Arithmetical Dynamics: Part 6

Arithmetical Dynamics: Part 6

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

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

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...
Arithmetical Dynamics: Part 4

Arithmetical Dynamics: Part 4

\( P^m(z) = z \ and \ P^N(z)=z \ where \ m|N \Rightarrow (P^m(z) – z) | (P^N(z)-z) \) 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

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

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 \( z \in C_j \)...