Written by Colin+ in further pure 2.

A student asks:

I know the method for finding the hyperbolic arcosine1 - but I get two roots out of my quadratic formula. Why is it just the positive one?

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 sight3.

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.

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:

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

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 $\arcosh$ function so that it always gives a non-negative answer.

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.

Actually, no. No rigour in this section.

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.

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 defined5

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?