Solution of Term Exam 4
Problem and Solution 1. (Graded by Vicentiu Tipu)
Compute the following definite and indefinite integrals in elementary
terms:

.
 With we have and hence
. Thus
 With we have and and when , also .
Thus
 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.
 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.
 Compute the volume of .
 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.
 Compute the surface area of .
Solution. (Graded by Cristian Ivanescu)
 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
.
 Taking
in the above formula we get
 The area of the surface obtained by revolving the graph of a
function around the axis, between and , is
.
 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.
 Compute the Taylor polynomial of degree for
around 0.
 Write the corresponding remainder term using one of the formulas
discussed in class.
 Determine (with proof) if there is an interval around 0 on which
.
Solution. (Graded by Julian C.N. Hung)

,
,
and in general,
. Thus the Taylor
coefficients are
and
so
 Our first remainder formula says that there is some between 0
and so that (with )
 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
.
 Show that for every positive integer there is a
positive integer
so that
.
 Show that
.
 (5 points bonus, no partial credit)
Is it always true that also
?
Solution. (Graded by Vicentiu Tipu)
 For any fixed we have that
so by the convergence
we can find an for which
. Rename to be and you are done.

.
 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.
 Compute the first 5 partial sums of the series
.
 Prove that
.
Solution. (Graded by Cristian Ivanescu)

,
,
,
and
.
 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
Copyright 19882000 Wolfram Research, Inc.
 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 )
Dror BarNatan
20041018