Hyperbolic Sine Function

Introduction

1.The geometric construction of the hyperbolic sine, based on the area of a sector of the equilateral hyperbola, appears in hyperbolic sine and cosine. For real $x,$ the hyperbolic sine is the function defined by the exponential formula:

$$\sinh(x) = \frac{e^x - e^{-x}}{2}$$

2.The function $f(x) = \sinh(x)$ is half the difference of $e^x$ and $e^{-x}.$ It is defined for every real number and has range $\mathbb{R}.$ Its graph is symmetric about the origin and has tangent $y = x$ at the point $(0, 0).$ The function tends to $+\infty$ as $x \to +\infty$ and to $-\infty$ as $x \to -\infty,$ while the circular sine oscillates. For large positive $x,$ the term $e^{-x}$ is close to zero and $\sinh(x)$ is close to $e^x/2.$ By oddness, $\sinh(x)$ is close to $-e^{-x}/2$ for large negative $x.$

3.The hyperbolic sine gives the arc length of the catenary $y = \cosh(x),$ along which a uniform chain hangs under its own weight. For $a \geq 0,$ the length of the arc from the vertex to the point of abscissa $a$ is $\sinh(a).$

Properties

4.The function has the following properties.

6.The hyperbolic sine and the hyperbolic cosine satisfy the fundamental hyperbolic identity:

$$\cosh^2(x) - \sinh^2(x) = 1$$

7.Since $\cosh(x)$ is positive for every real $x,$ solving the hyperbolic identity for $\cosh(x)$ gives:

$$\cosh(x) = \sqrt{1 + \sinh^2(x)}$$

8.Every real value has exactly one preimage, so the hyperbolic sine is a bijection of $\mathbb{R}$ onto $\mathbb{R}.$

Limits, derivatives, and integrals of the hyperbolic sine function

9.The remarkable limit of the exponential function determines the behaviour near the origin. The relevant quotient is:

$$\frac{\sinh(x)}{x} = \frac{1}{2}\left(\frac{e^x - 1}{x} + \frac{e^{-x} - 1}{-x}\right)$$

10.As $x$ tends to zero, both quotients in parentheses tend to $1.$ Their half-sum tends to one:

$$\lim_{x \to 0} \frac{\sinh(x)}{x} = 1$$

11.Near the origin, both $\sinh(x)$ and $\sin(x)$ are close to $x.$ The limits at infinity and the comparison with $e^x/2$ at $+\infty$ are:

$$ \begin{align} \lim_{x \to +\infty} \sinh(x) &= +\infty \\[6pt] \lim_{x \to -\infty} \sinh(x) &= -\infty \\[6pt] \lim_{x \to +\infty} \left(\sinh(x) - \frac{e^x}{2}\right) &= 0 \end{align} $$

12.The equality $\sinh(x) - e^x/2 = -e^{-x}/2$ shows that the curve approaches $e^x/2$ from below at $+\infty.$ At $-\infty,$ the equality $\sinh(x) + e^{-x}/2 = e^x/2 \to 0$ shows that the curve approaches $-e^{-x}/2$ from above. The graph has no vertical asymptotes because it is continuous on $\mathbb{R}.$ Since $\dfrac{\sinh(x)}{x} \to +\infty$ as $x \to \pm\infty,$ it has no horizontal or oblique asymptotes.

13.The function $\sinh(x)$ is continuous and differentiable on $\mathbb{R}.$ Termwise differentiation of the exponential expression gives its derivative:

$$\frac{d}{dx}\sinh(x) = \frac{e^x + e^{-x}}{2} = \cosh(x)$$

14.A second differentiation returns the original function, so the derivatives alternate with period two:

$$\frac{d^2}{dx^2}\sinh(x) = \sinh(x)$$

15.The parity of $n$ determines the $n$-th derivative:

$$ \frac{d^n}{dx^n}\sinh(x) = \begin{cases} \sinh(x) & \text{if } n \text{ is even} \\[6pt] \cosh(x) & \text{if } n \text{ is odd} \end{cases} $$

16.

The hyperbolic sine is the unique solution of the differential equation $y'' = y$ with $y(0) = 0$ and $y'(0) = 1.$ The circular sine solves $y'' = -y$ with the same initial conditions.

17.The indefinite integral of the hyperbolic sine is:

$$\int \sinh(x) \ dx = \cosh(x) + c$$

18.Because the function is odd, its definite integral over an interval symmetric about the origin vanishes:

$$\int_{-a}^{a} \sinh(x) \ dx = 0$$

19.For the catenary $y = \cosh(x),$ the derivative satisfies $y'(x) = \sinh(x),$ and the fundamental hyperbolic identity gives $\sqrt{1 + [y'(x)]^2} = \cosh(x).$ The arc-length formula therefore yields $\int_0^a \cosh(x) \ dx = \sinh(a)$ for $a \geq 0.$

20.For integrals containing $\sqrt{1 + x^2},$ the substitution $x = \sinh(t)$ gives $\sqrt{1 + x^2} = \cosh(t)$ and $dx = \cosh(t) \ dt,$ so the radical disappears. The trigonometric substitution for $\sqrt{1 - x^2}$ is $x = \sin(t)$ with $-\pi/2 \leq t \leq \pi/2.$

Monotonicity and convexity

21.The derivative $\cosh(x)$ is at least $1$ at every point, so the hyperbolic sine is strictly increasing on $\mathbb{R}$ and has no stationary points. Equality holds only at $x = 0,$ where the tangent is the line $y = x.$ Thus the smallest slope is $1.$

22.Since $\sinh''(x) = \sinh(x)$ is negative for $x < 0$ and positive for $x > 0,$ the graph is strictly concave on $(-\infty, 0)$ and strictly convex on $(0, +\infty).$ The origin is an inflection point, where the second derivative vanishes and changes sign.

Inverse function

23.The hyperbolic sine is continuous and strictly increasing, with range $\mathbb{R}.$ It is a bijection of $\mathbb{R}$ onto $\mathbb{R},$ and its inverse function is denoted by:

$$\mathrm{arsinh} : \mathbb{R} \to \mathbb{R}$$

24.With $t = e^y,$ the equation $x = \sinh(y)$ becomes $2x = t - t^{-1},$ which is equivalent to the quadratic equation:

$$t^2 - 2xt - 1 = 0$$

25.Its roots are $t = x \pm \sqrt{x^2 + 1}.$ Since $\sqrt{x^2 + 1} > |x|,$ only the root with the plus sign satisfies $t = e^y,$ because $e^y$ is positive. Taking the natural logarithm gives:

$$\mathrm{arsinh}(x) = \ln\left(x + \sqrt{x^2 + 1}\right)$$

26.For every real $x,$ the positivity of $\cosh$ and the fundamental hyperbolic identity give $\cosh(\mathrm{arsinh}(x)) = \sqrt{1 + x^2},$ so the derivative of the inverse function is:

$$\frac{d}{dx}\mathrm{arsinh}(x) = \frac{1}{\sqrt{1 + x^2}}$$

27.On $\mathbb{R},$ the antiderivatives of $\dfrac{1}{\sqrt{1 + x^2}}$ are:

$$\int \frac{1}{\sqrt{1 + x^2}} \ dx = \ln\left(x + \sqrt{x^2 + 1}\right) + c$$

Maclaurin series

28.The Maclaurin series of the hyperbolic sine follows from the series of the exponential function. Subtracting the expansion of $e^{-x}$ from that of $e^x$ cancels the even powers and doubles the odd powers. Dividing by $2$ gives a power series that converges for every real number:

$$\sinh(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots$$

29.Near the origin, the first term gives the linear approximation $y = x.$

30.For complex $x,$ replacing $x$ by $ix$ multiplies each term of degree $2n + 1$ by $i^{2n+1} = i(-1)^n,$ so $\sinh(ix) = i\sin(x).$ Applying this identity to $ix$ and using oddness gives $\sin(ix) = i\sinh(x).$