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

Batman Equation in text form:

((x7)2√||x|−3||x|−3+(y3)2√|y+3√337|y+3√337−1)(|x2|−(3√33−7112)x2−3+√1−(||x|−2|−1)2−y)(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)(.5−x)(.5+x)−y)(6√107+(1.5−.5|x|)√||x|−1||x|−1−6√1014√4−(|x|−1)2−y)=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)2−1=0 looks like this:

So the curve (x7)2√||x|−3||x|−3+(y3)2√|y+3√337|y+3√337−1=0 is the above ellipse, in the region where |x|>3 and y>−3√33/7:

That's the first factor.

The second factor is quite ingeniously done. The curve |x2|−(3√33−7)112x2−3+√1−(||x|−2|−1)2−y=0 looks like:

This is got by adding y=|x2|−(3√33−7)112x2−3, a parabola on the positive-x side, reflected:

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

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|:

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:

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

The fifth factor 2.25√|(.5−x)(x+.5)|(.5−x)(x+.5)−y=0 is the line y=2.25 truncated to −0.5<x<0.5.

Finally, 6√107+(1.5−.5|x|)−(6√10)14√4−(|x|−1)2−y=0 looks like:

so the sixth factor 6√107+(1.5−.5|x|)√||x|−1||x|−1−(6√10)14√4−(|x|−1)2−y=0 looks like

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)

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