First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 158
In [1]: function gauss(f::Function)
0.2369268851*f(-0.9061798459)+
0.2369268851*f(0.9061798459)+
0.5688888889*f(0)+
0.4786286705*f(0.5384693101)+
0.4786286705*f(-0.5384693101)
end
Out[1]: gauss (generic function with 1 method)
Now we compute 1 4
∫ 1
−1
( t+3
) t+3
4
dt using the code:
4
In [2]: 0.25*gauss(t->(t/4+3/4)^(t/4+3/4))
Out[2]: 0.41081564812239885
The next theorem is about the error of the Gauss-Legendre rule. Its proof can be found in
Atkinson [3]. The theorem shows, in particular, that the degree of accuracy of the quadrature
rule, using n nodes, is 2n − 1.
Theorem 80. Let f ∈ C 2n [−1, 1]. The error of Gauss-Legendre rule satisfies
∫ b
a
f(x)dx −
n∑
w i f(x i )=
i=1
2 2n+1 (n!) 4 f (2n) (ξ)
(2n +1)[(2n)!] 2 (2n)!
for some ξ ∈ (−1, 1).
Using Stirling’s formula n! ∼ e −n n n (2πn) 1/2 , where the symbol ∼ means the ratio of the
two sides converges to 1 as n →∞, it can be shown that
2 2n+1 (n!) 4
(2n +1)[(2n)!] 2 ∼ π 4 n .
This means the error of Gauss-Legendre rule decays at an exponential rate of 1/4 n as opposed
to, for example, the polynomial rate of 1/n 4 for composite Simpson’s rule.
Exercise 4.3-1: Prove that the sum of the weights in Gauss-Legendre quadrature is 2,
for any n.