Limit set

Problem:

Solution:

Consider $ x'^2 + y'^2 = (x - y - x\sqrt{x^2 + y^2})^2 + (x + y - y\sqrt{x^2 + y^2})^2 $

This can be expanded as follow:

$ x'^2 + y'^2 = (x - y)^2 - 2x(x - y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + (x + y)^2 - 2y(x + y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x(x-y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2y(x+y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x^2\sqrt{x^2 + y^2} + 2xy\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2xy\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

The whole point of the expansion is actually to simplify, let's group like terms and cancel first:

$ x'^2 + y'^2 = x^2 + y^2 - 2x^2\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2  +y^2 - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 + y^2 + x^2  + y^2  + x^2(x^2 + y^2) + y^2(x^2 + y^2) - 2x^2\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} $

And then factorize:

$ x'^2 + y'^2 = (x^2 + y^2)(2 + x^2 + y^2) - 2(x^2 + y^2)\sqrt{x^2 + y^2} $

Now we realize it can all be written in terms of the norm of the vectors, so we will do it.

$ a'^2 = a^2(2 + a^2) - 2a^3 $

If it were to converge, the after update norm should be the same as before update norm, so we solve 

$ a^2 = a^2(2 + a^2) - 2a^3 $

$ 1 = 2+a^2 - 2a $

Therefore we get $ a = 1 $, in other words, if the system converge at all, it must converge on the unit circle!