

Iterated analytic function problem
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 realvalued 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 fixedpoint 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 fixedpoint 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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Update: June 2012

