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Ivars · (This post was last modified: 03/07/2008, 11:06 PM by Ivars.)

bo198214 Wrote:
Ivars Wrote: $lim (n->infinity) {\frac{\theta n}{n} = 1$

What do you mean by $\theta n$ ? Btw. $\lim_{n\to\infty}$ has tex code
Code:
\lim_{n\to\infty}
.

Thanks. $\theta n$ is n-th iteration of $\theta$ , starting with $\theta=0$ , when n=0 .

It is the same as $\theta$ in iteration formula (map) , while $\theta'$ is next iteration result =angle $\theta$ n+1, I just did not know how to add indexes in tex.

This limit is called rotation or winding number.

In my simulation, when n=1829, $\theta n$ =1825.78027070570366694723301.... so

$\theta n$ /1829=0.998239623130510479468142710945

***bo198214*** · (This post was last modified: 03/09/2008, 11:47 AM by bo198214.)

Ivars Wrote:Thanks. $\theta n$ is n-th iteration of $\theta$ , starting with $\theta=0$ , when n=0 .

Ok that means $\theta_{n+1}=\theta_n + A + B\sin(2\pi\theta_n)$ where $A=\Omega$ and $B=\frac{1}{4\Omega}$ .

I put some effort into investigating this sequence $\theta_n$ . For the case of $A$ and $B$ that you describe one can see numerically that in the limit $\theta_n = \alpha+n$ for some constant $\alpha$ . By this observation we put up:

$\theta_n = \alpha + n + \epsilon_n$
$\theta_{n+1} = \alpha+n+\epsilon_n + A+B\sin\left(2\pi(\alpha+n+\epsilon_n)\right)$
$\alpha+n+1+\epsilon_{n+1}=\alpha+n+\epsilon_n + A+B\sin\left(2\pi(\alpha+\epsilon_n)\right)$
$1+\epsilon_{n+1}=\epsilon_n + A+B\sin\left(2\pi\alpha+2\pi\epsilon_n\right)$
(1) $\epsilon_{n+1}=\epsilon_n + A-1 + B\sin(2\pi\alpha)\cos(2\pi\epsilon_n)+B\cos(2\pi\alpha)\sin(2\pi\epsilon_n)$

If $\epsilon_n$ converges we can choose $\alpha$ so that $\epsilon_n$ converges to 0. In this case the equation has to be satisfied in the limit too:
$0= A-1 + B\sin(2\pi\alpha)$
So that if $\epsilon_n\to 0$ then
$\alpha_1=\frac{1}{2\pi}\arcsin\frac{1-A}{B}$ , $-\pi<\alpha_0\le\pi$ , where $|1-A|<|B|$ has to be satisfied. Let $\alpha_2=\frac{1}{2}-\alpha_1$ then all possible solutions are
$\alpha_1+N$ and $\alpha_2+N$ for integer $N$ . However only one of them is the actual value in $\alpha+n$ .

For example in our case $\alpha_0=\frac{1}{2\pi}\arcsin(4\Omega(1-\Omega))\approx 0.2197292942$ and $\alpha=0.5-\alpha_0-2\approx -1.7197292942$ .

If vice versa $\alpha$ is given as above, what conditions are to be satisfied that $\epsilon_n\to 0$ ?

We put $\alpha$ into equation (1), first:
$\sin(2\pi\alpha)=\frac{1-A}{B}$ , for simplicity let $0<1-A\le B$ ,
$B\sin(2\pi\alpha)=1-A$
$B\cos(2\pi\alpha)=\pm\sqrt{B^2-\left(B\sin(2\pi\alpha)\right)^2}=\pm\sqrt{B^2-(1-A)^2}$ where we get $+$ for solutions $\alpha=\alpha_1+N$ and $-$ for solutions $\alpha=\alpha_2+N$ .
$\delta:=\sin(2\pi\epsilon_n)$
now into (1)
$\begin{align*}\epsilon_{n+1}<br /> &=\epsilon_n + A-1 + (1-A)\sqrt{1-\delta^2}\pm\sqrt{B^2-(1-A)^2}\delta\\<br /> &=\epsilon_n - (1-A)(1-\sqrt{1-\delta^2})\pm\sqrt{B^2-(1-A)^2}\delta\end{align*}<br />$

The interesting cases is (a) if $\epsilon_n>0$ and $\epsilon_{n+1}<\epsilon_n$ and (b) if $\epsilon_n<0$ and $\epsilon_{n+1}>\epsilon_n$ together with c) $|\epsilon_{n+1}|<|\epsilon_n|$ . We assume in the following that $-\frac{1}{2}<\epsilon_n\le\frac{1}{2}$ and only consider the case $\alpha=\alpha_2+N$ .

If $\epsilon_n>0$ then $\delta>0$ and only positive values are subtracted from $\epsilon_n$ , so $\epsilon_{n+1}<\epsilon_n$ satisfies case (a).

If $\epsilon_n<0$ then $\delta<0$ and the condition $(1-A)(1-\sqrt{1-\delta^2}) \le -\sqrt{B^2-(1-A)^2}\delta$ is equivalent to $\epsilon_{n+1}>\epsilon_n$ .

In this case $-\delta > 0$ and as both side are positive we can quadrate getting equivalently:
$(1-A)^2(1-2\sqrt{1-\delta^2}+1-\delta^2) \le (B^2-(1-A)^2)\delta^2$
$2(1-A)^2(1-\sqrt{1-\delta^2}) \le B^2\delta^2$
$1-\sqrt{1-\delta^2} \le \frac{\delta^2}{2} \left(\frac{B}{1-A}\right)^2$ , let $c:=\frac{1-A}{B}$
$\left(1-\frac{\delta^2}{2}/c^2\right)^2 \le 1-\delta^2$
$1 - \delta^2/c^2 + \frac{\delta^4}{4}/c^4\le 1-\delta^2$
$-1/c^2 +\frac{\delta^2}{4}/c^4\le 1$
$\frac{\delta^2}{4}\le c^4+c^2$
so case (b) is occurs exactly for
$0>\delta\ge -2c\sqrt{1+c^2}$
So in the case $2c\sqrt{1+c^2}>1$ it is this satisified for all $\epsilon_n$ otherwise for
$0>\epsilon_n\ge -\frac{1}{2\pi}\arcsin(2c\sqrt{1+c^2})$ .

As a last step we give conditions for c) $|\epsilon_{n+1}|\le |\epsilon_n|$ .

Let $X=(1-A)(1-\sqrt{1-\delta^2})$ and $Y=\sqrt{B^2-(1-A)^2}|\delta|$ . In the case that $\epsilon_n>0$ we have to assure that the subtractions of $X$ and $Y$ do not lead below $-\epsilon_n$ , i.e. at most $X+Y<2\epsilon_n$ .

In the case $\epsilon_n<0$ , $\delta<0$ and $0>\epsilon_n\ge -\frac{1}{2\pi}\arcsin(2c\sqrt{1+c^2})$ we showed already that $X\le Y$ , so that the subtraction of $X$ and the addition of $Y$ increases $\epsilon_n$ . We shall ascertain that $Y\le 2|\epsilon_n|$ .

The condition $Y<2|\epsilon_n|$ is easy to check:
$\sqrt{B^2-(1-A)^2}|\sin(2\pi\epsilon_n)|<2|\epsilon_n|$
$2\pi\sqrt{B^2-(1-A)^2}<2\frac{2\pi\epsilon_n}{\sin(2\pi\epsilon_n)}$
Now we know that $x>sin(x)$ for $0<x<\pi$ , hence the condition is satisfied if we demand:
$\sqrt{B^2-(1-A)^2}<\frac{1}{\pi}$ .

The condition $X+Y \le 2\epsilon_n>0$ , let $p=1-A$ and $q=\sqrt{B^2-(1-A)^2}$ :
$p(1-\cos(2\pi\epsilon_n))+q\sin(2\pi\epsilon_n)\le 2\epsilon_n$
$2\pi p\frac{1-\cos(2\pi\epsilon_n)}{\sin(2\pi\epsilon_n)} + 2\pi q\le 2\frac{2\pi\epsilon_n}{\sin(2\pi\epsilon_n)}$ .
Similarly to the previous case the condition is satisfied for:
$p\frac{1-\cos(2\pi\epsilon_n)}{\sin(2\pi\epsilon_n)} + q\le \frac{1}{\pi}$ .

We know that $f(x)=\frac{1-\cos(x)}{\sin(x)}$ is striclty increasing for $0<x<\pi$ and $\lim_{x\downarrow 0} f(x)=0$ (graph it!). Hence there is an $\Delta$ such that $f(x)<\frac{\frac{1}{\pi}-q}{p}$ (which is equivalent to $X+Y<2 x$ ) for all $0<x<\Delta$ .

Now going backwards we get

Proposition. Let $0<1-A<B$ , $\sqrt{B^2-(1-A)^2}<\frac{1}{\pi}$ , $c=\frac{1-A}{B}$ , $2c\sqrt{1+c^2}>1$ , $\alpha_2=\frac{1}{2}-\frac{1}{2\pi}\arcsin ( c )$ and let $0<\Delta<\frac{1}{2}$ be the solution of $(1-A)\frac{1-\cos(2\pi\Delta)}{\sin(2\pi\Delta)}+\sqrt{B^2-(1-A)^2}=\frac{1}{\pi}$ .
If there exists an $n$ and integer $N$ such that $-\frac{1}{2}<\theta_n-(\alpha_2+N)<\Delta$ , then $\lim_{n\to\infty} \theta_n - (\alpha_2+N+n)=0$ , particularly $\lim_{n\to\infty} \frac{\theta_n}{n} = 1$ .

Verify the conditions for your choice:
$1-A=1-\Omega=0.4328567096>0$ , $B=\frac{1}{4\Omega}=0.4408057085>1-A$ , $\Delta=0.1583064982$ , $c=0.9819671144$ , $2c\sqrt{1+c^2}=2.752493874>1$ .

The last condition that $-\frac{1}{2}<\theta_n-(\alpha_2+N)<\Delta$ , or even the stronger condition that $-\frac{1}{2}<\theta_n-(\alpha_2+N)\le 0$ is quite probable for any sufficient chaotic choice of $A$ and $B$ .

So for your choice but also many other choices for $A$ and $B$ indeed $\lim_{n\to\infty} \frac{\theta_n}{n} = 1$ .

PS: By the length of the derivations I can not exclude that somewhere got an error into the computations, so dont take the result as granted.

Ivars · (This post was last modified: 03/09/2008, 12:21 PM by Ivars.)

bo198214 Wrote:Ok that means $\theta_{n+1}=\theta_n + A + B\sin(2\pi\theta_n)$ where $A=\Omega$ and $B=\frac{1}{4\Omega}$ .

I put some effort into investigating this sequence $\theta_n$ . For the case of $A$ and $B$ that you describe one can see numerically that in the limit $\theta_n = \alpha+n$ for some constant $\alpha$ .

Thanks,

It will .... take me some time to digest. I do not quite understand the first assumption,

Quote: one can see numerically that in the limit $\theta_n = \alpha+n$ for some constant $\alpha$ .

As far as I did it (1850 terms), nothing is constant-You mean you replaced small difference of sin from -1 with an argument?
But as sin argument nears n*(3pi/2), can You do it?

As I see it since teta(n+1)-teta(n) = 1 when n-> infinity, there shall be no differences between teta(n) and n integer part.

There can not be one also in reals, since limit n->infinity is 1.

So this constant seems suspicious to me-or may be I misunderstood something.

Ivars

***bo198214*** · 03/09/2008, 01:16 PM

Ivars Wrote:
Quote: one can see numerically that in the limit $\theta_n = \alpha+n$ for some constant $\alpha$ .

As far as I did it (1850 terms), nothing is constant-You mean you replaced small difference of sin from -1 with an argument?
But as sin argument nears n*(3pi/2), can You do it?

As I see it since teta(n+1)-teta(n) = 1 when n-> infinity, there shall be no differences between teta(n) and n integer part.

There can not be one also in reals, since limit n->infinity is 1.

So this constant seems suspicious to me-or may be I misunderstood something.

Whats not clear about this?
Just compute $\theta_n - (\alpha + n)$ with $\alpha=\frac{1}{2}-\frac{1}{2\pi}\arcsin\left(\frac{1-A}{B}\right)-2$ . You see numerically that $\theta_n - (\alpha + n)\to 0$ and that means that $\theta_n = \alpha+n$ in the limit.

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