By Mathematical Association of America
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Additional info for The American Mathematical Monthly, Number 3, March 2011
Example text
To check (S3) we note first that for j < n we have x j ∈ S ⊂ M, and x j therefore switches sides. Finally, x j+1 = σ (x j ) ∈ O f (x j ) by the definition of σ . 4 give the following main case of the Sharkovsky Theorem. 3. If an m-cycle O with m ≥ 2 contains a point that does not switch sides, then for each l m there is an elementary, O-forced l-loop of O-intervals, and hence an l-cycle. 6. PROOF OF THE SHARKOVSKY FORCING THEOREM. 3. We use the fact that the left and right halves of such a cycle are cycles for f 2 of half the length.
Burns, A simple special case of Sharkovskii’s theorem, Amer. Math. Monthly 107 (2000) 932–933. 2307/2695586 3. L. Block, J. Guckenheimer, M. -S. Young, Periodic points and topological entropy of one-dimensional maps, in Global Theory of Dynamical Systems, Z. Nitecki and C. , Lecture Notes in Mathematics, vol. 819, Springer Verlag, Berlin, 1980, 18–34. 4. U. Burkart, Interval mapping graphs and periodic points of continuous functions, J. Combin. Theory Ser. B 32 (1982) 57–68. 1016/0095-8956(82)90076-4 5.
Of course, cn = 1. Proof. Inversion in the circle of radius r is the map z → r 2 /¯z . This transforms f (z) into f (r 2 /¯z ), which has the same set of roots as g(z) = zn · f c¯0 r2 z¯ = zn + c¯1r 2 n−1 c¯2r 4 n−2 r 2n z + z + ··· + . c¯0 c¯0 c¯0 Since f and g are monic and have the same degree, the inversive stability of f is equivalent to the condition f = g. Equating coefficients and taking conjugates produces (5). When k = n, (5) becomes r 2n = c0 c¯0 = |c0 |2 , or r = |c0 |1/n . Taking moduli in (5) gives the following weaker condition of Lehmer [6]: r k |ck | = r n−k |cn−k |, k = 0, 1, .