Equivalence of Definitions of Real Inverse Hyperbolic Tangent

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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:

$\forall x \in S: \tanh^{-1} \left({x}\right) := \dfrac 1 2 \operatorname{ln} \left({\dfrac {1 + x} {1 - x} }\right)$

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:

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

$\blacksquare$


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