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Error Estimation for Simpsons Method

  • #1

Homework Statement



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].

Homework Equations



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


The Attempt at a Solution



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

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

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

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

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

Answers and Replies

  • #2
Looks like I made a mistake, the error estimation formula for Simpson's Method is:

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

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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