Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: | (75) |
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**Problem 1. ** In a very
condensed form, the definition of integration is as follows: For
bounded on and
a partition of
set
,
,
and
. Then set
and
. Finally, if we say that
`` is integrable on '' and set
.

From this definition alone, without using *anything* proven in class
about integration, prove that the function given below is integrable on
and compute its integral
:

**Solution. ** (Graded by Cristian Ivanescu) Let
be an arbitrary partition of . Then
for any the infimum
is 0 and so
. Thus
.
At the same time, for any the supremum
is , and hence
and so
. Now let
be given and let
be
the partition
. Then
while and so
. Thus
. But this is true for any
and we
already know that
. So it must be that . Thus
and hence is integrable on and its integral
is
.

**Problem 2. ** Prove that the function

**Solution. ** (Graded by Julian C.-N. Hung) Differentiate
using the first fundamental theorem of calculus. The first summand
yields
. The second summand is the first summand
pre-composed with the function
. So by the chain rule,
the derivative of the second summand is
.
is the sum of these two terms,
. Hence is a constant.

**Problem 3. ** In class we have proven that a
twice-differentiable function satisfying the equation is
determined by and . Use this fact and the known formulas for
the derivatives of and to derive a formula for
in terms of
, ,
and .

**Solution. ** (Graded by Julian C.-N. Hung)
Let be a constant and consider the
functions
and
. Then
and
so both and
satisfy . We also have
and
. So by what we have proven in
class or
.

**Problem 4. **
The function is defined by .

- Compute for all .
- Explain why has a differentiable inverse for .
- Let be the inverse function of (with the domain of considered to be ). Compute and simplify your result as much as you can. Your end result may still contain in it, but not , or .

**Solution. ** (Graded by Vicentiu Tipu)

- .
- For we have that and hence and . So is increasing on . It is also differentiable on that interval, so by a theorem proven in class, it has a differentiable inverse.
- .

**The results. ** 82 students took the exam; the average
grade was 69.3, the median was 78 and the standard deviation
was 26.76.

The generation of this document was assisted by
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Dror Bar-Natan 2004-10-18