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Find inverse hyperbolic function

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    find f-1(t) and compute f-1(0) where
    f(t) = sinh(t)


    2. Relevant equations
    sinh(t)=( et- e-t)/2


    3. 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: May 4, 2017
  2. jcsd
  3. Nov 23, 2009 #2

    lanedance

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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
     
  4. Nov 23, 2009 #3
    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.
     
  5. Nov 23, 2009 #4
    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!
     
  6. Nov 23, 2009 #5
    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: Nov 23, 2009
  7. Nov 23, 2009 #6

    lanedance

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    Homework Helper

    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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