Lecture 4 - Computer Science Department - University of Kentucky

CS321
Numerical Analysis
Lecture 4
Numerical Integration
Professor Jun Zhang
Department of Computer Science
University of Kentucky
Lexington, KY 40206‐0046
October 1, 2010

Definite Integral
A definite integral has an interval for integration. For a fixed integration
interval, the result is a number
π
2
∫ sin x dx = 1
0
An indefinite integral does not have an integration interval. The result
of an indefinite integral (antiderivative) is a class of functions
∫
sin x dx = − cos x + C
Numerical integration is for computing definite integrals
Fundamental Theorem of Calculus:
∫
b
a
x
∫ a
F'(
t
) dt = F
( x
) − F
( a
)
a
F'(
x)
dx = F(
b)
− F(
a)
2

Numerical Integration

Partition
The definite integral of a function can be viewed as the area under a
curve. This point of view lends us means to compute dfii definite integral
Let P be a partition of the interval of [a,b] as
P
{ a = x < x < x < < x < x b}
= 0 1 2 L n −1 n =
We have n subintervals as [x i ,x i+1 ]. Let m i be the greatest lower bound of
(a nonnegative function) f(x) on [x i ,x i+1 ] as
m
i
{ f ( x)
: x ≤ x ≤ x }
= i
inf i
+
and M i as the least upper bound on the same subinterval
M
i
{ f ( x)
: x ≤ x ≤ x }
= i
sup i
+
1
1
4

Partition

Lower and Upper Sums
The lower sums and upper sums of f corresponding to the given partition
P is
n−11
L(
f ; P)
= m x − x
U
(
f
;
P
) =
∑
i=
0
n−1
∑
=
i=
0
i
M
( )
i+
1
( x
x
)
i+ 1 −
If we consider the definite integral of a nonnegative f as the area under
the curve, we have
L ( f ; P)
f ( x)
dx ≤ U(
f ; P)
for all partitions P
∫
≤ b a
i
i
i
If f is continuous on [a,b], then the above inequality defines the definite
integral. The integral also exists it if f is monotone (ith (either increasing i or
decreasing) on [a,b]
6

Upper and Lower Bounds

Riemann‐Integrable Functions
If the greatest lower bound equals the least upper bound for all partitions of
[a,b], i.e.,
inf U(
f ; P)
= sup L(
f , P)
Then f is said to be Riemann‐integrable
P
Every continuous function defined on a closed and bounded interval of the
real line is Riemann‐integrable
We have
lim
n→∞
L(
f ; P
n
)
= ∫
b
a
P
f ( x)
dx =
lim U(
f ; P
n→∞
where P 0 , P 1 ,.., P n ,… are a sequence of partitions such that the length of the
largest subinterval in P n converges to 0 as n → ∞
We can construct nested (refined) partitions
n
)
8

Computation
1.) need a procedure to evaluate f(x)
2.) determine a partition (how many subintervals) of the interval [a,b]
3.) compute m i and M i on each subinterval
4.) compute the sums L(f; P) and U(f; P)
5.) an approximate value is obtained
b
a
[ U( f ; P) L( f P)
]
1
∫ f ( x)
dx ≈ + ;
2
6.) the error of this approximation is bounded by
[ U ( f
;
P
) L
( f
;
P
)
]
1
−
2
9

Trapezoid Rule
A strategy that is better than estimating both the upper and the lower
bounds of the area beneath the curve is to use trapezoids
The interval [a,b] is first divided into subintervals [x i ,x i+1 ], 0 ≤ i ≤ n –1. A
typical trapezoid has the subinterval [x i,x i+1 1] as its base, and the two
vertical sides are f(x i ) and f(x i+1 ). The area is given by the base times the
average height. The basic trapezoid rule for the subinterval [x i ,x i+1 ] is
x i
[ f (x ) + f (x )
](x
− x )
∫ + 1
f (x) dx ≈ 1
i i+
1 i+
x
2
1
i
The total area under the curve is
b
∫ f ( x)
dx ≈ T(
f ; P)
= ∑ − a
i=
n
1
A i
0
i
=
−1
1
1 n ∑ ( xi+
1 − xi
)[ f ( xi
) + f ( xi+
1)]
2 i=
0
10

Trapezoid Rule

Uniform Spacing
The lengths of the subintervals in a partition can be different. For fast
computation, a uniform partition of the interval may be advantageous
Let n be the number of subintervals, then h = (b – a)/n is the uniform
interval spacing. The nodal points are x i = a + i h, i = 0, 1,…, n. Hence
the composite trapezoid rule is
n
∑ − 1
h
T(
f ; P)
= [ f ( xi ) + f ( x + 1
2
)] i
2 i=0 0
Note that the end point of an interval is the starting point of the next
interval. This fact can save almost half of the computation,
∫
b
a
f
( x)
dx
≈
⎧
h⎨
⎩
n−1
∑
i=
1
f
( x
) +
1
⎫
[ f ( x0 ) + f ( )] ⎬
2
⎭
i x n
12

Trapezoid Rule with Uniform Spacing

Error Analysis (1)
If f”(x) exists and is continuous on [a,b], the error of the composite
trapezoid rule T is
for some ξ in (a,b)
b
b−a
2
2
∫ f ( x)
dx − T = − h f "( ξ ) = O(
h )
12
a
Proof. We first prove the result for a=0, b= 1 and h=1. That is
∫
1
0
[ 1
f
(0) +
f
(1) ] = −
f "(
ξ
)
f ( x
)
dx −
1
2
+
12
This simplified formula will be proved with the help of polynomial
interpolation
Define a polynomial of degree one that interpolates f at 0 and 1
p ( x
)
=
f
(0)
+
[
f
(1) −
f
(0
)]
x
14

Error Analysis (2)
It follows that
∫
1
0
p(
x)
dx
=
=
f (0) +
1
2
1
[
2
[ f (0) + f
f (1) −
(1)]
f
(0)]
Using the error formula for the polynomial interpolation, we have
Integrate it on both sides,
∫
1
0
f(x)
− p(x)
=
1 f"[
(x)]x(x
( −
2
ξ 1
)
2
1
1 1
f ( x)
dx −∫
p(
x)
dx = ∫ f "[ ξ(
x)]
x(
x −1)
dx
0
2
0
15

Error Analysis (3)
Using the Mean‐Value Theorem for Integrals
So we have
∫
1
0
∫
1
f "[ ξ ( x)]
x(
x −1)
dx =
f "[ ξ ( s)]
0 0
f ( x)
dx −
1
2
[
f
(0)
+
= −
We then do a change of variable, and let
f
1
6
(1)]
∫
1
f "( ζ )
=
x(
x −1)
dx
−
1
12
f "( ζ )
g( t)
= f [ a + t(
b − a)],
x = a + ( b − a)
t
dx
= ( b
− a)
dt,
g'(
t)
'[ a
g"(
t
) =
f
"[ a
+
t
(
b − a
)](
b − a
)
=
f
+ t(
b
2
− a)](
b
−
a)
16

Error Analysis (4)
Then using the result for the special case,
∫
a
b 1
f (x)dx
=
(b −
a)
= (b − a)
=
∫
∫
1
0
0
f[a
+
t(b −
a)]dt
g(t)dt
⎧
1
1
⎫
= (b − a) ⎨ [g( 0)
+ g( 1)]
− g"( ζ ) ⎬
⎩2
12 ⎭
3
b − a
(b − a)
[f (a)
+
f (b)] − f"(
ξ
)
2
12
This is the error formula for the trapezoid rule with only one subinterval.
Let [a,b] be divided into n equal subintervals by points
x
, x1,...,
xn,
with subinterval[
xi
,
xi
1
]
0 +
17

Let h be the interval length
Error Analysis (5)
x i h
∫
+ f
(x)dx =
[f (x
)
+
f (x
x i 2
1
)] −
h
12
=
i i+1
f"(
ξ
)
1 3
Sum over all subintervals to get the composite trapezoid rule
b
−1
x
−1
−1
i
∫ = ∑ ∫
+ 1
f ( x)
dx f ( x)
x = ∑ [ f ( xi
) + f ( xi
+ 1)]
− ∑
a
n
i=
0
x
i
h
2
n
i=
0
3
h
12
n
i=
0
f "( ξ )
i
Note that h=(b-a)/n, we use Intermediate-Value Theorem of Continuous
Functions,
3 1
1
h n −
n−
b − a 2 ⎡1
− ∑ f "( ξi
) = − h
12 i
12
⎢ ∑
= 0 ⎣n
i
=
0
⎤
f "( ξi
) ⎥ ⎦
i i ⎦
= −
b − a
12
h
2
f
"( ζ )
18

Example 1
Show that
a
h
∫
+ f ( x)
dx
a
h
2
=
3
h
[ f a)
+ f ( a + h)
] − f "( a)
+ L
(
12
Need to define
F ( t)
f ( x)
= ∫
t a
We can expand F(a+h) using Taylor series
dx
2
3
h h
F
( a+ h
) = F
( a
) + hF'(
a
) + F"(
a
) + F"'(
a
) +L
2 3!
Then by the Fundamental Theorem of Calculus, we have F’(t) = f(t).
Note that F(a) = 0, F”(t)=f’(t), F”’(t)=f”(t), and so on.
19

We have
Example 2
+ 2
3
a h
h h
∫ f
( x
) dx = hf
( a
) +
f '(
a
) +
f "(
a
) +L
3!
(1)
a
2
We can also apply the Taylor series directly on f(t) as
2
3
(2)
h h
f
( a
+ h
) =
f
( a
)
+
hf
'( a
)
+
f
"( a
)
+
f
"'( a
)
+L
2 3!
Adding f(a) on both sides of (2) and multiplying it by h/2, we obtain
(3)
h
[
2
2
3
h
h
f ( a)
+ f ( a+
h)]
= hf(
a)
+ f '( a)
+ f "( a)
+L
2 4
Subtracting (3) from (1), we finally get
a
∫ + h h
1 3
f ( x)
dx−
[ f ( a)
+ f ( a+
h)]
=− h f"(
a)
+L
a
2
12
20

Estimate Grid Spacing
Example. If the composite trapezoid rule is used to compute
1 2
−
∫ 0
e x
dx
with an error of at most 0.5 × 10 -4 , what is the uniform grid spacing h?
From the graph of the second derivative (a decreasing function)
f "( x)
= (4x
2 − x
− 2) e
2
We find that
f "(
x)
≤ f "(0) =
2
We need
−
b − a
12
h
2
f "( ξ )
≤
1
6
h
2
<
0.5×
10
−4
It follows that h ≤ 0.01732. The number of subintervals is n ≥ [1/h] = 58
21

Recursive Trapezoid Idea
What can we do if the initial partition of interval is not fine enough?
22

Recursive Trapezoid Formula
Given a parameter n, dividing [a,b] into 2 n equally spaced subintervals, we
have
n 1
h
T(
f ; P)
= h f ( xi ) + [ f ( x0 ) + f ( x n )]
2
∑ −
i=
1
n
= h∑ − f ( a + i h)
+ [ f ( a)
+ f ( b)
]
i=
1
2
Note that n = 2 n and h = (b – a)/2 n
Note that n 2 and h (b a)/2
R(
n,0)
=
h
n
2
∑ − 1
i=
1
1
f ( a +
h
h
i h)
+
2
[ f ( a)
+ f ( b)
]
Notice that R(n,0) can be viewed as dividing each subinterval of
R(n –1,0)into two equal sub‐subintervals. If we already computed
R(n –1,0), how can we compute R(n,0) cheaply?
23

Recursive Formula
If R(n –1,0)is available, R(n,0) can be computed as
R(n, 0)
1
= R(n −1,
0)
+ h
2
n−
2
1
∑
For n ≥ 1 using h = (b – a)/2 n . Initial starting value is
−
1
k=
1
f[a + ( 2k
−1)h]
1
R ( 0,
0)
= 2
(b − a)[f (a) + f (b)]
The trick is to only sum the function values at every other grid points
Proof. Note that
R
(
n,0)
=
h
2
n
1
∑
− f
(
a
+
i
h
)
+
C
i=
1
with C = h[f(a) + f(b)]/2 and
R(
n − 1,0) = 2h
−
2
n 1
∑ − 1
f ( a + 2 j h)
+ 2C
j=
1
24

Recursive Formula
Hence, we have
R(
n,0)
−
1
2
R(
n −1,0)
=
h
n
n−11
2 −
1
2 −
1
∑
i=
1
f ( a + ih)
− h
∑
j=
1
f
( a + 2
jh)
=
h
Each term in the first sum that corresponds to an even value of l is
cancelled by a term in the second term. The final result has only the odd
values of l
2
n−1
∑
k = 1
f [ a + (2k
−1)
h]
We can use the recursive trapezoid formula to compute a sequence of
approximations to a definite integral using the trapezoid rule, without
recomputing the values at points that have already been computed in the
previous step
25

Two Dimensional Integration
For one dimensional numerical integration on [0,1], using uniform space
h = 1/n
n−1
1 1
⎛ i ⎞
∫
f
(
x
)
dx
≈
[
f
(0)
+
2
∑
f
⎜
⎟ +
f
(1)]
0
2h
⎝ n ⎠
i=
1
n
⎛ i ⎞
= ∑ Ai
f ⎜ ⎟
i= 0
⎝
n
⎠
For two dimensional integration on a unit square
1 1
∫ ∫
0 0
f
(
x
,
y
)
dx
dy ≈
=
≈
1
n
∫ ∑
0
i=
1
n
∑
i
=
0
n
∑
i=
0
= n
A
A
n
A
i
∑∑
i= 0 j=
0
∫
i
1
0
n
∑
i
j=
0
A
⎛ i
f
⎜
,
⎝ n
i
f
⎛ ⎜
⎝
A
A
j
j
i
n
f
f
⎞
y
⎟
dy
⎠
⎞
, y⎟
dy
⎠
⎛
⎜
⎝
⎛
⎜
⎝
i
n
i
n
,
,
j
n
j
n
⎞
⎟
⎠
⎞
⎟
⎠
26

Romberg Algorithm
Recursive composite trapezoid method
For h = 1/2 n and n ≥ 1
1 R ( 0,0) 0)
=
(
b
−
a
)[
f
(
a
)
+
f
(
b
)]
2
n−
1
R(n, 0)
=
2
Using Richardson extrapolation, we can have
2 1 − 1
R(n −1,
0)
+ h ∑f[a
+ ( 2k
−1
2 k=1
R(
i,
j)
= R(
i,
j − 1)
+ −
j
1
j
[ R(
i,
j − 1) − R(
i − 1, −
4 1
For i ≥ j and j ≥ 1. This is the Romberg algorithm, which may yield better
approximate values for larger j
1)]
)h]
27

Deriving Romberg Algorithm
Composite trapezoid rule on 2 n-1 subintervals
∫
b
a
f
( x
)
dx =
R
( n −
1,0)
+ a
h
+ a
h
+ a
h
2
4
6
+ L
with h = (b – a)/2 n-1 and the coefficients a i depend on f but not on h
After one refinement and replacing n –1 with n and h with h/2, we have
b
a
∫ f ( x)
dx
=
1 1
1
R( n,0)
+ a h + a h + a h
4
2
Subtracting the 1 st equation from 4 times the 2 nd equation
where for n ≥ 1
∫
R
a
b
2
16
2
4
4
1
4
4
64
4
6
6
16
6
+ L
4 5
f (x) dx = R(n, 1)
− a h − a h
1
( n,1)
= R(
n,0)
+ [ R(
n,0)
− R(
n −
3
6
6
1,0)]
+L
28

More Romberg Algorithm
We could apply the extrapolation idea repeatedly to get
where
∫
b
a
1 6 21
f ( x)
dx = R(
n,2)
+ a h + a h
4
3
6
4
5
8
8
+L
1
R( n,2)
= R(
n,1)
+
15
[ R(
n,1)
− R(
n −1,1)]
This time, the truncation error is of sixth order
A few steps of extrapolation may generate very accurate
approximations
Too many extrapolations may make the computation tedious
29

General Extrapolation
Extrapolation processes can be applied in more general cases where the error
term can be represented as
E
=
a
h
α
+ b h
β
+ c h
γ
+ L
with 0 < α < β < γ, we show how the first term of the error expansion is
annihilated. Let
α β γ
L = φ( h)
+ a h + b h + c h + L (1)
Replacing h by h/2 yields
h
h α
h β
h γ
L = φ ( ) +
a
( ) +
b
( ) +
c
( ) + L
(2
)
2 2 2 2
Multiplying (2) by 2 α
α
α
h α α h β α h γ
( ) + a h + 2 b ( ) + 2 c ( ) + (3)
2 L = 2 φ
2
2
2
L
30

General Extrapolation – Cont.
Subtracting the previous two equations, we can remove the h α term
α
α
⎛ h⎞
⎞
(2 − 1) L = 2 φ⎜
⎟ − φ(
h)
⎝ 2 ⎠
+ (2
α −β
− 1) b h
β
+ (2
α −γ
We can write the new approximation formula as
α
2 ⎛ h⎞
1
L = φ⎜
⎟ − φ(
h)
+ bh
α
α
2 − 1
⎝
2
⎠
2 −
1
− 1) c h
β
γ
+ ch
+ L
γ
+ L
This approximation formula raises the order of truncation error from
O(h α ) to O(h β ) with α < β
Please read the book on p. 217 for a concrete example to show how the
approximation accuracy is improved using extrapolation
31

Basic Simpson’s Rule
Simple trapezoid rule uses two points for approximations
Can we get more accurate approximation using more points?
32

Basic Simpson’s Rule
A three point numerical integration rule using the middle point of the
interval is known as the Simpson’s s rule with different weights for each point
a
a
2h
[ f ( a)
+ 4 f ( a + h)
+ f ( a 2h
]
∫ + h
f ( x)
dx ≈
+
3
)
Using Taylor’s expansion, we can find the error term of this approximation
is
5
h (4)
− f ( ξ )
90
For some point ξ in (a,a+2h). This should be compared to the error term of
the simple trapezoid rule O(h 3 )
It is desirable to subdivide the interval adaptively so that refinement is only
placed in the area of large fluctuation of function value
33

Basic Simpson’s Rule
Comparing composite trapezoid rule and Simpson’s rule
34

Basic Simpson’s Rule
Using Taylor series for f(x) at a, we have
1 2
1 3
1 4 (4)
f(
a+ h)
= f(
a)
+ hf'(
a)
+ h f"(
a)
+ h f"'(
a)
+ h f ( a)
+L
2! 3! 4!
If we replace the interval size h by 2h, we obtain
4
2 4
3 2
4
(4)
f
( a+ 2 h
) =
f
( a
) +
2
hf
'( a
) +
2
h
f
"( a
) +
h F
"'( a
) +
h
f
( a
) +L
3 4!
By combining these two expansions, we get
2
3 20 4 (4)
f(
a)
+ 4f(
a+
h)
+ f(
a+
2 h)
= 6f(
a)
+ 6hf'(
a)
+ 4h
f"(
a)
+ 2h
f"'(
a)
+ h f ( a)
+L
4!
35

On the other hand, define
We expand F(a+2h) as
Basic Simpson’s Rule
F (
x
)
= x f
(
t
)
dt
a
∫
5
2 4
3 2
4
(4) 2
5
(5)
F ( a
+ 2 h
) =
F ( a
)
+
2
hF '( a
)
+
2
h F
"( a
)
+
h F
"'( a
)
+
h
f
( a
)
+
h F
( a
)
+L
3 3 5!
Note that F’=f, F(a)=0, F”=f’, F”’=f”, we have
∫ + 5
a 2h
2 4 3 2 4 2 5 (4)
f(
x)
dx=
2hf(
a)
+ 2h
f'(
a)
+ h f"(
a)
+ h f"'(
a)
+ h f ( a)
+L
a
3 3 5⋅4!
From the previous page, we have
h
[ f ( a)
+ 4 f ( a + h)
+ f ( a + 2h)]
=
3
2 4 3
2hf
( a)
+ 2h
f '(
a)
+ h f "( a)
3
+ 2 4 20 5 (4
h f "'(
a)
+ h f
) ( a)
+L
3 3⋅
4!
36

Basic Simpson’s Rule (1)
Subtracting the previous two terms, we have
a
∫ + 2h
a
5
h
h (4)
f ( x)
dx=
[ f ( a)
+ 4f
( a+
h)
+ f ( a+
2h)]
− f −L
3
90
We have the Simpson’s rule as
a
+ 2h
b−a
⎛ a+
b⎞
∫
f ( x ) dx≈
[
f
( a
) + 4
f
⎜
⎟+
f
( b
)]
a
6
⎝ 2
⎠
The error term of the Simpson’s rule us
5
1 (4)
⎛b−a⎞
− ⎜ ⎟ 90 ⎝
2
⎠
f
( ξ)
37

Adaptive Simpson’s Algorithm
Reduce the size of the intervals to get more accurate approximations
38

Adaptive Simpson’s Algorithm
Given an interval [a,b], we can use the basic Simpson’s rule to compute
an approximation to the integral as
where the approximation part is
I ∫ f ( x)
dx = S(
a,
b)
+ E(
a,
b)
≡ b a
( b − a)
⎡ ⎛ a + b ⎞ ⎤
S( a,
b)
= ⎢ f ( a)
+ 4 f ⎜ ⎟ + f ( b)
6
⎥
⎣ ⎝ 2 ⎠ ⎦
and the error term is
E(
a,
b)
= −
b−a
5 (4)
( ) f ( )
1
ξ
90 2
For simplicity, we assume f (4) (x) remains constant on (a,b). Let h = b – a,
we have
(1) (1)
I = S + E
For the first step approximation with
(1)
S = S(
a,
b)
39

Adaptive Simpson – (II)
And
E
(1)
=
−
1 ⎛ h⎞
⎜
⎟
90
⎝
2
⎠
5
f
(4)
We then subdivide the interval [a,b] and apply the basic Simpson’s rule
on the subintervals [a,c] and [c,b] respectively. We have a new
approximation on [a,b] as the sum of two separate approximations
where c = (a + b)/2 with
(2)
(2)
I = S +
S = S(
a,
c)
+
E
(2)
S(
c,
b)
and
5
5
(2) 1 ⎛ h / 2 ⎞ (4) 1 ⎛ h / 2 ⎞ (4) 1 (1)
E = − ⎜ ⎟ f − ⎜ ⎟ f = E
90 ⎝ 2 ⎠ 90 ⎝ 2 ⎠ 16
This is certainly til a better btt approximation since the subintervals are smaller
than the original interval
20

Adaptive Simpson – (III)
Subtracting the two approximations yields
(2)
(1)
(1)
(2)
S −
S
=
E −
E
=
15E
Hence the numerical integration can be
(2)
(2)
(2)
I =
S
+
E =
S
+
(2)
(2)
1
(
S −
S
15
The error term is then computable and can be used for building the
adaptive process
1
(2)
(1)
S − S < ε
15
If this test shows that the error is larger than ε, the interval [a,b] can be
split into two subintervals [a,c] and [c,b] with c = (a + b)/2. The
previously described procedure is replaced by ε/2 to make sure that the
error sum is smaller than ε
(1)
)
41

Adaptive Simpson’s Algorithm
Refine the intervals at the places where the function changes quickly
42

Adaptive Simpson – (IV)
Numerical integration on subintervals
b
c
b
I = ∫ f (x) dx = ∫ f (x) dx + ∫ f (x) dx = IL
+ I
a
(2)
L
a
Let S be thesumof S on[
a,c]
and S on[
c,b]
,wehave
I
If we want to have
−
S
=
I
I
L
L
1
15
It is more than enough to have
and
≤
=
S
+
I
−
R
(2)
S L
(2)
L −
−
I
S
− S
c
(2)
R
S
+
(1)
L
(2)
L
I
R
+
≤ ε
−
+
1
15
S
S
1 (2) (1)
S − ≤
ε
15 L S L 2
1 (2)
(1)
S −
≤
ε
15 R S R 2
(2)
R
(2)
R
(2)
R
S
−
S
(1)
R
R
43

Adaptive Simpson’s Algorithm
One decision to make is to choose where to refine the interval
44

Computational Procedure
The interval [a,b] is divided into four subintervals of equal length. Two
Simpson approximations are computed using two double‐width subintervals
and four single‐width subintervals
S
S
1
=
h
⎡
⎛ a
+
b
⎞
⎢ f ( a)
+ 4 f
⎜
⎟ +
6 ⎣ ⎝ 2 ⎠
h ⎡
=
12
⎢
⎣
⎤
f ( b)
⎥
⎦
⎛ a + c ⎞
⎛ c + b ⎞
f
( a
) + 4
f
⎜
⎟ + 2
f
( c
) + 4
f
⎜
⎟ +
⎝
2
⎠
⎝
2
⎠
2 f
⎤
( b
)
⎥
⎦
If |S 2 -S 1 | ≤15ε, we have done, and set
1
S =
[ 16
S2
−
S1
]
15
Otherwise the interval [a,b] is divided in half and the recursive procedure is
applied on the two subintervals [a,c] and [c,b], until either the error tolerance
is satisfied or the maximum number of subdivisions is reached
45

Adaptive Simpson’s Algorithm
Which sub‐interval or both to divide for refinement
46

Gaussian Quadrature Formulas
A general numerical integration formula is
∫
b
a
f x
) dx ≈ A
f
( x
) + A
f
( x
) + L+
( 0 0
1
1 A
n
f
( x
n
It suffices to know the nodes x 0 , x 1 ,…, x n and the weights A 0 , A 1 ,…, A n .
For important special functions, they are listed in some reference books
Suppose a set of nodes is given, how to find the weights. This can be
done using Lagrange interpolation polynomial as
n
p(
x)
= f ( x ) l ( x)
With
l
(
x
)
i
=
∑
i=
0
n
∏
j=
o,
j≠1
⎛
⎜
⎝
i
i
i
x −
x − x
x i
j
⎟⎞
⎟
⎠
)
If p is a good approximate to f, we anticipate
approximate to ∫ b f ( x
) dx
a
∫ b a
p ( x)
dx
is a good
47

We integrate over p(x) as
∫
b
a
f ( x
)
dx ∫ p
( x
)
dx
where we can compute
Gaussian Quadrature
b
≈ ∫ a
n
n
b
= ∑ f ( xi
) ∫ l =
a i ( x)
dx ∑
i=
0
i=
1
i
∫
= b l
a i
A ( x)
dx
A
i
f ( x
Note that the polynomial interpolation is exact for a polynomial of
degree at most n. It follows that the integration will be exact for such
polynomials
If the nodes can be chosen carefully, it is possible to increase the order
of polynomial with the exact integration remarkably. This was
discussed by Karl Gauss
i
)
48

An Example
Determine a quadrature formula when the interval is [‐2,2] and the
nodes are ‐1, 0, and 1.
We first need to compute the cardinal functions
2 ⎛ x − x ⎞
j 0 1 1
( ) ⎜ ⎟
⎛ x − ⎞⎛
x − ⎞
l0 x = ∏ = ⎜ ⎟⎜ ⎟ = x(
x −1)
j
=
1 ⎝ x
0
− x
j
⎠
⎝
( −1)
− 0
⎠⎝
−1−1
⎠ 2
2 ⎛ x − x ⎞
j ⎛ x + 1 ⎞⎛
x −1⎞
1 2
l ( x)
= ∏⎜
⎟ = ⎜ ⎟ ⎟ = − x
1 ⎜
j 1 x1
x
= ⎝ −
j ⎠ ⎝ 0 − ( −1)
⎠⎝
0 −1⎠
2 ⎛ x − x ⎞
1) ⎞
− 0
1
⎛ x − ( −
⎛ x ⎞
2
( ) = ∏ ⎜ j
l x
⎟ =
⎜
⎟
⎜
⎟ = x
( x
+
1)
j=
1 ⎝ x2
− x
j ⎠ ⎝ 1−
( −1)
⎠⎝
1−
0 ⎠ 2
The weights are computed by integrating these functions
A
2
2 1 2
1
⎡1
3 1 2
⎤
= ∫ l ( x)
dx = ( −1)
=
=
0 2 0
2
2
2 ⎢
−
− ∫ x x dx x x
−
⎣3
2 ⎥
⎦ −2
8
3
49

Similarly, we have
A
An Example
2 2
2
2
1
3
( )
(1 )
⎤
1 2 1
2 ⎢
⎡ = ∫ l x dx = − = − =
3 ⎥
⎣ ⎦
−
− ∫ x dx x x
−
−2
2
=
2
2 1 2
1 ⎡1
3 1 2 ⎤
A = ∫ l
( x
) dx =
+ 1) =
x(
x
dx
x + x
2 2 2
2
2
2
⎢
⎣
3
2
⎥ − ∫ −
⎦
So the quadrature formula defined on the interval [‐2,2] and using the
node ‐1, 0, 1, is
2
∫ − 2
8
4
8
f ( x)
dx≈
f ( −1)
− f (0) +
3 3 3
f (1)
It can be verified that this formula gives exact values for the three
functions
2
f ( x)
= 1, x,
x
−2
8
3
4
3
50

Gaussian Quadrature Theorem
Let q be a nontrivial polynomial of degree n + 1 such that
∫
b
a
x
k
q( x)
dx = 0 (0 ≤ k ≤ n)
Let x 0 , x 1 ,…, x n be zeroes of q. Then we define the formula
b
∑
∫
f
(
x
)
dx ≈
A i
f
(
x i
),
A i
=
∫
l
(
x
)
dx
a
≈ n
i=
0
With these x i ’s as nodes, the approximation will be exact for all
polynomials of degree at most 2n+1. All these nodes lie in the open
interval (a,b)
We can first figure out the quadrature for
∫
1
−1
f
( t
)
dt ≈
Then use the transformation t = [2x –(b – a)]/(b – a) for a Gaussian
quadrature on the general interval [a,b]
n
∑
i=
0
A
f
(
i t i
)
b
a i
51

The transformed integral is
∫ a f ( x
) dx
∫ b
=
Gaussian Theorem Proof
1
[
1
1
( b − a)
∫ −
f ( b − a)
t + ( b + a ]dt
1 2
1 2
2
)
Proof of Gaussian Quadrature Theorem:
Let f be any polynomial of degree at most (2n +1) 1). Dividing f by q with
quotient p and a remainder r
f = p q +
Both q and r are of degree at most n
r
By hypothesis, we have
∫ b a
Since x i are roots of q, we have
q( x)
p(
x)
dx =
f ( xi
) = p(
xi
) q(
xi
) + r(
xi
) = r(
xi
)
0
52

Gaussian Theorem Proof (II)
Since the degree of r is at most n, the integration
∫ b a
r ( x)
dx
is exact
∫
b
a
f ( x)
dx
=
∫
b
a
=
∫
r
(
x
)
dx
b
a
p(
x)
q(
x)
dx +
=
n
∑
A r
(
x
)
=
∫
b
a
n
∑
i
i
i= 0 i=
0
r(
x)
dx
A
i
f
(
x
Gaussian Quadrature Theorem guarantees high accuracy numerical
integration with just a few nodes. However, finding these nodes is not an
easy task. The roots of Legendre polynomials are the nodes for Gaussian
quadrature on the interval [-1,1]. With q 0 (x) = 1, q 1 (x) = x, we have for
n ≥ 2
q
n
⎛ 2n
− 1⎞
⎛ n − 1⎞
( x)
= ⎜ ⎟xqn−1(
x)
− ⎜ ⎟qn−2(
x)
⎝ n ⎠ ⎝ n ⎠
i
)
53