First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 149
Exercise 4.2-1: Show that the quadrature rule in Example 74 corresponds to taking
n =4in the composite Simpson’s formula (4.4).
Exercise 4.2-2: Show that the absolute error for the composite trapezoidal rule decays
at the rate of 1/n 2 , and the absolute error for the composite Simpson’s rule decays at the
rate of 1/n 4 , where n is the number of subintervals.
Example 75. Determine n that ensures the composite Simpson’s rule approximates
∫ 2
1 x log xdx with an absolute error of at most 10−6 .
Solution. The error term for the composite Simpson’s rule is b−a
180 h4 f (4) (ξ) where ξ is some
number between a =1and b =2,andh =(b − a)/n. Differentiate to get f (4) (x) = 2 x 3 . Then
b − a
180 h4 f (4) (ξ) = 1
180 h4 2 ξ 3 ≤ h4
90
where we used the fact that 2
ξ 3
than 10 −6 , that is,
≤ 2 1
=2when ξ ∈ (1, 2). Now make the upper bound less
h 4
90 ≤ 10−6 ⇒ 1
n 4 (90) ≤ 10−6 ⇒ n 4 ≥ 106
90 ≈ 11111.11
which implies n ≥ 10.27. Since n must be even for Simpson’s rule, this means the smallest
value of n to ensure an error of at most 10 −6 is 12.
Using the Julia code for the composite Simpson’s rule that will be introduced next,
we get 0.6362945608 as the estimate, using 10 digits. The correct integral to 10 digits is
0.6362943611, which means an absolute error of 2 × 10 −7 , better than the expected 10 −6 .
Julia codes for Newton-Cotes formulas
We write codes for the trapezoidal and Simpson’s rules, and the composite Simpson’s rule.
Coding trapezoidal and Simpson’s rule is straightforward.