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Just out of curiosity: Is it possible to parametrize a full circle or part of one with elementary functions but without using trigonometric functions? If so, what are advantages/disadvantages compared to the standard parametrizations using $\cos(t)$ and $\sin(t)$?

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3 Answers 3

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You may use the fact that $(1+it)/(1-it)$ has abs value 1. So: $$ x(t) = \frac{1-t^2}{1+t^2}\ \ , \ \ \ y(t) = \frac{2t}{1+t^2} $$ gives a parametrization for $S^1$ minus one point $(-1,0)$ (the limit of $t\rightarrow \infty$).

There is a non-trivial application in taking a skew-symmetric (or more generally anti self adjoint matrix, or operator) $S$ on a Hilbert space $H$ and produce the following orthogonal/unitary matrix: $$ U = (1+S) (1-S)^{-1}$$ It is also used in the so-called Cayley transform (see wiki) to analyse e.g. unbounded selfadjoint operators, with a factor of $i$ as well: $V=(1+iA)(1-iA)^{-1}$.

It is also used in numerical analysis, when you use a finite difference method for the wave equation and want to preserve e.g. the $L^2$ (discrete) norm.

Later edit: If you want the whole circle to be covered you may take a square before splitting into real/imag parts: $t \in {\Bbb R} \mapsto \frac{(1+it)^2}{(1-it)^2}\in {\Bbb C}$ covers the circle twice (although (1,0) only once). This gives: $$ x(t) = \frac{1-6t^2+t^4}{1+2t^2+t^4} , \ \ \ y(t) = \frac{4t -4t^3}{1+2t^2+t^4} $$

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    $\begingroup$ This one is also given in Mariano's link in the comments...very nice $\endgroup$
    – draks ...
    Oct 7, 2016 at 21:50
  • $\begingroup$ Thank you. Is there some situation where this parametrization would be more useful than the one with trigonometric functions? $\endgroup$
    – Viktor K.
    Oct 7, 2016 at 21:51
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    $\begingroup$ Yes, Victor, it leads to a formula for Pythagorean triples. $\endgroup$ Oct 7, 2016 at 21:55
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    $\begingroup$ @Algorithms, see this or search the site for "pythagoran triple rational" for more. $\endgroup$ Oct 8, 2016 at 4:15
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    $\begingroup$ I think the parametrization you give misses $(-1,0)$, not $(1,0)$. $\endgroup$
    – Wojowu
    Oct 8, 2016 at 9:55
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What about $f(x,\pm)=\pm\sqrt{1-x^2}$, where $f(\cdot,\cdot)$ has a discrete and continous parameter defined in $[-1,1]$...

You may also use $e^{it}=\cos(t)+i\sin(t)$ to represent a circle in the complex plane. With this calculating Fourier transforms becomes handy...


Just a comment to H.H. Rugh answer that needs graphical support:

His parametrisation is the stereograhic projection which has an application in Photography:

$\hskip1.5in$enter image description here

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  • $\begingroup$ Yes. But is there any reason to use any parametrization over another? $\endgroup$
    – Viktor K.
    Oct 7, 2016 at 21:47
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    $\begingroup$ Depends on your problem. Choose the one that suits you best! $\endgroup$
    – draks ...
    Oct 7, 2016 at 21:48
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    $\begingroup$ Thank you draks ... . It's too bad I can't accept two answers, as your answer is correct and useful too. $\endgroup$
    – Viktor K.
    Oct 7, 2016 at 22:01
  • $\begingroup$ @ViktorKaspervich no problem. Cheers... $\endgroup$
    – draks ...
    Oct 7, 2016 at 22:04
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This is an old question but I think the following is perhaps somewhat interesting:

A simple parametrization of the circle is $$ q\in[-1,1]:\quad x(q)=2|q|-1\quad y(q)=-\mathrm{sign}(q)\sqrt{1-x(q)^2}=-2\,\mathrm{sign}(q)\sqrt{|q|(1-|q|)} $$ This is pretty ugly in my opinion because of the $||$ and $\mathrm{sign}$ functions. However, we can bijectively map $[-1,1]$ to itself using the map $q=\mathrm{sign}(t)t^2$ so that $|q|=t^2$ and $\mathrm{sign}(q)=\mathrm{sign}(t)$. Then $$ t\in[-1,1]:\quad x(t)=2t^2-1\quad y(t)=-2\,\mathrm{sign}(t)\sqrt{t^2(1-t^2)}=-2t\sqrt{1-t^2} $$ is a parametrisation of the circle as $\mathrm{sign}(t)\sqrt{t^2}=\mathrm{sign}(t)|t|=t$.
However I still think there is a flaw with this $t$ parameter because $y(t)$ is not differentiable (or at least its one-sided derivatives do not exist) at the endpoints $t=\pm 1$. To remedy this, I tried to find another bijection from $[-1,1]$ to itself which has zero derivative at $\pm 1$ as this may cancel out the infinity from the square root (think $\sqrt[3]{x}$ and $x^3$). The simplest one I could find is: $$ t=\frac{3}{2}u-\frac{1}{2}u^3 $$ Let's analyse what happens just to $\sqrt{1-t^2}$: $$ \sqrt{1-t^2}=\sqrt{1-\left(\frac{3}{2}u-\frac{1}{2}u^3\right)^2}=\frac{1}{2}\sqrt{(4-u^2)(1-u^2)^2}=\frac{1-u^2}{2}\sqrt{4-u^2} $$ Luckily $1-u^2\geq 0$ so that $\sqrt{(1-u^2)^2}=1-u^2$. Anyway, the last expression above is clearly smooth on $[-1,1]$. Plugging $t(u)$ into the last parametrisation we had, this pops out: $$ u\in[-1,1]:\quad x(u)=\frac{1}{2}(u^2-2)(u^4-4u^2+1)\quad y(u)=-\frac{1}{2}u(1-u^2)(3-u^2)\sqrt{4-u^2}$$ So $(x(u),y(u))$ is a smooth parametrisation of the circle which is also bijective (with the exception of the point $(1,0)$ as the circle is a closed curve).
However, I don't really see any reason why you would want to use this parametrisation, it's more of a curiosity.

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