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

Batman logo

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Why don't you just try it? – Jonas Teuwen Jul 29 '11 at 21:19
@Jim: If you mouse over the downvote button, you see: "This question does not show any research effort; it is unclear or not useful." I downvoted because the OP was too lazy to type in the equation himself to any plotting program or calculator, which would have immediately shown that the equation is "for real". If the OP were asking for an explanation of how such an equation might be derived, as ShreevatsaR has done, that would be an appropriate question. – Zev Chonoles Jul 30 '11 at 17:08
@Zev Chonoles: i do not know of any web-site that can plot that equation. i wouldn't know where to begin. Also i don't understand how any solver could plot such a diagram. (Hence the question). – Ian Boyd Jul 30 '11 at 19:35
I don't understand why this question has so many upvotes. – Jonas Teuwen Aug 3 '11 at 18:36
@Jacob – splattne Aug 31 '11 at 14:40

11 Answers 11

up vote 946 down vote accepted

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 $\alpha < 0$.)

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 $y > -3\sqrt{33}/7$:

ellipse cut

That's the first factor.

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:

second factor

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:

second factor first term

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

second factor second term

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|:

Third factor without cut

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

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 $y = 3|x| + 0.75$:

fourth factor without cut

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

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 $-0.5 < x < 0.5$.

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:

sixth factor without cut

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

sixth factor

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,, chokes a bit on the third factor, and entirely on the fourth)

Wholly Batman

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I tip my hat to you for this comprehensive dissection. – Willie Wong Jul 30 '11 at 14:06
Holy upvotes Batman! The number of upvotes on this answer scares me. All I did was plot a bunch of equations. :-) – ShreevatsaR Jul 31 '11 at 0:41
Enjoy your gold badge, dude. ;) – J. M. Jul 31 '11 at 2:50
"I don’t know how ShreevatsaR did it but he sure brought a large luggage when intelligence showered the Earth." – Jonas Meyer Aug 3 '11 at 0:15
@Jonas Meyer: LOL, that's embarrassing! :P But then again, if Batman is what it takes for someone to appreciate mathematics a little, well good for Batman. :-) – ShreevatsaR Aug 3 '11 at 10:46

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):

enter image description here

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Geometer's Sketchpad? – Isaac Aug 3 '11 at 18:28
@Isaac: Probably right. – Tim Seguine Sep 6 '11 at 12:27
to see the graph just google the equation: 2*sqrt(-abs(abs(x)-1)*abs(3-abs(x))/((abs(x)-1)*(3-abs(x))))(1+abs(abs(x)-3)/(ab‌​s(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+‌​abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2‌​)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqr‌​t(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9 – Helder Velez Aug 20 '12 at 18:47
what program is that? – Joao Apr 4 '14 at 7:19
@HelderVelez 2014: "'abs' (and any subsequent words) was ignored because we limit queries to 32 words." - Google – Baby Nov 28 '14 at 8:11

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

enter image description here

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What if Commissioner Gordon uses Mathematica ???? – The Chaz 2.0 Jul 30 '11 at 5:25
@The Chaz: Then Commissioner Gordon should support the Mathematica SE Site Proposal on Area 51. – Isaac Jul 30 '11 at 19:22
touché ....$$$$ – The Chaz 2.0 Jul 30 '11 at 19:24
Here's Mathematica code. See Heike's post. I tried it on M8 and it works fine.… – Sol Aug 5 '11 at 22:48
@Sol: You can do better; see my answer. – J. M. Aug 9 '11 at 7:56

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.

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Yeah, the equation looks too contrived to me. :) A parametric form (it's just quadratic and linear arcs sewn together, it looks) would still be messy, but not as messy. (Probably a good job for splines...) – J. M. Jul 30 '11 at 2:43
+1 i was wondering how they split it up into sections. – Ian Boyd Jul 30 '11 at 19:38
" My guess is that if you try to plot this in C2 instead of R2 you will get all kinds of awful." What did you expect? The analytic continuation of the Batman symbol?? – jwg Aug 1 '13 at 8:43

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 = (1/2)*(3*(Abs[x - 1/2] + Abs[x + 1/2] + 6) - 11*(Abs[x - 3/4] + Abs[x + 3/4])), 
    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)*
     ((x + 4)/Abs[x + 4] - (x - 4)/Abs[x - 4]) - 3*Sqrt[1 - (x/7)^2]},
 {x, -7, 7}, AspectRatio -> Automatic, 
 Axes -> None, Frame -> True, PlotStyle -> GrayLevel[0]]

Mathematica graphics

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

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The following is what I got from the equations using MATLAB: enter image description here

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])

hold off
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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:


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"


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


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...

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...and that's half an hour of my life that I'll never get back... ;P – J. M. Jul 30 '11 at 18:47
Okay, make that an hour. Sheesh. *facepalm* – J. M. Jul 30 '11 at 19:56
Great, this would be much clearer. BTW, I think it's safer to leave it at "can be expressed using B-splines" than to actually try it out: who knows how many hours that will waste, right? :-) – ShreevatsaR Jul 31 '11 at 3:20

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 ] */
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
/* [wxMaxima: input   end   ] */

/* [wxMaxima: input   start ] */
/* [wxMaxima: input   end   ] */

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

The resulting plot is: enter image description here

(Note that this is, more or less, a copy of the entry I made on

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I really think that an indeterminate value multiplied by zero equals zero, so it seems to be legit. Is there any reason 0 * 0/0 should not be defined to be zero? – dbanet Jun 15 '15 at 19:45
@dbanet: What are you referring to? The issue above is that the original equations rely on the plotting software ignoring undefined values, which is peculiar, to say the least. The expression $\sqrt{\frac{|x|}{x}}$ (with $x$ being replaced by some expression) is what I referred to and it appears without being multiplied by $x$. – copper.hat Jun 15 '15 at 19:54
@copper-hat: The function $f(x)=\sqrt{\frac{|x|}{x}}$ appears only in boolean expressions $F:x\to \{\text{True},\text{False}\}$ of form $f(x)g(x)=0$, so if $g(x)$ is defined as $g:x\in\mathbb{C}\to{0}$, I would rather evaluate $F(0)$ to $\text{True}$ than to $\text{False}$, as $\forall{x}:f(x)g(x)=0\Longleftrightarrow \Big(f(x)=0\lor g(x)=0\Big)$. – dbanet Jun 15 '15 at 21:57
@dbanet: I'm really not sure what you are getting at. Look at the expressions in the question. They rely on the expression $\sqrt{\frac{|x|}{x}}$ returning zero for $x \le 0$, which is strange (look at Willie's answer My answer plots level sets, which avoids this whole issue. – copper.hat Jun 15 '15 at 22:10
@copper-hat: Why is it strange? For $x<0:\operatorname{Im}\left(\sqrt{\frac{|x|}{x}}\right)\neq{0}$. For $x=0$ it is indeterminate but that does not matter as long as any other factor evaluates to zero, so that the whole boolean expression holds. – dbanet Jun 15 '15 at 22:16

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



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The multiple "=0" lines are different than the mulitiplications in the original, there are some backslashes in there that throw things off, and the formatting is hard to read. The pastebin is better. – nealmcb Jul 30 '11 at 16:42
I've made a meta post about this answer. – Zev Chonoles Jul 30 '11 at 19:21
I would have commented it but I don't have enough rep to comment :P – stoicfury Jul 31 '11 at 4:45

Sorry but this is not the answer but too long for a comment: Probably the easiest verification is to type the equation on Google you'l be surprised : The easiest way is to Google :2 sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqrt(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9

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@copper, it is just because of the algorithm that draw uses for drawing implicit functions. You need to setup the variables ip_grid and ip_grid_in, that are the sampling values in your region. For example draw2d(ip_grid=[60,60], ip_grid_in=[20,20], implicit(y^2=x^3-2*x+1, x, -4,4, y, -4,4) ); – nicoguaro Oct 16 '14 at 19:21

Yes, this batman equation is real. For context, the batman curve is a piecewise curve in the shape of the logo of the Batman superhero originally posted on on Jul. 28, 2011.

It can written as two functions, one for the upper part and the other for the lower part, as

$$f(x) = (h-l) \,H \,(x+1)+(r-h) \, H\,(x-1)+(l-w)\,H\,(x+3)+(w-r)\,H\,(x-3)+w$$

$$g(x) = \frac 12\left[\left| \frac{1}{2x}\right|+ \sqrt {(1-(\lvert \lvert x\rvert -2\rvert -1)^2}-\frac{1}{112}(3 \sqrt {33}-7)x^2+3\sqrt{1-\left(\frac {1}{7}x\right)^2}-3\right]×\left(\frac{x+4}{\left|x+4\right|}-\frac{x-4}{\lvert x-4 \rvert}\right)-3 \sqrt{1-\left( \frac 17x \right)^2},$$

where $H(x)$ is the Heaviside step function and

$$w = 3 \sqrt{1-\left( \frac x7\right)^2}$$

$$l = \frac 12(x+3)- \frac37 \sqrt{10} \sqrt{4-(x+1)^2}+ \frac67 \sqrt{10}$$

$$h = \frac 12 \left[f\left(\left|x+ \frac12\right|+ \left|x-\frac 12\right|+6\right)-11\left(x+\frac 34\right)+\left|x- \frac 34\right|\right]$$

$$r = \frac 12(3-x)- \frac 37 \sqrt{10} \sqrt{4-(x-1)^2}+ \frac 67 \sqrt{10}$$

Its area can then be computed exactly as

$$A= \frac{955}{48}-\frac 27\left[2\sqrt{33}+7\pi+3 \sqrt{10}(\pi-1)\right]+21\left[cos^{-1}\left(\frac 37\right)+cos^{-1}\left(\frac 47\right)\right].$$

To illustrate this then on a graph:

batman graph

$\mathcal VOILA!!!$ The batman equation is indeed real!

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protected by Zev Chonoles Jul 30 '11 at 19:53

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