Is this Batman equation for real? [closed]

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

Batman logo

Batman Equation in text form:
((x7)2||x|3||x|3+(y3)2|y+3337|y+33371)(|x2|(3337112)x23+1(||x|2|1)2y)(3|(|x|1)(|x|.75)|(1|x|)(|x|.75)8|x|y)(3|x|+.75|(|x|.75)(|x|.5)|(.75|x|)(|x|.5)y)(2.25(x.5)(x+.5)(.5x)(.5+x)y)(6107+(1.5.5|x|)||x|1||x|1610144(|x|1)2y)=0

Answer

As Willie Wong observed, including an expression of the form |α|α is a way of ensuring that α>0. (As |α|/α is 1 if α>0 and non-real if α<0.)


The ellipse (x7)2+(y3)21=0 looks like this:

ellipse

So the curve (x7)2||x|3||x|3+(y3)2|y+3337|y+33371=0 is the above ellipse, in the region where |x|>3 and y>333/7:

ellipse cut

That's the first factor.


The second factor is quite ingeniously done. The curve |x2|(3337)112x23+1(||x|2|1)2y=0 looks like:

second factor

This is got by adding y=|x2|(3337)112x23, a parabola on the positive-x side, reflected:

second factor first term

and y=1(||x|2|1)2, the upper halves of the four circles (||x|2|1)2+y2=1:

second factor second term


The third factor 9(|(1|x|)(|x|.75)|)(1|x|)(|x|.75)8|x|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|x|+.75(|(.75|x|)(|x|.5)|(.75|x|)(|x|.5))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|(.5x)(x+.5)|(.5x)(x+.5)y=0 is the line y=2.25 truncated to 0.5<x<0.5.


Finally, 6107+(1.5.5|x|)(610)144(|x|1)2y=0 looks like:

sixth factor without cut

so the sixth factor 6107+(1.5.5|x|)||x|1||x|1(610)144(|x|1)2y=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, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)

Wholly Batman

Attribution
Source : Link , Question Author : a_hardin , Answer Author : ShreevatsaR

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