<s>[INST] Is this Batman equation for real? 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}
 [/INST] 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"
$$U(x)=\frac{x+|x|}{2x}$$

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

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...
</s>