Error Estimation for Simpsons Method

1. Mar 7, 2010

3.141592654

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution

The tricky part about this problem is finding $$f^{(4)}_{max}(x)$$:

$$f(x)=(1+x^2)^\frac{1}{4}$$

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

$$f''(x)=(\frac{1}{2}x*\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)+((1+x^2)^\frac{-3}{4}*\frac{1}{2})$$

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

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

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

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

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

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

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

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. Mar 7, 2010

3.141592654

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

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

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