(09/12/2009, 08:07 AM)mike3 Wrote: Hmm. However in the emails, you mentioned the use of a "multiplier" that works similar to the derivative at the fixed point but for a cycle.
Generally the multiplier of a cycle \( p_1,\dots,p_n \) (i.e. \( p_{k+1}=f(p_k) \) and \( f(p_n)=p_1 \)) is defined as:
\( f'(p_1)\cdot f'(p_2)\dots f'(p_n) \).
This is equal to the multiplier of \( f^{\circ n} \) at any point of the cycle (by the chain rule).
Example n=2
\( f(f(x))'=f'(f(x))\cdot f'(x) \). If you now plug in \( x=p_1 \) you get \( f'(p_2)\cdot f'(p_1) \) and if you plug in \( x=p_2 \) you get the same result \( f'(p_1)\cdot f'(p_2) \).
If you would depict the tangents at the left and right fixed point in the graph before, they would be parallel.
But this approach to consider \( f^{\circ 2} \) already failed.
