Finding the analytic expression for arcsinh

[Mr Davis 97]

Homework Statement

Given that $\sinh x = \frac{e^x-e^{-x}}{2}$, find an expression for $arcsinh x$

Homework Equations

The Attempt at a Solution

We can proceed by the normal procedure for finding inverses of well-defined functions, solve $x = \frac{e^y - e^{-y}}{2}$ for y. After doing some algebra and using the quadratic formula, we find that $y = \log (x \pm \sqrt{x^2 + 1})$. How do I know which root to take? It would seem that there are two inverse functions of sinhx

 

[ehild]

Homework Statement

Given that $\sinh x = \frac{e^x-e^{-x}}{2}$, find an expression for $arcsinh x$

Homework Equations

The Attempt at a Solution

We can proceed by the normal procedure for finding inverses of well-defined functions, solve $x = \frac{e^y - e^{-y}}{2}$ for y. After doing some algebra and using the quadratic formula, we find that $y = \log (x \pm \sqrt{x^2 + 1})$. How do I know which root to take? It would seem that there are two inverse functions of sinhx

What is the argument of the logarithm if you take the negative sign in $y = \log (x \pm \sqrt{x^2 + 1})$?

 

[Stephen Tashi]

The argument of $log(.)$ can't be a negative number.

 

[Mr Davis 97]

So since $e^y$ is always positive, and $x - \sqrt{x^2 + 1}$ is negative for all x, do we consider it an extraneous "solution", and thus take the other root? Why does $x - \sqrt{x^2 + 1}$ show up if it can't actually be a solution?

 

[Mark44]

So since $e^y$ is always positive, and $x - \sqrt{x^2 + 1}$ is negative for all x, do we consider it an extraneous "solution", and thus take the other root? Why does $x - \sqrt{x^2 + 1}$ show up if it can't actually be a solution?

If you had $u^2 - 2yu - 1 = 0$ and your goal was to solve for u, you would get $u = y \pm \sqrt{y^2 + 1}$. Here there is no problem with u being negative.

However, in your equation, you have $e^{2x} - 2ye^x - 1 = 0$. Solving for $e^x$ gives the same solution as for u, above. This time around, $e^x$ must be nonnegative, so we have to discard the solution $y - \sqrt{y^2 + 1}$. You weren't solving an actual quadratic equation -- instead, you were solving one that was only quadratic in form. Since $u = e^x$, u can only take on nonnegative values.

 

[Stephen Tashi]

Why does $x - \sqrt{x^2 + 1}$ show up if it can't actually be a solution?

Because our beloved algebraic manipulations are not completely reliable. We do them without keeping track of the assumptions we use when we apply them.

To repeat what Mark44 said:

From $2x = e^y + \frac{1}{e^y}$ we can multiply both sides by $e^y$ provided we assume $e^y$ is a number. From the properties of the function $e^y$, we must assume $e^y$ is a positive number.

Later where our work leads to the conclusion: $e^y = x + \sqrt{x^2 + 1}$ or $e^y = x - \sqrt{x^2 + 1}$ this work takes place under the assumption $e^y > 0$. So if we were using the appropriate words to keep track of our assumptions, we would see the alternative $e^y = x - \sqrt{x^2 + 1}$ is not viable.

It's similar to evaluating an expression in proposition logic like $A \land (B \lor \lnot A)$ to imply $A \land B$. When we do the algebraic manipulations we forget the "$A \land$" prefixes the work that leads to $(B \lor \lnot A)$ and then we wonder why the alternative $\lnot A$ doesn't work.