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Inverse hyperbolic functions (logarithmic form)

  1. Jun 18, 2015 #1
    To express the ##\cosh^{-1}## function as a logarithm, we start by defining the variables ##x## and ##y## as follows:
    $$y = \cosh^{-1}{x}$$
    $$x = \cosh{y}$$
    Where ##y ∈ [0, \infty)## and ##x ∈ [1, \infty)##.
    Using the definition of the hyperbolic cosine function, rearranging, and multiplying through by ##e^y##, we get:
    $$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 -1})}$$
    How exactly do we get rid of the minus sign?
     
  2. jcsd
  3. Jun 18, 2015 #2

    SteamKing

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    You haven't shown all the steps in your derivation of the inverse cosh function, especially how you came to have a "±" in your logarithmic expression.
     
  4. Jun 18, 2015 #3
    $$x = \cosh{y} = \frac{e^y + e^{-y}}{2}$$
    $$e^y - 2x + e^{-y} = 0$$
    $$e^{2y} - 2x e^y + 1 = 0$$
    $$e^y = \frac{2x \pm \sqrt{4x^2 - 4}}{2}$$
    $$e^y = x \pm \sqrt{x^2 - 1}$$
    $$y = \cosh^{-1}{x} = \ln{(x \pm \sqrt{x^2 - 1})}$$
     
  5. Jun 18, 2015 #4

    SteamKing

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  6. Jun 18, 2015 #5
    If ##x > \sqrt{x^2 - 1}## for all ##x \geq 1##, why should we ignore the second solution (##e^y = x - \sqrt{x^2 - 1}##)?


    image.jpg
     
  7. Jun 18, 2015 #6

    SteamKing

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  8. Jun 19, 2015 #7
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