MATH 247 — FALL 2000 — FINAL EXAM — BRIEF SOLUTIONS
NAME:
Total: 200 points. Do 8 out of 12 questions.
You MUST indicate which 8 questions
are to be graded; otherwise, just the first 8 problems will be graded.
EXPLAIN every answer. No books, notes, calculators or computers allowed on this exam.
1
(25 points)
.
(a) [8 points] A function
f
(
x
) on [
a, b
] is called
bounded
if there exists
M
∈
R
such that

f
(
x
)
 ≤
M
for all
x
∈
[
a, b
].
Negate this, so obtaining the definition of an
unbounded
function.
Solution.
For all
M
∈
R
there exists
x
∈
[
a, b
] such that

f
(
x
)

> M
.
(b) [8 points] Define what it means to say that “
a
n
converges to
L
”.
Solution.
For all
ε >
0 there exists
N
∈
N
such that for all
n
≥
N
we have

a
n

L

< ε
.
(c) [9 points] Negate your answer in part (b), thus obtaining a definition of “it is false that
a
n
converges to
L
”.
Solution.
There exists
ε >
0 such that for each
N
∈
N
there exists
n
≥
N
such that

a
n

L
 ≥
ε
.
1
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2
(25 points)
.
Consider a function
f
:
Z
→
R
such that
f
(1) = 2,
f
(
m
)
>
0 for all
m
∈
Z
,
and
f
(
j

k
) =
f
(
j
)
f
(
k
)
for all
j, k
∈
Z
.
Using these properties, find a formula for
f
(
m
)
, m
∈
Z
.
(Hint:
play around to guess a
formula, and then use induction ideas to give a proper proof.)
Solution.
The formula
f
(
j

k
) =
f
(
j
)
/f
(
k
) looks like the law of exponents, and so we guess
f
(
m
) =
a
m
for some
a
. Then since
f
(1) = 2 we guess
a
= 2.
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 Spring '09
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 Math, Rational Zeros Theorem, Continuous function, Prime number, 1 K, 2m, Chinese Remainder

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