August 14, 2021

Consider the following set

$S_1 = \{(x,y) | x=\frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1-1)$

For all $(x, y_1), (x, y_2) \in S_1$ , we have

$\frac{e^{y_1}-e^{-y_1}}{2}=x, \frac{e^{y_2}-e^{-y_2}}{2}=x.$

That is,

$(x, y_1), (x, y_2) \in S_1 \implies \frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0.\quad\quad\quad(1-2)$

Since $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$ (Exercise-1),

$\frac{e^{y_1}-e^{-y_1}}{2}-\frac{e^{y_2}-e^{-y_2}}{2}=0 \implies y_1 = y_2.\quad\quad\quad(1-3)$

From (1-2) and (1-3), we conclude:

$\forall (x, y_1), (x, y_2) \in S_1 \implies y_1 = y_2.$

i.e., $S_1$ defines a function.

Alternatively, (1-1) can be expressed as

$S_1 = \{(x, y) | x = \sinh(y)\}.$

It shows that $S_1$ is the inverse of $\sinh(x)$ . Therefore, we name the function defined by (1-1) $\sinh^{-1}$ and write it as

$y = \sinh^{-1}(x).$

Let’s look at another set:

$S_2 = \{(x,y)|x=\frac{e^y+e^{-y}}{2}\}.\quad\quad\quad(2-1)$

For a pair $(x, y_1>0) \in S_2$ , we have

$x=\frac{e^{y_1}+e^{-y_1}}{2}.\quad\quad\quad(2-2)$

For another pair $(x, y_2=-y_1)$ ,

$\frac{e^{y_2} + e^{-y_2}}{2} = \frac{e^{-y_1}+e^{-(-y_1)}}{2} = \frac{e^{-y_1} + e^{y_1}}{2} \overset{(2-2)}{=} x\implies (x, y_2=-y_1) \in S_2.$

Since $y_1>0, y_2 = -y_1 \neq y_1, (x, y_1), (x, y_2) \in S_1$ does not implie $y_1 = y_2$ . It means $S_2$ does not define a function.

However, modification of $S_2$ gives

$S_3 = \{(x, y) | x=\frac{e^y+e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2-3)$

It defines a function.

Rewrite (2-3) as

$S_3 = \{(x, y) | x = \cosh(y), y \ge 0\}.\quad\quad\quad(2-4)$

Then $\forall (x, y_1), (x, y_2) \in S_3$ , we have

$x=\cosh(y_1), x=\cosh(y_2)\implies \cosh(y_1) = \cosh(y_2).\quad\quad\quad(2-5)$

Notice that by (2-3), $y_1, y_2 \ge 0.$

Cleary, (2-5) is true if and only if $y_1 =y_2$ . For if $y_1 \ne y_2$ , by $(\star)$ , $\cosh(y_1) \ne \cosh(y_2).$

Therefore,

$(x, y_1), (x, y_2) \in S_3 \implies y_1 = y_2.$

We name the function defined by (2-3) $\cosh^{-1}$ as (2-4) shows that it is the inverse of $\cosh(x).$

See “Deriving Two Inverse Functions” for the explicit expressions of $\sinh^{-1}$ and $\cosh^{-1}$ .

Prove

$t_1 \ge 0, t_2 \ge 0, t_1 \ne t_2 \implies \cosh(t_1) \ne \cosh(t_2).\quad\quad\quad(\star)$

Without loss of generality, we assume that $t_2 > t_1.$ By Lagrange’s mean-value theorem (see “A Sprint to FTC“),

$\cosh(t_2) -\cos(t_1) =\cosh'(\xi) (t_2-t_1)$

where $\xi \in (t_1, t_2).$

We have

$t_2 > t_1 \implies t_2 - t_1 >0$

and

$\forall t > 0, (\cosh(t))' = (\frac{e^t+e^{-t}}{2})' = \frac{e^t-e^{-t}}{2} \overset{t>0\implies t >-t \implies e^t >e^{-t}}{>} 0.$

Consequently,

$\cosh(t_2) - \cosh(t_1) = \cosh'(\xi) (t_2-t_1) > 0.$

i.e.,

$\cosh(t_2) \ne \cosh(t_1).$

Exercise-1 Show that $\frac{e^{y_1}-e^{-y_1}}{2}- \frac{e^{y_2}-e^{-y_2}}{2}= 0$ if and only if $y_1 = y_2$ .

Day: August 14, 2021

“Instrumental Flying”