Download Final Exam Solutions - Intermediate Numerical Analysis 1 | MATH 4650 and more Exams Mathematics in PDF only on Docsity! Math 4650 Final Exam Solutions 1. True/False, and explain your answer in one sentence. Half credit for the correct circling, half credit for a correct justification. (a) To start an Adams-Bashforth 3-step method, we would typically use Euler’s method to get w1 and w2, then use the algorithm to get w3 and beyond. T F False. Euler’s method is only first-order while the 3-step method is supposed to be third-order; bad initial data will make the whole thing only first order. (b) If y′ = λy for some λ < 0, then all solutions calculated by the implicit Euler method will decay to 0. T F True. The implicit Euler method is A-stable. (c) If w (A) i is a differential equation method with local truncation error 6f (4)(ξ1)h 3 and w (B) i is a method with local truncation error −5f (6)(ξ2)h5, we can use w (A) i to roughly estimate the error in w (B) i without any other information. T F False. You can use a high-order method to estimate the error in a lower-order method, but you can’t do it the other way around. (d) If you are solving the same linear system Ax = b many times (for the same A and different b), the Gauss-Jordan method may be faster than Gaussian elimination. T F False. Gaussian elimination (to find the LU decomposition) is always faster than the Gauss-Jordan method. (e) It is often more efficient to calculate the determinant of a matrix to see if it’s nonsingular before going through Gaussian elimination to solve Ax = b. T F False. It is pretty much never more efficient to compute a determinant; in fact usually the determinant is computed by doing Gaussian elimination first. 2. Consider the differential equation dy dt = 2y − 1 1 + t , y(0) = 1, with exact solution y(t) = 1 2 t2 + t+ 1. Use the following methods to estimate y(1) with one step (h = 1). 1 (a) Second-order Taylor method. Solution: We need to compute y′′(0). We have y′′(t) = 2y′(t) 1 + t − 2y(t)− 1 (1 + t)2 = 2(2y − 1) (1 + t)2 − 2y − 1 (1 + t)2 = 2y − 1 (1 + t)2 , so that y′′(0) = 2·1−1 (1+0)2 = 1. Thus the Taylor method predicts y(1) ≈ w1 = y(0) + hy′(0) · h+ h 2 2 y′′(0) = 1 + 1 + 1 2 = 5 2 . (b) Midpoint method. Solution: The Midpoint method says y(1) ≈ w1 = y(0) + hf(0 + h/2, 1 + h/2f(0, 1)) = 1 + f(12 , 3 2 ) = 1 + 2 3 2 = 7 3 . (c) One-step Adams-Moulton (aka Implicit Trapezoid) method. Solution: The method says y(1) ≈ w1 = 1 + h2 ( f(0, 1) + f(h,w1) ) , which gives the equation w1 = 1 + 1 2 ( 1 + 2w1 − 1 2 ) , which reduces to w1 = 5 4 + 1 2 w1, and thus w1 = 5 2 . Which does the worst against the actual value? Solution: The exact value is y(1) = 5 2 , so both Taylor’s method and the Trapezoid method get it exactly, while the Midpoint method is off by 7 3 − 5 2 = −1 6 . 3. Consider the following symmetric matrix: A = 2 6 −86 25 −10 −8 −10 55 . (a) Determine the LDLt factorization. Solution: The multipliers in the first column are m21 = 3 and m31 = −4, which reduces the matrix to 2 6 −80 7 14 0 14 23 . 2
