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Taylor series power question

  1. Jan 13, 2010 #1
    when i develop the series of a cosine i have a (-1) member
    i wanted to represent the series as a sum
    so i need to take only the odd members so the power of -1 is 2k+1 i got
    but the solution says that the power of -1 is equal (-1)^{k-1}

    is it the same??
    why they have such an expression
    (they use n istead of k)
    http://i45.tinypic.com/6sszue.jpg

    how they got the power?
     
  2. jcsd
  3. Jan 13, 2010 #2

    Mark44

    Staff: Mentor

    You want the signs of your Taylor's series to alternate, right? (-1)k - 1 gives you that sign alternation. If you had (-1)2k + 1, the sign would always be negative, since you have odd powers of -1.
     
  4. Jan 14, 2010 #3
    you are correct
    how to get this expression?
     
  5. Jan 14, 2010 #4

    Mark44

    Staff: Mentor

    What exactly are you asking? Are you asked to find the Taylor's series for cos z at z = 2? There is a standard technique for finding the coefficients of this series.
     
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