Hyperbolic Trigonometry

Remember that ordinary trig functions are defined as follows:

$$\cos\theta=\frac 1 2(e^{i\theta}+e^{-i\theta})\qquad\qquad\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$$

and it is useful sometimes to use the extra functions

$${\rm sec} \theta\equiv \frac 1 {\cos\theta} \qquad\qquad{\r... ...ta} \qquad\qquad {\rm cot} \theta\equiv \frac 1 {\tan\theta}$$

Hyperbolic trig functions are defined similarly:

$$\cosh x={\textstyle \frac 1 2}(e^{x}+e^{- x})\qquad\qquad\sinh x=\frac{1}{2}(e^{x}-e^{- x})$$

$${\rm sech} x\equiv \frac 1 {\cosh x} \qquad\qquad {\rm cos... ...{\sinh x} \qquad\qquad {\rm coth} x\equiv \frac 1 {\tanh x}$$

From the definitions above it is easy to show that

$${{\rm d}\cosh x\over {\rm d}x}=\sinh x \qquad\qquad {{\rm d}... ...\qquad\qquad {{\rm d}\tanh x\over {\rm d}x}={\rm sech}^2\! x.$$

Often we are interested in the small- or large-$x$ limits of these functions. What we want is to find a simple function which approximates to a more complicated one in these limits. So while it is true that as $x\to 0$, $\sinh x\to 0$, that is not usually what we want; what we want is how it tends to zero.

From the small-$x$ expansion of the exponential $e^x=1+x+{\textstyle \frac 1 2}x^2+\ldots$ we get

$$\sinh x\stackrel{x\to 0}{ \longrightarrow} x \qquad\qquad... ...ckrel{x\to 0}{ \longrightarrow} 1+{\textstyle \frac 1 2}x^2$$

The limit of $\cosh x$ often causes problems; whether we keep the $x^2$ term depends on the context, given that we want to be able to say more than ``tends to 0'' or ``tends to $\infty$''. It may be useful to remember instead

$$\cosh x \stackrel{x\to 0}{ \longrightarrow} 1 \qquad\qqua... ...tackrel{x\to 0} { \longrightarrow}{\textstyle \frac 1 2}x^2$$

The same is true of the exponential:

$$e^x\stackrel{x\to 0}{ \longrightarrow}1 \qquad\qquad\hbox{but}\qquad\qquad e^x-1\stackrel{x\to 0}{ \longrightarrow} x.$$

In a particular problem we find that the energy of a system is

$$\left\langle E \right\rangle =\frac {\hbar \omega}{e^{\hbar\omega\beta}-1}$$

Naively we would say that at high temperatures, as $\beta\to0$, the denominator vanishes and the energy tends to infinity. That is true but not very helpful. If we are more sophisticated we see that the denominator actually tends to $\hbar\omega\beta$ and $\left\langle E \right\rangle \to 1/\beta=k_{\scriptscriptstyle B}T$. That is a much more useful prediction, since it can be verified experimentally.

The high-$x$ limits are easier; $e^{-x}\to 0$ and so

$$\sinh x\stackrel{x\to \infty}{ \longrightarrow}{\textstyl... ...uad\qquad \tanh x\stackrel{x\to \infty}{ \longrightarrow} 1$$