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Help with Taylor's Theorem to obtain error of approximation.

  1. Mar 27, 2006 #1
    I need to use Taylor's Theorem to obtain the upper bound for the error of the approximation on the following

    e^{\frac{1}{2}} \approx 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}}

    Here is an example problem in the textbook I am following.
    http://img157.imageshack.us/img157/8564/100027rh.jpg [Broken]

    Here is my work. Am I on the right track?

    e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}} + R_3 \left( {\frac{1}{2}} \right)

    e^{\frac{1}{2}} = 1 + \frac{1}{2} + \frac{{\left( {\frac{1}{2}} \right)^2 }}{{2!}} + \frac{{\left( {\frac{1}{2}} \right)^3 }}{{3!}} + \frac{{f^4 (z)}}{{4!}}\left( {\frac{1}{2}} \right)^4

    where [tex]0 < z < \frac{1}{2}[/tex]

    [tex]f^4 (z) = e^z [/tex]

    \frac{{e^0 }}{{4!}}\left( {\frac{1}{2}} \right)^4 < \frac{{e^z }}{{4!}}\left( {\frac{1}{2}} \right)^4 < \frac{{e^{\frac{1}{2}} }}{{4!}}\left( {\frac{1}{2}} \right)^4

    So the upper bound for the error on the approximation of [tex]\[
    e^{\frac{1}{2}} [/tex] is

    \frac{{e^{\frac{1}{2}} }}{{4!}}\left( {\frac{1}{2}} \right)^4 = \frac{{e^{\frac{1}{2}} }}{{384}} \approx 0.004
    Last edited by a moderator: May 2, 2017
  2. jcsd
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