4 hypothesis about iterated functions

[Shanghai46]

Wrong ?
Here you left out some conditions compared to the original I think.
Anyway remember there are no cyclic points by definition since all converge to the fixpoint.

Notice that if the function has at its fixpoint the derivative -1 or so then more likely there are cyclic points.

Im curious for your counterexample.


regards

tommy1729

I didn't forget any restrictions here. Let's take the function \(f(x)=-0.5x^2+1\). \(\tau=\sqrt3-1\). If \(x_0=1.4\), then all iterations of it belongs to the interval \([0,\sqrt2]\), which is the biggest monotomic interval so that for all \(x\) inside, it converges and all iterations of it stays inside this interval.
So with \(x_0=1.4\), all conditions are met, since it converges and always stays in the monotonic interval. (unlike for \(x_0=2\) because its first iteration is negative, exiting the monotonic part of the function that contains \(\tau\), yet it converges.) 
\(|f(1.4)-\tau|=0.71 > |x-\tau| = 0.67\).
The property doesn't work. Tho I managed to demonstrate it if and only if \(0<f'(\tau)<1\).

- sqrt 2 and sqrt 2 are the zero's of f(x).
And in that interval you have f ' (y) = 0 AND that is a local maximum so the function is not strictly monotone

I took the interval \([0,\sqrt2]\), NOT \([-\sqrt2,\sqrt2]\). In \([0,\sqrt2]\) f is monotonic. For all \(x\) inside, it converges towards \(\tau\) (\(\sqrt3-1\)). All positive iterations of \(f\) with the starting number being inside that interval is also inside that interval. I broke no conditions.

Oh like that.

I misunderstood you.

Congrats.

But then it is simply because f(x)  > x for say 1.4


regards

tommy1729

Nah no problem.