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Math Help - Percentage Errors using the Maclaurin Series

  1. #1
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    Percentage Errors using the Maclaurin Series

    Hi, I'm having trouble trying to find the percentage errors of P3(x) and P4(x).

    Here's the question;
    When x=0.7891.

    f(x)=x/(sin(x)+2)
    f(x)=0.2912

    P3(x)=1/2*x-1/4*x^2
    P3(x) =0.2389

    P4(x)=1/2*x-1/4*x^2+1/8*x^3
    P4(x) =0.3003

    Find the percentage errors in using P3(x), and P4(x) to approximate f(x).

    Thanks, DLL.
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  2. #2
    Senior Member Spec's Avatar
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    The formula for relative error is simply: \eta=\left | \frac{\Delta f(x)}{f(x)}\right | where \Delta f(x)=f(x)-f(x)_{approximation} for some value of x.
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  3. #3
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    I'm not sure how to use this formula as I have never come across it before. I was thinking of the Maclarin Remaider Therom. I'm not sure how I use these errors to approximate f(x) however.
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  4. #4
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    I don't need to use the relative error formula. I need to find the percentage errors of in usig P3(x) and P4(x) to approximate f(x).

    f(x)=x/(sins(x)=2) and the Maclurin polynomials are P3(x)=1/2*x-1/4*x^2 and P4(x)=1/2*x-1/4*x^2+1/8*x^3, when x=0.7891.

    I'm not sure how I use the percentage errors or use them to approximate f(x).
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  5. #5
    Senior Member Spec's Avatar
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    If you express the error as percentages, then it's relative to the true value. You can find the error of using a Maclaurin series to approximate a function, by using the Lagrange remainder theorem, but the answer you get is going to be the absolute error.

    The remainder is r(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} for some \xi \in [0,x]

    If you want to the find the absolute error by using the Lagrange remainder theorem, then you first need to find f^{4}(\xi) for the first polynomial ( P_3) and f^{5}(\xi) for the second ( P_4).
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  6. #6
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    Okay, thanks Spec.

    I used the relative error as you said and I got reasonable anwers. I think what I did was right so thanks for the insight on that.
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