A student asks:

I know the method for finding the hyperbolic arcosine ((OK, I confess: the student said “$arcosh$”, but I knew what he meant)) - 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 ((That joke never gets old.)) 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 ((they’re all hidden away)).

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:

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

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

$z - 2x + \frac1z = 0$ - 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:

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

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

By convention, we define the $\arcosh$ 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 ((they’re equal when $x=1$.))

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.


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

Do you know of any others?