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Homework Help: Taylor Series Question

  1. Dec 5, 2009 #1
    Trying to find the Taylor Series for cos(x) where x0 is PI.
    I've gotten

    Code (Text):

    cos(x) -1
    -sin(x) 0
    -cos(x) 1
    sin(x) 0
    cos(x) -1
    It's clearly 0 every other term so I need 2k or 2k-1. But the -1 term switches between -1 and 1
    How in world do I deal with this? xD

    Thanks for any suggestions. I am assuming I've made a mistake somewhere.
    I've asked my fellow students and they're cheating with k+1 which does work.

    Thanks! :)
  2. jcsd
  3. Dec 5, 2009 #2


    Staff: Mentor

    Why would you think k + 1 is cheating? I'm assuming you're talking about the exponent on -1 to give sign alternation.
  4. Dec 5, 2009 #3
    Is it correct?
  5. Dec 5, 2009 #4


    User Avatar
    Homework Helper

    Is what correct?
  6. Dec 6, 2009 #5


    User Avatar
    Science Advisor

    So far, you have, for the Taylor's series of cosine(x), about [itex]\pi[/itex],
    -1 + (1/2)x2- (1/4!)x4+ ...

    First, it should be clear that you will only have even powers of x and even numbers can be written "2n". That is, (1/(2(2))!)x4= (1/(2!) x2(2) and (1/6!)x6 = 1/(3(2))! x2(3). The only problem, then, is the power of -1.

    The first term is n= 1 and that is positive. (-1)n would give (-1)1= -1. To fix that you could just multiply the entire sum by -1:
    (-1)(-1+ (1/2!)x2- (1/6!)x6+ ...+ (-1)^n/(2n)! x2n+ ...)

    That is the same as multiplying the -1 in each term:
    1- (1/2!)x2- (1/6!)x6+ ...+ (-1)(-1)n/(2n)! x2n+...

    Which can also be written with (-1)n+1:
    1- (1/2!) x2- (1/6!)x6+ ...+ (-1)n+1/(2n)! x2n+ ...

    Nothing "cheating" about that.
  7. Dec 6, 2009 #6


    Staff: Mentor

    Since this is a Taylor's series about [itex]\pi[/itex], all of the terms should be powers of [itex]x - \pi[/itex], not powers of x.
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