Batman logo in equation form

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31 Responses to “Batman logo in equation form”

  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. Wonderful batman, is that a special formular or what?

  3. nffcnnr says:

    If this guy’s so smart, why didn’t he think of putting it on a larger piece of graph paper?

  4. Andrew Singleton says:

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

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

  5. unwesen says:

    That must be the ultimate nerdgasm: comics and mathematics combined.

  6. awjt says:

    I = b(a*t)+ma^n

  7. anne speck says:

    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.

    • john aguirre says:

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

  8. eaddict says:

    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.

  9. Ambiguity says:

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

  10. A school in Bogota Colombia did a whole project similar to this earlier this year using the educational math software Cabri and Derive.

    Memorable works include the Metallica logo, Adidas logo, Thundercats logo

    You can view the works in:

    http://proyectosilueta.blogspot.com/p/bocetos.html

  11. Sorry this is a better page to view all the works in the same page:

    http://proyectosilueta.blogspot.com/p/la-propuesta.html

  12. Jacob Ewing says:

    Ok, that beats my mammary math by a long shot.

  13. anne speck says:

    @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. 

  14. Joel Phillips says:

    sqrt{frac{||x| – 1|}{|x|-1}} is cheating.  

    • Haakon IV says:

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

  15. omems says:

    How long until DC issues a DMCA takedown notification?

  16. What is the reason for the paper being labeled in units of 1.4?  How is that generally more useful than integers?
     

    • sarah michel says:

      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.

  17. Chris Brown says:

    How long before someone posts a link to wolfram alpha?

    • hughstimson says:

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

  18. kumar mayank says:

    can someone type the text so that I can plot on Wolfram Alpha and confirm?

  19. hypnosifl says:

    I don’t think Wolfram Alpha’s equation plotter allows you to plot implicit functions, you’d have to solve for y if you wanted to use it. There’s an applet for implicit functions here, haven’t tried it though.

    • hypnosifl says:

      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

  20. Paul Munro says:

    It looks like the Grapher application, in the Applications/Utilities folder of EVERY Macintosh running OS X…

    • hypnosifl says:

      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…

  21. Hee S. Lee says:

    can somebody come up with the equations for wu-tang logo?

  22. 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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