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Math Help - Hyperbolic functions

  1. #1
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    Hyperbolic functions

    Find an explicit expression for tanh^-1x in terms of the logarithm function ln.

    Could anyone explain how to solve this. Cheers thanks v much.
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  2. #2
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    Quote Originally Posted by Haris View Post
    Find an explicit expression for tanh^-1x in terms of the logarithm function ln.
    Let y = \tanh x. Then y = \frac{e^x-e^{-x}}{e^x+e^{-x}}. Multiply out the fraction, and rearrange the equation so as so express it as a quadratic in e^x. Solve the quadratic to get a formula for e^x in terms of y, then take the logarithm of both sides.
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  3. #3
    Senior Member DeMath's Avatar
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    Can find this method

    Let {\tanh ^{ - 1}}\left( x \right) = y.

    Then x = \tanh y = \frac{{\sinh y}}{{\cosh y}} = \frac{{{e^y} - {e^{ - y}}}}{{{e^y} + {e^{ - y}}}} = \frac{{{e^{2y}} - 1}}{{{e^{2y}} + 1}} \Rightarrow x\left( {{e^{2y}} + 1} \right) = {e^{2y}} - 1

    \Rightarrow {e^{2y}} - x{e^{2y}} = 1 + x \Rightarrow \left( {1 - x} \right){e^{2y}} = 1 + x \Rightarrow {e^{2y}} = \frac{{1 + x}}{{1 - x}} \Rightarrow y = \frac{1}{2}\ln \frac{{1 + x}}{{1 - x}}.

    So {\tanh ^{ - 1}}\left( x \right) = \frac{1}{2}\ln \frac{{1 + x}}{{1 - x}}.
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