1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

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

    In an earlier part of the problem, I found

    [tex]
    f\left( {\frac{1}{3}} \right) = e^{3\left( {\frac{1}{3}} \right)}
    [/tex]

    [tex]
    \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
    [/tex]

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

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

    [tex]
    \begin{array}{l}
    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} \\
    \end{array}
    [/tex]

    [tex]
    f^{\left( 4 \right)} (z) = 27e^{3z}
    [/tex]

    So the bounds on my error

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

    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?

    Thanks

    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
    below.

    [tex]
    f^{\left( 4 \right)} (z) = 81e^{3z}
    [/tex]

    [tex]
    \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
    [/tex]

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    (e - 2.67) is approx. 0.05 not 0.5
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook