Inverse Hyperbolics as Logarithms

What arsinh x means

arsinh x is the number whose sinh is x. Call that number y. Then y = arsinh x says exactly the same as x = sinh y, and sinh has a definition in exponentials: $x=\frac{e^y-e^{-y}}{2}$x = (eʸ − e⁻ʸ)/2.

That turns the question into solving an equation for y. The aim is a formula for y in terms of x, and since y sits in an exponent, a logarithm will be needed at the end to bring it down.

Clear the negative power

Multiply both sides by 2: $2x=e^y-e^{-y}$2x = eʸ − e⁻ʸ. The term $e^{-y}$e⁻ʸ is $\frac{1}{e^y}$1/eʸ, so multiplying every term by $e^y$eʸ clears it, because $e^y\times e^{-y}=e^0=1$eʸ × e⁻ʸ = e⁰ = 1. The left side becomes $2x{e}^y$2x eʸ and the right side becomes $e^{2y}-1$e²ʸ − 1.

Collect everything on one side: $e^{2y}-2x{e}^y-1=0$e²ʸ − 2x eʸ − 1 = 0.

A quadratic in $e^y$eʸ

Write u for $e^y$eʸ. Then $e^{2y}={\left(e^y\right)}^2=u^2$e²ʸ = (eʸ)² = u², and the equation reads $u^2-2xu-1=0$u² − 2x u − 1 = 0, a quadratic in u with leading coefficient 1, middle coefficient −2x and constant −1.

The quadratic formula gives $u=\frac{2x\pm \sqrt{4x^2+4}}{2}$u = (2x ± √(4x² + 4))/2. Take 4 out of the root as 2: $\sqrt{4x^2+4}=2\sqrt{x^2+1}$√(4x² + 4) = 2√(x² + 1). Dividing by 2 leaves $e^y=x\pm \sqrt{x^2+1}$eʸ = x ± √(x² + 1).

The two roots as x varies. The gold curve, $x+\sqrt{x^2+1}$x + √(x² + 1), stays above the x-axis for every x; the plain curve, $x-\sqrt{x^2+1}$x − √(x² + 1), stays below it. At x = 0.75 the roots are 2 and −0.5, and at x = 0 they are 1 and −1: each pair multiplies to −1.

Reject the negative root

The two roots cannot both be values of $e^y$eʸ. Since $x^2+1$x² + 1 is bigger than $x^2$x², its square root is bigger than $\sqrt{x^2}=|x|$√(x²) = |x|, and |x| is never less than x. So $\sqrt{x^2+1}>x$√(x² + 1) > x for every x, and the root $x-\sqrt{x^2+1}$x − √(x² + 1) is always negative.

An exponential is never negative, so $e^y$eʸ cannot equal that root. Another way to see it: the two roots of $u^2-2xu-1=0$u² − 2x u − 1 = 0 multiply to the constant term, −1, so one is positive and the other negative whatever x is. The positive one is $e^y=x+\sqrt{x^2+1}$eʸ = x + √(x² + 1).

Take logarithms

Taking the natural logarithm of both sides brings y down from the exponent: $y=\ln \left(x+\sqrt{x^2+1}\right)$y = ln(x + √(x² + 1)). So $\mathrm{arsinh}x=\ln \left(x+\sqrt{x^2+1}\right)$arsinh x = ln(x + √(x² + 1)), for every real x. The bracket is always positive, so the logarithm always exists.

Try x = 0.75. Then $x^2+1=1.5625$x² + 1 = 1.5625, whose square root is 1.25, so the bracket is 0.75 + 1.25 = 2 and arsinh 0.75 = ln 2. Check it the other way: $\sinh \left(\ln 2\right)=\frac{2-\frac{1}{2}}{2}=\frac{3}{4}$sinh(ln 2) = (2 − ½)/2 = 3/4, since e to the power ln 2 is 2 and e to the power −ln 2 is ½.

arcosh: one sign changes

For y = arcosh x, start from $x=\cosh y=\frac{e^y+e^{-y}}{2}$x = cosh y = (eʸ + e⁻ʸ)/2. The same steps give $2x{e}^y=e^{2y}+1$2x eʸ = e²ʸ + 1, so $e^{2y}-2x{e}^y+1=0$e²ʸ − 2x eʸ + 1 = 0. Only the constant has changed sign, and the quadratic formula gives $e^y=x\pm \sqrt{x^2-1}$eʸ = x ± √(x² − 1). The root needs $x\ge 1$x ≥ 1, which is where arcosh is defined.

This time neither root is negative. The two roots multiply to the constant, +1, so they are reciprocals, and their logarithms are a number and its negative. At x = 2 the roots are $2+\sqrt{3}=3.7321$2 + √3 = 3.7321 and $2-\sqrt{3}=0.2679$2 − √3 = 0.2679, with logarithms 1.3170 and −1.3170. These are the two inputs where cosh is 2, one on each side of the y-axis.

arcosh takes the non-negative input, which comes from the larger root. So $\mathrm{arcosh}x=\ln \left(x+\sqrt{x^2-1}\right)$arcosh x = ln(x + √(x² − 1)) for $x\ge 1$x ≥ 1, and $\mathrm{arcosh}2=\ln \left(2+\sqrt{3}\right)=1.3170$arcosh 2 = ln(2 + √3) = 1.3170.

artanh: no quadratic at all

For y = artanh x, start from $x=\tanh y=\frac{e^y-e^{-y}}{e^y+e^{-y}}$x = tanh y = (eʸ − e⁻ʸ)/(eʸ + e⁻ʸ). Multiply the top and the bottom by $e^y$eʸ: $x=\frac{e^{2y}-1}{e^{2y}+1}$x = (e²ʸ − 1)/(e²ʸ + 1).

Multiply out: $x{e}^{2y}+x=e^{2y}-1$x e²ʸ + x = e²ʸ − 1. Gather the $e^{2y}$e²ʸ terms on one side: $e^{2y}\left(1-x\right)=1+x$e²ʸ(1 − x) = 1 + x, so $e^{2y}=\frac{1+x}{1-x}$e²ʸ = (1 + x)/(1 − x). This is a linear equation in $e^{2y}$e²ʸ, with one solution and nothing to reject.

Take logarithms and halve: $y=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$y = ½ ln((1 + x)/(1 − x)). The fraction is positive exactly when −1 < x < 1, the inputs artanh accepts. At x = 0.6 the fraction is $\frac{1.6}{0.4}=4$1.6/0.4 = 4, so artanh 0.6 = ½ ln 4 = ln 2.

Checks

At zero the arsinh formula gives $\ln \left(0+\sqrt{1}\right)=\ln 1=0$ln(0 + √1) = ln 1 = 0, and sinh 0 is 0, so arsinh 0 = 0 as it should be. The arcosh formula at 1 gives ln(1 + 0) = 0, and cosh 0 = 1. The artanh formula at 0 gives ½ ln 1 = 0, and tanh 0 = 0.

Three values give ln 2: $\mathrm{arsinh}\left(\frac{3}{4}\right)$arsinh(3/4), $\mathrm{arcosh}\left(\frac{5}{4}\right)$arcosh(5/4) and $\mathrm{artanh}\left(\frac{3}{5}\right)$artanh(3/5). For arcosh, $\frac{5}{4}+\sqrt{\frac{25}{16}-1}=\frac{5}{4}+\frac{3}{4}=2$5/4 + √(25/16 − 1) = 5/4 + 3/4 = 2. Going back, e to the power ln 2 is 2 and e to the power −ln 2 is ½, so $\sinh \left(\ln 2\right)=\frac{2-\frac{1}{2}}{2}=\frac{3}{4}$sinh(ln 2) = (2 − ½)/2 = 3/4, $\cosh \left(\ln 2\right)=\frac{2+\frac{1}{2}}{2}=\frac{5}{4}$cosh(ln 2) = (2 + ½)/2 = 5/4 and $\tanh \left(\ln 2\right)=\frac{1.5}{2.5}=\frac{3}{5}$tanh(ln 2) = 1.5/2.5 = 3/5.

For large x, $\sqrt{x^2+1}$√(x² + 1) is very close to x, so arsinh x is close to ln 2x. At x = 10, $\mathrm{arsinh}10=\ln \left(10+\sqrt{101}\right)=\ln 20.0499=2.9982$arsinh 10 = ln(10 + √101) = ln 20.0499 = 2.9982, and ln 20 = 2.9957.

The point (cosh t, sinh t) on the hyperbola $x^2-y^2=1$x² − y² = 1, beside the point at angle t on the unit circle; each shaded sector has area $\frac{t}{2}$t/2. At t = 0.5 the point is (1.1276, 0.5211), and $1.1276+0.5211=1.6487=e^{0.5}$1.1276 + 0.5211 = 1.6487 = e^0.5. Drag t to 1 and the point is (1.5431, 1.1752), whose coordinates add to 2.7183, which is e. Since $\cosh t+\sinh t=e^t$cosh t + sinh t = eᵗ for every t, t = ln(x + y) recovers t from the point.

The usual mistakes

Keeping both roots. $e^y=x-\sqrt{x^2+1}$eʸ = x − √(x² + 1) is negative, and no exponential is, so for arsinh only the plus sign gives an answer.

Mixing up the signs under the root. arsinh has $x^2+1$x² + 1, because its quadratic has constant −1. arcosh has $x^2-1$x² − 1, because its quadratic has constant +1.

Taking the logarithm of x alone. $e^y$eʸ equals the whole of $x+\sqrt{x^2+1}$x + √(x² + 1), so the whole of it goes inside the logarithm: $\ln \left(x+\sqrt{x^2+1}\right)$ln(x + √(x² + 1)), not $\ln x+\sqrt{x^2+1}$ln x + √(x² + 1).

Losing the half, or turning the fraction upside down, in artanh. The equation gives $e^{2y}$e²ʸ, not $e^y$eʸ, so the logarithm is halved, and the fraction is $\frac{1+x}{1-x}$(1 + x)/(1 − x), which is bigger than 1 when x is positive.

A cable and two boosts

In the first application below, a chairlift cable hangs as a cosh curve, and arcosh written as a logarithm finds where it reaches a tower. In the second, the speed of a particle is c tanh w, and artanh written as half a logarithm turns each speed into the quantity w that adds.