642:527
SOLUTIONS: ASSIGNMENT 4
FALL 2007
Some of these solutions were written by Professor Dan Ocone.
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
.
(4.1)
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. Alternatively, however, for a given
c
, you can deduce that constants
K
and
T
exist making (4.1) true, if you can show
lim
t
→∞

f
(
t
)

e
−
ct
<
∞
,
(4.2)
and so by using (4.2) you needn’t determine
K
and
T
explicitly to prove exponential order.
Conversely, if
lim
t
→∞

f
(
t
)

e
−
ct
=
∞
for every positive
c
,
(4.3)
then
f
is
not
of exponential order as
t
→ ∞
.
(a) Of exponential type, since (4.1) holds with
c
= 4,
K
= 5, and
T
= 0.
(c) Of exponential type by (4.1), using sinh 2
t
= (
e
2
t

e
−
2
t
)
/
2
≤
e
2
t
/
2.
(g) Again by (4.1):

cos
t
3
 ≤
1 so (4.1) holds with
c
= 0,
K
= 1,
T
= 0.
(j) Now it is easiest to use (4.3): for any
c >
0
lim
t
→∞
e
−
ct
cosh(
t
4
) = lim
t
→∞
parenleftBigg
e
t
4
−
ct
+
e
−
t
4
−
ct
2
parenrightBigg
=
∞
,
because lim
t
→∞
t
4

ct
=
∞
for every
c
, so cosh(
t
2
) is not of exponential order as
t
→ ∞
.
6. The function
f
(
t
) =
t
−
2
/
3
is not continuous, or even piecewise continuous, for
t
≥
0,
since lim
t
→∞
t
−
2
/
3
=
∞
. This means that Theorem 5.2.1 does not apply. Nevertheless,
f
(
t
) has a Laplace transform, since for
s >
0,
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 Fall '07
 SPEER
 Laplace, exponential order

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