(04/22/2023, 09:22 AM)Shanghai46 Wrote:(04/21/2023, 11:23 PM)tommy1729 Wrote: 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 \(0<f'(\tau)<1\).
your f(x) is symmetric ( f(x) = f(-x) ) since (-x)^2 = x^2 , so your interval is at least between - sqrt 2 and sqrt 2.
- 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 there and you broke 2 conditions.
Also because of that , the inverse function has a derivative being infinity and you get 2 branches.
That is what I said , the derivatives matter.
regards
tommy1729
