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The relation between the natural log composed with hyperbolic tangent and this ratio

  1. Jul 12, 2010 #1
    Hello,

    Consider [tex] x \in (0,1) [/tex], that is x between 0 and 1. Can someone explain why the following is true:
    [tex]\frac{x-1}{x+1} = \tanh \left( \ln \left( \frac{x}{2} \right) \right)[/tex]
     
  2. jcsd
  3. Jul 12, 2010 #2

    Mute

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    Re: The relation between the natural log composed with hyperbolic tangent and this ra

    It's not true. That equality doesn't hold. The correct expression is

    [tex]\frac{x-1}{x+1} = \mbox{tanh}\left(\frac{\ln x}{2}\right)[/tex]

    This follows from the identity

    [tex]\mbox{artanh}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x}\right)[/tex]

    You can get from this to the other form by making the replacement [itex]y = (1+x)/(1-x)[/itex]. To derive this identity, solve the following for w:

    [tex]z = \mbox{tanh}(w) = \frac{e^w-e^{-w}}{e^w+e^{-w}}[/tex]
     
  4. Jul 12, 2010 #3
    Re: The relation between the natural log composed with hyperbolic tangent and this ra

    Ah thank you!

    Sorry about that. I made a typo.
     
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