# Hyperbolic functions

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• Jan 8th 2009, 12:49 PM
Haris
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.
• Jan 8th 2009, 01:14 PM
Opalg
Quote:

Originally Posted by Haris
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.
• Jan 8th 2009, 01:36 PM
DeMath
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 ${\tanh ^{ - 1}}\left( x \right) = \frac{1}{2}\ln \frac{{1 + x}}{{1 - x}}$.