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Maclaurin errors

  1. Apr 19, 2006 #1
    Here's the problem:

    Determine the degree of Maclaurin polynomial required for the error to be less than .0001 if f(x)=cos(2x) and you are approximating f(0.6)

    I really don't know what I am doing. Here's what I've tried to do:

    Rn(.6) = ( (f^(n+1)(z)) / (n+1)! ) (.6)^(n+1)

    I don't know where to go from here....
  2. jcsd
  3. Apr 20, 2006 #2
    You must find the form of [tex] f^{n+1}(0) [/tex] first, then substitute it to Rn(0.6).
    Solve the inequation : Rn(0.6) < 0.001 for n.
    n is the required degree of Maclaurin polynomial for the given error.
  4. Apr 20, 2006 #3
    Where did this come from: Rn(.6) = ( (f^(n+1)(z)) / (n+1)! ) (.6)^(n+1) ?

    I would do this by trial and error, i.e calculating a larger and larger Maclaurin series expansion of cos (2x) until f(0.6) is where you need it to be.
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