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Math Help - Upper Bound For Maclaurin Approximation

  1. #1
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    Upper Bound For Maclaurin Approximation

    There is a homework problem that I don't even understand what they're asking. Can any of you help clarify what they mean?

    f(x) = sin(x)

    Find an upper bound for |[f(x)-[p9(x)]]|
    -7 <= x <= 7

    p9(x) is the nth order Maclaurin approximaxion to f
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  2. #2
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    Quote Originally Posted by soma View Post
    There is a homework problem that I don't even understand what they're asking. Can any of you help clarify what they mean?
    Where should we start? Obviously this problem expects you to know what a "MacLaurin series" is. Do you know that? It is simply the Taylor's series at x= 0. Do you know what that is? If not look up "MacLaurin series" or "Taylor's series" in the index of your book.

    And, I strongly suspect that in the section where this problem is given, there is a formula for the error you get in using cutting a McLaurin or Taylor's polynomial of degree n (that is, cutting off the McLaurin or Taylor's series at the nth power). It should look something like this:
    E\le \frac{M_{n+1}}{(n+1)!}x^{n+1} where " M_{n+1}" is an upper bound on the absolute value of the n+1 derivative of the function.

    Since this problem asked about the "9th order MacLaurin series for sin(x)" for x between -7 and 7. Okay you need M_{10}. What is the 10th derivative of sin(x)? What is its largest value between -7 and 7 (radians: 2\pi is about 6.28 so this is from less than 2\pi to larger than 2\pi).
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  3. #3
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    Ok, so the tenth derivative of sin(x) is -sin(x), and the maximum bound of -sin(x) is when x = -pi/2

    -sin(-pi/2) = 1

    so

    E <= [1/(9+1)!]*[(-pi/2)^(9+1)]
    E <= .0000252020

    Am I understanding this problem correctly?
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