Find inverse hyperbolic function

  • Thread starter synergix
  • Start date
  • #1
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Homework Statement


find f-1(t) and compute f-1(0) where
f(t) = sinh(t)


Homework Equations


sinh(t)=( et- e-t)/2


The Attempt at a Solution



t= (ey- e-y)/2

ln2t = y - - y

y = (ln2t)/2

but wikipedia says
http://upload.wikimedia.org/math/8/e/d/8edd21edb4cd413e462fe82ef4ac249b.png" [Broken]

what silly mistake have i made this time?
 
Last edited by a moderator:

Answers and Replies

  • #2
lanedance
Homework Helper
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i would check the step where you take the logarithm

ony way that may work, would be to let y = f(t), x=e^t then try and solve for x(y) first
 
  • #3
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y = (et- e-t)/2
2y = et- e-t
2yet = e2t - 1 (multiply both sides by et)

Let x = et, then you have a quadratic equation to help you isolate the t.
 
  • #4
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yes! this is how my instructor showed me but we have covered soo much this past couple of weeks I completely forgot about that technique. Thanks!
 
  • #5
178
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I used z instead of x and I got to
z^2-2tz-1=0
z= t+- sqrt(t^2+1)
z = e^y

so ln both sides

how do i know whether it is positive or negative?

since i can see from the formulas on wikipedia I went ahead with the rest and f-1(0) = ln1 = 0
 
Last edited:
  • #6
lanedance
Homework Helper
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so is [itex] t-\sqrt{t^2+1} [/itex] greater or less than zero? is this for all t? consider what happens when you take the logarithm, and which is actually a solution to the orginal equation

also, I find the way you swap y & t is quite confusing
you start with:
[itex] y(t) = sinh(t) [/itex]

so if I was you, I would solve for t(y)
 

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