Presented at Computer Science Department, Sharif University of Technology (Advanced Numerical Methods).

1. 1
Gaussian
Integration
M. Reza Rahimi,
Sharif University of Technology,
Tehran, Iran.

2. 2
Outline
• Introduction
• Gaussian Integration
• Legendre Polynomials
• N-Point Gaussian Formula
• Error Analysis for Gaussian Integration
• Gaussian Integration for Improper Integrals
• Legendre-Gaussian Integration Algorithms
• Chebyshev-Gaussian Integration Algorithms
• Examples, MATLAB Implementation and Results
• Conclusion

3. 3
Introduction
• Newton-Cotes and Romberg Integration usually use
table of the values of function.
• These methods are exact for polynomials less than N
degrees.
• General formula of these methods are as bellow:
b n
∫ f ( x)dx ≅ ∑w
a i =1
i f ( xi )
• In Newton-Cotes method the subintervals has the
same length.

4. 4
• But in Gaussian Integration we have the exact
formula of function.
• The points and weights are distinct for specific
number N.

5. 5
Gaussian Integration
• For Newton-Cotes methods we have:
b
b −a
1. ∫ f ( x )dx ≅ [ f (a) + f (b)].
a
2
b
b −a  a +b 
2. ∫
b
f ( x )dx ≅
6 
f (a ) + 4 f (
2
) + f (b) .

• And in general form:
b n
∫ f ( x)dx ≅ ∑ w f ( x )i i xi = a + (i − 1)h i ∈ {1,2,3,..., n}
a i =1
n
b− a n t− j
wi = ∏ dt
n − 1 ∫ j =1, j ≠ i i − j
0

6. 6
• But suppose that the distance among points are not
equal, and for every w and x we want the integration
to be exact for polynomial of degree less than 2n-1.
n 1
1.∑ wi = ∫ dx
i =1 −1
n 1
2.∑ xi wi = ∫ xdx
i =1 −1
.............
n 1
2n.∑ x 2 n −1 wi = ∫ x 2 n −1 dx
i =1 −1

7. 7
• Lets look at an example:
n =2 w1 , w2 , x1 , x 2 .
 .w1 + w2 = 2
1
2.x w + x w = 0
 1 1

2 2
1
2 ⇒2,4 ∴x 1 = x 2 ⇒3 ∴x 1 = .
2 2 2
 2
 .x 1 w1 + x 2 w2 = 3
2
3 3
 3
4.x 1 w1 + x 3 2 w2 = 0

1
x1 = −x 2 = , w1 = w2 =1.
3
• So 2-point Gaussian formula is:
1
1 −1
−
∫
1
f ( x ) dx ≅ f (
3
)+ f(
3
).

8. 8
Legendre Polynomials
• Fortunately each x is the roots of Legendre Polynomial.
1 d 2
PN ( x) = ( x − 1) n . n = 0,1,2,.....
2 n n! dx
• We have the following properties for Legendre
Polynomials.
1.Pn ( x) Has N Zeros in interval(-1,1).
2.( n +1) Pn +1 ( x ) = ( 2n +1) xPn ( x ) − nPn −1 ( x).
1
2
3. ∫ Pn ( x ) Pm ( x) dx = δmn
−1
2mn +1
1
4. ∫ x k Pn ( x) dx = 0 k = 0,1,2......, n - 1
−1
2 n +1 ( n!) 2
1
5.∫ x Pn ( x ) dx =
n
-1
(2n +1)!

9. 9
• Legendre Polynomials make orthogonal bases in (-1,1)
interval.
• So for finding Ws we must solve the following
equations:
n 1
1.∑ wi = ∫ dx = 2
i =1 −1
n 1
2.∑ wi x 2
i = ∫ xdx = 0
i =1 −1
....................
....................
n 1
1
n.∑ wi x n −1i = ∫ x n −1 dx = (1 − (−1) n )
i =1 −1
n

10. 10
• We have the following equation which has unique
answer:
... x1   w1   
n −1 T 2
1 x1  
   
1 x2 ... x 2   w2  
n −1 0 
   . = . 
 ... ... ... ...    
1 
1 ... x n   wn   n (1 − (− 1) ) 
n −1 n
 xn     
• Theorem: if Xs are the roots of legendre polynomials
1
and we got W from above equation then ∫P ( x)dx is −
1
exact for P ∈Π2 n −1 .

11. 11
• Proof:
p ∈ Π 2 n −1 ⇒ p( x) = q ( x) Pn ( x) + r ( x).
n −1 n −1
q( x) = ∑ q j Pj ( x) ; r ( x) = ∑ r j Pj ( x).
j =0 j =0
1 1 1 n −1 n −1
∫ p( x)dx = ∫ (q( x) P ( x) + r ( x))dx = ∫ ( P ( x)∑ q P ( x) + ∑ r P ( x))dx =
-1 −1
n
−1
n
j =0
j j
j =0
j j
n −1 1 n −1 1
∑ q ∫ P ( x) P ( x)dx + ∑ r ∫ P ( x) P (x)dx = 2r .
j =0
j j n
j =0
j 0 j 0
−1 −1
⇒
n n n
∑ w p( x ) = ∑ w (q( x ) P
i =1
i i
i =1
i i N ( x) + r ( xi )) = ∑ wi r ( xi )
i =1
n n −1 n −1 n n −1 1
= ∑ wi ∑ r j Pj ( xi ) = ∑ r j ∑ wi Pj ( x) = ∑ r j ∫ Pj ( x)dx = 2r0 .
i =1 j =0 j =0 i =1 j =0 −1

12. 12
Theorem:
1 n (x − x j )
wi = ∫ [ Li ( x )] dx ∏ (x
2
Li ( x ) =
−1 j = , j ≠i
1 i −xj )
Proof:
1 2
[ Li ( x)] 2 ∈ Π 2 n−2 ⇒ ∫ [ Li ( x)] 2 = ∑ w j [ Li ( x j )]
n
= wi .
−1 j =1

13. 13
Error Analysis for
Gaussian Integration
• Error analysis for Gaussian integrals can be derived
according to Hermite Interpolation.
b
Theorem : The error made by gaussian integration in approximation the integral ∫ f ( x )dx is ::
a
(b − a ) 2 n +1 ( N !) 4
EN ( f ) = f (2n)
(ξ ) ξ ∈ [ a, b].
(2n + 1)((2n)!) 3

14. 14
Gaussian Integration for Improper
Integrals
• Suppose we want to compute the following integral:
1
f ( x)
∫
−1 1−x2
dx
• Using Newton-Cotes methods are not useful in here
because they need the end points results.
• We must use the following:
1 1−ε
f ( x) f ( x)
∫
−1 1− x 2
dx ≅ ∫ε
−1+ 1− x 2
dx

15. 15
• But we can use the Gaussian formula because it does
not need the value at the endpoints.
• But according to the error of Gaussian integration,
Gaussian integration is also not proper in this case.
• We need better approach.
Definition : The Polynomial set { Pi } is orthogonal in (a, b) with respect to w(x) if :
b
∫ w( x) P ( x)P
a
i j ( x) dx = 0 for i ≠ j
then we have the following approximation :
b n
∫ w( x) f ( x)dx ≅ ∑ wi f ( xi )
a i =1
where xi are the roots for Pn and
b
wi = ∫ w( x)[ Li ( x)] dx
2
a
will compute the integral exactly when f ∈ Π 2 n −1

16. 16
Definition : Chebyshev Polynomials Tn ( x ) is defined as :
n 
2 
 
n 
Tn ( x ) = ∑ x n −2 k ( x 2 −1) k
 
k =0  2 k 
Tn ( x ) = 2 xTn ( x) − Tn −1 ( x), n ≥ 1, T0 ( x) = 1, T1 ( x ) = x.
If - 1 ≤ x ≤ 1 then :
( 2i −1)π 
Tn ( x ) = cos( n arccos x). roots xi = cos  .
 2n 
1
1
∫
−1 1−x 2
Ti ( x )T j ( x ) dx = 0 if i ≠ j.
• So we have following approximation:
1
1 π n  (2i − 1)π 
∫ f ( x)dx ≅ ∑ f ( xi ), xi = cos 
n i =1  2n   i ∈ {1,2,3,..., n}.
−1 1− x2

17. Legendre-Gaussian Integration
17
Algorithms
a,b: Integration Interval,
N: Number of Points,
f(x):Function Formula.
Initialize W(n,i),X(n,i).
Ans=0;
b−a b−a a+b
A( x ) = f( x+ ).
2 2 2
For i=1 to N do:
Ans=Ans+W(N,i)*A(X(N,i));
Return Ans;
End
Figure 1: Legendre-Gaussian Integration Algorithm

18. 18
a,b: Integration Interval,
tol=Error Tolerance.
f(x):Function Formula.
Initialize W(n,i),X(n,i).
Ans=0;
b −a b −a a +b
A( x ) = f( x+ ).
2 2 2
For i=1 to N do:
If |Ans-Gaussian(a,b,i,A)|<tol then return Ans;
Else
Ans=Gaussian(a,b,i,A);
Return Ans;
End
Figure 2: Adaptive Legendre-Gaussian Integration Algorithm.
(I didn’t use only even points as stated in the book.)

19. 19
Chebychev-Gaussian Integration
Algorithms
a,b: Integration Interval,
N: Number of Points,
f(x):Function Formula.
(b − a ) a +b a −b
A( x) = 1 − x 2 f( + x)
2 2 2
For i=1 to N do:
Ans=Ans+ A(xi); //xi chebyshev
roots
Return Ans*pi/n;
End
Figure 3: Chebyshev-Gaussian Integration Algorithm

20. 20
a,b: Integration Interval,
tol=Error Tolerance.
f(x):Function Formula.
(b − a ) a + b a − b
A( x ) = 1 − x 2 f( + x)
2 2 2
For i=1 to N do:
If |Ans-Chebyshev(a,b,I,A)|<tol then return Ans;
Else
Ans=Chebyshev(a,b,I,A);
Return Ans;
End
Figure 4: Adaptive Chebyshev-Gaussian Integration Algorithm

21. 21
Example and MATLAB
Implementation and Results
Figure 5:Legendre-Gaussian Integration

22. 22
Figure 6: Adaptive Legendre-Gaussian Integration

23. 23
Figure 7:Chebyshev-Gaussian Integration

24. 24
Figure 8:Adaptive Chebyshev-Gaussian Integration

25. 25
Testing Strategies:
• The software has been tested for
polynomials less or equal than 2N-1
degrees.
• It has been tested for some random inputs.
• Its Result has been compared with MATLAB
Trapz function.

26. 26
Examples:
Example 1:Gaussian-Legendre
1
1 π
∫ 2
−1 1 + x
dx exact → Arc tan(1) − Arc tan(−1) = ≅ 1.5707.

2
1 − (−1) 1 1
Trapezoid →(
  )( + ) = 1.0000.
2 1 + (−1) 2
1 + (1) 2
1 − (−1) 1 1 1
Simpson →(
 )( +4 + ) ≅ 1.6667.
6 1 + (−1) 2 1 + (0) 2 1 + (1) 2
2− Po int Gaussian → According To Software Resualt = 1.5000.
 
3−Po int Gaussian → According To Software Resualt = 1.5833.
 

27. 27
Example 2:Gaussian-Legendre
2
− e−x 3 − e −9 1
3
∫ xe
2
−x
dx  →(

exact
)0 = ( + ) ≅ 0.4999.
0
2 2 2
3− 0
Trapezoid →(
  )(0 + 3e −9 ) ≅ 0.0005.
2
3−0 2
Simpson → (
 )(0 + 1.5e −1.5 + 3e −9 ) ≅ 0.0792.
6
2− Po int Gaussian → ≅ 0.6494.
 
3− Po int Gaussian → ≅ 0.4640.
 
Example 3:Gaussian-Legendre
(b − a ) 2 n +1 ( n!) 4
En ( f ) = f 2n
(ξ ) ξ ∈[a, b].
( 2n +1)((2n)!) 3
π
(π − 0) 2 n +1 ( n!) 4
∫ sin( x)dx  → | (2n +1)((2n)!) 3 sin (ξ ) |≤ 5 ×10 ⇒ n ≥ 4.
−4
 2n
0
( 2 − 0) 2 n +1 (n!) 4 −ξ
2
∫ e dx  →| (2n +1)((2n)!) 3 e |≤ 5 ×10 ⇒ n ≥ 3.
−x −4

0

28. 28
Example 4:Gaussian-Legendre
3
x 1
∫0 1 + x 2 dx = ln(1 + x 2 ) ≅ 1.15129.
2
2 ⇒ ≅ 1.21622 ⇒ errora ≅ 0.06493.
3 ⇒≅ 1.14258 ⇒ errora ≅ 0.00871.
4 ⇒≅ 1.14902 ⇒ errora ≅ 0.36227.
5 ⇒≅ 1.15156 ⇒ errora ≅ 0.00027.
6 ⇒≅ 1.15137 ⇒ errora ≅ 0.00008.
Example 5:Gaussian-Legendre
3 2 3
e−x
∫ xe
2
−x
dx = ≅ 0.49994.
0
−2 0
2 ⇒≅ 0.64937 ⇒ errora ≅ 0.14943.
3 ⇒≅ 0.46397 ⇒ errora ≅ 0.03597.
4 ⇒≅ 0.50269 ⇒ errora ≅ 0.00275.
5 ⇒≅ 0.50007 ⇒ errora ≅ 0.00013.
6 ⇒≅ 0.49989 ⇒ errora ≅ 0.00005.

29. 29
Example 6:Gaussian-Legendre
π /2
∫ sin( x) dx :: Trapzoid :: 0.78460183690360
3.5
2
0 3
2 - Point ≅ 0.78539816339745.
2.5
3 - Point ≅ 0.78539816339745.
2
π
∫ sin( x)
2
dx :: Trapzoid :: 1.57079632662673 1.5
0
2 − Point ≅ 1.19283364797927. 1
3 - Point ≅ 1.60606730236915. 0.5
3π
2 0 -0.77 -0.57 0.57 0.77
∫ sin( x)
2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
dx ::Trapzoid :: 2.35580550989210
0
2 − Point ≅ 2.35619449019234. 4 - Point ≅ 3.53659228676239.
3 - Point ≅ 2.35619449019234. 5 - Point ≅ 3.08922572211956.
2π 6 - Point ≅ 3.14606122123817.
∫ sin( x)
2
dx ::Trapzoid :: 3.14159265355679
7 - Point ≅ 3.14132550162258.
0
2 − Point ≅ 5.91940603385020. 8 - Point ≅ 3.14131064749986.
3 - Point ≅ 1.47666903877755.

30. 30
Example 7:Adaptive Gaussian-Legendre
3
x
∫ 1 + x 2 dx ::
0
1)Adaptive Gaussian Integration :: error ≈ 5 ×10 -5 ⇒ 1.15114335351486.
2)Adaptive Gaussian Integration :: error ≈ 5 ×10 -4 ⇒ 1.15137188448013.
3
∫
2
xe x dx ::
0
1)Adaptive Gaussian Integration :: error ≈ 5 × 10 -5 ⇒ 0.49980229291620.
2)Adaptive Gaussian Integration :: error ≈ 5 × 10 -4 ⇒ 0.49988858784837.

31. 31
Example 8:Gaussian-Chebyshev
2 - Point Chebyshev Integration ≈ 0.48538619428604.
3 - Point Chebyshev Integration ≈ 1.39530571408271
2 - Point Chebyshev :: 0
3 - Point Chebyshev :: 0.33089431565488.

32. 32
Example 9:
1) w1 + w2 + w3 = 2
2) w1 x1 + w2 x 2 + w3 x3 = 0
2 2 2 2
3) w1 x1 + w2 x 2 + w3 x3 =
3
3 3 3
4) w1 x1 + w2 x 2 + w3 x3 =0
4 4 4 2
5) w1 x1 + w2 x 2 + w3 x3 =
5
5 5 5
6) w1 x1 + w2 x 2 + w3 x3 =0
w1 x1 + w2 x 2 1 
2,4 ⇒ 3 3
= 2 
w1 x1 + w2 x 2 x3  2 2 2 2 3 2 2 3 2 2
3 3  ⇒ w1 x1 ( x1 − x3 ) = w2 x 2 ( x3 − x 2 ), w1 x1 ( x3 − x1 ) = w2 x 2 ( x 2 − x3 )
w1 x1 + w2 x 2 1
4,5 ⇒ = 2
x3 
5 5
w1 x1 + w2 x 2 
⇒ x1 = − x 2
2 2 2 2
⇒ w1 x1 ( x1 − x3 ) = w2 (− x1 )( x3 − x1 ) ⇒ w1 = w2
w1 = w2 , x1 = − x 2 ⇒ 2) ⇒ w3 x3 = 0.
 
2 2 4 2
3,5 ⇒ 2w1 x1 = ,2 w1 x1 = . ⇒  x1 =
3
= − x2  ( w1 , w2 , w3 ) =  5 , 5 , 8 
 
3 5  5  9 9 9
2 5 8  3 3 
2
⇒ 2w1 x1 = ⇒ w1 = w2 = ⇒ 1 ⇒ w3 = ⇒ x3 = 0 ( x1 , x 2 , x3 ) = 
 ,− , 0 
3 9 9  5 5  

33. 33
Conclusion
• In this talk I focused on Gaussian Integration.
• It is shown that this method has good error
bound and very useful when we have exact
formula.
• Using Adaptive methods is Recommended Highly.
• General technique for this kind of integration
also presented.
• The MATLAB codes has been also explained.