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Laurent series

  1. Feb 13, 2009 #1
    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
     

    Attached Files:

  2. jcsd
  3. Feb 13, 2009 #2

    HallsofIvy

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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.
     
  4. Feb 13, 2009 #3
    Thanks
     
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