HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
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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 $y > -3\sqrt{33}/7$:
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:
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 $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:
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 $0.75 < |x| < 1$. 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 $0.5 < |x| < 0.75$. 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:
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
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)
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You may be able to see more easily the correspondences between the equations and the graph through the following graph which is from the link I got after a curious search on Google:
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Here's what I got from the equation using Maple...
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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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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... | |||||||||||||
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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:
This should work even for versions that do not have the | |||
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The following is what I got from the equations using MATLAB:
Here is the M-File (thanks to this link): | |||
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Here's the equations typed out if you want save time with writing it yourself. | |||||||||
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protected by Zev Chonoles♦ Jul 30 '11 at 19:53
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