As a matter of policy, I keep finals for one regular quarter. You may examine your final at any time in my office. If you wish to collect your final, you may do you in June.
Prob 1: (cf. Final Review bullet 4; Notes 4-23)
Prob 2: (cf. Mid I Review bullet 5, Final Review Bullet 1; Notes 2-4)
Prob 3: (cf. Mid I Review bullet 6; Notes 3-10, 3-11)
Prob 4: (cf. Mid I Review bullet 12; Notes 4-26A)
Prob 5: (cf. Mid I Review bullet 12)
Prob 6: (cf. Notes 6-27)
Prob 8: (cf. Final Review bullet 3; Notes 4-26A)
Prob 9: (cf. Mid II Review bullet 11, Final Review bullet 5; Notes 4-17)
I remember mentioning in class that "normal" equations are used in several ways in Numerical Analysis. This way is used to convert a non-symmetric system into a symmetric system. Some people had the basic idea of using AT, but both sides of the original matrix-vector equation must be PRE-multiplied by the transpose.
Prob 10: (cf. Final Review bullet 10; Notes 6-14)
Some had problems doing the simple evaluation of the function (i.e., the polynomial within the integral) correctly at 2 or 3 points to get the Gauss Quad values.
Prob 11: (cf. Final Review bullet 11)
This problem specified that we wanted an exact formula for a polynomial of degree 0.
A polynomial of degree 0 is a constant. Thus it can be taken out of the integral (on the left of the given equation) and "cancelled" with f(3/2) (which equals the same constant value) on the right. Then the left side of the initial equation can be evaluated exactly via the (integration by parts) formula given.
This was not intented so much as a "trick" question but as a question meant to see whether the reader took into account all the given information, which is very important in numerical problems.
A number of people tried to evaluate ln(3/2) which had nothing to do with the final value.
Prob 12: (cf. Final Review bullet 12; Notes 6-18)
This was basically the same procedure that needed to be done for the Runge-Kutta programming assignment. One introduces new variables which eliminate some of the derivatives of y and then one creates several equations with a derivative on the left side and an expression with no derivatives on the right. Since you are faced with a 3rd order DiffEq (i.e., one with a 3rd derivative in it), you should end up with 3 first order DiffEqs (with a derivative on the left and an expression without derivatives on the right) and three (new) initial conditions related to the old and new variables.
Prob 13: (cf. Final Review bullet 9; Notes 6-8)
One needs to use the error formula to determinine the width needed and thus also the number of panels. In the error formula the argument of f(4)(x) is chosen to maximize the value over the interval of integration. Thus x should be 3, since that where the 4th derivative is greatest on the interval given. For some reason, several evaluated the 4th derivative at the two end points and subtracted the values, which doesn't produce any usuable concept (in this case it is merely the difference of the max and min values of the 4th derivative on the interval).
One should use the error formula for the "composite" version given in the notes.
Since Simpson's rule demands that n be an even number, the minimum answer was 32 (the next higher even number greater than 30.006).
Prob 14: (cf. Final Review bullet 13; Notes 6-38A, 6-39)
Some had the correct discretization form for y'' but, unfortunately, made various algebraic errors (such as forgetting to correct for a minus sign inside parentheses) resulting in an incorrect final linear system.
A few used the discretized formula for the differential equation given in the notes as if it were a formula that could be used for any DiffEq.
Also, the question specifically asked for the linear system in "matrix form."
Prob 15: (cf. Final Review bullet 7; Notes 6-4)
Some used all 5 points for each approximation or varied the interval final point depending on the panel widths. In each case, one is supposed to compute the approximation to the same integral over the same interval (0,1) but using various widths for the panels. When the width h is 1, then there is only one panel (since the distance between the two limits of the integral is 1) and only the two endpoints are used. When the width is 1/2, there are two panels and the two endpoints and the middle point is used. When the width is 1/4, all 5 points are used.
Prob 16: (cf. Final Review bullet 8; Notes 6-11)
From information given in class, the Trapezoidal rule is of order 2 so the base value in the Richardson formula should be 2. Since the increase in panels form the first value (5) to the second value (10) is 2-fold, the exponent is also 2. Thus the denominator of the fraction is 22-1 or 3.
final nfinal 177 81 160 70 160 70 159 69 159 69 157 68 156 67 156 67 150 64 145 60 145 60 145 60 143 59 138 56 135 54 134 53 133 53 132 52 132 52 128 49 124 47 124 47 121 45 120 44 119 44 118 43 113 40 111 39 98 30 97 30 92 26 89 24 82 20 82 20 77 17 MAXIMUM 200 100
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 60- 80- 100- 120- 140- 160- 79 99 119 139 159 179 (1) (6) (4) (11) (10) (3)
This page is maintained by Dennis C. Smolarski, S.J.
Email: dsmolarski "at" scu.edu
© Copyright 2012 Dennis C. Smolarski, SJ, All rights reserved.
Last changed: 25 Mar 2012