Simpson's Rule demostration using Lagrange's polynomial approximation of degree 2

1. Centro de Investigaci´on y de Estudios Avanzados del Instituto Polit´ecnico
Nacional
Unidad Guadalajara
Sistemas el´ectricos de Potencia
M´etodos n´umericos y computaci´on: Investigaci´on
November 25, 2019
Nombre: Alexis Hern´andez San Germ´an
Derive the equation (1)
x2
x0
f(x)dx ≈
h
3
(f0 + 4f1 + f2) (1)
Start with the Lagrange polynomial PM (x) based on x0, x1, . . . , xM that can be used to
approximate f(x):
f(x) ≈ PM (x) =
M
k=0
fk LM,k(x) (2)
where fk = f(xk) for k = 0, 1, . . . , M. An approximation for the integral is obtained by
replacing the integrand f(x) with the polynomial PM (x). This is the general method for
obtaining a Newton-Cotes integration formula:
xM
x0
f(x)dx ≈
xM
x0
PM (x)dx
=
xM
x0
M
k=0
fk LM,k(x) dx =
M
k=0
xM
x0
fkLM,k(x)dx
=
M
k=0
xM
x0
LM,k(x)dx fk =
M
k=0
wkfk
(3)
The details for the general computations of the coefficients of wk in (3) are tedious. We
shall give a sample proof of Simpson’s rule, which is the case M = 2. This case involves the
approximating polynomial
P2(x) = f0
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
+ f1
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
+ f2
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
(4)
Since f0, f1, and f2 are constants with respect to integration, the relations in (3) lead to
x2
x0
f(x)dx ≈ f0
x2
x0
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
dx + f1
x2
x0
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
dx
+ f2
x2
x0
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
dx
(5)
We introduce the change of variable x = x0+ht with dx = h dt to assist with the evaluation
of the integrals in (5). The new limits of integration are from t = 0 to t = 2. The equal spacing
1

2. nodes xk = x0 + kh leads to xk − xj = (k − j)h and x − xk = h(t − k), which are used to
simplify (5) and get
x2
x0
f(x)dx ≈ f0
2
0
h(t − 1)h(t − 2)
(−h)(−2h)
h dt + f1
2
0
h(t − 0)h(t − 2)
(h)(−h)
h dt
+ f2
2
0
h(t − 0)h(t − 1)
(2h)(h)
h dt
= f0
h
2
2
0
(t2
− 3t + 2) dt − f1h
2
0
(t2
− 2t) dt + f2
h
2
2
0
(t2
− t) dt
= f0
h
2
t3
3
−
3t2
2
+ 2t
t=2
t=0
− f1h
t3
3
− t2
t=2
t=0
+ f2
h
2
t3
3
−
t2
2
t=2
t=0
= f0
h
2
2
3
− f1h
−4
3
+ f2
h
2
2
3
=
h
3
(f0 + 4f1 + f2)
(6)
References
[1] John H. Matthews, Kurtis K. Fink, Numerical Methods using MATLAB 4th Edition,
Pearson Education, 2010
2