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Taylor series derivative

  1. Sep 8, 2009 #1
    Is it correct to take the derivative of a taylor series the same as you would for a power series ie:
    [tex]
    sinx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}
    [/tex]
    [tex]
    \frac{d}{dx}(sinx)=cosx=\sum_{n=1}^{\infty}(-1)^n(2n+1)\frac{x^{2n}}{(2n+1)!}
    [/tex]
    it seems as if it wouldnt be
    [tex]
    cosx=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}
    [/tex]
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  2. jcsd
  3. Sep 8, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    have a look at your limits of summation
     
  4. Sep 9, 2009 #3
    Also, is this true?

    [tex]\frac{(2n+1)}{(2n+1)!} = \frac{1}{(2n)!}[/tex]

    for example

    [tex]\frac{7}{7!} = \frac{1}{6!}[/tex]
     
  5. Sep 9, 2009 #4
    Yup, that is true, you can prove it easily by factoring out (2n+1) from (2n+1)!
     
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