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Prove that sinhkt is of exponential order |k|?

  • #1

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 isnt bound. So c=/=0=/=t


oh boy... help?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
sinh(kt)=(exp(kt)-exp(-kt))/2. Think about it.
 
  • #3
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?
 
Last edited:
  • #4
1,174
5
Sigh...ok so,

OK....????? uhh... now what?
heh...its interesting how you were able to express emotion in your posts.
 
  • #5
heh...its interesting how you were able to express emotion in your posts.
Its late....and ive been bashing ny head against my desk thinking about this problem
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
Its late....and ive 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.
 

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