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Prove hyperbolic function

  1. Dec 3, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20111203/B1Ax1OcU.jpg

    Frankly, i've been sitting starring at that problem for long enough, and it just can't be solved through the direct use of the standard hyperbolic identities. I need a hint.
     
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  3. Dec 3, 2011 #2

    micromass

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    Use induction.
     
  4. Dec 3, 2011 #3

    sharks

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    That's a long shot... I just started this topic so i'm very new to it and i have no idea what induction means either. Can you elaborate a little?
     
  5. Dec 3, 2011 #4

    Curious3141

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    Really? Have you tried expressing cosh x and sinh x in terms of e^x and e^(-x)?

    No need for induction. This is a simple direct proof. And induction only covers non-negative integral n, whereas you can more easily prove the general result for all complex n directly.
     
    Last edited: Dec 3, 2011
  6. Dec 3, 2011 #5

    micromass

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    See our proofs FAQ: https://www.physicsforums.com/showthread.php?t=523874 [Broken] for a discussion on induction.

    To make it simpler, try and set n=2. Can you prove it for n=2?
     
    Last edited by a moderator: May 5, 2017
  7. Dec 3, 2011 #6

    Dick

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    No, no induction needed. Just put in the definition of cosh and sinh in terms of exponentials.
     
  8. Dec 3, 2011 #7

    dextercioby

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    Indeed, induction seems the most elegant proof.
     
  9. Dec 3, 2011 #8

    sharks

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    Alright, what is induction??
     
  10. Dec 3, 2011 #9

    micromass

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    Read the link I posted above. Or check your textbook.
     
  11. Dec 3, 2011 #10

    Curious3141

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    Have you tried the direct method? Start by expressing sinh and cosh as exponentials, then it's just algebra!
     
  12. Dec 3, 2011 #11

    sharks

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    First, i've expressed the L.H.S into two parts:
    (coshx + sinhx)^n and (coshx - sinhx)^n

    For the first one:
    (coshx + sinhx)^n = e^(nx)

    For the second one:
    (coshx - sinhx)^n = e^(-nx)

    I am aware that i somehow need to bring that power n down, but can't see how.
     
  13. Dec 3, 2011 #12

    micromass

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    Now do the same with the RHS
     
  14. Dec 3, 2011 #13

    sharks

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    For the R.H.S. i split it into 2 parts:

    (cosh nx + sinh nx) and (cosh nx - sinh nx)

    (cosh nx + sinh nx) = [e^(nx) + e^(-nx)]/2 + [e^(nx) - e^(-nx)]/2 = [e^(nx)]/2

    (cosh nx - sinh nx) = [e^(nx) + e^(-nx)]/2 - [e^(nx) - e^(-nx)]/2 = [e^(-nx)]/2

    OK, i see it now. They are both the same. It did not occur to me that i had to tinker with the R.H.S also, as normally, i should have been able to get the R.H.S. using only the L.H.S.
     
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