Prove hyperbolic function

  • Thread starter DryRun
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  • #1
DryRun
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Homework Statement
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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  • #2
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Use induction.
 
  • #3
DryRun
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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?
 
  • #4
Curious3141
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Homework Statement
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.

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.
 
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  • #5
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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?
 
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  • #6
Dick
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No, no induction needed. Just put in the definition of cosh and sinh in terms of exponentials.
 
  • #7
dextercioby
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Indeed, induction seems the most elegant proof.
 
  • #8
DryRun
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Alright, what is induction??
 
  • #9
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Alright, what is induction??

Read the link I posted above. Or check your textbook.
 
  • #10
Curious3141
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Alright, what is induction??

Have you tried the direct method? Start by expressing sinh and cosh as exponentials, then it's just algebra!
 
  • #11
DryRun
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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.
 
  • #12
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Now do the same with the RHS
 
  • #13
DryRun
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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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