Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (98) Next: Homework Assignment 24 Previous: Term Exam 4

# Solution of Term Exam 4

this document in PDF: Solution.pdf

Problem and Solution 1. (Graded by Vicentiu Tipu) Compute the following definite and indefinite integrals in elementary terms:

1. .
2. With we have and hence . Thus

3. With we have and and when , also . Thus

4. Using integration by parts with (thus ) and (thus say as well), we get

Problem 2. The unit ball'' in is the result of revolving the domain (for ) around the axis.

1. State the general Cosmopolitan Integral'' formula for the volume of a body obtained by revolving a domain bounded under the graph of a function around the axis.
2. Compute the volume of .
3. State the general Cosmopolitan Integral'' formula for the surface area of a body obtained by revolving a domain bounded under the graph of a function around the axis.
4. Compute the surface area of .

1. The volume of a body obtained by revolving a domain bounded under the graph of a function around the axis, between the vertical lines and , is .
2. Taking in the above formula we get

3. The area of the surface obtained by revolving the graph of a function around the axis, between and , is .
4. Taking and thus in the above formula we get

Problem 3. Let be a real number which is not a positive integer or 0, let and let be a positive integer.

1. Compute the Taylor polynomial of degree for around 0.
2. Write the corresponding remainder term using one of the formulas discussed in class.
3. Determine (with proof) if there is an interval around 0 on which .

Solution. (Graded by Julian C.-N. Hung)

1. , , and in general, . Thus the Taylor coefficients are and so

2. Our first remainder formula says that there is some between 0 and so that (with )

3. The remainder formula above is a product of three factors. The first in itself is a product of factors of the form for . If then , so the first factor is bounded by a constant times . For any given the second factor is bounded by which is independent of , and if then and so and so the third factor in the remainder formula is bounded by . Multiplying the three bounds we find that for the remainder is bounded by a constant times , and this goes to 0 when goes to . So at least on the interval the remainder goes to 0 and hence .

(Further analysis show that convergence occurs for all , but this doesn't concern us here).

Problem 4. Let be a sequence of sequences'' (an assignment of a real number to every pair of positive integers) and assume that is a sequence so that for every we have . Further assume that .

1. Show that for every positive integer there is a positive integer
so that .
2. Show that .
3. (5 points bonus, no partial credit) Is it always true that also ?

1. For any fixed we have that so by the convergence we can find an for which . Rename to be and you are done.
2. .
3. Take to be the identity matrix'': if though for all . But then for any fixed the sequence (regarded with varying) is eventually the constant 0, so and so . But .

Problem 5.

1. Compute the first 5 partial sums of the series .
2. Prove that .

1. , , , and .
2. Following this trend we guess that . This we prove by induction. There is no need to check low cases -- we've already done that. So all that remains is

But now,

Alternative Solution. Note that . Then use telescopic summation to find that

Now continue as at the end of the previous solution.

The results. 79 students took the exam; the average grade was 62.99, the median was 68 and the standard deviation was 23.32. The average is thus noticeably below the averages for the first three term exams, but still higher than last year's average. I don't know if the same factors from last year applied this time as well; but for what it's worth, see last year's handout What Went Wrong with Term Exam 4?''.

An unrelated computation.

drorbn@coxeter:~/classes/157AnalysisI:1 math
Mathematica 4.1 for IBM AIX
-- Motif graphics initialized --

In[1]:= D[ArcTan[x], {x, 10}]

9              7              5             3
-185794560 x    371589120 x    243855360 x    58060800 x    3628800 x
Out[1]= ------------- + ------------ - ------------ + ----------- - ---------
2 10            2 9            2 8            2 7          2 6
(1 + x )        (1 + x )       (1 + x )       (1 + x )     (1 + x )


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Dror Bar-Natan 2004-10-18