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# Gaussian Integration

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

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### Gaussian Integration

1. 1. 1 GaussianIntegration M. Reza Rahimi,Sharif University of Technology, Tehran, Iran.
2. 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. 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. 4• But in Gaussian Integration we have the exact formula of function.• The points and weights are distinct for specific number N.
5. 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. 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. 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. 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. 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. 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. 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. 12Theorem: 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. 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. 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. 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. 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. 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; EndFigure 1: Legendre-Gaussian Integration Algorithm
18. 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; EndFigure 2: Adaptive Legendre-Gaussian Integration Algorithm. (I didn’t use only even points as stated in the book.)
19. 19. 19Chebychev-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; EndFigure 3: Chebyshev-Gaussian Integration Algorithm
20. 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; EndFigure 4: Adaptive Chebyshev-Gaussian Integration Algorithm
21. 21. 21 Example and MATLABImplementation and Results Figure 5:Legendre-Gaussian Integration
22. 22. 22Figure 6: Adaptive Legendre-Gaussian Integration
23. 23. 23Figure 7:Chebyshev-Gaussian Integration
24. 24. 24Figure 8:Adaptive Chebyshev-Gaussian Integration
25. 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. 26Examples: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. 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. 28Example 4:Gaussian-Legendre3 x 1∫0 1 + x 2 dx = ln(1 + x 2 ) ≅ 1.15129. 22 ⇒ ≅ 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-Legendre3 2 3 e−x∫ xe 2 −x dx = ≅ 0.49994.0 −2 02 ⇒≅ 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. 29 Example 6:Gaussian-Legendreπ /2 ∫ sin( x) dx :: Trapzoid :: 0.78460183690360 3.5 2 0 32 - Point ≅ 0.78539816339745. 2.53 - Point ≅ 0.78539816339745. 2π∫ sin( x) 2 dx :: Trapzoid :: 1.57079632662673 1.502 − Point ≅ 1.19283364797927. 13 - Point ≅ 1.60606730236915. 0.53π 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 02 − 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 ≅ 3.14132550162258.02 − Point ≅ 5.91940603385020. 8 - Point ≅ 3.14131064749986.3 - Point ≅ 1.47666903877755.
30. 30. 30Example 7:Adaptive Gaussian-Legendre3 x∫ 1 + x 2 dx ::01)Adaptive Gaussian Integration :: error ≈ 5 ×10 -5 ⇒ 1.15114335351486.2)Adaptive Gaussian Integration :: error ≈ 5 ×10 -4 ⇒ 1.15137188448013.3∫ 2 xe x dx ::01)Adaptive Gaussian Integration :: error ≈ 5 × 10 -5 ⇒ 0.49980229291620.2)Adaptive Gaussian Integration :: error ≈ 5 × 10 -4 ⇒ 0.49988858784837.
31. 31. 31Example 8:Gaussian-Chebyshev2 - Point Chebyshev Integration ≈ 0.48538619428604.3 - Point Chebyshev Integration ≈ 1.395305714082712 - Point Chebyshev :: 03 - Point Chebyshev :: 0.33089431565488.
32. 32. 32Example 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. 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.