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Math Help - Function Approximation

  1. #1
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    Function Approximation

    I'm having trouble with the following question:

    Let f(x) = sinx and P5(x) be the fifth-order Maclaurin polynomial for f.

    a) What does Theorem 2 imply about the maximum approximation error committed by P5 over the interval [-2,2]?

    b) What is the maximum actual approximation error committed by P5 over the interval [-2,2]?

    Any help would be greatly appreciated .
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  2. #2
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    Let P_5(x) be the 5th degree Taylor polynomial, and f(x) = \sin x. Define R_5(x) = f(x) - P_5(x) then if x_0 is any non-zero point on (-2,2) it means by Lagrange's remainder there exists y between 0 and x_0 so that R_5(x_0) = \frac{f^{(6)}(y)}{6!} y^{6}. Now f^{(6)}(y) = -\sin y. This means |R_5(x_0)| = \left| \frac{-\sin y}{6!} y^6 \right| \leq \frac{2^6}{6!} = .0\bar 8, but maybe we can improve this estimate. Let us use Cauchy's remainder, R_5(x_0) = \frac{f^{(6)}(y)}{6!}(x_0 - y)^n x_0, note |x_0 - y|\leq 1 and |x_0|\leq 1 thus, |R_5(x_0)| \leq \frac{1}{6!} = .0013\bar 8.
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  3. #3
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    Quote Originally Posted by larson View Post
    I'm having trouble with the following question:

    Let f(x) = sinx and P5(x) be the fifth-order Maclaurin polynomial for f.

    a) What does Theorem 2 imply about the maximum approximation error committed by P5 over the interval [-2,2]?

    b) What is the maximum actual approximation error committed by P5 over the interval [-2,2]?

    Any help would be greatly appreciated .
    I not sure what theorem 2 is in your book, but I would guess it would be Taylor's Inequality...

    if
    |f^{n+1}(x)| \le M for |x-a|\le d , then the remainder R_n(x) of the Taylor series satisfies the inequality

    |R_n(x)|\le\frac{M}{(n+1)!}|x-a^{n+1}|

    Since the 6th derivative of Sin(x) is -Sin(x)

    Its max on [-2,2] is 1 so M =1. so in the formula a=0 because it is a Maclaurin series n=5 because it is 5th degree taylor poly.

    Plugging all that in gives us

    |R_n(x)| \le \frac{1}{(5+1)!}2^{5+1}=\frac{64}{720}=\frac{4}{45  }

    The alternating series estimation could also be used and is much easier as well.
    simplifying
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  4. #4
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    Thank you so so much.
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