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Thread: Error in Taylor Polynomial

  1. #1
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    Error in Taylor Polynomial

    Completely lost, all help appreciated


    1) Bound the error in the approximation

    sin(x) ≈ x

    for -pi/4 ≤ x ≤ pi/4

    --------------------------------------

    Let p(x) be the Taylor polynomial of degree n of the function
    f(x) = log(1-x) about a = 0.

    How large should n be to have

    | f(x)-p(x) | ≤ 10^(-4) for -1/2 ≤ x ≤ 1/2 ?


    For -1 ≤ x ≤ 1/2
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  2. #2
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    $\displaystyle \sin(x)\;=\;x - \frac{1}{6}x^{3} + OtherStuff$

    The alternating sign of the terms is VERY helpful.

    What is the size of the first term omitted on the desired interval?
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  3. #3
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    The error is less than the first term not used in the approximation

    |-x^3/3!| < pi^3/(4^3*3)= pi^3/384.

    So is that the answer??

    If they ask, "What is the taylor polynomial of degree 3?"

    Do you make sure that you have x^3 and no higher degree? Or do you keep 3 terms or? How do you know what's degree 3, etc. Thanks.
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  4. #4
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    Quote Originally Posted by amor_vincit_omnia View Post
    The error is less than the first term not used in the approximation

    |-x^3/3!| < pi^3/(4^3*3)= pi^3/384.
    That's the idea, but are you sure that's where the maximum value occurs, AT $\displaystyle \frac{\pi}{4}$? Also, 3! = 6, not 3. 3! = 3*2*1

    If they ask, "What is the taylor polynomial of degree 3?"
    Quit when you have a term in x^3 or of higher degree. For example, I dare you to find a degree 3 polynomial for f(x) = cos(x). Please don't count the number of terms while looking for degree.
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