(NASA)                           (OMCA)

      Saturn’s north pole has a hexagonal cloud boundary         Why we must go there

While deciding on our next home planet, let’s examine a mystery function $\boldsymbol{f(x,y)}$ discovered by a mathematical expedition to the hexagonal weather pattern on Saturn’s topside.

A positive test,  $f(x,y) = {\scriptsize\raise.4ex+}1 \kern1mu$, indicates an evenly triangular array of points above the clouds.


A subsurface probe,  $f(x,y) = -{\large\tfrac54} \kern1mu$, reveals a regular hexagonal array of points.   But we need a modulated survey,  $ f(x,y) = 1 {+} \, 3 \cos{\tiny\sqrt{\vphantom{\raise3.5ex~}\normalsize3}} \kern2mu r \kern1mu $, in order to account for the satellite view.


As this is part of a polar expedition,  $r = \sqrt{x^2 \! + y^2} \,$ and that last equation is the same as  $ f(x,y) = 1 {+} \, 3 \cos \! \sqrt{3 (x^2 \! + y^2)} \kern1mu \LARGE\raise-.4ex\strut $.   Its $y$- intercepts are difficult to specify but it crosses  $y = 0 \,$ at  $ \require{begingroup}\begingroup \def \3 {{ \tiny\sqrt{ \vphantom{\raise3ex~}\large 3 } }} \def \X#1{ {\pm}{ \large\tfrac{#1\pi}{3\3} \normalsize\,\raise.5ex{,} \, } } x = \X{ 2} \X{ 4} \X{ 8} \X{10} \X{14} \X{16} ~ \cdots \endgroup $

Laboratory measurements have determined that $f(x,y)$ contains 10 secret ingredients, each of which may be a variable, an operator, a trigonometric function, a decimal number, a symbolic constant, or a single bracket / fence such as left parenthesis, right brace, or absolute value bar. For example, $ 1 {+} \, 3 \cos \! \sqrt{3 (x^2 \! + y^2)} \, $ contains 13 ingredients.

So . . .        What is a 10 - ingredient formula for $~\boldsymbol{ f(x,y) }\,$?

In the spirit of scientific methodology, feel free to request a plot based on $f(x,y)$.   Another modulation,  $ f(x,y) = -\sin^2 \!{ \tfrac{\large 2} { \tfrac1{\large r} - \tfrac{1}{6\large\pi} } }\, $ for instance, can uncover some nuts and bolts.



Although $r$, $\theta$ and non-trigonometric functions are not available as ingredients within $f(x,y)$, any solutions that use them anyway would be more than welcome for the sake of interest and education.

The cartoonish “points” in plots of  $f(x,y) = {\scriptsize\raise.4ex+}1 \,$ and  $f(x,y) = -{\large\tfrac54} \,$ represent true points.

Complex numbers are allowed but not needed.

These implicit plots were made by EquationExplorer at KevinMehall.net.
Also good for implicit function plots is MathGrapher at eMathHelp.

  • $\begingroup$ Would you like to clarify whether those things that look like little circles in e.g. the plot of f(x,y)=1 are actually little (approximate) circles or single points? (The, er, point of my question is the point versus nontrivial-contour distinction; I'm not asking whether they're really circle-like versus e.g. figure-eight-like.) $\endgroup$
    – Gareth McCaughan
    Apr 3, 2017 at 13:18
  • $\begingroup$ Is $r$ available as a "cost 1" variable in the expression for $f$, and if so do we also have $\theta$? Or do we have only $x,y$? $\endgroup$
    – Gareth McCaughan
    Apr 3, 2017 at 13:25
  • $\begingroup$ Am I right in understanding that any number/constant costs us only 1 unit? E.g., if I wanted $\frac2{\sqrt3}+\frac{5\pi}{e^2}$ would that have cost 1? $\endgroup$
    – Gareth McCaughan
    Apr 3, 2017 at 13:33
  • $\begingroup$ Are there any particular rules on (1) what functions we're allowed to use (obviously trig, exp, log; what else?) and/or (2) when they need parens around their arguments? $\endgroup$
    – Gareth McCaughan
    Apr 3, 2017 at 15:29
  • $\begingroup$ If we want to say "absolute value of ...", can we do it with vertical bars |x| counting each bar as a sort of funny bracket? Or should we assume there's an abs function we can invoke? Or is it just not allowed? $\endgroup$
    – Gareth McCaughan
    Apr 3, 2017 at 15:42

1 Answer 1


The formula is

$f(x,y) = \cos y \cos \sqrt{3}x - \sin^2 y$.

As with @humn's similar find-the-formula puzzle from a few weeks ago, my path to the solution was rather ad hoc and unprincipled, though with hindsight I can see better routes I could have taken.

Obviously it's got to be

some sort of trigonometrical thing with appropriate symmetries. However, the "obvious" ways to construct these came up blank both because I couldn't get the right values and because I couldn't get them short enough.

So the next thing for this pure-mathematician-turned-hacker-and-engineer to do was

to figure out as many values of $f$ as possible from the diagrams, kludge together some sort of trigonometrical abomination, and see what happens.


first of all consider the line $y=0$. Obviously we have something here with period $\frac{2\pi}{\sqrt3}$ which I'll call $p$. We know that it's 1 at $x=0$, and from the first "modulated survey" we see that it's $-\frac12$ at $x=p/3$ and $2p/3$. This, together with known symmetries, is actually enough, but looking at the "nuts and bolts" plot shows us pretty well where the contours of $f=0$ and $f=-1$ are; in particular, $f(p/4)=f(3p/4)=0$ and $f(p/2)=-1$. [Here is one of those better routes: the contours of $f=-1$ are straight lines, and it's pretty obvious where they are, and I think following this line of thought would get us there much quicker than what I actually did.] And this is really enough even for an idiot like me to see that $f(x,0)$ is surely $\cos \sqrt{3}x$.

Now let's look at

what happens along the lines $y=\frac{\pi}3$ and $y=\frac{2\pi}3$. We expect to get something of the form $A(y)\cos\sqrt{3}x+B(y)$ where of course $A,B$ will themselves be trigonometric functions, with period $2\pi$ or some factor thereof. What we know here is that $f((1+2k)p/6,\pi/3)$ goes $-\frac12,-\frac54,-\frac12$ for $k=0,1,2$, and that's enough to give us $A=\frac12$ and $B=-\frac34$. Similarly, at $y=\frac{2\pi}3$ we get $A=-\frac12$ and $B=-\frac34$. At $y=0$ we already know $A=1$ and $B=0$; and at $y=\pi$ we readily see $A=-1$ and $B=0$. And now it's pretty clear that $A(y)=\cos y$ and $B(y)=\sin^2y$, and we're done.

The form of $f$ is a bit surprising at first glance. It's not immediately apparent that it has the symmetries we want. What's going on?

Well, taking a hint from the remark above about the contours of $f=-1$, let's rewrite it as $f(x,y)=\cos y\cos\sqrt3x-1+\cos^2y$. Now we can factorize: $f(x,y)=-1+\cos y(\cos\sqrt3x+\cos y)$. And now we can apply one of those sum-to-product rewrites we learned in secondary school and then forgot: $f(x,y)=-1+2\cos y\cos(y-\sqrt3x)\cos(y+\sqrt3x)$. And this does obviously have the symmetries we want.


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