(04/13/2015, 09:57 PM)sheldonison Wrote: But anyway, base b=0.1, find the fixed point, and find the multiplier \( \lambda \) at the fixed point, and from that the periodicity\( =\frac{2\pi i}{\ln(\lambda)} \); and that's a pretty darn good start, assuming you ever get that far .... The multiplier \( \lambda \) is defined where \( b^L=L \) and \( b^{L+\delta} \approx L + \lambda\delta \)
So then there is a formal Koenig solution that has \( S(\lambda z) = b^{S(z)}\;\;\;\exp_b^{\circ n} = S(\lambda^n) \). From that, you should be able to generate graphs, or a Taylor series, or whatever you like.
I had being reading about the Koenigs function, and I have not a clear understanding of it.
My function is \( \\[15pt]
{y(x)\,=\,^xb\,=\,^x0.01} \)
Do you mean finding a value \( \\[15pt]
{x_0} \) such that \( \\[15pt]
{0.01^{x_0}=\,x_0} \)? (x₀ would be the fixed point of \( \\[15pt]
{b^x} \))
Now, I need to find a function h(x), such that \( \\[20pt]
{h(0.01^x)=\lambda . h(x)} \), where \( \\[20pt]
{\lambda=\frac{\mathrm{d} (0.01^x)}{\mathrm{d} x}|_{x=x_0}} \)
I read that λ is the derivative of \( \\[15pt]
{\frac{\mathrm{d} (b^x)}{\mathrm{d} x}|} \)\( \\[5pt]
{_{x={\color{Red} 0}}} \), but you mean \( \\[15pt]
{\frac{\mathrm{d} (b^x)}{\mathrm{d} x}|} \)\( \\[0pt]
{_{x={\color{Red} x_0}}} \) (that makes more sense to me).
If I find h(x), then b˟ would be an eigenvector of h(x), and λ the eigenvalue of b˟.
Do \( \\[15pt]
{S(z)=h^{-1}(z)} \) ??
Is correct that \( \\[15pt]
{^nb\,=\,h^{-1}(\lambda^n)} \) ??
Edit: Can't be correct. I can't make it work.
