Batman logo in equation form

A Redditor called "i_luv_ur_mom" posted this math teacher's amusement, an equation that draws a lovely Bat-signal.

Do you like Batman? Do you like math? My math teacher is REALLY cool


  1. Brilliant!  That’s the kind of teacher who captures the students’ imaginations and makes a lasting impression on their lives.  I wish there were more teachers like him/her out there.

  2. Gonna have to agree. We need more teachers like this.

    Query: other than drawing the bat symbol… is it good for anything?

  3. It’s a step-wise polar equation. It probably doesn’t do anything other than make a space in minds which have been learning in the graphs=cartesian coordinates world for other kinds of plotting systems do exist. And why is it important to put it on larger graph paper? Take the formula and plug it into Geogebra and print it out as big as you want.

    1. That is a stepwise equation, but it is definitely not polar, it’s Cartesian.  Note that the variables are (x,y), not (r,theta).

  4. My daughter ‘s math class did that last year.  In fact, she did the Batman logo, others did the St Louis Cardinals letter logo, while others did various product logos.   Nice to see other teachers going the creative route.

  5. But the function isn’t analytic! I mean, look at those derivatives. Just look at ’em!

  6. @google-6e7df99bb682d2a2604859f86ea75e01:disqus D’oh! You’re right of course — polar would have made for a much cleaner equation. Apologies for my pre-coffee posting. 

    1. I have discovered a remarkable analytic function representing the bat signal, which this comment is unfortunately to small to contain.

    1. I think it’s probably in units of the square root of two, rather than units of 1.4, which would certainly be somewhat more significant.

    1. Are you suggesting Wolfram Alpha could be used to plot that beast? Awesome! And I nominate you sir!

    1. So for that implicit function applet I linked to, here’s the first of the six big expressions in parentheses, which gives the outer edge of the logo’s wings (set the range on the graph to xmin=-8, xmax=8, and also for ymin and ymax if you don’t want the shape distorted):

      ( ( ((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 ) = 0

    1. Grapher seems to have a kind of messed up way of handling expressions unless you type in a bunch of parentheses…for example, here are two sample inputs (with the first sqrt[…] being a term from the Batman equation) which give different responses:

      sqrt[abs[abs[x] – 3]/(abs[x] – 3)] + xy = 0 (gives “no valid operation found”)

      sqrt[(abs[abs[x] – 3])/(abs[x] – 3)] + xy = 0 (gives a working graph…only difference is parentheses around “abs[abs[x] – 3]”, shouldn’t the brackets be enough?)Anyway, that gripe aside, I tried taking a version of the equation posted by someone on the reddit thread and fixing it up a little to work in Grapher, Grapher was able to graph the 6 different parts of the equation individually (though with some it gave weird dotted lines at the end):(((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) = 0

      (abs[x/2] – ((3*sqrt[33] – 7)/112)*x^2 – 3 + sqrt[1 – (abs[abs[x] – 2] – 1)^2] – y) =0

      (9*sqrt[(abs[(abs[x] – 1)*(abs[x] – 3/4)])/((1 – abs[x])*(abs[x] – 3/4))] – 8*abs[x] – y) = 0

      (3*abs[x] + (3/4)*sqrt[(abs[(abs[x] – 3/4) (abs[x] – 1/2)])/((3/4 – abs[x])*(abs[x] – 1/2))] – y) = 0

      ((9/4)*sqrt[(abs[(x – 1/2)*(x + 1/2)])/((1/2 – x)*(1/2 + x))] – y) = 0

      ((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

      But when I tried multiplying them together into one giant equation and setting it all equal to zero, Grapher didn’t give an error but it also didn’t show anything on the graph. Maybe it’s too much for it to handle…

  7. Guys, the solution to the equation is complex (has imaginary part). The batman figure is one of the real solutions, that is what usually software will plot. The trick is done, for example, by the factor sqrt(abs( abs(x)-3 ) /abs(x)-3 ), that will be imaginary for x3, so the grapher will not be able to plot in the interval x<3, but the solution is still there!

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