Equivalence of Definitions of Real Area Hyperbolic Tangent
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Theorem
Let $S$ denote the open real interval:
- $S := \openint {-1} 1$
The following definitions of the concept of Real Area Hyperbolic Tangent are equivalent:
Definition 1
The inverse hyperbolic tangent $\artanh: S \to \R$ is a real function defined on $S$ as:
- $\forall x \in S: \map \artanh x := y \in \R: x = \map \tanh y$
where $\map \tanh y$ denotes the hyperbolic tangent function.
Definition 2
The inverse hyperbolic tangent $\artanh: S \to \R$ is a real function defined on $S$ as:
- $\forall x \in S: \map \artanh x := \dfrac 1 2 \map \ln {\dfrac {1 + x} {1 - x} }$
where $\ln$ denotes the natural logarithm of a (strictly positive) real number.
Proof
Definition 1 implies Definition 2
Let $x = \tanh y$.
Then:
| \(\ds x\) | \(=\) | \(\ds \frac {e^{2 y} - 1} {e^{2 y} + 1}\) | Definition 3 of Hyperbolic Tangent | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x e^{2 y} + x\) | \(=\) | \(\ds e^{2 y} - 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds e^{2 y} - x e^{2 y}\) | \(=\) | \(\ds 1 + x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds e^{2 y}\) | \(=\) | \(\ds \frac {1 + x} {1 - x}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 y\) | \(=\) | \(\ds \map \ln {\frac {1 + x} {1 - x} }\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\) |
$\Box$
Definition 2 implies Definition 1
Let $y = \dfrac {1 + x} {1 - x}$.
| \(\ds \map \tanh {\frac 1 2 \map \ln {\frac {1 + x} {1 - x} } }\) | \(=\) | \(\ds \map \tanh {\frac 1 2 \ln y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {e^{2 \paren {\frac 1 2 \ln y} - 1} } {e^{2 \paren {\frac 1 2 \ln y} + 1} }\) | Definition 3 of Hyperbolic Tangent | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {e^{\ln y} - 1} {e^{\ln y} + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {y - 1} {y + 1}\) | Exponential of Natural Logarithm | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\frac {1 + x} {1 - x} - 1} {\frac {1 + x} {1 - x} + 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {1 + x} - \paren {1 - x} } {\paren {1 + x} + \paren {1 - x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 x} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
$\Box$
Therefore:
| \(\text {(1)}: \quad\) | \(\ds x = y\) | \(\implies\) | \(\ds y = \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\) | Definition 1 implies Definition 2 | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds y = \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\) | \(\implies\) | \(\ds x = \tanh y\) | Definition 2 implies Definition 1 | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x = \tanh y\) | \(\iff\) | \(\ds y = \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }\) |
$\blacksquare$