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Math Help - taylor polynomial

  1. #1
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    taylor polynomial

    i cant believe i got this wrong but I did and I dont know why.

    find the 2nd order Taylor Polynomial of  f(x)= cos(x^2) centered at zero

    f(x) = cos x , then f(0) = 1
    f'(x) = -sinx, f'(0) =0
    f"(x) = -cosx f"(0) = -1

    second order TP of cosx is:
    1 + 0x - \frac{x^2}{2!}

    Replace x with x^2, second order TP of cos(x^2) :
    1 - \frac{x^4}{2!}

    But my book says that P"(x) = 1
    What gives?
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  2. #2
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    Re: taylor polynomial

    Quote Originally Posted by delgeezee View Post
    But my book says that P"(x) = 1
    What is P and at which x do you take the derivative?

    Your polynomial is correct; see WolframAlpha.
    Thanks from delgeezee
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  3. #3
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    Re: taylor polynomial

    Thanks for the Web site! This will help me study more efficiently.
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