Interesting value for W, h involving phi,Omega?

[Ivars]

To evaluate previous conjecture analytically, we need few more tools:

\( \lim_{n\to\infty}{\frac{\theta n}{n} = 1 \)

Let us put \( \theta(0) = 0 \)

\( \theta (1)= 0+\Omega-0=\Omega \)

\( \theta(2)= \theta(1)+\Omega-{\frac{1}{4*\Omega}}*sin(2\pi\theta(1)) = \Omega+\Omega+\frac{1}{4*\Omega}*sin(2\pi\Omega) \)

To evaluate \( sin(2\pi\Omega) \) we may use formula derived earlier:

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))),

so sin(2pi*Omega) = (-I/2)*(Omega^(2pi*Omega/(I*Omega))-Omega^(-2pi*Omega/(I*Omega))) = (-I/2)*((Omega^(-I*2pi)-(Omega^(+I*2pi))


But \( \Omega^{-I*2\pi}= -0.9123233..-I*0.4094706... \)

module \( \Omega^{-I*2\pi} =1 \)
Arg \( \Omega^{-I*2\pi} =0.41287 rad = 24.17158.. degrees \)
\( \Omega^{I*2\pi} = -0.9123233..+I*0.4094706... \)

module \( \Omega^{I*2\pi} =1 \)

So \( sin(2\pi\Omega)= {\frac{-I}{2}}*-I*0.8189142..= - 0.4094706. \)

\( \frac{1}{4*\Omega}*sin(2\pi\Omega)= -0.180497 \)

\( \theta(2) = \Omega+\Omega+0.180497= 1.31478.. \)

Which coincides with numerical calculation.
Ivars