- Question: Where and when is the exam?
The exam will be in 303 SH starting at about 8:40 and ending
at 10 (no later).
- Question: What is on the exam?
There will be two questions taken from the Final Questions worksheet.
They may be slightly modified with an additional part.
Remember you must submit your solutions to the Final Questions at the
beginning of the Final Exam.
- Question: Is there a typo in Problem 2?
Yes!!!. You should replace -1/4 by +1/4.
- Question: Problem 1: first, I calculated the eigenvalue to be 1
(multiplicity 3). Next,I have tried to find a second answer of the form x
= bte^t + ne^t where b and n are eigenvectors but my calculations result
in an augmented matrix where 0*n1 + 0*n2 + 0*n3 = 2. I'm just wondering
if the point of the problem was that there is only one eigenvector
associated with the eigenvalue 1 (of multiplicity 3) or if my calculations
are wrong.
I do not believe that 1 is an eigenvalue of multiplicity 3. You need
to carefully calculate the determinant that fixes the eigenvalues.
- Question: if the importance of this final is to know the path at
which the answer was retrieved and not just memorize each problem, is
there any way you could post the answers. That way I would not be trying
to figure out a way to find the other eigenvectors in the above question
if my assumption is correct.
The importance is for you do work through each problem and review any
concepts that you are not sure of. The problems are no harder than the
homework in the book. The idea is for you to become an expert at each
problem and learn the technique --- not memorize. On the exam, I may
change constant values which defeats memorization. Remember that you
can always check your answers so you will know if your are on the right
path. Does this help?
- Question: Problem 1 - I get an eigenvalue of 1 (with multiplicity=2).
The eigenvector I get is e=(1, -3/2, 1)T. In calculating (A-I)n=e, the
matrix which I get for (A-I) has a first row that is all zero's.
However, the first value of e=(1, -3/2, 1)T isn't zero, which leads to a
contradictory equation of 0*n1+0*n2+0*n3=1. If I ignore this little
inconsistency, the problem seems to work out. What did I do wrong!
You computed the eigenvalues wrong! The sign is that you got a
contradiction in trying to find a second eigenvector for 1.
- Question: Is problem 4 on the final problem set in the form of
equation (1) on page 680?
It looks like it is. You should treat the problem as posed with the
definition of the differential operator as I gave it.
- Question: For problem 4, is q(x)=0 or q(x)=1?, with respect to
equation (1) on page 680?
This is the same as before. It does not matter if you choose q=0 or
q=1 as long as you the choose mu and r to obtain the equation on the
exam. You really do not need to use the equation in the text, the problem
is completely defined in the question.
- Question: both of my linearly independent solutions, y1 and y2, each
involve both a series for a0 and also a series for a1. Is this also
wrong?
If the point is a regular singular point, you need to first concentrate on
the largest root of the indicial equation. You should re-check your
calculation for this root.
- Question: The Green's function I constructed for problem 6 seems not
to be
integrable along with F(s)=s. Here is what I obtained:
int(s*cos(s)*sin(x)/cos(-s+x), s = (0 ..
x))+int(s*sin(s)*cos(x)/cos(-s+x), s = (x .. pi))
Is this function right?
This can not be correct since the Green's function is supposed to be
continuous at x=s. Try again.
- Typo I believe there is a typo in problem 7. The
transformation should be z = sqrt{lambda} x.
- Question: You show a correction to problem #2 which states that -1/4
should be changed to 1/4. This is a Bessel equation which has the general
formula of
x^2y" + xy' + (x^2 - a^2)y = 0
The exam problem is as follows (with no corrections)
x^2y" + xy' - (x^2 - 1/4)y = 0
Should the correction be made to the (-) sign to a (+) sign before (x^2 -
1/4)y instead of changing -1/4 to +1/4?
The correct equation is:
x^2y" + xy' - (x^2 + 1/4)y = 0
I never called it the standard Bessel's equation. For fun and
to understand the solution a bit more, you might make the change of
dependent variable y(x) = v(x)/sqrt(x). I still want you to solve the
original equation by a series method.
- Question: I understand that you want us to become "experts"
regarding the processess by which these problems are solved. However, I
think that what has been requested is not to be given the complete
solutions but just the end result answers.
Becoming an expert would assume that I arrived at the right answer and
have not gone astray. If my answers do not agree with yours, I would at
least know that I need to go back and find out where I went wrong. If I
think my answers are correct, and they are not, I will not have learned
the proper way to arrive at the correct answers. Keep in mind, we only
want the answers not the solution process by which the answers were
derviced.
I used the word expert in the student sense, which means I
am hoping you can solve all the problems. You can check you solutions for
almost each problem. You can use Maple to check solutions or try by hand.
Knowing the answer in advance is not a good learning tool. One of the
most important concepts is trying to decide if a solution is correct or
not. Don't ask your future boss for the answer in advance!!!
- Question: How do you know if L has an inverse?
L having an inverse is similar to whether a matrix A has an inverse
but you cannot use the determinant.
- Question: These problems are quite lengthy and complicated. I don't
know if you can expect us to regurgitate two of them in a little over an
hour without looking things up.
Hmmm! That's why the questions are in take-home form. I do not want
anyone to regurgitate during the exam. That would be most
unpleasant. I only expect you to be able to solve a couple of reasonable
questions. Remember, I can shorten them, add a new part, or change
constants.
- Question: There have been several more questions regarding problem 6.
Instead of posting each let me try to give you some hints. I hope I am
computing this correctly because my solution is not very complicated.
In the alternate formulation of the problem for the Green's function G, we
must solve the homogeneous eqn G'' + G = 0 with the two bc's but with
continuity and the jump condition. I am using G' to mean the partial with
repect to x. The general solution is
G = a1 cos(x) + a2 sin(x), x less s and b1 cos(x) + b2 sin(x), x greater s
We then apply the bc's to the appropriate function. So G(pi)=0 implies
that b1 cos(pi) + b2 sin(pi) =0 so -b1 = 0 since sin(pi)=0 -- OK! Next
the bc at x=0 implies that -a1 sin(pi) + a2 cos(pi) =0 so a2 = 0 -- OK!
So we now have
G = a1 cos(x), x less s and b2 sin(x), x greater s.
See if you can use continuity and the jump condition from here! I am
going to get some lunch!
- Signing off I am signing off for the night and I will see
everyone in the morning. I hope that everyone learned a lot from the
final questions and the course in general. We covered a substantial
amount of material and I hope you were able to absorb it, as well as see
the connections to other math courses that you have taken.