Equivalence of Definitions of Real Inverse Hyperbolic Cosine

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Theorem

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

Definition 1

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

$\forall x \in S: \cosh^{-1} \left({x}\right) := y \in \R_{\ge 0}: x = \cosh \left({y}\right)$

where $\cosh \left({y}\right)$ denotes the hyperbolic cosine function.

Definition 2

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

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

where:

$\sqrt {x^2 - 1}$ denotes the positive square root of $x^2 - 1$
$\ln$ denotes the natural logarithm of a (strictly positive) real number.


Proof

Definition 1 implies Definition 2

Let $x = \cosh y$, where $y > 0$.

Let $z = e^y$.

Then:

\(\displaystyle x\) \(=\) \(\displaystyle \frac {e^y + e^{-y} } 2\) Definition of Hyperbolic Cosine
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 x\) \(=\) \(\displaystyle e^y + e^{-y}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 x e^y\) \(=\) \(\displaystyle e^{2 y} + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^2 - 2 x z + 1\) \(=\) \(\displaystyle 0\) Power of Power
\(\displaystyle \leadsto \ \ \) \(\displaystyle z\) \(=\) \(\displaystyle \frac {2 x \pm \sqrt {\paren {-2 x}^2 - 4} } 2\) Quadratic Formula
\(\displaystyle \) \(=\) \(\displaystyle x \pm \sqrt{x^2 - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^y\) \(=\) \(\displaystyle x \pm \sqrt{x^2 - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \map \ln {x \pm \sqrt {x^2 - 1} }\)


Also, from Minimum of Real Hyperbolic Cosine Function:

$x = \cosh y \ge 1$


Also:

\(\displaystyle x\) \(=\) \(\displaystyle \sqrt {x^2}\) $x$ is positive
\(\displaystyle \) \(>\) \(\displaystyle \sqrt {x^2 - 1}\) Square Root is Strictly Increasing
\(\displaystyle \leadsto \ \ \) \(\displaystyle x - \sqrt {x^2 - 1}\) \(>\) \(\displaystyle 0\)


Thus $x - \sqrt {x^2 - 1}$ is (strictly) positive.


Aiming for a contradiction, suppose $x - \sqrt {x^2 - 1} > 1$.

Then:

\(\displaystyle x - \sqrt {x^2 - 1}\) \(>\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x - 1\) \(>\) \(\displaystyle \sqrt {x^2 - 1}\) Both sides are (strictly) positive
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x - 1}^2\) \(>\) \(\displaystyle x^2 - 1\) right hand side is (strictly) positive because $x \ge 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 - 2 x + 1\) \(>\) \(\displaystyle x^2 - 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2\) \(>\) \(\displaystyle 2 x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(<\) \(\displaystyle 1\)

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:

$y = \ln z$
\(\displaystyle \map \cosh {\map \ln {x + \sqrt {x^2 - 1} } }\) \(=\) \(\displaystyle \map \cosh {\ln z}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {e^{\ln z} + e^{-\ln z} } 2\) Definition of Hyperbolic Cosine
\(\displaystyle \) \(=\) \(\displaystyle \frac {z + \frac 1 z} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {z^2 + 1} {2 z}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {x + \sqrt {x^2 - 1} }^2 + 1} {2 x + 2 \sqrt {x^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2 + 2 x \sqrt {x^2 - 1} + \paren {x^2 - 1} + 1} {2 x + 2 \sqrt {x^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 x^2 + 2 x \sqrt {x^2 - 1} } {2 x + 2 \sqrt {x^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {2 x + 2 \sqrt {x^2 - 1} } x} {2 x + 2 \sqrt {x^2 - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle x\)


If $-1 < x < 1$, $z$ is not defined.

If $x \le -1$:

\(\displaystyle \sqrt {x^2 - 1}\) \(<\) \(\displaystyle \sqrt {x^2}\) Square Root is Strictly Increasing
\(\displaystyle \) \(=\) \(\displaystyle -x\) $x$ is negative
\(\displaystyle \leadsto \ \ \) \(\displaystyle x + \sqrt{x^2 - 1}\) \(<\) \(\displaystyle 0\)

If $x \ge 1$, $z \ge 1$.

Therefore, $y = \ln z \ge \ln 1 = 0$.

$\Box$


Therefore:

\(\text {(1)}: \quad\) \(\displaystyle y > 0 \land x = \cosh y\) \(\implies\) \(\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }\) Definition 1 implies Definition 2
\(\text {(2)}: \quad\) \(\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }\) \(\implies\) \(\displaystyle x = \cosh y \land y > 0\) Definition 2 implies Definition 1
\(\displaystyle \leadsto \ \ \) \(\displaystyle y > 0 \land x = \cosh y\) \(\iff\) \(\displaystyle y = \map \ln {x + \sqrt {x^2 + 1} }\)

$\blacksquare$


Also see