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Homework Help: Error Estimation for Simpsons Method

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the error estimation for the integral [tex]\int(1+x^2)^\frac{1}{4}[/tex] with limits [tex][0, 2][/tex] for [tex]n=8[/tex].

    2. Relevant equations

    [tex]\epsilon\leq \frac{|f^{(4)}_{max}(x)|(b-a)^5}{180n^4}[/tex]

    3. The attempt at a solution

    The tricky part about this problem is finding [tex]f^{(4)}_{max}(x)[/tex]:





    [tex]=\frac{1}{2} [\frac{-6}{4}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}][/tex]

    [tex]=\frac{1}{2} [\frac{-3}{2}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}][/tex]

    [tex]f'''(x)=\frac{1}{2} [(\frac{-3}{2}x^2*\frac{-7}{4}(1+x^2)^\frac{-11}{4}*2x)+((1+x^2)^\frac{-7}{4}*\frac{-6}{2}x)+(\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)][/tex]

    [tex]=\frac{1}{2} [\frac{42}{8}x^3(1+x^2)^\frac{-11}{4}-\frac{6}{2}x(1+x^2)^\frac{-7}{4}-\frac{6}{4}x(1+x^2)^\frac{-7}{4}][/tex]

    [tex]=\frac{1}{2} [\frac{21}{4}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}-\frac{3}{2}x(1+x^2)^\frac{-7}{4}][/tex]

    [tex]=\frac{3}{2} [\frac{7}{4}x^3(1+x^2)^\frac{-11}{4}-x(1+x^2)^\frac{-7}{4}-\frac{1}{2}x(1+x^2)^\frac{-7}{4}[/tex]

    [tex]\frac{3}{2} [\frac{7}{2}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}][/tex]

    I entered the derivatives I calculated along with the derivatives taken from an online derivative calculator into my graphing calculator to make sure I was coming out with the right answer. It appears I've made a mistake with the 3rd derivative but I can't find the error. Thanks for any help.
    Last edited: Mar 7, 2010
  2. jcsd
  3. Mar 7, 2010 #2
    Looks like I made a mistake, the error estimation formula for Simpson's Method is:


    So I'm trying to find the maximum of the fourth derivative. However, I still can't find the error in my third deriv.
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