Why is arcosh the positive root?

A student asks:

I know the method for finding the hyperbolic arcosine ¹ - but I get two roots out of my quadratic formula. Why is it just the positive one?

A quick refresher, in case you don’t know the method

Hyperbolic functions are the BEST FUNCTIONS IN THE WHOLE WIDE WORLD ² and I’ve loved them since the moment I realised you didn’t really have to faff around with minus signs like you do with the trig functions. In fact, via Osborn’s rule, they opened up a whole load of analogies with sine and cosine that made my life as an A-level student so much easier. (And that was for someone who already had a fairly easy life as an A-level student.)

In case you’re a bit rusty on your Further Pure, though, let’s quickly define the hyperbolic functions:

$\sinh (x):=\frac{e^x-e^{-x}}{2}$ $\cosh (x):=\frac{e^x+e^{-x}}{2}$

Easy enough, right? It’s the work of a few moments to see that $\sinh (x)$ differentiates to $\cosh (x)$ and vice-versa - no minus signs in sight ³.

There’s also $\tanh (x):=\frac{\sinh (x)}{\cosh (x)}=\frac{e^{2x}-1}{e^{2x}+1}$ (why?), but today, I don’t care about that.

Inverting hyperbolic functions

If you remember your C3, you’ll know that to invert a function, you find $y=f^{-1}(x)$ by applying $f$ to both sides of the equation to get $f(y)=x$ - and then solve for $y$. Let’s do that:

$y=\mathrm{arcosh}(x)$ $\cosh (y)=x$ $\frac{e^y+e^{-y}}{2}=x$. or $e^y+e^{-y}=2x$, or even $e^y-2x+e^{-y}=0$

It looks like we’re stuck, but let’s define $z:=e^y$:

$z-2x+\frac{1}{z}=0$ - it’s a disguised quadratic. $z^2-2xz+1=0$ (*)

We can solve it, either by the quadratic formula, or by completing the square (I quite like CTS here):

$(z-x{)}^2-x^2+1=0$ $(z-x{)}^2=x^2-1$ $z=x\pm \sqrt{x^2-1}$

And getting back into $y$, as we started with:

$y=\ln (x\pm \sqrt{x^2-1})$

TWO solutions?

In fact, that gives two valid solutions: as long as $\left|x\right|>1$, $\sqrt{x^2-1}$ is always defined and smaller than $x$ (why?) - so you’ll be taking the natural logarithm of a positive number whether you pick the positive or negative root.

The trouble is the shape of the graph $y=\cosh (x)$. We’ve committed one of the cardinal sins of inverting functions and tried to invert something that isn’t one-to-one. The hyperbolic cosine graph looks a bit like a quadratic (it gets a lot steeper a lot more quickly, but it’s still got that distinctive U-shape) - so any given $y$ value corresponds to two distinct $x$ values. ⁴

By convention, we define the $\mathrm{arcosh}$ function so that it always gives a non-negative answer.

Waving a hand at it

Three heuristic reasons that you need the positive root:

• Obviously the bigger root is the positive one, so - given that $\cosh (x)$ is symmetrical about the $y$-axis, we need the positive root.

• The roots $z_1$ and $z_2$ of the quadratic equation (*) have a product of 1 - so $z_1=\frac{1}{z_2}$ and $y_1=-y_2$ using the log rules. The root with the positive sign is bound to give us the positive value of $y$.

• If $x$ is a large positive number, the solutions for $z$ are approximately $z_1\simeq 2x$ and $z_2$ just barely above zero - clearly having a negative logarithm, meaning the positive root is the one we want.

A bit of rigour so that Michael Gove doesn’t explode

Actually, no. No rigour in this section.

A bit of rigour so that it looks like I know how to cook

That’s more like a good reason. What I really want to show is that $x-\sqrt{x^2-1}\le 1$, for all $x\ge 1$, so that the logarithm gives a negative answer.

Method 1: directly

This is quite neat. We could reformulate what we want to show as $x-1\le \sqrt{x^2-1}$.

$x-1$ is the same thing as $\sqrt{x-1}\sqrt{x-1}$, while $\sqrt{x^2-1}=\sqrt{x-1}\sqrt{x+1}$

Clearly, $\sqrt{x-1}\sqrt{x-1}\le \sqrt{x-1}\sqrt{x+1}$ as long as those square roots are defined ⁵

Method 2: sort-of-inductively

Alternatively, you can say that $x-\sqrt{x^2-1}\le 1$ is true when $x=1$ - the two are equal. Now differentiate the left-hand-side to get $1-\frac{x}{\sqrt{x^2-1}}$. Because the denominator of the second term is strictly smaller than $x$, the second term is strictly larger than 1, and the derivative is strictly negative. That means, for all valid values of $x$, the graph is going downhill - it can never get any higher than the value of 1 it starts from.

Phew!

So there you go: several different methods to show why the $\mathrm{arcosh}$ function’s inverse only uses the positive quadratic root.

Do you know of any others?

Footnotes:

1. OK, I confess: the student said “$arcosh$”, but I knew what he meant

2. That joke never gets old.

3. they’re all hidden away

4. That’s not true of $y=\sinh (x)$ or $y=\cosh (x)$, which are one-to-one on the real numbers.

5. they’re equal when $x=1$.