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Homework 1: Solutions and remarks on selected problems
Greenberg
§
5.2, # 1.
A function
f
has exponential order as
t
→ ∞
if there exist
constants
K
≥
0,
T
≥
0, and a constant
c
, such that

f
(
t
)
 ≤
Ke
ct
for all
t
≥
T
.
(1)
A straightforward way to show that a function has exponential order is to identify constants
K
,
c
, and
T
for which it can be shown that (1) holds. However, for a given
c
, you can deduce
that constants
K
and
T
exist making (1) true, if you can show
lim
t
→∞

f
(
t
)

e

ct
<
∞
,
(2)
and so by using (2) you needn’t determine
K
and
T
explicitly to prove exponential order.
Conversely, if
lim
t
→∞

f
(
t
)

e

ct
=
∞
for every positive
c
,
(3)
then
f
is
not
of exponential order as
t
→ ∞
. Thus for problem 5(e), observe that
lim
t
→∞
e

ct
sinh(
t
2
) = lim
t
→∞
(1
/
2)
±
e
t
2

ct

e

t
2

ct
²
=
∞
for every
c >
0 because lim
t
→∞
t
2

ct
=
∞
for every
c
. Thus sinh(
t
2
) is not of exponential
order as
t
→ ∞
.
For problem 5(h), repeated application of L’Hˆopital’s rule shows that for
any
c >
0,
lim
t
→∞
t
100
e

ct
= 0, and hence
t
100
is of exponential order as
t
→ ∞
. Pick a
c >
0. By using
the ﬁrst derivative test, one ﬁnds that
t
100
e

ct
, as a function on the domain 0
≤
t <
∞
,
achieves its maximum at
˜
t
= 100
/c
. Thus, for all
t
≥
0,
t
100
e

ct
≤
(100
/c
)
100
e

100
.
Hence
t
100
≤
(100
/c
)
100
e
100
e
ct
for all
t
≥
0, which establishes (1) with explicit
K
and
T
corresponding to
c
.
Greenberg,
§
5.2, #5,6.
Suppose
R
b
a
f
(
t
)
dt
is an improper integral. This happens if one or
both of the limits of integration are inﬁnite and/or if the function
f
has a vertical asymptote
at a ﬁnite limit of integration. The comparison test for improper integrals implies that if
f
(
t
)
≥
g
(
t
)
≥
0, and if
R
b
a
g
(
t
)
dt
diverges, then so does
R
b
1
f
(
t
)
dt
. On the other hand, if
R
b
a
f
(
t
)
dt
converges, then so does
R
b
a
g
(
t
)
dt
.
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This note was uploaded on 09/29/2009 for the course 650 527 taught by Professor Danocone during the Fall '06 term at Rutgers.
 Fall '06
 DanOcone

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