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I'm trying to find the equation of the inverse of the hyperbolic tangent as follows:

Take an $x \in ]-1,1[$ and define $y:=\text{arctanh}(x)$ so that $\tanh y=x$.
This means that $\frac{e^{2y}-1}{e^{2y}+1}=x$
This should be equivalent to $e^{2y}=\frac{1+x}{1-x}$

I can't seem to find how to go from one equation to the other. I've tried multiplying both sides by $\frac{e^{2y}}{x}$ but that gets me nowhere.

Joshua
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1 Answers1

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$$y=\text{arctanh}\, x:=\frac{e^x-e^{-x}}{e^x+e^{-x}}\stackrel{\cdot\frac{e^x}{e^x}}=\frac{e^{2x}-1}{e^{2x}+1}\implies$$

$$\implies (y-1)e^{2x}=-y-1\implies e^{2x}=\frac{1+y}{1-y}\implies x=\frac12\log\frac{1+y}{1-y}=\log\sqrt\frac{1+y}{1-y}$$

within the proper limits, of course.

Timbuc
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