Is this Batman equation for real?

Question

HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?

Batman Equation in text form: \begin{align} &\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\ &\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\ &\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\ &\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\ &\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0 \end{align}

Answer 1

As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)

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The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:

So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$:

That's the first factor.

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The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:

This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:

and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:

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The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|:

truncated to the region $0.75 < |x| < 1$.

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Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$:

truncated to the region $0.5 < |x| < 0.75$.

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The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.

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Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:

so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like

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As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)

Answer 2

Here's what I got from the equation using Maple...

Answer 3

Looking at the equation, it looks like it contains terms of the form $$ \sqrt{\frac{| |x| - 1 |}{|x| - 1}} $$ which evaluates to $$\begin{cases} 1 & |x| > 1\\ i & |x| < 1\end{cases} $$

Since any non-zero real number $y$ cannot be equal to a purely imaginary non-zero number, the presence of that term is a way of writing a piece-wise defined function as a single expression. My guess is that if you try to plot this in $\mathbb{C}^2$ instead of $\mathbb{R}^2$ you will get all kinds of awful.

Answer 4

Here's the equations typed out if you want save time with writing it yourself.

(x/7)^2*SQRT(ABS(ABS(x)-3)/(ABS(x)-3))+(y/3)^2\*SQRT(ABS(y+3*SQRT(33)/7)/(y+3*SQRT(33)/7))-1=0
ABS(x/2)-((3*SQRT(33)-7)/112)*x^2-3+SQRT(1-(ABS(ABS(x)-2)-1)^2)-y=0
9*SQRT(ABS((ABS(x)-1)*(ABS(x)-0.75))/((1-ABS(x))*(ABS(x)-0.75)))-8*ABS(x)-y=0
3*ABS(x)+0.75*SQRT(ABS((ABS(x)-0.75)*(ABS(x)-0.5))/((0.75-ABS(x))*(ABS(x)-0.5)))-y=0
2.25*SQRT(ABS((x-0.5)*(x+0.5))/((0.5-x)*(0.5+x)))-y=0
(6*SQRT(10))/7+(1.5-0.5*ABS(x))*SQRT(ABS(ABS(x)-1)/(ABS(x)-1))-((6*SQRT(10))/14)*SQRT(4-(ABS(x)-1)^2)-y=0

Also: http://pastebin.com/x9T3DSDp

Answer 5

In fact, the five linear pieces that consist the "head" (corresponding to the third, fourth, and fifth pieces in Shreevatsa's answer) can be expressed in a less complicated manner, like so:

$$y=\frac{\sqrt{\mathrm{sign}(1-|x|)}}{2}\left(3\left(\left|x-\frac12\right|+\left|x+\frac12\right|+6\right)-11\left(\left|x-\frac34\right|+\left|x+\frac34\right|\right)\right)$$

This can be derived by noting that the functions

$$\begin{cases}f(x)&\text{if }x<c\\g(x)&\text{if }c<x\end{cases}$$

and $f(x)+(g(x)-f(x))U(x-c)$ (where $U(x)$ is the unit step function) are equivalent, and using the "relation"

$$U(x)=\frac{x+|x|}{2x}$$

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Note that the elliptic sections (both ends of the "wings", corresponding to the first piece in Shreevatsa's answer) were cut along the lines $y=-\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)$, so the elliptic potion can alternatively be expressed as

$$\left(\left(\frac{x}{7}\right)^2+\left(\frac{y}{3}\right)^2-1\right)\sqrt{\mathrm{sign}\left(y+\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)\right)}=0$$

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Theoretically, since all you have are arcs of linear and quadratic curves, the chimera can be expressed parametrically using rational B-splines, but I'll leave that for someone else to explore...

Answer 6

You may be able to see more easily the correspondences between the equations and the graph through the following picture which is from the link I got after a curious search on Google(link broken now):

Answer 7

The following is what I got from the equations using MATLAB:

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Here is the M-File (thanks to this link):

clf; clc; clear all; 
syms x y

eq1 = ((x/7)^2*sqrt(abs(abs(x)-3)/(abs(x)-3))+(y/3)^2*sqrt(abs(y+3/7*sqrt(33))/(y+3/7*sqrt(33)))-1);
eq2 = (abs(x/2)-((3*sqrt(33)-7)/112)*x^2-3+sqrt(1-(abs(abs(x)-2)-1)^2)-y);
eq3 = (9*sqrt(abs((abs(x)-1)*(abs(x)-.75))/((1-abs(x))*(abs(x)-.75)))-8*abs(x)-y);
eq4 = (3*abs(x)+.75*sqrt(abs((abs(x)-.75)*(abs(x)-.5))/((.75-abs(x))*(abs(x)-.5)))-y);
eq5 = (2.25*sqrt(abs((x-.5)*(x+.5))/((.5-x)*(.5+x)))-y);
eq6 = (6*sqrt(10)/7+(1.5-.5*abs(x))*sqrt(abs(abs(x)-1)/(abs(x)-1))-(6*sqrt(10)/14)*sqrt(4-(abs(x)-1)^2)-y);


axes('Xlim', [-7.25 7.25], 'Ylim', [-5 5]);
hold on

ezplot(eq1,[-8 8 -3*sqrt(33)/7 6-4*sqrt(33)/7]);
ezplot(eq2,[-4 4]);
ezplot(eq3,[-1 -0.75 -5 5]);
ezplot(eq3,[0.75 1 -5 5]);
ezplot(eq4,[-0.75 0.75 2.25 5]);
ezplot(eq5,[-0.5 0.5 -5 5]);
ezplot(eq6,[-3 -1 -5 5]);
ezplot(eq6,[1 3 -5 5]);
colormap([0 0 1])

title('Batman');
xlabel('');
ylabel('');
hold off

Answer 8

Since people (not from this site, but still...) keep bugging me, and I am unable to edit my previous answer, here's Mathematica code for plotting this monster:

Plot[{With[{w = 3 Sqrt[1 - (x/7)^2], 
            l = 6/7 Sqrt[10] + (3 + x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x + 1)^2], 
            h = (3 (Abs[x - 1/2] + Abs[x + 1/2] + 6) -
                 11 (Abs[x - 3/4] + Abs[x + 3/4]))/2, 
            r = 6/7 Sqrt[10] + (3 - x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x - 1)^2]}, 
           w + (l - w) UnitStep[x + 3] + (h - l) UnitStep[x + 1] +
           (r - h) UnitStep[x - 1] + (w - r) UnitStep[x - 3]],
      1/2 (3 Sqrt[1 - (x/7)^2] + Sqrt[1 - (Abs[Abs[x] - 2] - 1)^2] + Abs[x/2] -
      ((3 Sqrt[33] - 7)/112) x^2 - 3) (Sign[x + 4] - Sign[x - 4]) - 3*Sqrt[1 - (x/7)^2]},
     {x, -7, 7}, AspectRatio -> Automatic,  Axes -> None, Frame -> True,
     PlotStyle -> Black]

This should work even for versions that do not have the Piecewise[] construct. Enjoy. :P

Answer 9

The 'Batman equation' above relies on an artifact of the plotting software used which blithely ignores the fact that the value $\sqrt{\frac{|x|}{x}}$ is undefined when $x=0$. Indeed, since we’re dealing with real numbers, this value is really only defined when $x>0$. It seems a little ‘sneaky’ to rely on the solver to ignore complex values and also to conveniently ignore undefined values.

A nicer solution would be one that is unequivocally defined everywhere (in the real, as opposed to complex, world). Furthermore, a nice solution would be ‘robust’ in that small variations (such as those arising from, say, roundoff) would perturb the solution slightly (as opposed to eliminating large chunks).

Try the following in Maxima (actually wxmaxima) which is free. The resulting plot is not quite as nice as the plot above (the lines around the head don’t have that nice ‘straight line’ look), but seems more ‘legitimate’ to me (in that any reasonable solver should plot a similar shape). Please excuse the code mess.

/* [wxMaxima batch file version 1] [ DO NOT EDIT BY HAND! ]*/
/* [ Created with wxMaxima version 0.8.5 ] */

/* [wxMaxima: input   start ] */
load(draw);
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
f(a,b,x,y):=a*x^2+b*y^2;
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
c1:sqrt(26);
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
draw2d(implicit(
f(1/36,1/9,x,y)
+max(0,2-f(1.5,1,x+3,y+2.7))
+max(0,2-f(1.5,1,x-3,y+2.7))
+max(0,2-f(1.9,1/1.7,(5*(x+1)+(y+3.5))/c1,(-(x+1)+5*(y+3.5))/c1))
+max(0,2-f(1.9,1/1.7,(5*(x-1)-(y+3.5))/c1,((x-1)+5*(y+3.5))/c1))
+max(0,2-((1.1*(x-2))^4-(y-2.1)))
+max(0,2-((1.1*(x+2))^4-(y-2.1)))
+max(0,2-((1.5*x)^8-(y-3.5)))
-1,
x,-6,6,y,-4,4));
/* [wxMaxima: input   end   ] */

/* Maxima can't load/batch files which end with a comment! */
"Created with wxMaxima"$

The resulting plot is:

(Note that this is, more or less, a copy of the entry I made on http://blog.makezine.com.)