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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
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

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Numerical_Methods_Simpson_Rule

  • 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