Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: (263) Next: Homework Assignment 24 Previous: Term Exam 4

Solution of Term Exam 4

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Problem 1. Is there a non-zero polynomial defined on the interval and with values in the interval so that it and all of its derivatives are integers at both the point 0 and the point ? In either case, prove your answer in detail. (Hint: How did we prove the irrationality of ?)

Solution. There isn't. Had there been one, we could reach a contradiction as in the proof of the irrationality of . Indeed we would have that , hence the integral is not an integer. But repeated integration by parts gives

The assumptions on and along with the fact that , , and are all integers imply that the boundary terms are all integers. If is large enough, and hence the remaining integral is 0. So is an integer, and that's a contradiction.

Problem 2. Compute the volume of the Black Pawn'' on the right -- the volume of the solid obtained by revolving the solutions of the inequalities and about the axis (its vertical axis of symmetry). (Check that and hence the height of the pawn is ).

Solution. This is the area of the rotation solid with radius bounded by and . Thus

Problem 3.

1. Compute the degree Taylor polynomial of the function around the point 0.
2. Write a formula for the remainder in terms of the derivative evaluated at some point .
3. Show that at least for very small values of , .

Solution.

1. , , and so it can be shown by induction that . Thus and hence .
2. Cauchy's formula for the remainder is for some .
3. If then and and hence and thus . Therefore , as required.

Problem 4.

1. Prove that if and the function is continuous at , then
2. Let be a number, and define a sequence via the relations and for . Assuming that this sequence is convergent to a positive limit, determine what this limit is.

Solution.

1. See the easy'' part of Theorem 1 of Spivak's Chapter 22.
2. Assume . Then . Using the first part of this question on the function , which is continuous at , we find that . Hence satisfies . Dividing by we get which is which along with implies that .

Problem 5. Do the following series converge? Explain briefly why or why not:

1. .

Solution. hence by the vanishing test the series cannot converge.

2. .

Solution. . The latter is a multiple of the harmonic series which doesn't converge, hence the original series doesn't converge either.

3. .

Solution. Ignoring the first two terms of the series, which don't change convergence anyway,

The latter sequence is summable as we have shown in class, hence the original series is convergent.

4. .

Solution. The function is positive at and simple differentiation shows that for , hence it is increasing, and hence it is positive for all . Thus which is summable as was shown in class.

5. .

Solution. That's a tough one. Here's a solution inspired by the solution to Problem 20 of Spivak's Chapter 23, which by itself is inspired by the proof of the divergence of the harmonic series:

If we replace each of the inner sums here by the number of terms in it times the smallest of those, which is the last of those, it only becomes smaller. Hence

The latter are partial sums of a divergent positive series, hence they approach infinity. Therefore the partial sums approach infinity and our series is divergent.

The results. 75 students took the exam; the average grade is 47.4, the median is 46 and the standard deviation is 23.55.

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Dror Bar-Natan 2004-03-19