Proof:
(a,b) values  = reset |
(xn,yn) values |
When you click on the circle situated on top of the kidney-like black area, and further a number of times on the figure on the right, then you will see that you create the sequence of points (x0,y0), (x1,y1), etc. These points approach 3 points in the plane (3 "attracting" limits).
On the other hand, when you do not click inside the circle, but yust below that circle within the big black area, then the points creep slowly from the first three starting values away, ("repulsive") toward a common limit.
When a point (xn,yn) resides in a vicinity of such a limit (x,y), then the next point (xn+1,yn+1) is even closer to (x,y). (x,y) is "attracting" and not "repulsive".
Now assume that (xn,yn) is close to (x,y).
For convenience we assume (xn,yn) = (x+h,y) where h is a small number.
Then (xn+1,yn+1) = ((x+h)2-y2+a,2(x+h)y+b).
We demand that for small values of h:
((xn+1-x)2 + (yn+1-y)2) < ((xn-x)2 + (yn-y)2)or that (xn+1-x)2 + (yn+1-y)2 < (xn-x)2 + (yn-y)2
or that ((x+h)2-y2+a-x)2 + (2(x+h)y+b-y)2 < h2
or that (x2-y2+a-x+h2+2hx)2 + (2xy+b-y+2hy)2 < h2
Now x = x2-y2+a and y = 2xy+b, so this yields
(h2+2hx)2 + (2hy)2 < h2
Divide by h2, then (h+2x)2 + (2y)2 < 1
As this equation is true for very small values of h, and thus also for the limit with h to 0, this yields:
(2x)2 + (2y)2 ≤ 1 or x2 + y2 ≤ 1/4