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4th order Maclaurin Polynomial - Help with error approximation

  1. Mar 16, 2006 #1
    I have the the follow 3rd order polynomial approximation for e^3x

    f(x) = e^{3x} \approx 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3

    In an earlier part of the problem, I found

    f\left( {\frac{1}{3}} \right) = e^{3\left( {\frac{1}{3}} \right)}

    \approx 1 + 3\left( {\frac{1}{3}} \right) + \frac{9}{2}\left( {\frac{1}{3}} \right)^2 + \frac{9}{2}\left( {\frac{1}{3}} \right)^3 = \frac{8}{3} \approx 2.667

    I need to find the upper bound for the error of the above approximation.

    f(x) = e^{3x} = 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 + R_3 (x)

    R_3 \left( {\frac{1}{3}} \right) = \frac{{f^{\left( 4 \right)} \left( z \right)}}{{4!}}\left( {\frac{1}{3}} \right)^4 \\
    {\rm{where }}0 < z < \frac{1}{3} \\

    f^{\left( 4 \right)} (z) = 27e^{3z}

    So the bounds on my error

    \frac{{27e^{3(0)} }}{{4!}} < \frac{{27e^{3z} }}{{4!}} < \frac{{27e^{3\left( {\frac{1}{3}} \right)} }}{{4!}}

    Which means my error for this function when evaluated at 1/3 is
    somewhere between 1.125 and ~3.058. This must be wrong
    because f(1/3) is approx. 2.67.
    (e - 2.67) is approx 0.5.

    I beleive the bounds that I found are saying that my error must be
    greater than 1.125 and less than 3.058. Where did I go wrong?


    EDIT: I found some errors while looking over the problem more carefully.
    Please let me know if it looks correct now. Thanks. My corrections are

    f^{\left( 4 \right)} (z) = 81e^{3z}

    \frac{{81e^{3(0)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \frac{{81e^{3(z)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \frac{{81e^{3\left( {\frac{1}{3}} \right)} }}{{4!}}\left( {\frac{1}{3}} \right)^4

    \approx 0.04167 < \frac{{81e^{3(z)} }}{{4!}}\left( {\frac{1}{3}} \right)^4 < \approx 0.11326
    Last edited: Mar 16, 2006
  2. jcsd
  3. Mar 16, 2006 #2


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    This still looks wrong.
  4. Mar 16, 2006 #3

    (e - 2.67) is approx. 0.05 not 0.5
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