# Do “Parabolic Trigonometric Functions” exist?

The parametric equation

\begin{align*} x(t) &= \cos t\\ y(t) &= \sin t \end{align*}

traces the unit circle centered at the origin ($x^2+y^2=1$). Similarly,

\begin{align*} x(t) &= \cosh t\\ y(t) &= \sinh t \end{align*}

draws the right part of a regular hyperbola ($x^2-y^2=1$). The hyperbolic trigonometric functions are very similar to the standard trigonometric function.

Do similar functions exist that trace parabolas (because it is another conic section) when set up as parametric equations like the above functions? If so, are they also similar to the standard and hyperbolic trigonometric functions?

• I think you probably have to consider ellipses, not just the circle. There are a couple of ways of looking at the process of moving from circle to ellipse to parabola to hyperbola. There is a very geometric way of taking a right circular cone extended in both directions with axis vertical. A horizontal plane will cut the cone in a circle. Rotate the plane until it is vertical, and the section will change as indicated. Algebraically this is changing the eccentricity of the conic. The parabola is a limiting/special case, so you might expect to see something "in between". – Mark Bennet Apr 21 '12 at 21:28
• – hmakholm left over Monica Apr 22 '12 at 0:03
• @J.M. I agree with the rest of your edits, but I don't think it was necessary to replace the British spelling of "centred" with the American one. – user856 Jul 7 '12 at 4:14
• @Rahul, sorry. Old habits... – J. M. isn't a mathematician Jul 7 '12 at 4:17

Indeed there are, but they're not usually called that. What ordinary trigonometric and hyperbolic functions have in common is that they are solutions to the differential equation $$f''(t) = af(t)$$ When $a$ is negative, the solutions are ordinary sines and cosines, scaled horizontally by a factor that depends on $a$. If you take a solution $f$ and draw the parametric plot $(x,y)=(f'(t), f(t))$, the result is an ellipse whose eccentricity depends on $a$. For $a=-1$ the ordinary sine and cosine are solutions, and you get a circle.

On the other hand, when $a$ is positive the solutions are either hyperbolic sines or hyperbolic cosines, again with a horizontal scaling factor that depends on $a$. A plot of $(x,y)=(f'(t), f(t))$ is one arm of a hyperbola with a central angle that depends on $a$. For $a=1$ the hyperbolic sine and cosine are solutions, and the hyperbola is right-angled.

Intuitively, then, since a parabola is the limiting case between an ellipse and a hyperbola, we should expect to get a "parabolic function" by setting $a=0$. Unfortunately the differential equation then becomes $$f''(t)=0$$ whose solutions are first-degree polynomials, and it is hard to make those create a parabola. However, there's a way out (many thanks to Qiaochu Yuan for pointing this out!): Instead of $f''(t)=af(t)$ we can take the basic differential equation to be $$f'''(t)=af'(t)$$ In the $a\ne 0$ case all this changes is to allow us to add a constant term to solutions, which just moves the conic about in the plane. But for $a=0$, the solutions are now all the polynomials of degree $\le 2$. And when we take any quadratic polynomial $f$ and plot $(x,y)=(f'(t), f(t))$, what we get is indeed a parabola centered around the $y$-axis!

If we take $f$ to be a first-degree polynomial, the parametric plot is just a straight (vertical) line, another limiting case of conic sections.

In all of the above cases, plotting $(f_1(t),f_2(t))$ for two unrelated solutions (for the same $a$) generally produces a conic of the same general kind, but perhaps moved and rotated. And the dependency on $a$ of the eccentricity/angle disappears; that was mediated through the derivative in the $x$ position.

So a "parabolic function" is simply another (redundant) term for a quadratic polynomial. It is not quite clear which should be counted as the parabolic sine and cosine, though. Cases could be made for either $\operatorname{sinp}(t) = t$ and $\operatorname{cosp}(t) = 1+\frac12 t^2$ or the other way around -- but worrying too much about that is just silly.

• @Qiaochu: Thanks for the salvaging suggestion! – hmakholm left over Monica Apr 21 '12 at 21:52
• No problem. The issue here is that if you restrict yourself to things centered at the origin, then an ellipse or a hyperbola naturally degenerates into a pair of lines rather than a parabola, which is why you were getting linear rather than quadratic functions. The "center" of a parabola is at infinity so you need to let the center vary as well. (Note that a pair of lines is also a conic section!) – Qiaochu Yuan Apr 21 '12 at 22:07
• @QiaochuYuan: Makes sense. That also explains why in the $(f'(t),f(t))$ case it is enough to allow vertical translations. – hmakholm left over Monica Apr 21 '12 at 22:11
• Somewhat interestingly, $\operatorname{cosp}(x+y)-\operatorname{cosp}(x-y)= 2\operatorname{sinp}x\operatorname{sinp}y$. – sdcvvc Apr 21 '12 at 22:39
• @sdcvvc: Nice observation. Unfortunately, the other trigonometric sum/product identities cannot generalize to any nontrivial polynomials, since the degrees wouldn't match up. – hmakholm left over Monica Apr 21 '12 at 23:06

The problem of generalized forms of trigonmetry has been touched in the past by several authors, E. Ferrari (Rome university) proposed different forms and proved the link with elliptic functions.

Dattoli, Migliorati and Ricci used the Ferrari's approach to study the parabolic trigonometric functions and the relevant link with Chebyshev polynomials. The relevant papers have appeared on arXiv:

You could write down some equivalent functions, but they wouldn't be very useful. For instance, for $y = x^2 -1$: $$x=\frac{-1\pm\sqrt{1+4\tan^2 (t)}}{2\tan (t)}$$ $$y=\frac{1\mp\sqrt{1+4\tan^2 (t)}}{2\tan^2 (t)}$$

Why these functions are less useful than the usual hyperbolic and trigonometric functions, I don't know.

$\DeclareMathOperator{\cosp}{cosp}$ $\DeclareMathOperator{\sinp}{sinp}$ $\DeclareMathOperator{\tanp}{tanp}$ $\DeclareMathOperator{\cotp}{cotp}$ $\DeclareMathOperator{\secp}{secp}$ $\DeclareMathOperator{\cscp}{cscp}$

This is similar to another post, so I'll repeat my answer:

Defining $$\cosp u = \cosh 2u\quad\text{ and }\quad\sinp u = \sqrt{2}\sinh u$$ gives the identities $$\cosp u - \sinp^2 u = 1,\quad \cotp u - \sinp u = \cscp u,\quad\text{ and }\quad\cscp u - \tanp u = \secp u \cscp u,$$ 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, $x^2-y^2 = 1$.

Let's think about trig functions as being convenient ways of going from cartesian to polar coordinates. Converting a "unit" parabola to polar, you would get something like

$$r = \frac{\sin\theta}{\cos^2\theta} = \frac{\sin\theta}{1 - \sin^2\theta}$$

where one could go on ad nauseum with trig identities

• Welcome to MathSE! Your posts will be better received if you follow a few guidelines. Learning $\LaTeX$ will make your equations much more clear. Also, you should answer the question as explicitly as possible. Your answer does not seem to address the issue of functions other than trigonometric ones relating to parabolas. Could you do further clarification in your post? – Rory Daulton Dec 21 '14 at 23:29