Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The three-point quadrature rule with error term is given by $$\int_{-1}^1f(x)dx=\frac59f\left(\frac{-\sqrt{15}}5\right)+\frac89f(0)+\frac59f\left(\frac{\sqrt{15}}5\right)+kf^{(6)}(c).$$ Find $k$.

After using Lagrange's interpolation (interpolating $f$ at $\frac{-\sqrt{15}}5,0,\frac{\sqrt{15}}5$), I found that the error term should be of the form $$\int_{-1}^1\frac{f'''(c_x)}6x(x^2-\frac53)dx.$$ However, I can't use the mean value theorem, because $x(x^2-\frac53)$ changes sign in $[-1,1]$. So how to continue from here?

I think this is somewhat related to the Simpson's error terms. But the textbook I'm using (Sauer) omitted the proof.

share|improve this question
add comment

1 Answer 1

If you can take it as given that the error term has this form, then you can calculate $k$ by substituting a function with constant sixth derivative, $f(x)=x^6$. Then

$$\frac27=\int_{-1}^1x^6\,\mathrm dx=2\cdot\frac59\left(\frac{\sqrt{15}}5\right)^6+k\cdot6!\;,$$

and thus

$$k=\frac1{6!}\left(\frac27-\frac{10}9\left(\frac{\sqrt{15}}5\right)^6\right)=\frac1{15750}\;.$$

share|improve this answer
    
But I think $c$ depends on $x$. –  asepe Apr 15 '12 at 17:57
    
@asepe: I don't understand what that means. The only equation in which you use a variable called $c$ is the first one, and in that equation there's no free variable $x$; the variable $x$ occurs in that equation only as an integration variable. Thus it's not clear to me what it would mean for $c$ to depend on $x$. –  joriki Apr 15 '12 at 18:19
    
@asepe: Sorry, there was an error in the numbers; I've corrected it. –  joriki Apr 21 '12 at 7:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.