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Taylor series of sinz-sinhz

  1. Feb 19, 2012 #1
    I have to find the first three non zero terms of this series by hand. I know the answer and it is

    -(z^3/3) - z^7/2520 - z^11/19958400

    Which will take ages to get to by brute force. Is there a quicker way?
     
  2. jcsd
  3. Feb 19, 2012 #2

    tiny-tim

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    hi connor415! :smile:

    but that's just (n!)/2 …

    what's brutish about that? :confused:
     
  4. Feb 19, 2012 #3
    The method I used was brutish. I didnt use the generalized form of the series for sin and sinh that I think youre referring to. I now am, and its faster. Thanks
     
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