Az Egri Ho Si Minh Tanárképző Főiskola Tud. Közleményei. 1984. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 17)

II. TANULMÁNYOK A TERMÉSZETTUDOMÁNYOK KÖRÉBŐL - Szepessy Bálint: Megjegyzések a valós függvények iterálásához III. (A tetszőleges magasrendű ciklusokról)

Remarks 011 iteration oi real functions III. BY BÁLINT SZEPESSY (Summary) Let f(x) be a real valued function defined on the closed interval [a, b]. If f(x) satisfies the conditions (i) f(x) is a continuous function at every inside points of the interval [a, b]

; furthermore f(x) is continuous on the right and on the left at point a and b respectively; (ii) f(x) maps the interval [a, b] onto itself; (iii) there is no subinterval of the interval [a, b] where f(x) is a constant function

; then f(x) is called iterational basic function. For i = 0, 1, 2, ... the function fi(x), defined by f

0(x) = x and fj(x) = f(fi_

x(x)) for i > 0, is called i

t

h iterated function of f(x). We say a real number c is a fix point of f(x) of order one if f(c) = c, furthermore c .is a fix point of order r if f

r(c) = c but f

n(c) ^ c for n = 1, 2, . . ., r — 1. If c is a fix point of f(x) of order r, then the numbers f(e) = c

1

; f(c

t) = c

2, . . f(c

r_

1) = c are also fix points of order r and the fix points, c

1, c

2, ... c give a cycle of order r. In [10] we looken for conditions for function f(x) if f(x) has no fix point of order greater than two. In [9] we studied iterational basic functions for which the orders of the cycles have not upper bound. We have proved: If f(x) is a function satisfying the conditions (i), (ii), (iii) and there are two subinterval of [a, b] without oommon points which are mapped onto the interval [a, b] by f(x), then there are cycles of arbitrary order. In this paper we give some other sufficient conditions for f(x) having fix points of arbitrary order. Among others the following theorem is proved. Let a, b, c and d real numbers with conditions a < c < d < b and let f(x) be an iterational basic function defined on the interval [a, b]. If f(c) = c, f(d) = b and [a, 1>] has a subinterval which is mapped by f(x) onto the interval [a, b], then f(x) has fix points of order n for any natural number n. 843