1. ## Singularities

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

2. 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

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

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

3. Originally Posted by Moo
Hello,

cot=cos/sin

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

For any $\displaystyle x \in \mathbb{Z}$, there is a singularity, because thereafter, $\displaystyle \sin(\pi x)=0$
Now I believe there is a triple pole for 0, and simple pole (not too sure) for any integer in $\displaystyle \mathbb{Z}^\times$
you can check it by finding the Laurent series near these points.
The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow 0} x^m f(x) \neq \infty$ is $\displaystyle m = 3$ so $\displaystyle x = 0$ is a pole of order 3.

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

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

The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow n} x^m f(x) \neq \infty$ is $\displaystyle n = 1$ so $\displaystyle x = n$ is a simple pole (the more PC expression is pole of order 1).
And the limit has to be $\displaystyle \neq 0$, 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 $\displaystyle \lim_{x \to n} (x-n)^m f(x)$ ?

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

Actually, I couldn't be sure for the second poles, because I didn't study the limit
And I thought it was $\displaystyle \lim_{x \to n} (x-n)^m f(x)$ ? Mr F says: Yes, it is.
Whoops. Actually there were a couple of typos from my rushed cut and paste. Fixed now. Thanks.

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

thanks guys