Ingenious Batman Equation

Saw this on Facebook, and I thought it's fake.

But out of curiosity, I went to google 'Batman equation", and found that somebody actually solve it!!
These people are brilliant...
And the prof who came out with this equation is genius.


Solution taken from HERE:




The ellipse (x7)2+(y3)2−1=0$\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:

So the curve (x7)2∣∣∣∣x∣∣−3∣∣∣∣x∣∣−3−−−−−√+(y3)2∣∣y+333√7∣∣y+333√7−−−−−−√−1=0$\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where |x|>3$|x|>3$ and y>−333−−√/7$y > -3\sqrt{33}/7$:

That's the first factor.

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The second factor is quite ingeniously done. The curve ∣∣x2∣∣−(333√−7)112x2−3+1−(||x|−2|−1)2−−−−−−−−−−−−−−−√−y=0$\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:

This is got by adding y=∣∣x2∣∣−(333√−7)112x2−3$y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:

and y=1−(||x|−2|−1)2−−−−−−−−−−−−−−−√$y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles (||x|−2|−1)2+y2=1$\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:

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The third factor 9(∣∣(1−∣∣x∣∣)(∣∣x∣∣−.75)∣∣)(1−∣∣x∣∣)(∣∣x∣∣−.75)−−−−−−−−−−−−√−8|x|−y=0$9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|:

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

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Similarly, the fourth factor 3|x|+.75(∣∣(.75−∣∣x∣∣)(∣∣x∣∣−.5)∣∣(.75−∣∣x∣∣)(∣∣x∣∣−.5))−−−−−−−−−−−−−−√−y=0$3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines y=3|x|+0.75$y = 3|x| + 0.75$:

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

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The fifth factor 2.25∣∣(.5−x)(x+.5)∣∣(.5−x)(x+.5)−−−−−−−−−√−y=0$2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line y=2.25$y = 2.25$ truncated to −0.5<x<0.5$-0.5 < x < 0.5$.

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Finally, 610√7+(1.5−.5|x|)−(610√)144−(|x|−1)2−−−−−−−−−−−√−y=0$\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:

so the sixth factor 610√7+(1.5−.5|x|)∣∣∣∣x∣∣−1∣∣∣∣x∣∣−1−−−−−√−(610√)144−(|x|−1)2−−−−−−−−−−−√−y=0$\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like

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As a product of factors is 0$0$ iff any one of them is 0$0$, 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)