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Homework Help: Taylor series question

  1. Feb 11, 2009 #1
    how to find the taylor series for
    y(x)=\sin^2 x
    i need to develop a general series which reaches to the n'th member
    so i cant keep doing derivatives on this function till the n'th member

    how to solve this??
  2. jcsd
  3. Feb 12, 2009 #2


    Staff: Mentor

    Multiply the two Taylor series together (or Maclaurin series if that's what you're using).
    [tex]sin^2(x) = (x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)((x - x^3/3! + x^5/5! -+ ... + (-1)x^{2n + 1}/(2n + 1)! +...)[/tex]

    The first term will be x^2
  4. Feb 12, 2009 #3


    User Avatar
    Science Advisor

    Why can't you "keep doing derivatives on this function till the n'th member"?

    y= sin^2(x)
    y'= 2sin(x)cos(x)= sin(2x)
    and the rest is easy.
  5. Feb 14, 2009 #4
    what about the remainder ob the multiplication??
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