Problem:
Solution:
Consider x′2+y′2=(x−y−x√x2+y2)2+(x+y−y√x2+y2)2
This can be expanded as follow:
x′2+y′2=(x−y)2−2x(x−y)√x2+y2+x2(x2+y2)+(x+y)2−2y(x+y)√x2+y2+y2(x2+y2)
x′2+y′2=x2−2xy+y2−2x(x−y)√x2+y2+x2(x2+y2)+x2+2xy+y2−2y(x+y)√x2+y2+y2(x2+y2)
The whole point of the expansion is actually to simplify, let's group like terms and cancel first:
x′2+y′2=x2+y2−2x2√x2+y2+x2(x2+y2)+x2+y2−2y2√x2+y2+y2(x2+y2)
x′2+y′2=x2+y2+x2+y2+x2(x2+y2)+y2(x2+y2)−2x2√x2+y2−2y2√x2+y2
And then factorize:
x′2+y′2=(x2+y2)(2+x2+y2)−2(x2+y2)√x2+y2
Solution:
Consider x′2+y′2=(x−y−x√x2+y2)2+(x+y−y√x2+y2)2
This can be expanded as follow:
x′2+y′2=(x−y)2−2x(x−y)√x2+y2+x2(x2+y2)+(x+y)2−2y(x+y)√x2+y2+y2(x2+y2)
x′2+y′2=x2−2xy+y2−2x(x−y)√x2+y2+x2(x2+y2)+x2+2xy+y2−2y(x+y)√x2+y2+y2(x2+y2)
x′2+y′2=x2−2xy+y2−2x2√x2+y2+2xy√x2+y2+x2(x2+y2)+x2+2xy+y2−2xy√x2+y2−2y2√x2+y2+y2(x2+y2)
x′2+y′2=x2+y2−2x2√x2+y2+x2(x2+y2)+x2+y2−2y2√x2+y2+y2(x2+y2)
x′2+y′2=x2+y2+x2+y2+x2(x2+y2)+y2(x2+y2)−2x2√x2+y2−2y2√x2+y2
And then factorize:
x′2+y′2=(x2+y2)(2+x2+y2)−2(x2+y2)√x2+y2
Now we realize it can all be written in terms of the norm of the vectors, so we will do it.
a′2=a2(2+a2)−2a3
If it were to converge, the after update norm should be the same as before update norm, so we solve
a2=a2(2+a2)−2a3
1=2+a2−2a
Therefore we get a=1, in other words, if the system converge at all, it must converge on the unit circle!
