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 7:
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