# Laurent series

1. Feb 13, 2009

### asi123

1. The problem statement, all variables and given/known data

Hey guys.
I need to develop this function into Laurent series.
I used the Sin Taylor series and got what I got.
Now, is there a trick or something to get the z-2 inside series or is this enough?

Thanks.

2. Relevant equations

3. The attempt at a solution

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2. Feb 13, 2009

### HallsofIvy

Staff Emeritus
Expand sin(z) about z= 2, not 0! And that can be done relatively easily by writing sin(z) as sin(u+ 2) where u= z- 2. sin(u+2)= cos(u)sin(2)+ sin(u)cos(2) so
$$sin(z)= \sum_{n=0}^\infty \frac{(-1)^n sin(2)}{(2n)!}(z-2)^{2n}+ \frac{(-1)^n cos(2)}{(2n+1)!}(z-2)^{2n+1}$$
Where I have expanded sin(u) and cos(u) in the usual series around u= 0.

3. Feb 13, 2009

Thanks