The Batman equation
Moderators: gmalivuk, Moderators General, Prelates
Re: The Batman equation
Someone type up the equation in either latex prettyness or standard typed out form so it can be copied/verified.
Also, the picture is huge and cut off for me so I don't actually get to see the whole equation without jumping through certain hoops.
edit:
Warning: All of the 2's that were supposed to be squaring are lacking ^'s below, so you'll need to replace them appropriately.
((x/7)2 Sqrt[Abs[Abs[x]  3]/(Abs[x]  3)] + (y/3)2 Sqrt[Abs[y + (3 Sqrt[33])/7]/(y + (3 Sqrt[33])/7)]  1) (Abs[x/2]  ((3 Sqrt[33]  7)/112) x2  3 + Sqrt[1  (Abs[Abs[x]  2]  1)2 ]  y) (9 Sqrt[Abs[(Abs[x]  1) (Abs[x]  3/4)]/((1  Abs[x]) (Abs[x]  3/4))]  8 Abs[x]  y) (3 Abs[x] + .75 Sqrt[Abs[(Abs[x]  3/4) (Abs[x]  1/2)]/((3/4  Abs[x]) (Abs[x]  1/2))]  y) (9/4 Sqrt[Abs[(x  1/2) (x + 1/2)]/((1/2  x) (1/2 + x))]  y) ((6 Sqrt[10])/7 + (3/2  Abs[x]/2) Sqrt[Abs[Abs[x]  1]/(Abs[x]  1)]  (6 Sqrt[10])/14 Sqrt[4  (Abs[x]  1)2 ]  y) == 0
Should be mathematica friendly.
Also, the picture is huge and cut off for me so I don't actually get to see the whole equation without jumping through certain hoops.
edit:
Warning: All of the 2's that were supposed to be squaring are lacking ^'s below, so you'll need to replace them appropriately.
((x/7)2 Sqrt[Abs[Abs[x]  3]/(Abs[x]  3)] + (y/3)2 Sqrt[Abs[y + (3 Sqrt[33])/7]/(y + (3 Sqrt[33])/7)]  1) (Abs[x/2]  ((3 Sqrt[33]  7)/112) x2  3 + Sqrt[1  (Abs[Abs[x]  2]  1)2 ]  y) (9 Sqrt[Abs[(Abs[x]  1) (Abs[x]  3/4)]/((1  Abs[x]) (Abs[x]  3/4))]  8 Abs[x]  y) (3 Abs[x] + .75 Sqrt[Abs[(Abs[x]  3/4) (Abs[x]  1/2)]/((3/4  Abs[x]) (Abs[x]  1/2))]  y) (9/4 Sqrt[Abs[(x  1/2) (x + 1/2)]/((1/2  x) (1/2 + x))]  y) ((6 Sqrt[10])/7 + (3/2  Abs[x]/2) Sqrt[Abs[Abs[x]  1]/(Abs[x]  1)]  (6 Sqrt[10])/14 Sqrt[4  (Abs[x]  1)2 ]  y) == 0
Should be mathematica friendly.
Last edited by Dopefish on Sat Jul 30, 2011 4:57 am UTC, edited 1 time in total.
 agelessdrifter
 Posts: 225
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Re: The Batman equation
It's really counterintuitive to me that something like that could be defined without going piecewise.
And furthermore how was anyone able to extract the formula?
And furthermore how was anyone able to extract the formula?
 skeptical scientist
 closedminded spiritualist
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Re: The Batman equation
agelessdrifter wrote:It's really counterintuitive to me that something like that could be defined without going piecewise.
And furthermore how was anyone able to extract the formula?
Not really. It makes heavy use of things like x/x, which basically let you define piecewise functions with a single formula. Any piecewise function where the pieces are intervals can be written as a sum of the functions defining pieces times the characteristic function of the corresponding interval, and the function (1+ax/(ax)*xb/(xb))/2 is basically the characteristic function of (a,b) (it's undefined at a and at b, but that doesn't really matter to a calculator trying to draw the graph).
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
Re: The Batman equation
Well, I can't exactly replicate all of it, but enough to make it look legit. Used your corrected mathematica friendly post.
((x/7)2 Sqrt[Abs[Abs[x]  3]/(Abs[x]  3)] + (y/3)2 Sqrt[Abs[y + (3 Sqrt[33])/7]/(y + (3 Sqrt[33])/7)]  1) (Abs[x/2]  ((3 Sqrt[33]  7)/112) x^2  3 + Sqrt[1  (Abs[Abs[x]  2]  1)2 ]  y) (9 Sqrt[Abs[(Abs[x]  1) (Abs[x]  3/4)]/((1  Abs[x]) (Abs[x]  3/4))]  8 Abs[x]  y) (3 Abs[x] + .75 Sqrt[Abs[(Abs[x]  3/4) (Abs[x]  1/2)]/((3/4  Abs[x]) (Abs[x]  1/2))]  y) (9/4 Sqrt[Abs[(x  1/2) (x + 1/2)]/((1/2  x) (1/2 + x))]  y) ((6 Sqrt[10])/7 + (3/2  Abs[x]/2) Sqrt[Abs[Abs[x]  1]/(Abs[x]  1)]  (6 Sqrt[10])/14 Sqrt[4  (Abs[x]  1)2 ]  y) == 0
((x/7)2 Sqrt[Abs[Abs[x]  3]/(Abs[x]  3)] + (y/3)2 Sqrt[Abs[y + (3 Sqrt[33])/7]/(y + (3 Sqrt[33])/7)]  1) (Abs[x/2]  ((3 Sqrt[33]  7)/112) x^2  3 + Sqrt[1  (Abs[Abs[x]  2]  1)2 ]  y) (9 Sqrt[Abs[(Abs[x]  1) (Abs[x]  3/4)]/((1  Abs[x]) (Abs[x]  3/4))]  8 Abs[x]  y) (3 Abs[x] + .75 Sqrt[Abs[(Abs[x]  3/4) (Abs[x]  1/2)]/((3/4  Abs[x]) (Abs[x]  1/2))]  y) (9/4 Sqrt[Abs[(x  1/2) (x + 1/2)]/((1/2  x) (1/2 + x))]  y) ((6 Sqrt[10])/7 + (3/2  Abs[x]/2) Sqrt[Abs[Abs[x]  1]/(Abs[x]  1)]  (6 Sqrt[10])/14 Sqrt[4  (Abs[x]  1)2 ]  y) == 0
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 Screen shot 20110730 at 12.04.33 AM.png (10 KiB) Viewed 33356 times
 gmalivuk
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Re: The Batman equation
It is, but it's also wrong, since everything that should be squared is multiplied by 2.Dopefish wrote:Should be mathematica friendly.
Re: The Batman equation
Yup, I noticed that was happening in one place, and corrected it. Apparently the same mistake is in multiple places.
 skeptical scientist
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Re: The Batman equation
Doing find and replace ")2" > ")^2" should catch most of them.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson
Re: The Batman equation
Oh bother, sorry about that. I'll fix it later when I have time, but I'll add a warning to my earlier post pointing that out until then.
Re: The Batman equation
Well, I have managed to clean my code a little (I can't seem to just use a contour plot or any similar mathematica option. I am numerically solving the equation and plotting the different solutions)
 gmalivuk
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Re: The Batman equation
ImplicitPlot should work.

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Re: The Batman equation
I tried it in Maxima and got much less successful results. Maybe it's a limitation of Maxima, but I also could very well not be using it correctly. It was almost the Batman logo, it was just deformed.
Re: The Batman equation
I tried it in Mac OS X's Grapher... It didn't render anything for the full equation. But if I split it into 6 separate equations the results look pretty good.
I'm not sure if it's because I'm new here and don't have permission to attach images, or if I'm just being an idiot and not seeing how to attach images... but a link will have to suffice:
http://www.macopz.com/crosis/BatmanEqn.png
I'm not sure if it's because I'm new here and don't have permission to attach images, or if I'm just being an idiot and not seeing how to attach images... but a link will have to suffice:
http://www.macopz.com/crosis/BatmanEqn.png
Re: The Batman equation
Tried plotting using matlab, with very limited success:
http://tinypic.com/r/161ykuc/7
Any thoughts on where I went wrong? For those playing at home, use the following to see the equation in a somewhat more readable form (after first running the script above):
Code: Select all
batman = '((x/7)^2 * sqrt(abs(abs(x)3)/(abs(x)3)) + (y/3)^2 * sqrt(abs(y+ 3*sqrt(33)/7)/(y+3*sqrt(33)/7))1) * (abs(x/2)(3*sqrt(33)7)/112*x^23+sqrt(1(abs(abs(x)2)1)^2)y) * (9*sqrt(abs((abs(x)1)*(abs(x)0.75))/((1abs(x))*(abs(x)0.75)))8*abs(x)y) * (3*abs(x)+0.75*sqrt(abs((abs(x)0.75)*(abs(x)0.5))/((0.75abs(x))*(abs(x)0.5)))y) * (2.25*sqrt(abs((x0.5)*(x+0.5))/((0.5x)*(0.5+x)))y) * (6*sqrt(10)/7+(1.50.5*abs(x))*sqrt(abs(abs(x)1)/(abs(x)1))6*sqrt(10)/14*sqrt(4(abs(x)1)^2)y)';
figure(1);
ezplot(batman, [7.2 7.2 3.5 3.5])
title('Batman')
xlabel(' ')
ylabel(' ')
http://tinypic.com/r/161ykuc/7
Any thoughts on where I went wrong? For those playing at home, use the following to see the equation in a somewhat more readable form (after first running the script above):
Code: Select all
bat = syms(batman)
syms x y
pretty(bat)
Re: The Batman equation
The form of the overall equation is a product of six terms. Since we are dealing, presumably, with complex variables, we are in a field, and all fields are integral domains, hence a product equals zero exactly when at least one of its multiplicands is zero. Thus we may consider each of the six terms independently. After the first term, the remaining five, when set equal to zero, are trivial to solve for y as a function of x. Thus, for any arbitrary x that does not cause division by zero, we will get at least five values for y, although occasionally some of these might coincide. We may get even more y values from the first term.
We may view this socalled “Batman equation” as a multivalued map from ℂ\{a finite number of bad points, plus a finite number of branch cuts unless we take the multivalued square root operator} to ℂ\{possibly a finite number of bad points plus possibly a finite number of branch cuts}. Indeed, by inspection and a bit of reasoning, we can see that each of the terms considered alone produces a surface that is continuous everywhere except possibly at the bad points. The overal equation is just the union of all such surfaces, possibly with additional holes punched in at the points that are bad on any other such surface. We also see that each of the bad points is basically of the form √(z/z), which gets no worse than a step discontinuity along any path into the bad point. So we have no poles.
So, viewed as a set in ℂ×ℂ, the graph of the Batman equation is a collection of a finite number of surfaces continuous except at a few bad points (and possibly branch cuts). When we take the 2dimensional slice of ℂ×ℂ corresponding to ℝ×ℝ, then we do in fact get see the curves presented. However, that image does no justice to the full graph of the equation, which naturally lies in what may be considered a Euclidean 4space.
We may view this socalled “Batman equation” as a multivalued map from ℂ\{a finite number of bad points, plus a finite number of branch cuts unless we take the multivalued square root operator} to ℂ\{possibly a finite number of bad points plus possibly a finite number of branch cuts}. Indeed, by inspection and a bit of reasoning, we can see that each of the terms considered alone produces a surface that is continuous everywhere except possibly at the bad points. The overal equation is just the union of all such surfaces, possibly with additional holes punched in at the points that are bad on any other such surface. We also see that each of the bad points is basically of the form √(z/z), which gets no worse than a step discontinuity along any path into the bad point. So we have no poles.
So, viewed as a set in ℂ×ℂ, the graph of the Batman equation is a collection of a finite number of surfaces continuous except at a few bad points (and possibly branch cuts). When we take the 2dimensional slice of ℂ×ℂ corresponding to ℝ×ℝ, then we do in fact get see the curves presented. However, that image does no justice to the full graph of the equation, which naturally lies in what may be considered a Euclidean 4space.
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Re: The Batman equation
rjb, that's kindasorta similar to what I got in Maxima, except yours is closer to correct.
This code requires loading the draw package.
Resulted in
Update: I tried it again, this time typing out the equation from the original photo, using this command:
With this result:
Everything's there, but there's a lot of stuff that we don't want there, too. Makes me wonder if there's a domain limitation or something that we're missing, but the fact that other people are getting better results makes me think that something's wrong on my end (either PEBKAC or Maxima).
Code: Select all
draw2d(ip_grid=[100,100],ip_grid_in=[20,20],
implicit(
((x/7)^2 * sqrt(abs(abs(x)  3)/(abs(x)  3)) + (y/3)^2 * sqrt(abs(y + (3 * sqrt(33))/7)/(y + (3 * sqrt(33))/7))  1) * (abs(x/2)  ((3 * sqrt(33)  7)/112) * x*2  3 + sqrt(1  (abs(abs(x)  2)  1)^2 )  y) * (9 * sqrt(abs((abs(x)  1) * (abs(x)  3/4))/((1  abs(x)) * (abs(x)  3/4)))  8 * abs(x)  y) * (3 * abs(x) + .75 * sqrt(abs((abs(x)  3/4) * (abs(x)  1/2))/((3/4  abs(x)) * (abs(x)  1/2)))  y) * (9/4 * sqrt(abs((x  1/2) * (x + 1/2))/((1/2  x) * (1/2 + x)))  y) * ((6 * sqrt(10))/7 + (3/2  abs(x)/2) * sqrt(abs(abs(x)  1)/(abs(x)  1))  (6 * sqrt(10))/14 * sqrt(4(abs(x)1)^2)y)=0,
x,9,9,y,4,4));
This code requires loading the draw package.
Resulted in
Update: I tried it again, this time typing out the equation from the original photo, using this command:
Code: Select all
draw2d(ip_grid=[100,100],ip_grid_in=[20,20],line_type=dots,
implicit(
((x/7)^2*sqrt((abs(abs(x)3)/(abs(x)3)))+(y/3)^2*sqrt((abs(y+3*sqrt(33)/7))/(y+3*sqrt(33)/7))1)
*
(abs(x/2)(3*sqrt(33)7)/(112)*x^23+sqrt(1(abs(abs(x)2)1)^2)y)
*
(9*sqrt((abs((abs(x)1)*(abs(x).75)))/((1abs(x))*(abs(x).75)))8*abs(x)y)
*
(3*abs(x)+.75*sqrt((abs((abs(x).75)*(abs(x).5)))/((.75abs(x))*(abs(x).5)))y)
*
(2.25*sqrt((abs((x.5)*(x+.5)))/((.5x)*(.5+x)))y)
*
(6*sqrt(10)/7+(1.5.5*abs(x))*sqrt((abs(abs(x)1))/(abs(x)1))6*sqrt(10)/14*sqrt(4(abs(x)1)^2)y)
=0,
x,9,9,y,4,4))$
With this result:
Everything's there, but there's a lot of stuff that we don't want there, too. Makes me wonder if there's a domain limitation or something that we're missing, but the fact that other people are getting better results makes me think that something's wrong on my end (either PEBKAC or Maxima).
Last edited by adanedhel728 on Sat Jul 30, 2011 7:43 pm UTC, edited 1 time in total.

 Posts: 9
 Joined: Fri Jul 29, 2011 6:58 pm UTC
Re: The Batman equation
Dopefish wrote:Someone type up the equation in either latex prettyness or standard typed out form so it can be copied/verified.
I think this should be right:
[math]\left(
\left( \frac{x}{7} \right) ^2
\sqrt{ \frac{ x 3  }{ x3 } }
+
\left( \frac{y}{3} \right) ^2
\sqrt{ \frac{ \left y + \frac{3\sqrt{33}}{7} \right }{ y + \frac{3\sqrt{33}}{7} } }
 1
\right)
\cdot
\left(
\left \frac{x}{2} \right

\left( \frac{ 3\sqrt{33} 7 }{112} \right) x^2
 3 +
\sqrt{ 1  (x  2  1)^2 }
 y
\right)
\cdot
\left(
9 \sqrt{ \frac{
 (x1)(x.75) 
}{
(1x)(x.75)
} }
 8xy
\right)
\cdot
\left(
3x+.75 \sqrt { \frac {
 (x.75)(x.5) 
}{
(.75x)(x.5)
} }  y
\right)
\cdot
\left(
2.25 \sqrt{ \frac{
 (x.5)(x+.5) 
}{
(.5x)(.5+x)
}}  y
\right)
\cdot
\left(
\frac { 6 \sqrt{10} }{ 7 }
+ (1.5  .5x) \sqrt{ \frac{ x1 }{ x1 } }
 \frac{ 6\sqrt{10} }{ 14 } \sqrt{ 4(x  1)^2 }  y
\right)
=0[/math]
The forum cuts it off for being wide, though, so here's the latex source:
Code: Select all
\left(
\left( \frac{x}{7} \right) ^2
\sqrt{ \frac{ x 3  }{ x3 } }
+
\left( \frac{y}{3} \right) ^2
\sqrt{ \frac{ \left y + \frac{3\sqrt{33}}{7} \right }{ y + \frac{3\sqrt{33}}{7} } }
 1
\right)
\cdot
\left(
\left \frac{x}{2} \right

\left( \frac{ 3\sqrt{33} 7 }{112} \right) x^2
 3 +
\sqrt{ 1  (x  2  1)^2 }
 y
\right)
\cdot
\left(
9 \sqrt{ \frac{
 (x1)(x.75) 
}{
(1x)(x.75)
} }
 8xy
\right)
\cdot
\left(
3x+.75 \sqrt { \frac {
 (x.75)(x.5) 
}{
(.75x)(x.5)
} }  y
\right)
\cdot
\left(
2.25 \sqrt{ \frac{
 (x.5)(x+.5) 
}{
(.5x)(.5+x)
}}  y
\right)
\cdot
\left(
\frac { 6 \sqrt{10} }{ 7 }
+ (1.5  .5x) \sqrt{ \frac{ x1 }{ x1 } }
 \frac{ 6\sqrt{10} }{ 14 } \sqrt{ 4(x  1)^2 }  y
\right)
=0
and a PNG rendering at (I'm apparently not allowed to use [url]) <http://i.imgur.com/vtwYh.png>.
 phlip
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Re: The Batman equation
It's 6 terms, multiplied together, and called equal to zero... you might have better luck plotting it as 6 separate curves, one for each term, equal to zero. Might make life a bit easier for the plotting tool...
Code: Select all
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 SSJ3 Mario Brothers
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 Joined: Thu Apr 03, 2008 2:06 am UTC
Re: The Batman equation
phlip wrote:It's 6 terms, multiplied together, and called equal to zero... you might have better luck plotting it as 6 separate curves, one for each term, equal to zero. Might make life a bit easier for the plotting tool...
That works to an extent, however the problem is when you set one of the six terms equal to zero, you miss out on the implied bounds imposed by the other five. Which is where all those extra lines come from.
I have so far been unsuccessful in plotting it in MATLAB; I know I could do it by manually calculating the bounds for each of the six parts, but where's the fun and elegance in that?

 Posts: 272
 Joined: Thu Jul 09, 2009 9:26 am UTC
Re: The Batman equation
SSJ3 Mario Brothers wrote:phlip wrote:It's 6 terms, multiplied together, and called equal to zero... you might have better luck plotting it as 6 separate curves, one for each term, equal to zero. Might make life a bit easier for the plotting tool...
That works to an extent, however the problem is when you set one of the six terms equal to zero, you miss out on the implied bounds imposed by the other five. Which is where all those extra lines come from.
I'm not sure I understand what you mean by missing out on the implied bounds, as far as I can see splitting it into 6 separate graphs should work.
Don't have access to Mathematica or anything, but surprisingly wolfram alpha can handle most bits of it.
[url=http://www.wolframalpha.com/input/?i=plot+((x/7)^2*sqrt((abs(abs(x)3)/(abs(x)3)))%2B(y/3)^2*sqrt((abs(y%2B3*sqrt(33)/7))/(y%2B3*sqrt(33)/7))1)%3D0++x%3D9..9,+y%3D9..9]Part 1[/url]
[url=http://www.wolframalpha.com/input/?i=plot+(abs(x/2)(3*sqrt(33)7)/(112)*x^23%2Bsqrt(1(abs(abs(x)2)1)^2)y)%3D0+x%3D9..9,+y%3D9..9]Part 2[/url]
Part 3
Part 4
Part 5
[url=http://www.wolframalpha.com/input/?i=plot+(6*sqrt(10)/7%2B(1.5.5*abs(x))*sqrt((abs(abs(x)1))/(abs(x)1))6*sqrt(10)/14*sqrt(4(abs(x)1)^2)y)%3D0+x%3D9..9,+y%3D9..9]Part 6[/url]
You might have to click on the 'try again with more time' button for it to work but it seems to have had a good go at everything except part 4 for me.
Edit: That's weird. Not sure why some of the URL tags haven't worked, but you can copy & paste the URL I guess.

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Re: The Batman equation
There's a nice deconstruction over here:
http://math.stackexchange.com/questions/54506/isthisbatmanequationforreal
http://math.stackexchange.com/questions/54506/isthisbatmanequationforreal
An apple every eight hours will keep three doctors away
Re: The Batman equation
The equations are definitely correct. Graphing it implicitly doesn't work because when the square root of a negative number is taken, most graphing applications return "not a number".
I rewrote the equation as two separate piecewise functions, and I was able to graph it using my graphing calculator application on the iPhone.
y1(x)=(.75+3*abs(x))*(abs(x)≤.75)*(abs(x)≥.5)+(98*abs(x))*(abs(x)≤1)*(abs(x)≥.75)+2.25*(abs(x)≤.5)+((6√(10))(1.5√(4(abs(x)1)^2))/7+(3abs(x))/2)*(abs(x)≥1)*(abs(x)≤3)+3√(1x^2/49)*(abs(x)≥3)
y2(x)=((73√(33))*x^2/112+abs(x/2)3+√(1(abs(abs(x)2)1)^2))*(abs(x)≤4)+(3√(1x^2/49))*(abs(x)≥4)
It looks like I'm not allowed to post images. You can see the result at the following URL: http://iphonecalc.com/batman.png
I rewrote the equation as two separate piecewise functions, and I was able to graph it using my graphing calculator application on the iPhone.
y1(x)=(.75+3*abs(x))*(abs(x)≤.75)*(abs(x)≥.5)+(98*abs(x))*(abs(x)≤1)*(abs(x)≥.75)+2.25*(abs(x)≤.5)+((6√(10))(1.5√(4(abs(x)1)^2))/7+(3abs(x))/2)*(abs(x)≥1)*(abs(x)≤3)+3√(1x^2/49)*(abs(x)≥3)
y2(x)=((73√(33))*x^2/112+abs(x/2)3+√(1(abs(abs(x)2)1)^2))*(abs(x)≤4)+(3√(1x^2/49))*(abs(x)≥4)
It looks like I'm not allowed to post images. You can see the result at the following URL: http://iphonecalc.com/batman.png
Re: The Batman equation
This function is very delightful for drawing such a graph, but write it down in latex is very tedious.
((x7)2⋅x−3(x−3)+(y3)2⋅y+3⋅337y+3⋅337−1)⋅(x2−((3⋅33−7)112)⋅x2−3+1−(x−2−1)2−y)⋅(9⋅(x−1)⋅(x−0.75)((1−x)∗(x−0.75))−8⋅x−y)⋅(3⋅x+0.75⋅(x−0.75)⋅(x−0.5)((0.75−x)⋅(x−0.5))−y)⋅(2.25⋅(x−0.5)⋅(x+0.5)((0.5−x)⋅(0.5+x))−y)⋅(6⋅107+(1.5−0.5⋅x)⋅x−1x−1−(6⋅1014)⋅4−(x−1)2−y)=0
As a product of factors is 0 if and only if any one of them is 0, multiplying these six factors puts the curves together. This graph is no more than the combination of six curves.
All these six curves are very simple, for instance, the first factor is the ellipse (x7)2+(y3)2, in the region where x>3 and y>−333/7 ; the region were restricted by x−3(x−3) and y+3⋅337y+3⋅337−1
Here’s what I got from the equation using ggplot2 (R Language)…
Image: http://www.cloudst.at:8080/tmp/file3ff25618.png
Source Code:
Source: The Batman Equation: http://cloudst.at/index.php?do=/kafechew/blog/thebatmanequation/
((x7)2⋅x−3(x−3)+(y3)2⋅y+3⋅337y+3⋅337−1)⋅(x2−((3⋅33−7)112)⋅x2−3+1−(x−2−1)2−y)⋅(9⋅(x−1)⋅(x−0.75)((1−x)∗(x−0.75))−8⋅x−y)⋅(3⋅x+0.75⋅(x−0.75)⋅(x−0.5)((0.75−x)⋅(x−0.5))−y)⋅(2.25⋅(x−0.5)⋅(x+0.5)((0.5−x)⋅(0.5+x))−y)⋅(6⋅107+(1.5−0.5⋅x)⋅x−1x−1−(6⋅1014)⋅4−(x−1)2−y)=0
As a product of factors is 0 if and only if any one of them is 0, multiplying these six factors puts the curves together. This graph is no more than the combination of six curves.
All these six curves are very simple, for instance, the first factor is the ellipse (x7)2+(y3)2, in the region where x>3 and y>−333/7 ; the region were restricted by x−3(x−3) and y+3⋅337y+3⋅337−1
Here’s what I got from the equation using ggplot2 (R Language)…
Image: http://www.cloudst.at:8080/tmp/file3ff25618.png
Source Code:
Code: Select all
#an image with an equation which apparently draws the Batman logo.
require(ggplot2)
f1 < function(x) {
y1 < 3*sqrt(1(x/7)^2)
y2 < 3*sqrt(1(x/7)^2)
y < c(y1,y2)
d < data.frame(x=x,y=y)
d < d<d$y > 3*sqrt(33)/7,>
return(d)
}
x1 < c(seq(3, 7, 0.001), seq(7, 3, 0.001))
d1 < f1(x1)
p1 < ggplot(d1,aes(x,y)) + geom_point(color="red")
x2 < seq(4,4, 0.001)
y2 < abs(x2/2)(3*sqrt(33)7)*x2^2/1123 + sqrt(1(abs(abs(x2)2)1)^2)
p2 < p1 + geom_point(aes(x=x2,y=y2), color="yellow")
x3 < c(seq(0.75,1,0.001), seq(1,0.75,0.001))
y3 < 98*abs(x3)
p3 < p2+geom_point(aes(x=x3,y=y3), color="green")
x4 < c(seq(0.5,0.75,0.001), seq(0.75,0.5,0.001))
y4 < 3*abs(x4)+0.75
p4 < p3+geom_point(aes(x=x4,y=y4), color="steelblue")
x5 < seq(0.5,0.5,0.001)
y5 < rep(2.25,length(x5))
p5 < p4+geom_point(aes(x=x5,y=y5))
x6 < c(seq(3,1,0.001), seq(1,3,0.001))
y6 < 6 * sqrt(10)/7 +
(1.5  0.5 * abs(x6)) * sqrt(abs(abs(x6)1)/(abs(x6)1)) 
6 * sqrt(10) * sqrt(4(abs(x6)1)^2)/14
p6 < p5+geom_point(aes(x=x6,y=y6), colour="blue")
p < p6+theme_bw()
print(p)
Source: The Batman Equation: http://cloudst.at/index.php?do=/kafechew/blog/thebatmanequation/
Re: The Batman equation
Nice one! I will make sure to use one of these on the upcoming test for my students. Thanks a lot
Jacob
Jacob
 Proginoskes
 Posts: 313
 Joined: Mon Nov 14, 2011 7:07 am UTC
 Location: Sitting Down
Re: The Batman equation
jacobs wrote:Nice one! I will make sure to use one of these on the upcoming test for my students. Thanks a lot
Jacob
This finally made it to school.failblog.org, where someone posted:
For extra credit, have them convert it to polar coordinates.
Spoiler:
Re: The Batman equation
Proginoskes wrote:Spoiler:
That's an odd way of spelling three.
my pronouns are they
Magnanimous wrote:(fuck the macrons)
 Proginoskes
 Posts: 313
 Joined: Mon Nov 14, 2011 7:07 am UTC
 Location: Sitting Down
Re: The Batman equation
eSOANEM wrote:Proginoskes wrote:Spoiler:
That's an odd way of spelling three.
Yes, but there are two for some values of theta as well.
Nyah!
Re: The Batman equation
There are some places around 40 degrees where there are two distinct values, but I think that, given the behaviour of the graph nearby it would be better to view it as a single repeated value (like the root of x^2=0)
my pronouns are they
Magnanimous wrote:(fuck the macrons)
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