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Prove that sinhkt is of exponential order |k|?

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    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


    2. Relevant equations
    |f(t)|= Me^(ct), t>T


    3. 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?
     
  2. jcsd
  3. Jul 30, 2008 #2

    Dick

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    sinh(kt)=(exp(kt)-exp(-kt))/2. Think about it.
     
  4. Jul 30, 2008 #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: Jul 30, 2008
  5. Jul 30, 2008 #4
    heh...its interesting how you were able to express emotion in your posts.
     
  6. Jul 30, 2008 #5
    Its late....and ive been bashing ny head against my desk thinking about this problem
     
  7. Jul 30, 2008 #6

    Dick

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    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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