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Inverse Hyberbolic Functions

  1. May 17, 2005 #1
    Just a quick question

    Can anyone give a method to derive arcsinh(x) from the definition of sinh(x)?

    Thanks
     
  2. jcsd
  3. May 17, 2005 #2
    [tex]\sinh{x} = \frac{e^x - e^{-x}}{2}[/tex].

    Assuming the existence of arcsinh, for every x we must have:

    sinh(arcsinh(x)) = x.

    For simplicity, let arcsinh(x) = z, so that

    [tex]\sinh(z) = x[/tex]

    <=>

    [tex]e^z - e^{-z} = 2x[/tex]

    <=>

    [tex](e^z)^2 - 1 = e^z \cdot 2x[/tex]

    That's a quadratic equation in e^z, which can be easily solved.
     
  4. May 17, 2005 #3
    thanks

    haven't done that since a-level and had forgotten it completely!
     
  5. May 17, 2005 #4

    dextercioby

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    Since it's a quadratic equation,u'll need to specify the domain.Note that the direct function is defined on all [itex] \mathbb{R} [/itex],while i'm sure u can't say the same about its inverse.

    Daniel.
     
  6. May 17, 2005 #5
    I think arcsinh is ok on all of [itex] \mathbb{R} [/itex]
    With arccosh x has to be greater than or equal to one, but I can't remember the conditions for arctanh
     
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