# Thus z b a f x w x dx n x j 0 a j f x j 825 this

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Chapter 4 / Exercise 7
Numerical Analysis
Burden/Faires
Expert Verified
Thus, Z b a f ( x ) w ( x ) dx = n X j =0 A j f ( x j ) . (8.25) This proves that the Gaussian quadrature has degree of precision k = 2 n +1. Now suppose that the interpolatory quadrature (8.18) has maximal degree of precision 2 n + 1. Take f ( x ) = P ( x )( x - x 0 )( x - x 1 ) · · · ( x - x n ) where P is a polynomial of degree n . Then, f is a polynomial of degree 2 n + 1 and Z b a f ( x ) w ( x ) dx = Z b a P ( x )( x - x 0 ) · · · ( x - x n ) w ( x ) dx = n X j =0 A j f ( x j ) = 0 . Therefore, the polynomial ( x - x 0 )( x - x 1 ) · · · ( x - x n ) of degree n + 1 is orthogonal to all polynomials of degree n . Thus, it is a multiple of ψ n +1 . Example 24. Consider the interval[-1,1]and the weight functionw1.The corresponding orthogonal the Legendre Polynomials1, x, x2-13, x3- 3 5 x, · · · . Take n = 1 . The roots of ψ 2 are x 0 = - q 1 3 and x 1 = q 1 3 . There- fore, the corresponding Gaussian quadrature is Z 1 - 1 f ( x ) dx A 0 f r - 1 3 ! + A 1 f r 1 3 ! (8.26)
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Chapter 4 / Exercise 7
Numerical Analysis
Burden/Faires
Expert Verified
118 CHAPTER 8. NUMERICAL INTEGRATION where A 0 = Z 1 - 1 l 0 ( x ) dx, (8.27) A 1 = Z 1 - 1 l 1 ( x ) dx. (8.28) We can evaluate evaluate these integrals directly or use the method of un- determined coefficients to find A 0 and A 1 . The latter is generally easier and we illustrate it now. Using that the quadrature has to be exact for 1 and x we have 2 = Z 1 - 1 1 dx = A 0 + A 1 , (8.29) 0 = Z 1 - 1 xdx = - A 0 r 1 3 + A 1 r 1 3 . (8.30) Solving this 2 × 2 linear system we get A 0 = A 1 = 1 . So the Gaussian quadrature for n = 1 in [ - 1 , 1] is Q 1 [ f] =f-T[ r 1 3 ! + f r 1 3 ! (8.31) Let us compare this quadrature to the elementary Trapezoidal Rule. Take f ( x ) = x 2 . The Trapezoidal Rule, f ] , gives T [ x 2 ] = 2 2 [ f ( - 1) + f (1)] = 2 (8.32) whereas the Gaussian quadrature Q 1 [ f ] yields the exact result: Q 1 [ x 2 ] = - r 1 3 ! 2 + r 1 3 ! 2 = 2 3 . (8.33) Example 25. Let us take again the interval [ - 1 , 1] but now w ( x ) = 1 1 - x 2 . As we know (see 2.4 ), ψ n +1 ( x ) = T n +1 ( x ) , i.e. the ( n + 1) st Chebyshev Polynomial. Its zeros are x j = cos[ 2 j +1 2( n +1) π ] for j = 0 , . . . , n . For n = 1 we have x 0 = cos π 4 = r 1 2 , (8.34) x 1 = cos 5 π 4 = - r 1 2 . (8.35)
8.3. GAUSSIAN QUADRATURES 119 We can use again the method of undetermined coefficients to find A 0 and A 1 : π = Z 1 - 1 1 1 1 - x 2 dx = A 0 + A 1 , (8.36) 0 = Z 1 - 1 x 1 1 - x 2 dx = - A 0 r 1 2 + A 1 r 1 2 , (8.37) which give A 0 = A 1 = π 2 . Thus, the corresponding Gaussian quadrature to approximate R 1 - 1 f ( x ) 1 1 - x 2 dx is Q 1 [ f ] = π 2 " f - r 1 2 ! + f r 1 2 !# . (8.38) 8.3.1 Convergence of Gaussian Quadratures Let f C [ a, b ] and consider the interpolation quadrature (8.18). Can we guarantee that the error converges to zero as n → ∞ , i.e., Z b a f ( x ) w ( x ) dx - n X j =0 A j f ( x j ) 0 , as n → ∞ ? The answer is no. As we know, convergence of the interpolating polynomial to f depends on the smoothness of f and the distribution of the nodes. However, if the interpolatory quadrature is a Gaussian one the answer is yes. This follows from the following special properties of the quadrature weights A 0 , A 1 , . . . , A n in the Gaussian quadrature.