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Taylor Series Remainder

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    What degree Taylor Polynomial around a = 0(MacLaurin) is needed to approximate cos(0.25) to 5 decimals of accuracy?


    2. Relevant equations
    taylor series....to complicated to type out here

    remainder of nth degree taylor polynomial = |R(x)| <= M/(n+1)! * |x - a|^(n+1)
    where a = 0 in this case
    and
    M >= |f^(n+1)(t)|



    3. The attempt at a solution
    I don't really get this question at all. I know that |R(0.25)| = 0.00001 <= M/(n+1)! * |x - a|^(n+1)
    But how do I get M when |f^(n+1)(t)| is unknown? I don't even know what |f^(n+1)(t)| means!
     
  2. jcsd
  3. Nov 10, 2009 #2
    f(n+1)(t) is the n+1:th derivative of f(t). So is your plan to find the lowest upper bound for Mn? It might be easier (and more likely to be correct too) if you just calculated enough terms from the series until you have the desired accuracy.
     
  4. Nov 10, 2009 #3
    But we aren't marked on that...it has to be through the remainder method.

    Anyways, I know what f^(n+1)(t). I just don't know what to plug in for t. And after that, doesn't it just become a plug-and-check game for n until I get less than 0.00001?
     
  5. Nov 10, 2009 #4

    HallsofIvy

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    All derivatives of cosine are [itex]\pm cosine[/itex] or [itex]\pm sine[/itex]. What is the largest possible value of a sine or cosine?
     
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