Developing a Laurent Series: Seeking Assistance

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SUMMARY

The discussion focuses on developing the Laurent series for the function sin(z) expanded around z=2. The user initially attempted to use the Taylor series but was advised to express sin(z) as sin(u+2), where u=z-2. The resulting series expansion is given as sin(z) = ∑_{n=0}^∞ [(-1)^n sin(2)/(2n)!](z-2)^{2n} + [(-1)^n cos(2)/(2n+1)!](z-2)^{2n+1}, utilizing standard series expansions for sin(u) and cos(u) around u=0.

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Homework Statement



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.


Homework Equations





The Attempt at a Solution

 

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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
[tex]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}[/tex]
Where I have expanded sin(u) and cos(u) in the usual series around u= 0.
 
HallsofIvy said:
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
[tex]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}[/tex]
Where I have expanded sin(u) and cos(u) in the usual series around u= 0.

Thanks
 

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