Equivalence of Definitions of Real Area Hyperbolic Cosine
Theorem
The following definitions of the concept of Real Area Hyperbolic Cosine are equivalent:
Definition 1
The inverse hyperbolic cosine $\cosh^{-1}: S \to \R$ is a real multifunction defined on $S$ as:
- $\forall x \in S: \map {\cosh^{-1} } x := \set {y \in \R: x = \map \cosh y}$
where $\map \cosh y$ denotes the hyperbolic cosine function.
Definition 2
The inverse hyperbolic cosine $\cosh^{-1}: S \to \R$ is a real multifunction defined on $S$ as:
- $\forall x \in S: \map {\cosh^{-1} } x := \map \ln {x \pm \sqrt {x^2 - 1} }$
where:
- $\ln$ denotes the natural logarithm of a (strictly positive) real number.
- $\sqrt {x^2 - 1}$ denotes the square root of $x^2 - 1$
Proof
Definition 1 implies Definition 2
Let $x = \cosh y$, where $y > 0$.
Let $z = e^y$.
Then:
| \(\ds x\) | \(=\) | \(\ds \frac {e^y + e^{-y} } 2\) | Definition of Hyperbolic Cosine | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 x\) | \(=\) | \(\ds e^y + e^{-y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 x e^y\) | \(=\) | \(\ds e^{2 y} + 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z^2 - 2 x z + 1\) | \(=\) | \(\ds 0\) | Power of Power | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \frac {2 x \pm \sqrt {\paren {-2 x}^2 - 4} } 2\) | Quadratic Formula | ||||||||||
| \(\ds \) | \(=\) | \(\ds x \pm \sqrt{x^2 - 1}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds e^y\) | \(=\) | \(\ds x \pm \sqrt{x^2 - 1}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \map \ln {x \pm \sqrt {x^2 - 1} }\) |
Also, from Minimum of Real Hyperbolic Cosine Function:
- $x = \cosh y \ge 1$
Also:
| \(\ds x\) | \(=\) | \(\ds \sqrt {x^2}\) | $x$ is positive | |||||||||||
| \(\ds \) | \(>\) | \(\ds \sqrt {x^2 - 1}\) | Square Root is Strictly Increasing | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - \sqrt {x^2 - 1}\) | \(>\) | \(\ds 0\) |
Thus $x - \sqrt {x^2 - 1}$ is (strictly) positive.
Aiming for a contradiction, suppose $x - \sqrt {x^2 - 1} > 1$.
Then:
| \(\ds x - \sqrt {x^2 - 1}\) | \(>\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - 1\) | \(>\) | \(\ds \sqrt {x^2 - 1}\) | Both sides are (strictly) positive | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x - 1}^2\) | \(>\) | \(\ds x^2 - 1\) | right hand side is (strictly) positive because $x \ge 1$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x^2 - 2 x + 1\) | \(>\) | \(\ds x^2 - 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2\) | \(>\) | \(\ds 2 x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(<\) | \(\ds 1\) |
and a contradiction is deduced.
Therefore:
- $x - \sqrt {x^2 - 1} < 1$
From Logarithm is Strictly Increasing:
- $y = \map \ln {x - \sqrt {x^2 - 1} } < \ln 1 = 0$
Since $y$ is (strictly) positive from the first definition of real inverse hyperbolic cosine:
- $y = \map \ln {x + \sqrt {x^2 - 1} }$
$\Box$
Definition 2 implies Definition 1
Let $z = x + \sqrt {x^2 - 1}$.
Then:
- $y = \ln z$
| \(\ds \map \cosh {\map \ln {x + \sqrt {x^2 - 1} } }\) | \(=\) | \(\ds \map \cosh {\ln z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {e^{\ln z} + e^{-\ln z} } 2\) | Definition of Hyperbolic Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {z + \frac 1 z} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {z^2 + 1} {2 z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {x + \sqrt {x^2 - 1} }^2 + 1} {2 x + 2 \sqrt {x^2 - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x^2 + 2 x \sqrt {x^2 - 1} + \paren {x^2 - 1} + 1} {2 x + 2 \sqrt {x^2 - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 x^2 + 2 x \sqrt {x^2 - 1} } {2 x + 2 \sqrt {x^2 - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {2 x + 2 \sqrt {x^2 - 1} } x} {2 x + 2 \sqrt {x^2 - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
If $-1 < x < 1$, $z$ is not defined.
If $x \le -1$:
| \(\ds \sqrt {x^2 - 1}\) | \(<\) | \(\ds \sqrt {x^2}\) | Square Root is Strictly Increasing | |||||||||||
| \(\ds \) | \(=\) | \(\ds -x\) | $x$ is negative | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x + \sqrt{x^2 - 1}\) | \(<\) | \(\ds 0\) |
If $x \ge 1$, $z \ge 1$.
Therefore, $y = \ln z \ge \ln 1 = 0$.
$\Box$
Therefore:
| \(\text {(1)}: \quad\) | \(\ds y > 0 \land x = \cosh y\) | \(\implies\) | \(\ds y = \map \ln {x + \sqrt {x^2 + 1} }\) | Definition 1 implies Definition 2 | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds y = \map \ln {x + \sqrt {x^2 + 1} }\) | \(\implies\) | \(\ds x = \cosh y \land y > 0\) | Definition 2 implies Definition 1 | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y > 0 \land x = \cosh y\) | \(\iff\) | \(\ds y = \map \ln {x + \sqrt {x^2 + 1} }\) |
$\blacksquare$