First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 146
Remark 71. Both closed and open Newton-Cotes formulas using odd number of nodes (n
is even), gains an extra degree of accuracy beyond that of the polynomial interpolant on
which it is based. This is due to cancellation of positive and negative error.
There are some drawbacks of Newton-Cotes formulas:
• In general, these rules are not of the highest degree of accuracy possible for the number
of nodes used.
• The use of large number of equally spaced nodes may incur the erratic behavior associated
with high-degree polynomial interpolation. Weights for a high-order rule may
be negative, potentially leading to loss of significance errors.
• Let I n denote the Newton-Cotes estimate of an integral based on n nodes. I n may not
converge to the true integral as n →∞for perfectly well-behaved integrands.
Example 72. Estimate ∫ 1
0.5 xx dx using the midpoint, trapezoidal, and Simpson’s rules.
Solution. Let f(x) =x x . The midpoint estimate for the integral is 2hf(x 0 ) where h =
(b − a)/2 =1/4 and x 0 =0.75. Then the midpoint estimate, using 6-digits, is f(0.75)/2 =
0.805927/2 =0.402964. The trapezoidal estimate is h [f(0.5) + f(1)] where h =1/2, which
2
results in 1.707107/4 =0.426777. Finally, for Simpson’s rule, h =(b − a)/2 =1/4, and thus
the estimate is
h
1
[f(0.5) + 4f(0.75) + f(1)] = [0.707107 + 4(0.805927) + 1] = 0.410901.
3 12
Here is a summary of the results:
Midpoint Trapezoidal Simpson’s
0.402964 0.426777 0.410901
Example 73. Find the constants c 0 ,c 1 ,x 1 so that the quadrature formula
∫ 1
has the highest possible degree of accuracy.
0
f(x)dx = c 0 f(0) + c 1 f(x 1 )
Solution. We will find how many of the polynomials 1,x,x 2 , ... the rule can integrate exactly.
If p(x) =1, then
If p(x) =x, we get
∫ 1
0
∫ 1
p(x)dx = c 0 p(0) + c 1 p(x 1 ) ⇒ 1=c 0 + c 1 .
0
p(x)dx = c 0 p(0) + c 1 p(x 1 ) ⇒ 1 2 = c 1x 1