Here is a good discussion of the batman logo over on Math Stack
Exchange:
http://math.stackexchange.com/questions/54506/is-this-batman-equation-for-real
Dana
On Jul 29, 10:06 pm, Matt Rissler <discn...@gmail.com> wrote:
> var('x y')
> f1(x,y)=((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)
> f2(x,y)=(abs(x/2)-((3*sqrt(33)-7)/112)*x^2-3+sqrt(1-
> (abs(abs(x)-2)-1)^2)- y)
> f3(x,y)=(9*sqrt(abs((abs(x)-1)*(abs(x)-3/4))/((1-
> abs(x))*(abs(x)-3/4)))-8*abs(x)-y)
> f4(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)
> f5(x,y)=(9/4*sqrt(abs((x-1/2)*(x+1/2))/((1/2-x)*(1/2+x)))-y)
> f6(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)
>
> p1=implicit_plot(f1==0,(-8,8),(-3,3),plot_points=200)
> p2=implicit_plot(f2==0,(-8,8),(-3,3),plot_points=200)
> p3=implicit_plot(f3==0,(-8,8),(-3,3),plot_points=200)
> p4=implicit_plot(f4==0,(-8,8),(-3,3),plot_points=200)
> p5=implicit_plot(f5==0,(-8,8),(-3,3),plot_points=200)
> p6=implicit_plot(f6==0,(-8,8),(-3,3),plot_points=200)
> show(p1+p2+p3+p4+p5+p6)
>
> So the above works. I did a contour_plot for each factor and the
> intersection of the domains for all of the factors looks like the
> empty set, so I'm guessing the original plot is a little fishy
> (batty?).
>
> Matt
>
> On Jul 29, 3:26 pm, Rob Beezer <goo...@beezer.cotse.net> wrote:
>
>
>
>
>
>
>
> > Cute. ;-)
>
> > What happens if you try building it up one factor at a time?
>
> > On Jul 29, 11:36 am, "D.C. Ernst" <ernst.tr...@gmail.com> wrote:
>
> > > A student of mine just sent me the following batman logo:
>
> > >http://i.imgur.com/CNy9J.jpg
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