Hyperbolic functions

The first known calculation of a hyperbolic trigonometry problem is attributed to Gerardus Mercator when issuing the Mercator map projection circa 1566. It requires tabulating solutions to a transcendental equation involving hyperbolic functions. The first to suggest a similarity between the sector of the circle and that of the hyperbola was Isaac Newton in his 1687 Principia Mathematica. Roger Cotes suggested to modify the trigonometric functions using the imaginary unit i=\sqrt{-1} to obtain an oblate spheroid from a prolate one. Lambert credited Riccati for the terminology and names of the functions, but altered the abbreviations to those used today. == Notation ==

With hyperbolic angle u, the hyperbolic functions sinh and cosh can be defined with the exponential function eu. In the figure A =(e^{-u}, e^u), \ B=(e^u, \ e^{-u}), \ OA + OB = OC . Exponential definitions of and of and • Hyperbolic sine: the odd part of the exponential function, that is, \sinh x = \frac {e^x - e^{-x}} {2} = \frac {e^{2x} - 1} {2e^x}. • Hyperbolic cosine: the even part of the exponential function, that is, \cosh x = \frac {e^x + e^{-x}} {2} = \frac {e^{2x} + 1} {2e^x}. • Hyperbolic tangent: \tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}} = \frac{e^{2x} - 1} {e^{2x} + 1}. • Hyperbolic cotangent: for , \coth x = \frac{\cosh x}{\sinh x} = \frac {e^x + e^{-x}} {e^x - e^{-x}} = \frac{e^{2x} + 1} {e^{2x} - 1}. • Hyperbolic secant: \operatorname{sech} x = \frac{1}{\cosh x} = \frac {2} {e^x + e^{-x}} = \frac{2e^x} {e^{2x} + 1}. • Hyperbolic cosecant: for , \operatorname{csch} x = \frac{1}{\sinh x} = \frac {2} {e^x - e^{-x}} = \frac{2e^x} {e^{2x} - 1}. Differential equation definitions The hyperbolic functions may be defined as solutions of differential equations: The hyperbolic sine and cosine are the solution of the system \begin{align} c'(x)&=s(x),\\ s'(x)&=c(x),\\ \end{align} with the initial conditions s(0) = 0, c(0) = 1. The initial conditions make the solution unique; without them any pair of functions (a e^x + b e^{-x}, a e^x - b e^{-x}) would be a solution. and are also the unique solution of the equation , such that , for the hyperbolic cosine, and , for the hyperbolic sine. Complex trigonometric definitions Hyperbolic functions may also be deduced from trigonometric functions with complex arguments: • Hyperbolic sine: \sinh x = -i \sin (i x). • Hyperbolic cosine: \cosh x = \cos (i x). • Hyperbolic tangent: \tanh x = -i \tan (i x). • Hyperbolic cotangent: \coth x = i \cot (i x). • Hyperbolic secant: \operatorname{sech} x = \sec (i x). • Hyperbolic cosecant:\operatorname{csch} x = i \csc (i x). where is the imaginary unit with . The above definitions are related to the exponential definitions via Euler's formula (See below). == Characterizing properties==

Hyperbolic cosine It can be shown that the area under the curve of the hyperbolic cosine (over a finite interval) is always equal to the arc length corresponding to that interval: \text{area} = \int_a^b \cosh x \,dx = \int_a^b \sqrt{1 + \left(\frac{d}{dx} \cosh x \right)^2} \,dx = \text{arc length.} Hyperbolic tangent The hyperbolic tangent is the (unique) solution to the differential equation , with . ==Useful relations==

The hyperbolic functions satisfy many identities, all of them similar in form to the trigonometric identities. In fact, '''Osborn's rule''' \operatorname{cosh}(t) \leq e^{t^2 /2}. It can be proved by comparing the Taylor series of the two functions term by term. ==Inverse functions as logarithms==

\begin{align} \operatorname {arsinh} (x) &= \ln \left(x + \sqrt{x^{2} + 1} \right) \\ \operatorname {arcosh} (x) &= \ln \left(x + \sqrt{x^{2} - 1} \right) && x \geq 1 \\ \operatorname {artanh} (x) &= \frac{1}{2}\ln \left( \frac{1 + x}{1 - x} \right) && | x | 1 \\ \operatorname {arsech} (x) &= \ln \left( \frac{1}{x} + \sqrt{\frac{1}{x^2} - 1}\right) = \ln \left( \frac{1+ \sqrt{1 - x^2}}{x} \right) && 0 ==Derivatives==

\begin{align} \frac{d}{dx}\sinh x &= \cosh x \\ \frac{d}{dx}\cosh x &= \sinh x \\ \frac{d}{dx}\tanh x &= 1 - \tanh^2 x = \operatorname{sech}^2 x = \frac{1}{\cosh^2 x} \\ \frac{d}{dx}\coth x &= 1 - \coth^2 x = -\operatorname{csch}^2 x = -\frac{1}{\sinh^2 x} && x \neq 0 \\ \frac{d}{dx}\operatorname{sech} x &= - \tanh x \operatorname{sech} x \\ \frac{d}{dx}\operatorname{csch} x &= - \coth x \operatorname{csch} x && x \neq 0 \end{align} \begin{align} \frac{d}{dx}\operatorname{arsinh} x &= \frac{1}{\sqrt{x^2+1}} \\ \frac{d}{dx}\operatorname{arcosh} x &= \frac{1}{\sqrt{x^2 - 1}} && 1 ==Second derivatives==

Each of the functions and is equal to its second derivative, that is: \frac{d^2}{dx^2}\sinh x = \sinh x \frac{d^2}{dx^2}\cosh x = \cosh x \, . All functions with this property are linear combinations of and , in particular the exponential functions e^x and e^{-x} . ==Standard integrals==

\begin{align} \int \sinh (ax)\,dx &= a^{-1} \cosh (ax) + C \\ \int \cosh (ax)\,dx &= a^{-1} \sinh (ax) + C \\ \int \tanh (ax)\,dx &= a^{-1} \ln (\cosh (ax)) + C \\ \int \coth (ax)\,dx &= a^{-1} \ln \left|\sinh (ax)\right| + C \\ \int \operatorname{sech} (ax)\,dx &= a^{-1} \arctan (\sinh (ax)) + C \\ \int \operatorname{csch} (ax)\,dx &= a^{-1} \ln \left| \tanh \left( \frac{ax}{2} \right) \right| + C = a^{-1} \ln\left|\coth \left(ax\right) - \operatorname{csch} \left(ax\right)\right| + C = -a^{-1}\operatorname{arcoth} \left(\cosh\left(ax\right)\right) +C \end{align} The following integrals can be proved using hyperbolic substitution: \begin{align} \int {\frac{1}{\sqrt{a^2 + u^2}}\,du} & = \operatorname{arsinh} \left( \frac{u}{a} \right) + C \\ \int {\frac{1}{\sqrt{u^2 - a^2}}\,du} &= \sgn{u} \operatorname{arcosh} \left| \frac{u}{a} \right| + C \\ \int {\frac{1}{a^2 - u^2}}\,du & = a^{-1}\operatorname{artanh} \left( \frac{u}{a} \right) + C && u^2 a^2 \\ \int {\frac{1}{u\sqrt{a^2 - u^2}}\,du} & = -a^{-1}\operatorname{arsech}\left| \frac{u}{a} \right| + C \\ \int {\frac{1}{u\sqrt{a^2 + u^2}}\,du} & = -a^{-1}\operatorname{arcsch}\left| \frac{u}{a} \right| + C \end{align} where C is the constant of integration. ==Taylor series expressions==

It is possible to express explicitly the Taylor series at zero (or the Laurent series, if the function is not defined at zero) of the above functions. \sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} + \cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} This series is convergent for every complex value of . Since the function is odd, only odd exponents for occur in its Taylor series. \cosh x = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \frac {x^6} {6!} + \cdots = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} This series is convergent for every complex value of . Since the function is even, only even exponents for occur in its Taylor series. The sum of the sinh and cosh series is the infinite series expression of the exponential function. The following series are followed by a description of a subset of their domain of convergence, where the series is convergent and its sum equals the function. \begin{align} \tanh x &= x - \frac {x^3} {3} + \frac {2x^5} {15} - \frac {17x^7} {315} + \cdots = \sum_{n=1}^\infty \frac{2^{2n}(2^{2n}-1)B_{2n} x^{2n-1}}{(2n)!}, \qquad \left |x \right | where: • B_n is the nth Bernoulli number • E_n is the nth Euler number ==Infinite products and continued fractions==

The following expansions are valid in the whole complex plane: :\sinh x = x\prod_{n=1}^\infty\left(1+\frac{x^2}{n^2\pi^2}\right) = \cfrac{x}{1 - \cfrac{x^2}{2\cdot3+x^2 - \cfrac{2\cdot3 x^2}{4\cdot5+x^2 - \cfrac{4\cdot5 x^2}{6\cdot7+x^2 - \ddots}}}} :\cosh x = \prod_{n=1}^\infty\left(1+\frac{x^2}{(n-1/2)^2\pi^2}\right) = \cfrac{1}{1 - \cfrac{x^2}{1 \cdot 2 + x^2 - \cfrac{1 \cdot 2x^2}{3 \cdot 4 + x^2 - \cfrac{3 \cdot 4x^2}{5 \cdot 6 + x^2 - \ddots}}}} :\tanh x = \cfrac{1}{\cfrac{1}{x} + \cfrac{1}{\cfrac{3}{x} + \cfrac{1}{\cfrac{5}{x} + \cfrac{1}{\cfrac{7}{x} + \ddots}}}} ==Comparison with circular functions==

area and hyperbolic functions depending on hyperbolic sector area . The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. Both types depend on an argument, either circular angle or hyperbolic angle. Since the area of a circular sector with radius and angle (in radians) is , it will be equal to when . In the diagram, such a circle is tangent to the hyperbola at . The yellow sector depicts an area and angle magnitude. Similarly, the yellow and red regions together depict a hyperbolic sector with area corresponding to hyperbolic angle magnitude. The legs of the two right triangles with the hypotenuse on the ray defining the angles are of length times the circular and hyperbolic functions. The hyperbolic angle is an invariant measure with respect to the squeeze mapping, just as the circular angle is invariant under rotation. The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic functions that does not involve complex numbers. The graph of the function is the catenary, the curve formed by a uniform flexible chain, hanging freely between two fixed points under uniform gravity. ==Relationship to the exponential function==

The decomposition of the exponential function in its even and odd parts gives the identities e^x = \cosh x + \sinh x, and e^{-x} = \cosh x - \sinh x. Combined with Euler's formula e^{ix} = \cos x + i\sin x, this gives e^{x+iy}=(\cosh x+\sinh x)(\cos y+i\sin y) for the general complex exponential function. Additionally, e^x = \sqrt{\frac{1 + \tanh x}{1 - \tanh x}} = \frac{1 + \tanh \frac{x}{2}}{1 - \tanh \frac{x}{2}} ==Hyperbolic functions for complex numbers==

Since the exponential function can be defined for any complex argument, we can also extend the definitions of the hyperbolic functions to complex arguments. The functions and are then holomorphic. Relationships to ordinary trigonometric functions are given by Euler's formula for complex numbers: \begin{align} e^{i x} &= \cos x + i \sin x \\ e^{-i x} &= \cos x - i \sin x \end{align} so: \begin{align} \cosh(ix) &= \frac{1}{2} \left(e^{i x} + e^{-i x}\right) = \cos x \\ \sinh(ix) &= \frac{1}{2} \left(e^{i x} - e^{-i x}\right) = i \sin x \\ \tanh(ix) &= i \tan x \\ \cosh(x+iy) &= \cosh(x) \cos(y) + i \sinh(x) \sin(y) \\ \sinh(x+iy) &= \sinh(x) \cos(y) + i \cosh(x) \sin(y) \\ \tanh(x+iy) &= \frac{\tanh(x) + i \tan(y)}{1 + i \tanh(x) \tan(y)} \\ \cosh x &= \cos(ix) \\ \sinh x &= - i \sin(ix) \\ \tanh x &= - i \tan(ix) \end{align} Thus, hyperbolic functions are periodic with respect to the imaginary component, with period 2 \pi i (\pi i for hyperbolic tangent and cotangent). ==See also==