $\neq 0$

**pols89** · January 5th 2009, 06:46 AM

thanks guys

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