Select Page Consider fix point of $$R(z) = z^2 – z$$ . Which is the solution of $$R(z) = z \\ \Rightarrow z^2 – z =z \\ \Rightarrow z^2 – 2z =0 \\ \Rightarrow z(z -2) =0$$ Now , consider the fix point of $$R^2(z)$$ . $$\\$$ $$i.e. R^2(z) = R . R(z) =... ## Arithmetical Dynamics: Part 5 And suppose that R has no periodic points of period n . Then (d, n) is one of the pairs $$(2,2) ,(2,3) ,(3,2) ,(4 ,2)$$ , each such pair does arise from some R in this way . The example of such pair is$$ 1. R(z) = z +\frac {(w-1)(z^2 -1)}{2z} ; it \ has \ no \... ## Arithmetical Dynamics: Part 0

Rational function $$R(z)= \frac {P(z)}{Q(z)}$$ ; where P and Q are polynimials . There are some theory about fixed points . Theorem: Let $$\rho$$ be the fixed point of the maps R and g be the Mobius map . Then $$gRg^{-1}$$ has the same number of fixed points at... $$P^m(z) = z \ and \ P^N(z)=z \ where \ m|N \Rightarrow (P^m(z) – z) | (P^N(z)-z)$$ The proof of the theorem in Part 0 : Let , P be the polynomials satisfying the hypothesis of theorem 6.2.1 . Let , $$K = \{ z \in C | P^N(z) =z \} \\$$ and let $$M =\{ m \in... ## Arithmetical Dynamics: Part 3 Theory: Let \( \{ \zeta_1 , ……., \zeta_m \}$$ be a ratinally indifferent cycle for R and let the multiplier of $$R^m$$ at each point of the cycle be $$exp \frac {2 \pi i r}{q}$$ where $$(r,q) =1$$ . Then $$\exists \ k \in Z$$ and$$mkq$$ distinct... ## Arithmetical Dynamics: Part 2

The lower bound calculation is easy . But for the upper bound , observe that each $$z \in K$$ lies in some cycle of length m(z) and we these cycles by $$C_1 , C_2 …..,C_q$$ . Further , we denote the length of the cycle by $$m_j$$ , so , if $$z \in C_j$$...