MthSc 365: Numerical methods for engineers
Lecture 17
Pietro Belotti
Dept. of Mathematical Sciences
Clemson University
December 4, 2012
Numerical integration
Rectangle 1: compute f on lower bound of [xk , xk+1 ], i.e., xk .
Z b
N
X
f (x)dx ≈
f (xk )(xk+1 − xk ).
a
k=1
Rectangle 2: compute f on upper bound of [xk , xk+1 ], i.e., xk+1 .
Z b
N
X
f (xk+1 )(xk+1 − xk ).
f (x)dx ≈
a
k=1
Rectangle 3: compute f on mid bound of [xk , xk+1 ], i.e.,
Z b
N
X
xk+1 + xk
f
f (x)dx ≈
(xk+1 − xk ).
2
a
k=1
Trapezoidal method:
Z b
N
X
f (xk+1 ) + f (xk )
(xk+1 − xk ).
f (x)dx ≈
2
a
k=1
xk+1 +xk
.
2
Gauß quadrature
Let’s generalize the above methods.
Rb
In general, a f (x)dx is approximated as follows:
1. Define a set of points x1 , x2 . . . , N
2. Compute appropriate coefficients Ck , for k = 1, 2, . . . , N
3. Compute the integral as
Z
b
f (x)dx =
a
N
X
Ck f (xk )
k=1
Numerical integration
Z
Rectangle 1:
b
a
f (x)dx ≈
PN
k=1 f (xk )(xk+1
N
X
Ck f (xk )
k=1
− xk ) means Ck = xk+1 − xk = h
Rectangle 2 and 3: Same as R1 (different choice of xk ): Ck = h.
Trapezoidal method:
◮
C1 = CN = 2h .
◮
Ck = h for k = 2, 3, . . . , N − 1.
Simpson’s 1/3 method:
h
3
◮
C1 = CN =
◮
For k even, Ck = 43 h
◮
For k odd and 1 6= k 6= N, Ck = 32 h
Gauß quadrature
Now the question is: “Are there other ways to select Ck ’s?”
Problem: Find x1 , x2 , . . . , xN and C1 , C2 , . . . , CN to approximate
Z
b
a
f (x)dx ≈
N
X
Ck f (xk )
k=1
This clearly depends on f , but we can’t adapt this for every f .
◮
Idea: f can be approximated by a Taylor series expansion
1
f (x) = f (x0 ) + f ′ (x0 )(x − x0 ) + f ′′ (x0 )(x − x0 )2 + . . .
2
⇒ Find the appropriate xk , Ck for 1, x, x2 , x3 . . .
Gauß quadrature
Solve a slightly simpler problem:
Z
1
f (x)dx
−1
And set N = 2. Then find
Z 1
f (x)dx = C1 f (x1 ) + C2 f (x2 ),
−1
at least for the following four functions:
f (x) = 1
f (x) = x2
f (x) = x
f (x) = x3
Gauß quadrature
R1
f (x) dx = C1 f (x1 ) + C2 f (x2 )
−1
R1
1
R−1
1
x
−1
R1
x2
−1
R1
3
x
−1
⇐⇒
=
=
=
=
C1
C1
C1
C1
1
x1
x1 2
x1 3
2 = C1
0 = C1
2
3 = C1
0 = C1
1
x1
x1 2
x1 3
dx
dx
dx
dx
+
+
+
+
+
+
+
+
C2
C2
C2
C2
C2
C2
C2
C2
1
x2
x2 2
x2 3
1
x2
x2 2
x2 3
A system of 4 nonlinear equations in 4 variables C1 , C2 , x1 , x2 .
Gauß quadrature
We need four functions because there are four variables.
Its solution is C1 = C2 = 1, x1 = − √13 , x2 =
√1
3
Hence from now on we can approximate
Z
1
−1
1
f (x)dx = 1 · f − √3 + 1 · f √13
for any function f . (Results may vary.)
Gauß quadrature
For N = 3, then we have six variables ⇒ six functions:
f (x) = 1
f (x) = x3
R1
R−1
1
−1
R1
R−1
1
R−1
1
R−1
1
−1
1
x
x2
x3
x4
x5
dx
dx
dx
dx
dx
dx
f (x) = x2
f (x) = x5
f (x) = x
f (x) = x4
=
=
=
=
=
=
C1
C1
C1
C1
C1
C1
+
+
+
+
+
+
1
x1
x1 2
x1 3
x1 4
x1 5
C2
C2
C2
C2
C2
C2
1
x2
x2 2
x2 3
x2 4
x2 5
+
+
+
+
+
+
C3
C3
C3
C3
C3
C3
1
x3
x3 2
x3 3
x3 4
x3 5
Gauß quadrature
2 = C1
0 = C1
2
3 = C1
0 = C1
2
5 = C1
0 = C1
Solution:
◮
◮
+
+
+
+
+
+
1
x1
x1 2
x1 3
x1 4
x1 5
5 8 5
9, 9, 9
C2
C2
C2
C2
C2
C2
1
x2
x2 2
x2 3
x2 4
x2 5
+
+
+
+
+
+
C3
C3
C3
C3
C3
C3
1
x3
x3 2
x3 3
x3 4
x3 5
(C1 , C2 , C3 ) =
q
q (x1 , x2 , x3 ) = − 35 , 0, 35
We can approximate
q Z 1
f (x)dx = 59 · f − 35 +
−1
8
9
· f (0) +
5
9
·f
q 3
5
Gauß quadrature for the general case: [a, b]
Easy! Just create a variable t such that
◮
◮
x = a ⇔ t = −1
x = b ⇔ t = +1
⇒ x = a + 12 (b − a)(t + 1)
◮
b−a
2 dt
Note: dx =
From Calculus we know that
Z 1 Z b
b−a
1
dt
f a + (b − a)(t + 1)
f (x)dx =
2
2
−1
a
So we simply create g(t) = f a + 12 (b − a)(t + 1)
Z
b
f (x)dx =
a
Z
1
g(t)dt =
−1
5
9
q · g − 35 +
8
9
b−a
2
· g (0) +
5
9
and . . .
·g
q 3
5
Errors in numerical integration
Consider the first technique we’ve seen to estimate an integral:
Z
b
a
f (x)dx ≈
N
X
k=1
f (xk )(xk+1 − xk ).
Let’s restrict to the k-th interval:
Z xk+1
f (x)dx ≈ f (xk )(xk+1 − xk ) = f (xk )(xk+1 − xk ).
xk
The error of this estimate is
Z xk+1
f (x)dx =
xk
where ξk ∈ [xi , xk+1 ]
f (xk )h +
1 ′
f (ξk )h2
2
ξi who?
So
1
2
f ′ (ξi )h2 is the error for some ξi ∈ [xi , xi+1 ].
We don’t care what ξi is. The total error is
E=
N X
1
k=1
2
′
2
f (ξi )h
Note: The average f ′ on [a, b] is f ′ =
N
h2 X ′
=
f (ξi )
2
k=1
1
N
1
E = h2 Nf ′
2
Now h =
b−a
N ,
so E =
1 b−a
2 N
· Nhf ′ =
PN
k=1 f
b−a ′
2 hf
′ (ξ
i ).
Hence,
∈ O(h)
Errors in numerical integration
For the rectangular method that uses f (xi+1 ), the error is similar.
However, for the mid-point this changes substantially:
E=
b − a ′′ 2
f h ∈ O(h2 ),
24
much better (h2 tends to zero much faster for h ↓ 0). Similarly
for the trapezoidal method:
E=
b − a ′′ 2
f h ∈ O(h2 ).
12
Simpson’s 1/3 and 3/8 methods behave better:
E1/3 =
E3/8 =
b−a IV 4
180 f h
b−a IV 4
80 f h
∈ O(h4 )
∈ O(h4 )
Simpson’s 3/8 is better than 1/3: h3/8 =
b−a
3N
and h1/3 =
b−a
2N .