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Are there parabolic and elliptical functions analogous to the circular and hyperbolic functions sin(h),cos(h), and tan(h)?

Also, in matters of conic sections, are there other properties such that it helps to group the circle and hyperbola in one, and the parabola and ellipse in the other?

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    $\begingroup$ The "corresponding" functions for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ are just scalar multiples of the circular functions: the coordinates of the ellipse are parametrized by $(a\cos t, b\sin t)$, so the "elliptical" functions would just be $a\sin$, $b\cos$, and $\frac{a}{b}\tan$. For the parabola $y=x^2$, the functions are $(t,t^2)$, so "parabolic cosine" would be the identity, the "parabolic sine" would be the square; not very interesting. $\endgroup$ – Arturo Magidin Aug 26 '11 at 20:15
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    $\begingroup$ Note also that there is a "natural" choice for the circular functions, the circle $x^2+y^2=1$, and a "natural" choice for the hyperbolic functions, the hyperbola $x^2-y^2=1$. But there is no 'natural' choice for the ellipse (in fact, a circle is a kind of ellipse, so the circular functions "are" the 'natural' case of the elliptical ones). (cont) $\endgroup$ – Arturo Magidin Aug 26 '11 at 20:52
  • $\begingroup$ erased continuation comment. $\endgroup$ – Arturo Magidin Aug 26 '11 at 21:05
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    $\begingroup$ The hyperbola $x^2-y^2=1$ is parametrized by $(\cosh,\sinh)$. If you're willing to go one degree beyond quadratics, nonsingular cubics can be transformed into elliptic curves, which can be parametrized with Weierstrass $\wp$-functions and their first derivatives. $\endgroup$ – anon Aug 26 '11 at 21:23
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    $\begingroup$ @Arturo: With no other mission but to parameterize $y=x^2$, I could easily take "parabolic cosine" to be a fairly exotic bijection, not necessarily the identity; then "parabolic sine" would be the square of that function, not necessarily the square function itself. In this case, I'll grant that the relationship between "cosp" and "sinp" is "not very interesting", but the individual functions could be interesting in themselves. Besides, who says that we'd have to parameterize the specific parabolic incarnation $y=x^2$? The standard trigs don't parameterize conics that pass through the origin. $\endgroup$ – Blue Aug 29 '11 at 8:17
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Ellipses and circles are "really" the same: a circle is an ellipse in which the focal distance is $0$. Or, put another way, from among all the ellipses $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ if you are going to choose a "canonical" one to define some functions, the natural choice is to use $a=b=1$, which just leads you back to the unit circle. That is, trying to do "elliptic trigonometric functions" pretty soon either drops you into either an arbitrary choice of parameters, or the regular circular trigonometric functions.

Now, one way in which we can "unify" the circular and the hyperbolic functions is through the complex exponential: the circular trigonometric functions correspond to certain exponentials with purely imaginary arguments, while the hyperbolic ones correspond to purely real arguments: for $t$ a real number, $$\begin{align*} \cos t&= \frac{e^{it}+e^{-it}}{2} &\qquad \sin t &= \frac{e^{it}-e^{-it}}{2}\\ \cosh t &= \frac{{e^t}+e^{-t}}{2} & \sinh t &= \frac{e^t-e^{-t}}{2} \end{align*}$$

Where do these come from? One place to find them is the differential equation $$y'' + \lambda y = 0.$$ If $\lambda\gt 0$, the solution set is spanned by $\cos(\sqrt{\lambda}\;t)$ and $\sin(\sqrt{\lambda}\;t)$. You get the standard functions by normalizing with $\lambda=1$. If $\lambda\lt 0$, then the solution set is spanned by $\cosh(\sqrt{|\lambda|t})$ and $\sinh(\sqrt{|\lambda|}t)$, and you get the standard functions by normalizing with $\lambda=-1$.

This would suggest looking for any putative "parabolic trigonometric functions" in the only remaining case: $\lambda=0$ (this corresponds to the fact that if you view conic sections as being given by slicing a cone with a plane, you obtain the parabola in the boundary between ellipses and hyperbolas). But the differential equation $y''=0$ has solution space spanned by $1$ and $t$, so that the natural "parabolic functions" are just $1$ and $t$.

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[Figures would really help this answer. I'll see if I can add some later.]

According to one definition, the circular and hyperbolic functions parameterize the circle and hyperbola (respectively) according to an appropriate definition of "radian" measure of an angle.

Let $\angle IOP$ have its vertex at the origin, point $I$ at $(1,0)$, and point $P$ somewhere above the $x$-axis. For convenience, let $P$ lie on the unit circle ($x^2+y^2=1$), and let ray $OP$ determine the point $P^\prime$ on the "unit hyperbola" ($x^2-y^2=1$). Then the (circular) radian measure of $\angle IOP$ is twice the area of circular sector $IOP$; and the hyperbolic radian measure is twice the area of the hyperbolic sector $IOP^\prime$ (provided that the angle is no greater than half a right angle). Thus,

  • We define the circular trig functions by calling $T(\cos(t), \sin(t))$ the point on the unit circle such that $\angle IOT$ has (circular) radian measure $t$.

  • We define the hyperbolic trig functions by calling $T\,^\prime(\cosh(t), \sinh(t))$ the point on the unit hyperbola such that $\angle IOT\,^\prime$ has hyperbolic radian measure $t$.

(It is a worthwhile exercise to show that the hyperbolic functions defined in this way have standard explicit representations in terms of $t$. Here's a sketch: Rotating the hyperbola $45$ degrees and scaling appropriately yields the curve $y=1/x$. From there, the definition of the natural logarithm allows for determining the area of the hyperbolic sector from $T\,^\prime$, so that the exponential appears when reversing the process.)

It is not at all unreasonable to seek "parabolic" and "elliptical" (and, let's not forget, "not-rectangularly-hyperbolic"!) variants of these trig functions. (In fact, I made the very same investigation as a high school student.) The key is to decide upon exactly what curve to use, and how to assign a "generalized radian measure" to a given angle. I'll describe an approach that generalizes circular trig functions to conics of every eccentricity, $e$; the downside is that it doesn't match the standard hyperbolic trig functions for $e=\sqrt{2}$.

We simply generalize the centered unit circle to a conic with focus at the origin and with "semi-latus rectum" length $1$. More specifically, letting the "other" focus drift down the negative $x$-axis as $e$ approaches $1$ and drift in from the positive $x$-axis as $e$ exceeds $1$; that is, we'll concern ourselves with the (branch of) the conic with closest vertex to the right of the origin, "opening" to the left. Such a conic has equation

$$x^2( 1 - e^2) + 2 e x + y^2 = 1$$

which is the template for the generalized "Pythagorean relation" of our new functions.

The vertex, $V$, of (the preferred branch of) our conic has $x$-coordinate $v := \frac{1}{1+e}$. Let $P(p,q)$ be a point on the curve, with $q\ge 0$. Then we define the generalized radian measure ("grm") of $\angle VOP$ as twice the area of sector $VOP$. The area of the sector is equal to the full area under the curve from $P$ to $V$, adjusted by the area ($\frac{1}{2}|p|q$) of the right triangle with hypotenuse $VP$ (subtracting the triangle if $p<0$ and adding otherwise, so it turns out that removing the absolute value sign from $p$ will "do the right thing"). That is,

$$\begin{eqnarray*} \mathrm{grm}(\angle VOP) &:=& 2 \left( \frac{1}{2} p q + \int_{p}^{v} \sqrt{1-x^2(1-e^2)-2 e x} \; dx\right) \\\\ &=& p \sqrt{1-p^2(1-e^2)-2ep} + 2\int_{p}^{v} \sqrt{1-x^2(1-e^2)-2 e x} \; dx \end{eqnarray*}$$

(This explains the disconnect with the standard hyperbolic trig functions. We're defining our radians from "convex" sectors based on a focus, whereas the standard hyperbolic functions define them from "concave" sectors based on the center.)

And now, with vigorously-waving hands ... Writing "$t$" for "$\mathrm{grm}(\angle VOP)$", the above gives a formula for $t$ in terms of $p$. Inverting gives, in terms of $t$, a formula for $p$ ... which we interpret as "the generalized cosine of $t$", or "$\cos_e t $". We solve for the "generalized sine of $t$" from the generalized Pythagorean relation

$$( 1 - e^2 )\cos_e^2 t + 2 e \cos_e t + \sin_e^2 t = 1$$

thereby defining $P(\cos_e t,\sin_e t)$ as the point on our generalized conic such that $VOP$ has generalized radian measure $t$.

General(ised)ly speaking, of course.

Details (and roadblocks) are, as they say, "left to the reader". (I suggest tackling the case $e=1$ first.)

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    $\begingroup$ Note that the conics being considered by Don here can be represented in polar coordinates as $r=\frac1{1+e\cos\,\theta}$; that is another way to generate the expression for $\mathrm{grm}$. $\endgroup$ – J. M. is a poor mathematician Aug 31 '11 at 13:49
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On the other hand, if you bring the Jacobian elliptic functions into the fray, it becomes possible to consider elliptic analogs (though the "elliptic" in "elliptic functions" is not due to these functions being useful in parametrizing the ellipse; it's a long tale for another day).

Briefly: if you consider the function pair $\left(a\operatorname{sn}\left(u\mid \frac{a^2}{b^2}\right),\quad a\operatorname{dn}\left(u\mid \frac{a^2}{b^2}\right)\right)$, they parametrize the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ by virtue of the identity $\operatorname{dn}^2(u\mid m)+m\operatorname{sn}^2(u\mid m)=1$.

In any event: I'd like to suggest that you take a look at A.I. Markushevich's The Remarkable Sine Functions; there is a lot of material on the functions that have been studied as generalizations of the usual circular sine.

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Well, if we define cosp u = cosh 2u and sinp u = sqrt(2)*sinh u then we have the identities cosp u - sinp^2 u = 1, cotp u - sinp u = cscp u, and cscp u - tanp u = secp u cscp u, corresponding to the parabola x - y^2 = 1. These are analogous to the circular functions with defining equation x^2 + y^2 =1 or the hyperbolic functions, x^2-y^2 = 1.

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