Ocio: La ecuación de Batman
Función matemática que al graficarla nos revela el símbolo de este superheroe:
((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
Visto en: http://www.reddit.com/r/pics/comments/j2qjc/do_you_like_batman_do_you_like_math_my_math/
((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
Visto en: http://www.reddit.com/r/pics/comments/j2qjc/do_you_like_batman_do_you_like_math_my_math/
