y=sech⁡xy = \operatorname{sech} x

Graph of the Hyperbolic Secant y=sech⁡xy = \operatorname{sech} x

The hyperbolic secant function y=sech⁡xy = \operatorname{sech} x is defined as the reciprocal of the hyperbolic cosine1.

sech⁡x=1cosh⁡x=2ex+e−x\operatorname{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}

It is the hyperbolic counterpart of the ordinary secant sec⁡x\sec x.

Domain and range

  • The domain is all real numbers
  • The range is the half-open interval (0,1](0, 1]
  • The maximum 11 is attained at x=0x = 0
  • It is an even function

The denominator cosh⁡x\cosh x is always at least 11 and never 00, so the function is defined for every real number. Since cosh⁡x≥1\cosh x \geq 1, the value stays at 11 or below and never reaches 00. Where sec⁡x\sec x had asymptotes and satisfied ∣y∣≥1|y| \geq 1, this function is the exact opposite: bounded and smooth.

Symmetry

Since cosh⁡x\cosh x is even, its reciprocal sech⁡x\operatorname{sech} x is even too, and the graph is symmetric about the yy-axis.

Monotonicity and maximum

The derivative is ddxsech⁡x=−sech⁡xtanh⁡x\dfrac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x \tanh x. As sech⁡x\operatorname{sech} x is positive, the sign comes from −tanh⁡x-\tanh x: the function increases for x<0x < 0 and decreases for x>0x > 0. The maximum point is (0,1)(0, 1).

Asymptote and decay

As x→±∞x \to \pm\infty we have cosh⁡x→∞\cosh x \to \infty and hence sech⁡x→0\operatorname{sech} x \to 0, so the xx-axis is a horizontal asymptote. The values are always positive, so the curve approaches 00 from above.

Far out cosh⁡x≈e∣x∣2\cosh x \approx \dfrac{e^{|x|}}{2}, so the decay is exponential.

sech⁡x≈2e−∣x∣\operatorname{sech} x \approx 2e^{-|x|}
xxsech⁡x\operatorname{sech} x
0011
11≈0.6481\approx 0.6481
22≈0.2658\approx 0.2658
33≈0.0993\approx 0.0993

The graph is a bell shape, symmetric, raised in the middle and falling away smoothly on both sides. It resembles the density of a normal distribution, but its tails decay exponentially rather than as steeply as a Gaussian.

An identity

Dividing both sides of cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1 by cosh⁡2x\cosh^2 x gives the following.

1−tanh⁡2x=sech⁡2x1 - \tanh^2 x = \operatorname{sech}^2 x

The right-hand side is exactly the derivative of tanh⁡x\tanh x. It corresponds to 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x for trigonometric functions, differing only in a sign.

Applications

The smooth bell shape appears in the soliton solutions of equations describing nonlinear waves2. The profile of a solitary wave is written with sech⁡\operatorname{sech} or sech⁡2\operatorname{sech}^2.

  • Solitons of the KdV equation and of the nonlinear Schrödinger equation
  • Light pulses that travel along an optical fiber without changing shape
  • Solitary waves on shallow water
  1. Hyperbolic functions, Wikipedia
  2. Soliton, Wikipedia