Deriving Two Inverse Functions

July 22, 2021September 9, 2021 / Michael Xue / Leave a comment

In “Instrumental Flying“, we defined function $y=\sinh^{-1}(x)$ as

$\{(x, y) | x = \frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1)$

From $x = \frac{e^y-e^{-y}}{2}$ , we obtain

$(e^y)^2-2x\cdot e^y-1=0.$

It means either $e^y = x-\sqrt{x^2+1}$ or $e^y = x+\sqrt{x^2+1}.$

But $e^y = x-\sqrt{x^2+1}$ suggests $e^y < 0$ (see Exercise-1), contradicts the fact that $\forall t \in R, e^t > 0$ (see “Two Peas in a Pod, Part 2“).

Therefore,

$e^y = x+\sqrt{x^2+1} \implies y = \log(x + \sqrt{x^2+1}).$

i.e.,

$\sinh^{-1}(x) = \log(x + \sqrt{x^2+1}), \;\;x \in (-\infty, +\infty).$

We also defined a non-negative valued function $y = \cosh^{-1}(x)$ :

$\{(x, y) | x = \frac{e^y + e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2)$

Simplifying $x=\frac{e^y+e^{-y}}{2}$ yields

$(e^y)^2-2x\cdot e^y+1=0.$

It follows that either $e^y = x-\sqrt{x^2-1}$ or $e^y=x+\sqrt{x^2-1}.$

For both expressions’ right-hand side to be valid requires that $x \ge 1$ . However, when $x = 2$ ,

$e^y = x-\sqrt{x^2-1} = 2 - \sqrt{3} < 1$

suggests that $y < 0$ (see Exercise-2,3), contradicts (2).

Hence,

$e^y = x+\sqrt{x^2-1} \implies y = \log(x+\sqrt{x^2-1}).$

i.e.,

$\cosh^{-1}(x) = \log(x+\sqrt{x^2-1}), \;\;x \in [1, +\infty).$

See also “A Relentless Pursuit“.

Exercise-1 Show that $\forall x \in R, x-\sqrt{x^2+1} < 0.$

Exercise-2 Without calculator or CAS, show that $2-\sqrt{3} < 1.$

Exercise-3 Prove $\forall t \ge 0. e^t \ge 1$ (hint: see “Two Peas in a Pod, Part 2“)