4 hypothesis about iterated functions

[Shanghai46]

I'm working on my iteration formula, and I made a few hypothesis based on intuition and tests. These seem really useful in order to find the final interval restrictions of it. 

If you have an answer "it's true", could you link to a demonstration, or to a property that demonstrates that, and if it's false, could you provide a counter example, as well as, if possible, put some restrictions to make it work, then giving the demonstration? Thank you.

ONE : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations of \(f(x_0)\) are inside the interval \(I\).
is \(\forall n\in\mathbb{N}, |f^{\circ n+1}(x_0)-\tau| < |f^{\circ n}(x_0)-\tau|\) true? (kinda answered by Tommy but without evidence..... So......)

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 \(I\)?

THREE : 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 THE BIGGEST monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\). 
are ALL positive iterations of \(f(x_0)\) inside \(I\)?

FOUR : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations are inside the interval \(I\).
is \(\forall r\in\mathbb{R}, arg(f^{\circ r}(x_0)-\tau)=arg(f'(\tau)^r) \) true? (\(arg\) is the argument in the complex plane)


Here they are. Thank you.

[JmsNxn]

I'm working on my iteration formula, and I made a few hypothesis based on intuition and tests. These seem really useful in order forme to find the final interval restrictions of it. 

If you have an answer "it's true", could you link to a demonstration, or to a property that demonstrates that, and if it's false, could you provide a counter example, as well as, if possible, put some restrictions to make it work, then giving the demonstration? Thank you.

ONE : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations of \(f(x_0)\) are inside the interval \(I\).
is \(\forall n\in\mathbb{N}, |f^{\circ n+1}(x_0)-\tau| < |f^{\circ n}(x_0)-\tau|\) true? (kinda answered by Tommy but without evidence..... So......)

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 \(I\)?

THREE : 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 THE BIGGEST monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\). 
are ALL positive iterations of \(f(x_0)\) inside \(I\)?

FOUR : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations are inside the interval \(I\).
is \(\forall r\in\mathbb{R}, arg(f^{\circ r}(x_0)-\tau)=arg(f'(\tau)^r) \) true? (\(arg\) is the argument in the complex plane)


Here they are. Thank you.

LMAO!

Yes this is true! See my post on your other thread!

GOOD JOB IDENTIFYING THIS!

This is a deep result, which appears more often in abel function analysis!

GOOD JOB!

You've rediscovered something that is unbelievably hard to prove!

[Shanghai46]

LMAO!

Yes this is true! See my post on your other thread!

GOOD JOB IDENTIFYING THIS!

This is a deep result, which appears more often in abel function analysis!

GOOD JOB!

You've rediscovered something that is unbelievably hard to prove!

 

so they all are correct, but hard to prove? Technically I don't need to demonstrate them if they already are in a theorem, which I'd need the name. But yeah if not I'll need to demonstrate them.

[Shanghai46]

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?

[Shanghai46]

ONE : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations of \(f(x_0)\) are inside the interval \(I\).
is \(\forall n\in\mathbb{N}, |f^{\circ n+1}(x_0)-\tau| < |f^{\circ n}(x_0)-\tau|\) true? (kinda answered by Tommy but without evidence..... So......)

Also wrong, I found a counter example

[tommy1729]

ONE : let's assume there's a function \(f\) that has an attractive fixed point \(\tau\). Let's consider the real interval \(I\) which is the biggest monotomic interval so that \(\forall x_0\in I\), the infinite iteration of \(f(x_0)\) converges towards \(\tau\), and that all positive iterations of \(f(x_0)\) are inside the interval \(I\).
is \(\forall n\in\mathbb{N}, |f^{\circ n+1}(x_0)-\tau| < |f^{\circ n}(x_0)-\tau|\) true? (kinda answered by Tommy but without evidence..... So......)

Also wrong, I found a counter example

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

[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

[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 \(0<f'(\tau)<1\).

[tommy1729]

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

[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.