# Why is Romberg integration usually based on trapezoidal rule?

The wikipedia article on Romberg Integration says that it's simply Richardson Extrapolation applied to either the Trapezoidal Rule or the Midpoint Rule.

I'm reading out of a couple of textbooks on numerical analysis (Burden and Faires Numerical Analysis 9th Edition and Atkinson Intro to Numerical Analysis 2nd Edition) and both textbooks only mention Romberg in the context of the trapezoid rule.

I was wondering why we can't apply the same thing to, say Simpson's Rule?

I get that in order to apply Richardson Extrapolation (which is what Romberg integration is based on) you need the error in a very particular form, but Simpson's Rule seems to have that form.

• Some advice on picking better titles... in particular, the word "question" is redundant in titles. – user147263 Aug 11 '14 at 21:58
• Does Simpson's Rule have error of the form $c_1 h^5 + c_2 h^7 + ...$? – EpicMochi Aug 12 '14 at 6:52

## 2 Answers

You can apply it to Simpson's Rule. However, Simpson's Rule is obtained by one extrapolation step from the Trapezoidal Rule, so it makes no difference.

In my Applied Numerical Methods course, we are taught using Simpson's method also. Both works equally well. However, Trapezoidal rule works for any interval length as compared to Simpson's rule which needs even number of intervals. So if we have say 5 values of h, we can apply trapezoidal rule for all those 5 values. But Simpson's method may not apply on certain h values. Those will have to be evaluated by Trapezoidal rule. While coding these on any computer, it is easier then to use Trapezoidal rule since writing the algorithm would be easy. Hope this makes sense.