Functional super-iteration and hierarchy of functional hyper-iterations
#7
(05/02/2009, 08:32 AM)andydude Wrote: So do you mean the following?

\( {}^{\circ 2}f(x) = f^{f(x)}(x) \)
\( {}^{\circ 3}f(x) = f^{f^{f(x)}(x)}(x) \)
\( {}^{\circ 4}f(x) = f^{f^{f^{f(x)}(x)}(x)}(x) \)

Yes. EDIT 5/2/09: actually f^^3 (x) == f^[f^[f]](x), not [[f]^f]^f(x). Along the same lines as "functional root" \( \sqrt[n]{f}(x), \) (a function which, iterated n times, gives f(x)), the "functional logarithm" can be defined so that \( \operatorname{flog}_f (f^n(x)) = n \) for all n. So we should be able to define \( {}^{\circ0} f(x) = \operatorname{flog}_f [{}^{\circ 1} (f^1)] (x) = 1. \) <EDIT: of course I shouldn't call this logarithm, but "iteration number", perhaps?> There's not much else that I think we can do, however, because functions don't behave quite like numbers. A function might have no fixed points, and superiterations of a function may not even be continuous. I just wanna know what areas of mathematics do they use when doing real-valued iteraitons?
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Messages In This Thread
RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/02/2009, 06:14 PM

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