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Name: existing
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995


Question:
If you consider a point in a space to represent a real-valued function can one generate an operator which takes a point in the space to another point such that the orbit generated when the operator is iterated covers a ball of points. I am trying to generalize the concept of orbits from a simple case where the points are elements of the reals to the case where the points are elements of a space of functions. So in the simple case where the point is an element of the reals, say x, I want to have:

f(x) f, f(f(x)), f(f(f(x))), . . .

to form in the limit a continuous interval if more than a countable number of iterations are taken.


Replies:
This sounds like iterated analytic function theory, and I, too, would appreciate a good reference.

hawley


Maybe I am not completely understanding your question, but it sounds like you are asking about, e.g., operators in a Hilbert Space.

I am a little puzzled by some of what you describe, as with the use of the terms "ball of points" and "orbit." Given a point in a space, we do speak of a neighborhood (often called a ball) of that point; this implies that a "distance" measure, or metric, has been defined on the set (i.e., that we are talking about a metric space). And indeed, metrics may be specified on spaces of functions. The term "orbit" is generally used to refer to a graph of y'(x) versus y(x) for an ordinary differential equation. Perhaps you are thinking of a graph of f^n(x) versus n (where "f^n" means n iterations of f) for a given x? For a specified x, the sequence {f(x),f(f(x)), . . .} will not give a continuous interval. It may happen that this sequence converges, say, to a number A, in which case any interval around A will contain an infinite number of members of this sequence.

Yes, an operator maps a function to a function; the operator you describe is very specialized, in that its "domain" is f, f(f), f(f(f)), etc. Thus, a given function f completely specifies both the operator and its domain. I am not sure what one could do with such an operator.

I concur with the previous responses, a lot of what you describe falls under the category of fixed-point iteration. We can define a sequence of functions {f_i(x), i=1,2, . . .} by: f_1(x)=f(x), f_[n+1](x)=f(f_n(x)). The form in which fixed-point iteration is usually written is: x_[n+1] = f(x_n), with x_0 chosen by the user. This would be equivalent to your iteration (via the above sequence definition) x_n = f_n(x_0).

Your question touches on many subjects. Keep wondering!

rcwinther


Hey you guys, I do not think you read the question properly! "existing" really does come up with some profound ones!

Anyway, let us consider the iteration of a real function first. What kinds of orbits do we normally find? If there are fixed points, then of course those orbits have only one point in them. Neighbors may, upon iteration, be attracted to the fixed points. Similarly, orbits of size 2 are fixed points of the function f(f(x)), and orbits of neighbors may be attracted to these size 2 fixed points. The standard mapping of chaos theory exhibits orbits which double in length until a transition is reached, at which point the stable orbits (which others are attracted to) are in fact of infinite length, and cover the interval densely. Of course in a count- able number of iterations you cannot get an uncountable number of points, and I am really not sure what it would mean to apply an uncountable number of iterations - iterations by there very nature seem to be countable. Does anybody else have a better idea?

Actually, I am sure you could get whatever you wanted by allowing your function to be discontinuous, but maybe that is cheating. Anyway, to restate "existing's" real question, is there something equivalent to at least densely filling a region of function space by applying an operator that takes functions to other functions, presumably within some bounded region of function space at least? There ought to be! Maybe somebody has an example?

asmith



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