4 hypothesis about iterated functions

[tommy1729]

TWO : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\) so that \(f'(\tau)<0\). Let's consider the real interval \(I\) which is a monotomic interval (not the biggest) so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\). 
If \(x_0\) and \(f(x_0)\in I\), are ALL positive iterations of \(f(x_0)\) inside \(I\)?

FALSE. I found a counter example with \(f=-\frac{\tan(x)}{2}\) with the interval \([-\pi/2,\pi/2]\). With \(x_0=1.2\) 
\(x_0=1.2\) in the interval 
\(f(x_0)=-1.29\) in the interval 
\(f(f(x_0))=1.71\) NOT in the interval

I believe there's a link with what I call "pseudo repulsive fixed point". Because it seems that the allowed interval is between approximately \(-1.16\) and \(1.16\) , which exactly are the repulsive fixed points for minus \(f\). My hypothesis is that for a function that has a central symmetry at \(a\) (or a axial symmetry with the line \(x=a\)), the repulsive fixed point on that function are the pseudo repulsive fixed one on the function \(f(a-x)\).
Maybe H2 only works if there isn't any (pseudo) repulsive fixed point between x and tau
Is it right?

huh ?

your interval of values going to the fixpoint is larger than the one you gave !?

And your - tan(x)/2 is not continu everywhere.

And your fixpoint derivative is smaller than 0.

And some values your interval might not converge to the fixpoint.

1.83.. is a fixpoint and your value is 1.71 is close ...
suspicious.



regards

tommy1729