APPM/MATH Problem Set 6 Solutions

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1 APPM/MATH 460 Problem Set 6 Solutions This assignment is due by 4pm on Wednesday, November 6th You may either turn it in to me in class or in the box outside my office door (ECOT ) Minimal credit will be given for incomplete solutions or solutions that do not provide details on how the solution is found You may discuss the problems with your classmates, but all work (analysis and code) must be your own For problems that require you to implement a method, you may use code from our textbook website faires/numerical-analysis/programs/, although they may require some modification For your own personal understanding of the methods involved, it is highly recommended that you attempt your own implementations 1 (a) Determine the values of the weights, w, and w so that the quadrature formula has degree of precision at least f(x) dx f(/) + w f(0) + w f(1/) Solution: We need to select the weights so that the quadrature rule is exact for a basis of degree- polynomials, namely { 1, x, x } This yields the following system of equations + w + w + w 4 + w 4 dx x dx 0 x dx which can be written in matrix form as / 0 1/ 1/4 0 1/4 w w 0 / Solving the linear system yields [ w w ] T [ 4 by 4 ] T The quadrature rule is then give f(x) dx 4 f(/) f(0) + 4 f(1/) (b) Once the values of, w, and w have been computed, determine the overall degree of precision of the quadrature rule Solution: We know, by construction, that the quadrature rule is exact for polynomials of degree or less We need to find the first polynomial basis function that the rule fails to integrate exactly ( ) 4 + ( ) ( ) 0 x 0 ( 4 ) 1 + ( ) ( So the degree of precision of the quadrature rule is ) x 4

2 In this problem you will approximate the value of π by numerically computing the integral 1 + x π (a) The Composite Trapezoid Rule has the following error formula: I(f) T n (f) (b a) h f (ξ) 1 Determine the minimum number of subintervals necessary to approximate π to within an absolute error tolerance of 10 6 using the Composite Trapezoid Rule on the given integral Solution: We have h (b a) n n (b a) /n /n, so the error term becomes I(f) T n (f) n max f (x) x [,1] We have f (x) 4 ( x 1 ) (x + 1) which takes the maximum absolute value of 4 when x 0 So, on [, 1] we have I(f) T n (f) n To guarantee that the error bound is less than 10 6 we require that n 10 6 n 106 n So to guarantee a priori that the error will be less then the tolerance we should use at least 16 subintervals In practice, it is possible to reach the tolerance with fewer subintervals (b) Verify that the Composite Trapezoid Rule is second-order accurate by approximating the given integral on n 1,, 4,, and 16 subintervals Make a table with columns corresponding to the approximation, the absolute error in the approximation, and the relevant error ratios Solution: See table below n T n (f) e n e n /e n From the error ratios we see that the Composite Trapezoid has second-order accuracy (c) Approximate the integral using the Composite Simpson s Rule on n 1,, 4,, and 16 subintervals Make a table with columns corresponding to the approximation, the absolute error in the approximation, and the relevant error ratios What is the order of accuracy of the method? Does this agree with the theory we developed in class?

3 Solution: See table below n S n (f) e n e n /e n Since the error ratios eventually settle down to e n /e n 64 the Composite Simpson s Rule applied to this problem is sixth-order accurate The theory predicts that the CSR method will be at least fourth-order accurate We get the extra accuracy due to a special property of this particular integral Extra Credit: Thoroughly (ie with math) explain the weird result you observe in Part (c) Solution: Like the Composite Trapezoid Rule, the Composite Simpson Rule has an alternative error formula that depends on it s derivatives at the interval endpoints: I(f) S h (f) + E 4h 4 [f (b) f (a)] + E 6h 6 [f (b) f (a)] + 4! 6! We notice that the function f(x) 1 + x satisfies f (1) f () 0 which eliminates the fourth-order error term Then, the leading error term is O ( h 6) which explains the sixth-order accurate behavior we see from the code (d) Use Romberg Integration to approximate π to within an absolute error tolerance of 10 6 Print out the resulting integration table and corresponding error table Compare the total number of function evaluations used in the Romberg Integration to the number predicted for the Composite Trapezoid Rule in part (a) Solution: See table below n R k,1 R k, R k, R k,4 R k, R k, n e k,1 e k, e k, e k,4 e k, e k, e e e e 06 10e e 09 7e e 0 116e e 0 We also observe some unexpected behavior from Romberg Integration because of the special properties of the integrand Recall that the derivation of Romberg Integration assumed that the trend in the error of CTR has the form I(f) T h (f) + k h + k 4 h 4 + k 6 h 6 + But this turns out not to be true for this integrand Recall the special error formula we wrote down in class when discussing the application of CTR to periodic functions: I(f) T h (f)+ B h [f (b) f (a)]+ B 4h 4 [f (b) f (a)]+ + B kh k! 4! (k)! [ ] f (k) (b) f (k) (a)

4 But, since the third-derivative term is zero for this function the actual error trend is actually I(f) T h (f) + k h + k 6 h 6 + This suggests that the second column in the Romberg Integration table should have sixth-order accuracy In fact, if we compute the error ratios for the second column we find n e n /e n From the table we can see that the first approximation that falls below the 10 6 tolerance is R, which has an error of and requires a total of 17 function evaluations to compute This is significantly fewer function evaluations predicted for the standard Composite Trapezoid Rule predicted in Part (a) (a) Find the quadrature points x 1, x, x and associated weights, w, w for the Gaussian quadrature rule of the form f(x) dx w i f(x i ) i1 Solution: The optimal quadrature nodes are the roots of the degree- Legendre Polynomial P (x) x x : x 1 x 0 x There are numerous ways to compute the Gaussian Quadrature weights: Lagrange Basis Functions w w Legendre Polynomial Formula (x x ) (x x ) (x 1 x ) (x 1 x ) dx 9 (x x 1 ) (x x ) (x x 1 ) (x x ) dx 9 (x x 1 ) (x x ) (x x 1 ) (x x ) dx 9 Assuming that the Legendre Polynomials have been normalized so that P n (1) 1, we can use the following general formula to compute the weights: The Vandermonde Matrix w i (1 x i ) [P n(x i )] The final method for determining the weights is to enforce that the n-point Gaussian Quadrature rule be exact for all polynomials of degree (n 1), represented by the basis { 1, x, x,, x n} Let x 1, x,, x n be the quadrature points chosen as the roots of the n-degree Legendre Polynomial Assuming the quadrature rule has the form

5 this leads to the following system of equations: n I(f) w i f(x i ) i1 + w + + w n x 1 + x w + + x n w n x 1 + x w + + x nw n x n 1 + x n w + + x n n w n dx x dx 0 x dx x n dx This system of equations can be re-written in matrix form as x 1 x x x n x 1 x x x n x n 1 x n x n x n n w w w n f 1 f f f n and then solved for the quadrature weights This matrix is called the Vandermonde matrix and has numerous applications in numerical analysis It should be noted that use of Vandermonde matrix to compute quadrature weights is only recommended for very small degree quadrature rules as the matrix can become very ill-conditioned as n increases In fact, the Vandermonde matrix for a 10-point quadrature rule already has condition number κ (V ) 10 4 (b) Use the quadrature rule developed in part (a) to approximate the integral from Problem Give the absolute error in the approximation Solution: Using the weights and quadrature points computed in Part (a) we have I(f) ) ( ) ( 9 f + 9 f(0) + 9 f 166 which has absolute error π If we compare this to Simpson s Rule approximation (which uses the same number of quadrature points) we have S(f) 1 (f() + 4f(0) + f(1)) which has absolute error π 01917