04/14/2015, 02:36 PM
(This post was last modified: 04/14/2015, 02:38 PM by sheldonison.)
(04/14/2015, 01:43 AM)marraco Wrote: Do \( \\[15pt]Yes. For b=0.01; the primary fixed point, which is repelling is \( L\approx0.277987425;\;\;\; \lambda\approx -1.28017940 \)
{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.
\( {0.01}^{\left( z+0.277987425 \right)} \; ~= \;
L -1.28017940 \cdot z
+2.94772200\cdot z^2
-4.52492050\cdot z^3 ...
\)
\( S(z) = h^{-1}(z)=L + z + a_2 z^2 + a_3 z^3 + ... a_n\cdot z^n \)
There is a formal solution for a2, a3 etc in terms of the Taylor series of the desired function at its fixed point. I get a2~=1.00982628 ...
Normally, we would use \( f(z)=S(\lambda^z) \) for the superfunction, but since lambda is a negative number, perhaps one could use \( f(z)=S\left( (-\lambda)^z \cdot \sin(\pi z)\right) \) to generate the exponentially increasing pseudo 2-periodic solution the Op is looking for.
- Sheldon
