Laurent series

[Bella123456789]

How would i find the laurent series for cot(z)/z^4?
I've never done it using trigonometric functions, and powers of more than 2.

 

[galactus]

The Laurent Series for cot(z) about the origin is:

$\displaystyle cot(z)=\frac{1}{z}-\frac{1}{3}z+O(z^{3})$

$\displaystyle cot(z)=\frac{cos(z)}{sin(z)}=\left(1-\frac{z^{2}}{2}+O(z^{4})\right)\left(\frac{1}{z}+\frac{z}{6}+O(z^{3})\right)$

$\displaystyle =\frac{1}{z}-\frac{z}{3}+O(z^{3})$

Now, divide by z^4.

 

[Bella123456789]

is that meant to be an O, standing for any letter?

 

[galactus]

See here for Big Oh notation:

http://en.wikipedia.org/wiki/Big_O_notation

 

[Bella123456789]

The Laurent Series for cot(z) about the origin is:

$\displaystyle cot(z)=\frac{1}{z}-\frac{1}{3}z+O(z^{3})$

$\displaystyle cot(z)=\frac{cos(z)}{sin(z)}=\left(1-\frac{z^{2}}{2}+O(z^{4})\right)\left(\frac{1}{z}+\frac{z}{6}+O(z^{3})\right)$

$\displaystyle =\frac{1}{z}-\frac{z}{3}+O(z^{3})$

Now, divide by z^4.

So would the next terms be $\displaystyle =\frac{1}{z}-\frac{z}{3}+\frac{z^3}{5}-\frac{z^5}{7}$

Giving the laurent series as $\displaystyle =\frac{1}{z^5}-\frac{1}{3z^3}+\frac{1}{5z}$