5.5 Gaussian Quadrature
Now we come back to the numerical integration based on the polynomial interpolation
1
I   f ( x)dx .
1
The general formula for the numerical integration is
n
I  Ai f ( xi ) .
(1.1)
i 0
If we take
1
x  xj
1 j  i
xi  x j
Ai   
dx ,
(1.2)
Then the formula will be exact for all polynomials f ( x ) of degree  n . For example, the
trapezoid rule, where there are two points x0 and x1 , A0  A1 
x1  x0
, is exact for all
2
polynomials of degree at most 1; the Simpson’s rule, where there are three points
x0  1, x2  1, x1  0, A0 
x1  x0
x x
4( x1  x0 )
, A2  1 0 , A1 
, is exact for all polynomials
6
6
6
of degree at most 3.
Question: Can we select the nodes x 0 ,x 1 ,
,x n such that, the formula (1.1) and (1.2) is exact
for all polynomials of degree at most 2n  1 .
Example. n  0, we take x0  0, A0  2. Now the formula is
g0 ( f )  A0 f (0) .
(1.3)
We claim that (1.3) is exact for all polynomials of degree  1 .
Solution: Since any polynomials f ( x ) of degree  1 is a linear combination of 1 and x , we only
need to check that (1.3) is exact for the trial function 1, x in the following sense


1
1
1dx  A0 f (0)  2  g0 (1) ,
1
xdx  A0 f (0)  2  0  g0 ( x) .
1
1
Example. n  1, we take x0  
3
3
, x1 
, A0  A1  1. The formula for this case is
3
3
g1 ( f )  f (
3
3
) f ( ).
3
3
(1.4)
We say that (1.4) is exact for all polynomials of degree  3 .
Solution. Again, we only need to check it is exact for trial functions for 1, x, x 2 , x3 .
1
g1 (1)  2   1dx ,
1
1
g1 ( x)  0   xdx ,
1
g1 ( x 2 )  (
1
3 2
3
2
)  ( ) 2    x 2 dx ,
1
3
3
3
1
g1 ( x3 )  0   x3dx .
1
Question: What is the general rule?
q(x ) = (x - x 0 )(x - x 1 ) (x - x n ) .
Theorem (Theorem on Gaussian Quadrature)
Suppose q( x) is orthogonal to all polynomials p ( x ) of degree  n in the sense

1
q( x) p( x)dx  0 , for all polynomials p ( x ) of degree  n .
1
(1.5)
Then the formula (1.1) and (1.2) will be exact for all polynomials f ( x ) of degree  2n  1 .
Proof. Let f ( x ) be a polynomial of degree  2n  1 . Divided by q( x) , obtaining a quotient
p ( x ) and remainder r ( x ) . Then
f ( x)  q ( x) p ( x)  r ( x) ,
Where p ( x ) is a polynomial of degree  n , and r ( x ) is a polynomial of degree  n . Using (1.5)
and the fact that (1.1) and (1.2) is exact for all polynomial of degree  n , we have

1
1
n
f ( x)dx   q ( x) p ( x)dx   r ( x )dx   r ( x )dx  Ai r ( xi ) .
1
1
1
1
1
1
i 0
2
Note that
r ( xi )  f ( xi )  q( xi ) p( xi )  f ( xi ) .
Therefore,

1
1
n
f ( x)dx   r ( x)dx  Ai f ( xi ) .
1
1
i 0
Now it is the time to answer the Question before: Can we select the nodes x 0 ,x 1 ,
,x n such
that, the formula (1.1) and (1.2) is exact for all polynomials of degree at most 2n  1 . We
should choose the nodes such that they are the zeros of q(x ) , which is orthogonal to all
polynomials p ( x ) of degree  n . Furthermore, this can be achieved by the following so called
Legendre polynomials.
The Legendre polynomials of degree n
Ln ( x) 
1 dn
[( x 2  1) n ] .
2n n ! dx n
The properties
 2
, mn
2

(1)  Ln ( x) Lm ( x) dx 
.
 mn   2n  1
1
2n  1

mn
 0,
1
(2) The recursive formula
(n  1) Ln1 ( x)  (2n  1) xLn ( x)  nLn1 ( x) .
(3)
L0 ( x)  1
L1 ( x)  x
1
L2 ( x)  (3x 2  1)
2
1
L3 ( x)  (5 x3  3x)
2
1
L4 ( x)  (35 x 4  30 x 2  3)
8
1
L5 ( x)  (63x5  70 x3  15 x)
8
3
(4) Lk ( x)  (1) k Lk ( x) .
(5)

1
L ( x)q( x)dx  0 for all polynomials q( x) of degree  n 1.
1 n
(6) Ln ( x) has n zeros -1< x 0 < x 1 <
< x n-1 < 1.
The zeros for the Legendre polynomial:
(1) L0 ( x)  1 , no zeros.
(2) L1 ( x)  x , x0  0 is the unique zero.
(3) L2 ( x) 
3
3
1
.
, x1 
(3 x 2  1) , there are two zeros x0  
3
3
2
3
3
1
(5 x 3  3 x) , there are three zeros x0   , x1  0, x2 
.
2
5
5
1
(5) L4 ( x)  (35 x 4  30 x 2  3) , there are four zeros.
8
(4) L3 ( x) 
Suppose now the zeros of the Legendre polynomial Ln1 ( x) are given as x 0 ,x 1 ,
,x n . Then
the coefficients can be computed by the following definite integral
1
A j = ò l j (x )dx ,
-1
where
x  xi
.
i j x  x
j
i
i 0
n
l j ( x)  
By using the feature that the Gaussian quadrature is exact for all polynomials of degree  2n  1 .
One can prove (by using the integration by parts) that
Aj 
2wn2（1）
1
, where w n+1 (x ) = (x - x 0 )(x - x 1 )
2
(1  x j )( wn' 1 ( x j )) 2
(x - x n ) .
Example. Use the Gaussian quadrature to compute the approximation of the following definite
integral
1
I   cos xdx  2sin1  1.682941 .
1
4