# Equivalence of Definitions of Real Inverse Hyperbolic Tangent

## Theorem

Let $S$ denote the open real interval:

$S := \left({-1 \,.\,.\, 1}\right)$

The following definitions of the concept of Real Inverse Hyperbolic Tangent are equivalent:

### Definition 1

The inverse hyperbolic tangent $\tanh^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \tanh^{-1} \left({x}\right) := y \in \R: x = \tanh \left({y}\right)$

where $\tanh \left({y}\right)$ denotes the hyperbolic tangent function.

### Definition 2

The inverse hyperbolic tangent $\tanh^{-1}: S \to \R$ is a real function defined on $S$ as:

$\forall x \in S: \map {\tanh^{-1} } x := \dfrac 1 2 \map \ln {\dfrac {1 + x} {1 - x} }$

where $\ln$ denotes the natural logarithm of a (strictly positive) real number.

## Proof

### Definition 1 implies Definition 2

Let $x = \tanh y$.

Then:

 $\displaystyle x$ $=$ $\displaystyle \frac {e^{2 y} - 1} {e^{2 y} + 1}$ Definition of Hyperbolic Tangent $\displaystyle \implies \ \$ $\displaystyle x e^{2 y} + x$ $=$ $\displaystyle e^{2 y} - 1$ $\displaystyle \implies \ \$ $\displaystyle e^{2 y} - x e^{2 y}$ $=$ $\displaystyle 1 + x$ $\displaystyle \implies \ \$ $\displaystyle e^{2 y}$ $=$ $\displaystyle \frac {1 + x} {1 - x}$ $\displaystyle \implies \ \$ $\displaystyle 2 y$ $=$ $\displaystyle \ln \left({ \frac {1 + x} {1 - x} }\right)$ $\displaystyle \implies \ \$ $\displaystyle y$ $=$ $\displaystyle \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)$

$\Box$

### Definition 2 implies Definition 1

Let $y = \dfrac {1 + x} {1 - x}$.

 $\displaystyle \tanh \left({\frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)}\right)$ $=$ $\displaystyle \tanh \left({\frac 1 2 \ln y}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {e^{2 \left({\frac 1 2 \ln y}\right) - 1} } {e^{2 \left({\frac 1 2 \ln y}\right) + 1} }$ Definition of Hyperbolic Tangent $\displaystyle$ $=$ $\displaystyle \frac {e^{\ln y} - 1} {e^{\ln y} + 1}$ $\displaystyle$ $=$ $\displaystyle \frac {y - 1} {y + 1}$ Exponential of Natural Logarithm $\displaystyle$ $=$ $\displaystyle \frac {\frac {1 + x} {1 - x} - 1} {\frac {1 + x} {1 - x} + 1}$ $\displaystyle$ $=$ $\displaystyle \frac {\left({1 + x}\right) - \left({1 - x}\right)} {\left({1 + x}\right) + \left({1 - x}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {2x} 2$ $\displaystyle$ $=$ $\displaystyle x$

$\Box$

Therefore:

 $\text {(1)}: \quad$ $\displaystyle x = \tanh \left({y}\right)$ $\implies$ $\displaystyle y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)$ Definition 1 implies Definition 2 $\text {(2)}: \quad$ $\displaystyle y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)$ $\implies$ $\displaystyle x = \tanh \left({y}\right)$ Definition 2 implies Definition 1 $\displaystyle \implies \ \$ $\displaystyle x = \tanh \left({y}\right)$ $\iff$ $\displaystyle y = \frac 1 2 \ln \left({ \frac {1 + x} {1 - x} }\right)$

$\blacksquare$