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So, I was studying Numerical Integration from Hildebrand where I came across Simpson's Rule and 3/8 rule. Now, Simpson's rule has an error of order 5 and degree of precision 3, which is same as 3/8 rule. The difference lies in the Lagrange polynomials we use to interpolate.

In 3/8 rule we need an extra computation point, hence I feel it's computationally insufficient as compared to Simpson's rule. Or if we restate Simpson's rule is better. My question lies, does 3/8 rule have any advantage?

Simpson's Rule: $\int_a^b f(x) dx\approx \frac{b-a}{6}\Big{[}f(a)+4f(\frac{a+b}{2})+f(b)\Big{]}$ with error as $\frac{1}{90}\Big{(}\frac{b-a}{2}\Big{)}^5|f^{4}(z)|$

3/8 Rule: $\int_a^b f(x)dx\approx \frac{(b-a)}{8}\Big{[}f(a)+3f(\frac{2a+b}{3})+3f(\frac{a+2b}{3})+f(b)\Big{]}$ with error as $\Big{|}\frac{(b-a)^5}{6480}f^4(z)\Big{|}$

Reference: Wikipedia for Formula

Abhinav Jha
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    Does the central term in Simpson's rule need a factor of $4$ ? and there is a close bracket needed in the $3/8$ rule ... could someone edit ? – Donald Splutterwit Mar 17 '17 at 13:10
  • The advantage will be how accurate the result is ... the more terms & strips you use the better the approximation will be ... you have answered your own question ? ... the $3/8$ rule will be better with an error $72$ smaller than Simpson's rule. – Donald Splutterwit Mar 17 '17 at 13:14
  • But they both have the same Degree of Precision. Doesn't it mean that they will give me same answers? – Abhinav Jha Mar 17 '17 at 13:22
  • The numerical value returned will not be the same ... the level of accuracy will be of the same order. The more strips you use the better the approximation & the better interpolatating polynomial you use the better the approximation. (There are exceptions to this for very badly behaved integrands.) – Donald Splutterwit Mar 17 '17 at 13:28

2 Answers2

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There are two advantages of the 3/8ths rule: First, the error term is smaller than Simpson's rule.

The second more important use of the 3/8ths rule is for uniformly sampled function integration. Suppose you have a function known at equally spaced points. If the number of points is odd, then the composite Simpson's rule works just fine. If the number of points is even, then you have a problem at the end. One solution is to use the 3/8ths rule. For example, if the user passed 6 samples, then you use Simpson's for the first three points, and 3/8ths for the last 4 (the middle point is common to both). This preserves the order of accuracy without putting an arbitrary constraint on the number of samples.

user14717
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  • If you use the same number of intervals for both Simpson's 1/3 and 3/8, then Simpson's 1/3 gives a more accurate answer, Simpson's 3/8 error term is just the same as Simpson's 1/3 except that you multiply it by 2.25. – Malzahar Apr 03 '20 at 11:05
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For Newton-Cotes formulas, when 'n' is an even integer the degree of precision is n+1 and when n is odd degree of precision is'n'(Ref. Numerical Analysis by Burden). Could someone explain how?

Natasha
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