Inverse Hyperbolic Functions

Undoing a function

To undo a function is to start from an output and find the input that produced it. That works only if each output comes from exactly one input. If two inputs give the same output, the output alone cannot say which one it came from.

$\sinh x=\frac{e^x-e^{-x}}{2}$sinh x = (eˣ − e⁻ˣ)/2 passes this test. Its gradient is cosh x, which is never less than 1, so sinh rises everywhere and never comes back to a height it has already had. It also reaches every real number: as x grows, $\frac{e^x}{2}$eˣ/2 grows without bound and $\frac{e^{-x}}{2}$e⁻ˣ/2 shrinks toward nothing, and the same happens in the negative direction with the signs reversed.

So every real number is sinh of exactly one input. For example, sinh 1 = 1.1752 and sinh(−1) = −1.1752. A horizontal line at any height, such as y = 2, crosses the curve exactly once, here at x = 1.4436.

The curve y = sinh x, rising everywhere, with the points (1, 1.1752) and (−1, −1.1752). The line y = 2 crosses it once, at x = 1.4436.

Reflect it in y = x

The inverse, arsinh, sends each output of sinh back to its input: arsinh x is the number whose sinh is x. Every point (a, b) on y = sinh x becomes the point (b, a) on y = arsinh x. Swapping the two coordinates of every point is the same as reflecting the graph in the line y = x.

Since sinh 1 = 1.1752, arsinh 1.1752 = 1. Since sinh 1.4436 = 2, arsinh 2 = 1.4436. sinh takes every real value, so arsinh accepts every real number, and its outputs cover every real number too. sinh 0 = 0 gives arsinh 0 = 0, and arsinh(−2) = −1.4436, the same size as arsinh 2 with the sign changed.

The reflection also changes how fast the curve grows. sinh climbs faster and faster, so arsinh climbs slower and slower: arsinh 3 = 1.8184 and arsinh 10 = 2.9982.

The gold curve is y = arsinh x and the plain curve is y = sinh x, mirror images in the line y = x. The point (1, 1.1752) on sinh reflects to (1.1752, 1) on arsinh, and (1.4436, 2) reflects to (2, 1.4436).

cosh takes each value twice

$\cosh x=\frac{e^x+e^{-x}}{2}$cosh x = (eˣ + e⁻ˣ)/2 fails the test. Replacing x by −x swaps the two exponentials and leaves their sum alone, so cosh(−x) = cosh x: the graph is symmetrical about the y-axis. Its lowest point is cosh 0 = 1, and every height above 1 is reached twice, once on each side.

Height 2, for example, is reached at x = 1.3170 and at x = −1.3170. Reflecting the whole curve in y = x would send the input 2 back to both of those numbers, and a function must give one answer for each input. Heights below 1 are never reached at all, so nothing below 1 has an input to go back to.

The curve y = cosh x, lowest at (0, 1). The line y = 2 crosses it twice, at x = −1.3170 and x = 1.3170.

Keep the right half

The cure is to keep only part of the curve. For $x\ge 0$x ≥ 0, cosh x rises from 1 and never repeats a height, so this half can be reflected. The reflection is arcosh. It accepts every $x\ge 1$x ≥ 1, starts at arcosh 1 = 0, and is never negative.

Of the two inputs where cosh is 2, the reflection keeps the non-negative one: arcosh 2 = 1.3170, not −1.3170. This is called the principal value. The other input is still there to be found, as the negative of the principal value, when a problem needs it.

One consequence: cosh(arcosh x) = x for every $x\ge 1$x ≥ 1, but arcosh(cosh x) = x only when $x\ge 0$x ≥ 0. Start from −1.3170: cosh gives 2, and arcosh brings back 1.3170. In general arcosh(cosh x) = |x|.

The plain curve is y = cosh x for $x\ge 0$x ≥ 0 only, starting at (0, 1). The gold curve is its reflection in y = x, y = arcosh x, starting at (1, 0). The point (1.3170, 2) reflects to (2, 1.3170).

tanh stays between −1 and 1

tanh x = sinh x / cosh x rises everywhere, like sinh, so it repeats no value and needs no cut. But its values are trapped: tanh 1 = 0.7616, tanh 2 = 0.9640 and tanh 3 = 0.9951, closing in on 1 without ever reaching it, and the same toward −1 on the negative side.

So artanh accepts only inputs strictly between −1 and 1. Its outputs cover every real number: artanh 0.5 = 0.5493, artanh 0.9 = 1.4722 and artanh 0.99 = 2.6467. As x gets close to 1, artanh x grows without bound, so the lines x = 1 and x = −1 are vertical asymptotes. They are the reflections of tanh's horizontal asymptotes, y = 1 and y = −1.

The plain curve is y = tanh x, between the dashed lines y = 1 and y = −1. The gold curve is its reflection, y = artanh x, between the dashed lines x = 1 and x = −1. The point (1, 0.7616) reflects to (0.7616, 1).

The three side by side

arsinh x is defined for every real x, and its value can be any real number.

arcosh x is defined for $x\ge 1$x ≥ 1, and its value is never negative.

artanh x is defined for −1 < x < 1, and its value can be any real number.

In each case the inputs of the inverse are the outputs of the original function, and the outputs of the inverse are the inputs that were kept.

The gold curve is $y=x^2$y = x², with its reflection in y = x dashed. P at x = 1.5 and the point at x = −1.5 have the same height, 2.25, so the reflection sends 2.25 back to two places, as cosh does with every height above 1. Switch to $x^3$x³, which repeats no height, and the reflection is a function.

The usual mistakes

Reading arcosh as the reciprocal of cosh. They are different things: $1/\cosh 2=0.2658$1 / cosh 2 = 0.2658, while arcosh 2 = 1.3170, the number whose cosh is 2.

Giving the negative input as arcosh. cosh(−1.3170) is 2 as well, but arcosh 2 is the non-negative one, 1.3170.

Swapping the restrictions. $x\ge 1$x ≥ 1 belongs to arcosh, because cosh never goes below 1. −1 < x < 1 belongs to artanh, because tanh never leaves that interval. arsinh has no restriction at all.

Putting 1 or −1 into artanh. tanh never reaches 1, so artanh 1 has no value.

A cooling tower

In the application below, a cooling tower's outline is a hyperbola traced by x = 30 cosh u and y = 50 sinh u. A height gives sinh u, a radius gives cosh u, and arsinh and arcosh turn each back into u.