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)
\( \begin{align*}\epsilon_{n+1}
&=\epsilon_n + A-1 + (1-A)\sqrt{1-\delta^2}\pm\sqrt{B^2-(1-A)^2}\delta\\
&=\epsilon_n - (1-A)(1-\sqrt{1-\delta^2})\pm\sqrt{B^2-(1-A)^2}\delta\end{align*}
\)
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.