Natural Log Composed with Hyperbolic Tangent & this Ratio

[afbase]

Hello,

Consider $$x \in (0,1)$$, that is x between 0 and 1. Can someone explain why the following is true:
$$\frac{x-1}{x+1} = \tanh \left( \ln \left( \frac{x}{2} \right) \right)$$

 

[Mute]

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

$$\frac{x-1}{x+1} = \mbox{tanh}\left(\frac{\ln x}{2}\right)$$

This follows from the identity

$$\mbox{artanh}(x) = \frac{1}{2} \ln \left( \frac{1+x}{1-x}\right)$$

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

$$z = \mbox{tanh}(w) = \frac{e^w-e^{-w}}{e^w+e^{-w}}$$

 

[afbase]

Ah thank you!

Sorry about that. I made a typo.