y = \operatorname{arsech} x

The Inverse Hyperbolic Secant $y = \operatorname{arsech} x$

$\operatorname{arsech} x$ , the inverse hyperbolic secant, is the inverse of $\operatorname{sech} x = \dfrac{1}{\cosh x}$ ¹. Since $\operatorname{sech}$ is even it is not one-to-one on the whole line, so it is restricted to $x \geq 0$ , where the value falls monotonically from $1$ to $0$ , and the inverse is taken there.

Definition and closed form

The equation $\operatorname{sech} y = x$ is the same as $\cosh y = \dfrac{1}{x}$ , so $\operatorname{arsech} x = \operatorname{arcosh}\dfrac{1}{x}$ . Written with a logarithm it takes the following form.

\operatorname{arsech} x = \ln\frac{1 + \sqrt{1-x^{2}}}{x}

Domain and range

The domain is $0 < x \leq 1$
The range is $y \geq 0$
It decreases monotonically
It is neither even nor odd

The range of $\operatorname{sech}$ was $(0, 1]$ , and that interval has become the domain of the inverse. At $x = 1$ the value is $0$ , and as $x \to 0^{+}$ we have $\ln\dfrac{2}{x} \to +\infty$ , so the $y$ -axis is a vertical asymptote.

Monotonicity

The derivative is as follows.

\frac{d}{dx}\operatorname{arsech} x = -\frac{1}{x\sqrt{1-x^{2}}}

It is negative throughout the domain, so the function decreases monotonically. As $x \to 1^{-}$ the factor $\sqrt{1-x^{2}}$ tends to $0$ and the derivative diverges, so the tangent at $(1, 0)$ is vertical. The curve stands up at its right end and stretches away to the left along the $y$ -axis.

Concavity

The second derivative is $\dfrac{1-2x^{2}}{x^{2}(1-x^{2})^{3/2}}$ . The denominator is positive, so the sign comes from $1-2x^{2}$ , which changes at $x = \dfrac{1}{\sqrt{2}}$ . That single point is the inflection point, where the value is $\operatorname{arcosh}\sqrt{2} = \ln(1+\sqrt{2}) \approx 0.881$ . The curve is concave up to its left and concave down to its right.

Notable values

$x$	$\operatorname{arsech} x$
$0.1$	$\approx 2.9932$
$0.5$	$\ln(2+\sqrt{3}) \approx 1.3170$
$\dfrac{1}{\sqrt{2}}$	$\ln(1+\sqrt{2}) \approx 0.8814$
$1$	$0$

How it differs from the other inverse hyperbolic functions

Function	Domain
$\operatorname{arsinh} x$	all real numbers
$\operatorname{arcosh} x$	$x \geq 1$
$\operatorname{artanh} x$	$-1 < x < 1$
$\operatorname{arsech} x$	$0 < x \leq 1$

Only $\operatorname{arsech}$ takes charge of a bounded interval. That $\operatorname{sech}$ was even and bounded shows up on the inverse side as a narrow domain and a restriction of the values to the positive side.

Relation to the tractrix

The tractrix is given by the following equation, whose first term is exactly this function².

y = \ln\frac{1+\sqrt{1-x^{2}}}{x} - \sqrt{1-x^{2}} = \operatorname{arsech} x - \sqrt{1-x^{2}}

The path of an object dragged by a string of fixed length is the inverse hyperbolic secant with a semicircle subtracted. Together with the fact that the pseudosphere, obtained by revolving the tractrix, has constant negative curvature, this function also shows its face on the side of non-Euclidean geometry.

Inverse hyperbolic functions, Wikipedia
Tractrix, Wikipedia

The Inverse Hyperbolic Secant y=arsech⁡xy = \operatorname{arsech} xy=arsechx