Exam results for Math 421:03, fall 2004

The second exam

Extra credit
Max grade 12 22 16 12 18 12 8 5 93
Min grade 2 4 0 2 5 0 0 0 34
Mean grade 7.82 17.76 4.15 6.97 12.62 9.92 4.68 2.21 66.12
Median grade 7 20 2.5 7 13 11 4 0 68.5

34 students took this exam. Numerical grades will be retained for use in computing the final letter grade in the course. Here are approximate letter grade assignments for this exam:

Range[85,100][80,84][70,79] [65,69][60,64][55,59][0,54]

Discussion of the grading

Generally arithmetic errors will be penalized only minimally. If, however, your error makes the problem much simpler, more credit will be deducted. I wrote what I hoped students would accept as axioms for my exams, at least.


  1. Do my problems, not those you invent.
  2. I cannot read your mind or imagine your answers: what you write is what I will read (and grade).
  3. If you do extensive computation, you are probably not doing my problem (see 1) or you are doing my problem inefficiently.

Problem 1 (12 points)
6 points for each part. I will read what was written. I can not guess what you meant to write. I required that what was written be responsive to what was asked. In part a), an important "If ... then ..." needed to be inferred from what you wrote (for example, "If [a certain linear combination is 0] then [all the coefficients are 0]". Or "The only way a certain linear combination is 0 is if all the coefficients are 0". I also accepted statements about none of the vectors being linear combinations of the others. But I read what you wrote, and tried to read it carefully.
I took 2 points off in part b) if the definition used AX=X and did not specify that X must be non-zero. This specification is important. If a definition using the equation det(A-In)=0 was given, there was no such risk. Again, I read what students wrote, and did not guess at what they did not write.

Problem 2 (22 points)
a) (2 points) Take the determinant of A-I. You may compute this determinant in any way you like.
b) (2 points) You may have already simplified the characteristic polynomial in part a), or you can do it here. The roots should be obvious.
c) (4 points) You need to solve three (3) homogeneous systems of linear equations. But they all are rather simple.
d) (2 points) You are merely asked to write P and D, which you certainly should be able to do after parts b) and c). Verification of your statements occurs in the next few parts of the problem. If incorrect results from b) and c) are used correctly here, I gave full credit.
e) (3 points) You may find P-1 in any way you like. The answer is easy enough to check, so I gave 1 point out of 3 for an incorrect answer. The exam I handed out, by the way, had two parts labeled d. That should be fixed in the posted version.
f) (2 points) Compute the product requested.
g) (2 points) Compute the product requested. If you do not get a correct diagonal matrix, I gave no points.
h) (5 points) 2 points for setting up the requested relationship A6=PD6P-1. 1 point for computing D6 and 2 points for computing A6 (information was given allowing you to check your answer).

Problem 3 (16 points)
a) (8 points) A restatement of the definition in the language of the problem (that is, writing an arbitrary linear combination of the functions of the problem, setting this equal to 0, etc.) earns 2 points. The balance is earned when the answer contains verification that the coefficients of the linear combination must be 0.
b) (8 points) The correct answer (No) gets 2 points. The cover page states, "An answer alone may not receive full credit." Correct supporting evidence is needed for the other 6 points.

Problem 4 (12 points)
In this problem, I expected students to evaluate the determinant (8 points). A few students tried other strategies (using RREF), and one or two were successful. Students who "plugged in" values for a and b and c were simplifying the problem too much and can earn at most 3 of these 8 points.
I hoped the determinant evaluation would be combined with the knowledge that a matrix is singular (not invertible) exactly when the determinant is 0 (2 points). This then could be matched up with the perpendicularity condition to get the desired conclusion (2 points).

Problem 5 (18 points)
It is possible to make mistakes in a) and have serious effects on work in b) and c). If the result of a) were as complex as the correct answer, points were only taken off in a). A similar approach was followed for errors in b).
a) (5 points) One application of integration by parts. Keep track of the n's and the signs.
b) (4 points) Evaluate the antiderivative. Notice that sin(nPi) and sin(n 0) are 0 and that cos(nPi)=(-1)n and cos(n 0) is 1. I took off 2 points for errors which really fouled up the answers in c) (examples: omitting the sign "flip" or omitting the Pi).
c) (2 points) The points were earned if the result of b) was used correctly.
d) (7 points)
The left-hand graph (4 points)
In the graph of the partial sum, I looked for the following qualitative behavior:

  1. Continuity on [0,Pi] with matching values at 0 and Pi
  2. The value 0 at both 0 and Pi.
  3. Closeness (with "wiggling) to the line segment inside the interval.
  4. Gibbs phenomena (overshoot) at both ends
I took off a point for each feature that was missing.
The right-hand graph (3 points)
This is supposed to be the graph of the sum of the whole Fourier sine series. Here the behavior required was much simpler:
  1. Identical to x+1 except at the ends (1 point)
  2. 0 at both ends (1 point each)

Problem 6 (12 points)
With orthogonality, there is almost no computation in this problem (yes, other than small integers). I'll take off 1 point if the normalization constant is misquoted (I did this in my own solution of the problem, so I would have scored 99 at most). If the integrals are computed as the sum of squares of the coefficients with no constant (or, better, with the constant=1) I will take 2 points off.
Some students antidifferentiated instead of differentiating, and this will lose 2 points.
I took off 3 points for an error I didn't anticipate, an answer which essentially declares that the integral of (a negative number)2 is negative. I can't read people's minds, and I don't know on what level this error was made: through fatigue and nervousness under exam conditions, or because of serious misunderstanding.

Problem 7 (8 points)
I am not satisfied with the statement of this problem. The statement might have made the problem more difficult for students. I wanted to ask: what is the polynomial formula when x<0 for the odd (respectively, even) extension of the polynomial x+x4? I think now I should just have asked exactly that. The statement of the question(s) seemed to invite misinterpretation.
4 points for each part, with 2 points for the specification of x<0, 1 point for x>0, and 1 point for identifying which Fourier coefficients must be 0.
I'll give 1 point on each part to people who correctly write the Fourier sine (respectively, cosine) series for F(x) (respectively, G(x)).
I'll put a "cleaner" version of the question in the posted exam, but I'll also show the wording I actually used, since I believe that affected student work on the problem.