This section introduces a new class of functions: the hyperbolic functions.
Figure1.1.1.Definition of the Trigonometric Functions
The basic idea for hyperbolic functions comes from the construction of trigonometric functions. Recall how I defined the sine and cosine functions, as shown in Figure 1.1.1. For a circle with radius one, angle (in radians) can be defined to be the arc length of the inscribed arc or, equivalently, twice the shadded area. Then the \(x\) and \(y\) coordinates of a point on the edge of the unit circle, dependent on the angle, are given by the cosine and sine functions, respectively. The important observation is that there is an natural, intrinsic definition of angle, and that the trigonometric functions give cartesian coordinates based on that intrinsic angle.
Figure1.1.2.Definition of the Hyperbolic Functions
The unit circle is the locus of the equation \(x^2 + y^2 =
1\text{.}\) If I change this very slightly to \(x^2 - y^2 = 1\text{,}\) then I have the unit hyperbola instead. Hyperbolics come the hyperbola the same way the trigonometric functions come from the circle. First I have to define the hyperbolic equivalent of (radian) angle.
Definition1.1.3.
As seen in the shaded region in Figure 1.1.2, hyperbolic angle is defined to be the area bounded by the \(x\) axis, the hyperbola, and a line from the origin to the point on the hyperbola. Hyperbolic cosine and sine are the \(x\) and \(y\) coordinates of a point on the hyperbola as functions of hyperbolic angle.
Recall that the trigonometric angle is bounded between 0 and \(2\pi\text{.}\) Hyperbolic angle is unbounded: as the point keeps moving the point up the hyperbola, the inscribed area grows to infinity.
Now that I have defined a notion of hyperbolic angle, I can define the hyperbolic functions in parallel with the trig functions. They will both take hyperbolic angle as input; one will give the \(x\) coordinate and the other the \(y\) coordinate of a point on the hyperbola.
Definition1.1.4.
Let \((x,y)\) be a point on the hyperbola \(x^2 = y^2 -
1 \) and let \(\theta\) be the inscribed hyperbolic angle corresponding to that point. Then hyperbolic cosine is the unique function that gives \(x =
\cosh \theta \) and hyperbolic sine is likewise the unique function that gives \(y = \sinh \theta\text{.}\)
Since hyperbolic angle is unbounded, the hyperbolics are not periodic. You can also see this in the unbounded nature of the hyperbola: the \(x\) and \(y\) coordinates of points on the hyperbola are unbounded and the hyperbolics have to capture all of these possible values. Therefore, as the angle grows, I expect both \(\cosh \theta\) and \(\sinh
\theta\) to grow to infinity instead of oscillating like the trigonometric functions.
Subsection1.1.2Hyperbolic Identities
The fundamental trigonometric identity is \(\sin^2 \theta +
\cos^2 \theta =1\text{,}\) which comes from the fact that these are \(x\) and \(y\) coordinates and the circle is the locus of \(x^2 + y^2 = 1\text{.}\) For the hyperbolics, the equation is now \(x^2 - y^2 = 1\text{,}\) so the fundamental hyperbolic identity is
In parallel with the trigonometric functions, there are many other hyperbolic identities. In almost all cases, they are the same as the trigonometric identities excepts for differences in sign. I’ll give two examples here; the rest are found in the reference materials.
\begin{equation*}
\sinh (x+y) = \sinh x \cosh y + \cosh x \sinh y \hspace{2cm}
\tanh^2 x = 1 - \sech^2 x
\end{equation*}
Also in parallel with trigonometry, the remaining hyperbolic functions (tanh, coth, sech, csch) are defined in terms of hyperbolic sine and cosine. For example, since \(\tan x =
\frac{\sin x}{\cos x}\text{,}\) I also define \(\tanh x =
\frac{\sinh x}{\cosh x}\text{.}\)
Now I can move on to the most surprising fact about hyperbolics, a fact that doesn’t (at least at this point) have any parallel with trigonometry. The fact is this: hyperbolics are actually not new functions at all. They can be quite easily built out of exponentials. For the base functions \(\sinh
x\) and \(\cosh x\text{,}\) here are the exponential definition.
\begin{equation*}
\cosh x = \frac{e^x + e^{-x}}{2} \hspace{1cm} \text{ and }
\hspace{1cm} \sinh x = \frac{e^x - e^{-x}}{2}\text{.}
\end{equation*}
With this definition, the remaining four hyperbolics are defined in terms of exponentials as follows.
\begin{align*}
\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \amp
\hspace{2cm} \coth x = \frac{e^x + e^{-x}}{e^x -
e^{-x}}\\
\sech x = \frac{2}{e^x + e^{-x}} \amp \hspace{2cm} \csch x
= \frac{2}{e^x - e^{-x}}
\end{align*}
In particular, I know the asymptotic behaviour and asymptotic order of the hyperbolics. \(\cosh x\) and \(\sinh x\) both have the same asymptotic order as \(e^x\text{.}\)\(\sech
x\) and \(\csch x\) both have exponential decay to 0. \(\coth x\) and \(\tanh x\) both have horizontal asymptotes at \(y = \pm 1\text{.}\)
I can also reverse the exponential identities to give a pleasant description of the exponential function.
\begin{equation*}
e^x = \cosh x + \sinh x\text{.}
\end{equation*}
This leads us to an interesting question: if hyperbolics relate both to trigonometry (in definition, form and identities) and exponentials, is there also a hidden connection between trigonometry and exponentials? This is a question I will return to late in the course.
Subsection1.1.3Inverse Hyperbolics
For trigonometry, I used the notation of \(\arcsin x\) for the inverse of \(\sin x\text{;}\) I will use a similar notation for hyperbolics. Inverses will be indicated by a ‘arc’ prefix, so \(\arcsinh x\) is the inverse of the hyperbolic sine function.
The following table summarizes the domain restrictions required to define the inverse hyperbolics. Since the hyperbolics are not periodic, these domain restrictions are much less restrictive than they were for trigonometric functions. Since the inverse switches domain and range, the resulting range in the table will be the domain of the inverse.
\begin{equation*}
\begin{array}{llll}
\text{ Function } \amp \text{ Restricted Domain } \amp
\text{ Resulting Range } \amp \text{ Inverse Function } \\
\hline
\sinh x \amp \RR \amp \RR \amp \arcsinh x \\
\cosh x \amp [0, \infty) \amp [0, \infty) \amp \arccosh x \\
\tanh x \amp \RR \amp (-1,1) \amp \arctanh x \\
\sech x \amp [0, \infty) \amp (0, 1] \amp \arcsech x \\
\csch x \amp x \neq 0 \amp x \neq 0 \amp \arccsch x \\
\coth x \amp x \neq 0 \amp (-\infty,-1) \cup (1, \infty)
\amp \arccoth x
\end{array}
\end{equation*}
Since the hyperbolics have exponential descripions, I expect that their inverses will have logarithmic descriptions. This is true.
Now that I have defined these new functions, I’d like to know their limit, derivative and integral behaviour.
Limits are almost entirely the same as for other elementary function. The hyperbolics are continuous on their domains, so where there isn’t division by zero, I can just evaulate limits. Where there is division by zero, I can use the techniques I already know from limits: algebraic manipulation to adjust the limit or L’Höpital’s Rule. Much like trig, there is one special limit which is very useful.
Here are the derivatives of the remaining hyperbolics.
\begin{align*}
\frac{d}{dx} \tanh x \amp = \frac{d}{dx} \frac{\sinh
x}{\cosh x} = \sech^2 x\\
\frac{d}{dx} \coth x \amp = \frac{d}{dx} \frac{\cosh
x}{\sinh x} = -\csch^2 x\\
\frac{d}{dx} \csch x \amp = \frac{d}{dx} \frac{1}{\sinh x}
= - \csch x \ \coth x\\
\frac{d}{dx} \sech x \amp = \frac{d}{dx} \frac{1}{\cosh x}
= - \sech x \tanh x
\end{align*}
Again, there is a similarity in form to the trigonometric derivatives but with difference in sign. In particular, for sine and cosine, taking four derivatives returns to the original function. For hyperbolic sine and cosine, since the derivatives don’t introduce a negative sign, only two derivatives return to the original.
One way of thinking about exponentials, trigonometric functions and hyperbolics is in terms of solutions to differential equations. This table summarizies three of the most basic and most important DEs and their solutions (where \(a\) and \(b\) are real constants).
\begin{equation*}
\begin{array}{ll}
\text{ DE } \amp \text{ Solution } \\
\hline
\dfrac{df}{dx} = f \amp f = ae^x \\[1em]
\dfrac{d^2f}{dx^2} = -f \amp f = a \sin x + b \cos x \\[1em]
\dfrac{d^2f}{dx^2} = f \amp f = a \sinh x + b \cosh x
\end{array}
\end{equation*}
The derivatives of (most of) the inverse hyperbolic are as follows.
Like inverse trigonometric derivatives, the results of these derivatives are particularly interesting because they are special algebraic functions which don’t have algebraic anti-derivatives. If I reverse the direction, this means I need inverse hyperbolics to do the following integrals.
\begin{align*}
\int \frac{1}{\sqrt{1+x^2}} dx \amp = \arcsinh x + c\\
\int \frac{1}{\sqrt{1-x^2}} dx \amp = \arccosh x + c\\
\int \frac{1}{1 - x^2} dx \amp = \arctanh x + c \text{ or }
\arccoth x + c
\end{align*}
Notice the strangness with inverse hyperbolic cotangent and tangent. Both have the same derivatives and both solve the same integrals. This seems odd, since the solutions to integrals are unique up to a constant. The confusion is solved by realizing that \(\arccoth\) and \(\arctanh\) have mutually exclusive domains. Therefore, the ‘or’ in the solution is appropriate: the anti-derivative is either of the two functions depending on your location on the real number line.
