Solution of Term Exam 1
Problem 1.
All that is known about the angle is that
. Can you find
and
? Explain your reasoning in full detail.
Solution. (Graded by C.-N. (J.) Hung)
In class we wrote the formula
. Also
using
and taking
we get
Dividing the numerator and denominator by
this
becomes
Likewise using
we get
Problem 2.
- State the definition of the natural numbers.
- Prove that every natural number has the property that whenever
is natural, so is .
Solution. (Graded by V. Tipu)
- The set of natural numbers
is the smallest set of numbers
for which
Alternatively, the set of natural numbers
is the intersection of all
sets of numbers satisfying
- Let be the assertion ``whenever is natural, so is ''.
We prove by induction on :
- asserts that ``whenever is natural, so is ''. This is
true by the second bullet in the definition of
.
- Assume , that is, assume that whenever is natural, so is
. Let be a natural number. Then
is a natural
number because by the number is natural and because adding one
to a natural number gives a natural number by the second bullet in the
definition of
. So we have shown that whenever is natural so is
, and this is the assertion .
Problem 3.
Recall that a function is called ``even'' if
for all
and ``odd'' if
for all , and let be some
arbitrary function.
- Find an even function and an odd function so that .
- Show that if
where and are even and
and are odd, then and .
Solution. (Graded by C. Ivanescu)
- Set
and
. Then
while
(so is even)
and
(so is
odd).
- Assume where is even and is odd. Then
So necessarily
. Now if
as above, then both and can play the role of in this
argument, so they are both equal to
and in particular
they equal each other. Likewise,
and arguing like before,
.
Problem 4.
Sketch, to the best of your understanding, the graph of the function
(What happens for near 0? Near ? For large ? Is the graph
symmetric? Does it appear to have a peak somewhere?)
Solution. (Graded by C. Ivanescu)
If then and so
; furthermore, the
larger is (while ), the larger is and hence the
smaller
is. When approaches from above,
approaches 0 from above and hence
becomes larger and
larger. If the and so
. When , and when approaches
from below, approaches from below and approaches 0 from
below, and so
becomes more and more negative. In summary,
the graph looks something like:
Problem 5.
- Suppose that
for all , and that the limits
and
both exist. Prove that
.
- Suppose that for all , and that the limits
and
both exist. Is it always true
that
? (If you think it's always true,
write a proof. If you think it isn't always true, provide a
counterexample).
Solution. (Graded by C.-N. (J.) Hung)
- Let
and
and assume by contradiction that ; that is, that
. Use the existence of the two limits to find
and
so that
and
Now choose some specific for which both
and
. But then
and so
while
and so
. Therefore remembering that
for all we get
or
or
which is a contradiction. Thus the assumption that must be incorrect
and thus .
- Take for all and for all and
. Then for all but
. So it isn't always true that if
for all and the limits exist, then
.
The results. 105 students took the exam; the average
grade was 67.19, the median was 70 and the standard deviation
was 21.12.
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Dror Bar-Natan
2004-10-18