Practice Final Exam Solutions - Numerical Analysis | Math 105, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Practice Final Exam Solutions - Numerical Analysis | Math 105 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Jim Lambers Math 105A Summer Session I 2003-04 Practice Final Exam Solution 1. Use the Newton divided-difference formula to compute the cubic interpolating polynomial p3(x) for the function f(x) = ex, using interpolation points x0 = 0, x1 = 1, x2 = 2, and x3 = 3. What is an upper bound for the error |p3(0.5)− f(0.5)|? Solution The divided-difference table for the given data is i xi f [xi] 0 0 1 1.718282 1 1 2.718282 1.476246 4.670774 0.8455357 2 2 7.389056 4.012853 12.69648 3 3 20.08554 It follows that p3(x) = 1 + 1.718282x+ 1.476246x(x− 1) + 0.8455357x(x− 1)(x− 2). The error in this polynomial is f(x)− p3(x) = eξ 4! x(x− 1)(x− 2)(x− 3), since f (4)(x) = ex. It follows that |p3(0.5)− f(0.5)| = ∣∣∣∣eξ4! (0.5)(0.5− 1)(0.5− 2)(0.5− 3) ∣∣∣∣ = ∣∣∣∣ eξ24 1516 ∣∣∣∣ ≤ ∣∣∣∣ e324 1516 ∣∣∣∣ ≤ 0.7845913, since the maximum of eξ on [0, 3] occurs at ξ = 3. 1 2. Find the Hermite interpolating polynomial for the following data. i xi f(xi) f ′(xi) 0 −1 0 1 1 0 1 0 2 1 4 −1 Solution We define z0 = x0, z1 = x0, z2 = x1, z3 = x1, z4 = x2, z5 = x2. The divided-difference table for the given data is i zi f [zi] 0 −1 0 1 1 −1 0 0 1 −1 2 0 1 −1 1.5 0 2 −3 3 0 1 3 −4.5 3 −7 4 1 4 −4 −1 5 1 4 It follows that the Hermite interpolating polynomial is H5(x) = (x+ 1)− (x+ 1)2x+ 1.5(x+ 1)2x2 − 3(x+ 1)2x2(x− 1). 3. Find the cubic spline interpolant s(x) for the data i xi f(xi) 0 1 −1 1 3 3 2 5 0 that satisfies free, or natural, boundary conditions. Solution The spline s(x) is a piecewise polynomial consisting of the two pieces s0(x) = a0 + b0(x− 1) + c0(x− 1)2 + d0(x− 1)3, s1(x) = a1 + b1(x− 3) + c1(x− 3)2 + d1(x− 3)3. 2 From Taylor’s Theorem, f(x0 − h) = f(x0)− f ′(x0)h+ f ′′(x0) 2 h2 − f ′′′(ξ1) 6 h3, f(x0 + 2h) = f(x0) + f ′(x0)2h+ f ′′(x0) 2 4h2 + f ′′′(ξ2) 6 8h3, where ξ1 ∈ [x0 − h, x0] and ξ2 ∈ [x0, x0 + 2h]. Substituting these Taylor expansions into the differentiation rule yields −4f(x0 − h) + 3f(x0) + f(x0 + 2h) 6h = − 2 3h [ f(x0)− f ′(x0)h+ f ′′(x0) 2 h2 − f ′′′(ξ1) 6 h3 ] + 1 2h f(x0) + 1 6h [ f(x0) + f ′(x0)2h+ f ′′(x0) 2 4h2 + f ′′′(ξ2) 6 8h3 ] = f ′(x0)− 2 3h [ f ′′(x0) 2 h2 − f ′′′(ξ1) 6 h3 ] + 1 6h [ f ′′(x0) 2 4h2 + f ′′′(ξ2) 6 8h3 ] = f ′(x0) +O(h2), since the terms of O(h) cancel. 5. Use Romberg integration to compute an approximation to∫ 1 −1 ex cos 2x dx that is sixth-order accurate. Solution We begin by using the Trapezoidal Rule, or, equivalently, the Composite Trape- zoidal Rule∫ b a f(x) dx ≈ h 2 f(a) + n−1∑ j=1 f(xj) + f(b)  , h = b− a n , xj = a+ jh, with n = 1 subintervals. Since h = (b− a)/n = (1− (−1))/1 = 2, we have R1,1 = 2 2 [f(−1) + f(1)] = −1.28429624943104. 5 Next, we apply the Composite Trapezoidal Rule with n = 2 and h = (1− (−1))/2 = 1, which yields R2,1 = 1 2 [f(−1) + 2f(0) + f(1)] = 0.35785187528448. Now, we can use Richardson Extrapolation to obtain a more accurate approximation, R2,2 = R2,1 + R2,1 −R1,1 3 = 0.90523458352299, Because the error in the Composite Trapezoidal Rule satisfies ∫ b a f(x) dx = h 2 f(a) + n−1∑ j=1 f(xj) + f(b) +K1h2 +K2h4 +K3h6 +O(h8), where the constants K1, K2 and K3 depend on the derivatives of f(x) on [a, b] and are independent of h, we can conclude that R2,1 has fourth-order accuracy. We can obtain a second approximation of fourth-order accuracy by using the Composite Trapezoidal Rule with n = 4 to obtain a third approximation of second-order accuracy. We set h = (1− (−1))/4 = 1/2, and then compute R3,1 = 0.5 2 [f(−1) + 2[f(−0.5) + f(0) + f(0.5)] + f(1)] = 0.78818484680003. Now, we can apply Richardson Extrapolation to R2,1 and R3,1 to obtain R3,2 = R3,1 + R3,1 −R2,1 3 = 0.93162917063855. It follows from the error term in the Composite Trapezoidal Rule, and the formula for Richard- son Extrapolation, that R2,2 = ∫ 1 0 e−x 2 dx+ K̃2h4 +O(h6), R2,2 = ∫ 1 0 e−x 2 dx+ K̃2 ( h 2 )4 +O(h6). Therefore, we can use Richardson Extrapolation with these two approximations to obtain a new approximation R3,3 = R3,2 + R3,2 −R2,2 24 − 1 = 0.93338880977959. Because R3,3 is a linear combination of R3,2 and R2,2 in which the terms of order h4 cancel, we can conclude that R3,3 is of sixth-order accuracy. 6 6. Use Gaussian Quadrature with n = 3 to evaluate the integral∫ 1 0 sin3 x dx. What is the degree of this quadrature rule? Solution The particular Gaussian quadrature rule that we will use consists of 3 nodes x1, x2 and x3, and 3 weights w1, w2 and w3. To determine the proper nodes and weights, we use the fact that the nodes and weights of a 3-point Gaussian rule for integrating over the interval [−1, 1] are given by i Nodes r3,i Weights c3,i 1 0.7745966692 0.5555555556 2 0.0000000000 0.8888888889 3 −0.7745966692 0.5555555556 To obtain the corresponding nodes and weights for integrating over [−1, 1], we can use the fact that in general, ∫ b a f(x) dx = ∫ 1 −1 f ( b− a 2 t+ a+ b 2 ) b− a 2 dt, as can be shown using the change of variable x = [(b− a)/2]t+ (a+ b)/2 that maps [a, b] into [−1, 1]. We then have∫ b a f(x) dx = ∫ 1 −1 f ( b− a 2 t+ a+ b 2 ) b− a 2 dt ≈ 5∑ i=1 f ( b− a 2 r3,i + a+ b 2 ) b− a 2 c3,i ≈ 5∑ i=1 f(xi)wi, where xi = b− a 2 r3,i + a+ b 2 , wi = b− a 2 c3,i, i = 1, . . . , 5. In this example, a = 0 and b = 1, so the nodes and weights for a 3-point Gaussian quadrature rule for integrating over [0, 1] are given by xi = 1 2 r3,i + 1 2 , wi = 1 2 c3,i, i = 1, . . . , 3, which yields 7