Prove that sinhkt is of exponential order |k|?

[physicsfan24]

Homework Statement

Prove that sinh(kt) is of exponential order |k|.
Find M>0, c>=0 and T>0 to show that
|f(t)|= Me^(ct), t>T

Homework Equations

|f(t)|= Me^(ct), t>T

The Attempt at a Solution

I'm looking at the graph of sinhkt (i graphed a few values of k) and indeed it is of an exponential order.

Now,
|sinhkt|<=Me^(ct), t>T

sinhkt isn't bound. So c=/=0=/=t


oh boy... help?

 

[Dick]

sinh(kt)=(exp(kt)-exp(-kt))/2. Think about it.

 

[physicsfan24]

Sigh...ok so,

|e^kt-e^-kt|<=2Me^ct
|e^kt-1/e^kt|<=2Me^ct
|(e^3kt-e^kt)/(e^2kt)|<=2Me^ct
|e^3kt-e^kt|<=2Me^[t(c+2k)]
...

ln|e^3kt-e^kt|<=ln|2Me^[t(c+2k)]|
ln|e^2kt - 1|+ tk <= (c+2k)*t+ln|2M|
ln|[e^2kt - 1]/(2M)]| <= (c+2k)t+kt
ln|[e^2kt - 1]/(2M)]| <= (c+3k)t


OK...? uhh... now what?

 

[Gear300]

Sigh...ok so,

OK...? uhh... now what?

heh...its interesting how you were able to express emotion in your posts.

 

[physicsfan24]

heh...its interesting how you were able to express emotion in your posts.

Its late...and I've been bashing ny head against my desk thinking about this problem

 

[Dick]

Its late...and I've been bashing ny head against my desk thinking about this problem

All that head bashing in vain, sigh. When you are trying to prove 'order of' questions you are allowed to make generous estimates. All you need is SOME M,c,T. Not a set that is in any sense optimal. I'll walk you through it. |sinh(kt)|=sinh(|kt|) which is equal to sinh(|k|t) for t>0. Now that's |exp(|k|t)-1/exp(|k|t)|/2. Which is
(1/2)*exp(|k|t)*|(1-exp(-2|k|t)|. Since the last factor goes to 1 as t->infinity there is a value of T such that for t>T, (1-exp(-2|k|t))>(1/2). Can you explicitly find such a T? So for t>T, |sinh(kt)|>(1/4)exp(|k|t). All done.