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 Functional super-iteration and hierarchy of functional hyper-iterations Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/02/2009, 02:23 AM (This post was last modified: 05/12/2009, 02:12 AM by Base-Acid Tetration.) We all know the functional "exponentiation" or "power" for functional iteration, $f^{\circ n} (x).$ (I use F^circle n, f^n (x), or f^circle n (x) if clarity is needed, for iteration) That means we can do this for some function: $f^{g(x)}(x),$ where the function f is iterated g(x) times at x. We could investigate the properties and results of functional "tetration", or super-iteration. The "upper-left exponent" notation can be abuse, or use the circle notation like ${}^{\circ n} f(x)$ Initial Definition: (might have gotten my variables wrong) Given a $f : S \rightarrow S$ and a $n \in \mathbb{N},$ a functional super-iteration is defined by the recurrence relation: $(\forall n \, \forall x) \mbox{ so that the result of the iteration is defined, } {}^{\circ n+1} f(x) =[ f^ {{}^{\circ{n}} f(x)}(x) ] ,$ ${}^{\circ 1} f(x) = f(x).$ Already a hierarchy of operations on functions that is much like the arithmetic hyper-operations is evident: Function composition: $g \circ f (x) = g(f(x)), \, g(x) = c \rightarrow g(f(x)) = c.$ Function iteration: $f^{n+1}(x) = f (f^n (x)), \, f^0(x) = x.$ A function can be iterated a constant times, or by g(x) times. Functional superiteration: ${}^{\circ n+1} f(x) = \left [ {}^{\circ n} f^{\circ f(x)} \right ] (x)}, {}^{\circ 1} f(x) = f(x).$ bo198214 Administrator Posts: 1,402 Threads: 91 Joined: Aug 2007 05/02/2009, 07:48 AM (This post was last modified: 05/02/2009, 07:48 AM by bo198214.) Yes, some years ago I thought about a similar hierarchy, though in a different context of converting exponential binary trees into analytic functions. To give an impression of this operator I compute the first three functions starting with $f(x)=x+1$. To make the notation more extensible I use $I$ for the operator, that maps $f$ to $x\mapsto f^{\circ f(x)}(x)$. $I^0 f = f = x \mapsto x+1$, $(I^0f)^{\circ n}(x) = x + n$ $I^1 f = x \mapsto 2x+1$, $(I^1f)^{\circ n}(x) = 2^n x + 2^n - 1$ $I^2 f = x\mapsto 2^x(x+1) - 1$, $(I^2 f)^{\circ n} = ?$ And of course next we would put $f(x)$ into the exponent of the operator and so on, getting nice hierarchy of just too quickly growing functions. This hierarchy can already be built on positive integer functions, so no analytic something needed. I guess the growth of $I^n (x\mapsto x+1)$ is equal to the growth of $x\mapsto b [n] x$. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2009, 08:32 AM So do you mean the following? ${}^{\circ 2}f(x) = f^{f(x)}(x)$ ${}^{\circ 3}f(x) = f^{f^{f(x)}(x)}(x)$ ${}^{\circ 4}f(x) = f^{f^{f^{f(x)}(x)}(x)}(x)$ I know I've seen this somewhere... I want to say this paper but I can't find it. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2009, 09:23 AM Interesting, if $f(x) = f_1x + f_2x^2 + \cdots$ and $g(x) = g_1x + g_2x^2 + \cdots$ both have a (non-parabolic) fixed point at 0, then their supercomposition would have a parabolic fixed point. $f^{\circ g(x)}(x) = x + x^2 g_1 \ln(f_1) + x^3\ln(f_1)\left(\frac{f_2g_1}{f_1(f_1-1)} + \frac{g_1^2}{2}\ln(f_1) + g_2\right) + \cdots$ also, one could find the infinite superiterate by looking at the first n coefficients of ${}^{\circ n}f(x)$: ${}^{\circ(\infty)}f(x) = x + \ln(f_1)x^2 + \left(\frac{3}{2}\ln(f_1)^2 + \frac{f_2\ln(f_1)}{f_1(f_1-1)}\right)x^3 + \cdots$ tommy1729 Ultimate Fellow Posts: 1,507 Threads: 358 Joined: Feb 2009 05/02/2009, 12:22 PM (05/02/2009, 09:23 AM)andydude Wrote: Interesting, if $f(x) = f_1x + f_2x^2 + \cdots$ and $g(x) = g_1x + g_2x^2 + \cdots$ both have a (non-parabolic) fixed point at 0, then their supercomposition would have a parabolic fixed point. $f^{\circ g(x)}(x) = x + x^2 g_1 \ln(f_1) + x^3\ln(f_1)\left(\frac{f_2g_1}{f_1(f_1-1)} + \frac{g_1^2}{2}\ln(f_1) + g_2\right) + \cdots$ also, one could find the infinite superiterate by looking at the first n coefficients of ${}^{\circ n}f(x)$: ${}^{\circ(\infty)}f(x) = x + \ln(f_1)x^2 + \left(\frac{3}{2}\ln(f_1)^2 + \frac{f_2\ln(f_1)}{f_1(f_1-1)}\right)x^3 + \cdots$ to be honest , i dont have a clue what you are talking about , how the **** did you arrive at those coefficients ? im quite skeptical about " super-iteration " ... bo198214 Administrator Posts: 1,402 Threads: 91 Joined: Aug 2007 05/02/2009, 01:14 PM (05/02/2009, 12:22 PM)tommy1729 Wrote: im quite skeptical about " super-iteration " ... Ahm, skeptical about what? That it is useful, that it is possible, that it is correct? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/02/2009, 06:14 PM (This post was last modified: 05/03/2009, 02:34 AM by Base-Acid Tetration.) (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 ${}^{\circ0} f(x) = \operatorname{flog}_f [{}^{\circ 1} (f^1)] (x) = 1.$ 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? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2009, 06:38 PM (This post was last modified: 05/02/2009, 06:38 PM by andydude.) (05/02/2009, 12:22 PM)tommy1729 Wrote: how the **** did you arrive at those coefficients ? Regular iteration. Whenever a function $f(x)$ has a fixed point at zero, then its iterates can be described by polynomials or geometric series. For example, if $f(x) = f_1x + f_2x^2 + f_3x^3 + \cdots$ then $f(0) = 0$ is a fixed point of that function. The iterates of $f(x)$ will then be:$f^2(x) = f_1(f_1x + f_2x^2 + \cdots) + f_2(f_1x + f_2x^2 + \cdots)^2 + \cdots = f_1^2 x + (f_1 + f_1^2)f_2x^2 + \cdots$ $f^3(x) = f_1(f_1(f_1x + f_2x^2 + \cdots) + \cdots) + \cdots = f_1^3 x + (f_1^2 + f_1^3 + f_1^4)f_2x^2 + \cdots$ For $f_1 \ne 1$ this can be generalized to $f^t(x) = f_1^tx + \frac{f_1^{t-1}(f_1^{t} - 1)f_2}{f_1-1}x^2 + \cdots$ and substituing $t=g(x)$ in this formula, then expanding the series about x again will give the formulas in my last post. Andrew Robbins bo198214 Administrator Posts: 1,402 Threads: 91 Joined: Aug 2007 05/02/2009, 07:27 PM (This post was last modified: 05/02/2009, 07:30 PM by bo198214.) (05/02/2009, 06:14 PM)Tetratophile Wrote: 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? But before, one should really look at integer functions where your definition is unambigous. I am not especially aware that someone defined/used such a construct. If so I guess it belongs to recursion theory, like the Ackermann function. But as Andrew mentioned: As soon we get a parabolic fixed point, this fixed point stays forever. That means you can do regular iteration there and have even an analytic hierarchy. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/03/2009, 01:08 PM EDIT 5/2/09: For the hierarchy of iterations, or hyper-iterations, we can use the operator: $\operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace,$ where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property $f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x).$ So unless otherwise specified by the use of the down arrow like this, $f {\operatorname{It}\downarrow}_k ,$ super-iteration, like other hyper-iterations, is evaluated from the top down, like ordinary tetration, not from the bottom up as I said originally. I am sorry for any confusion the mistake in the original post may have caused. ANOTHER EDIT 5/2/09: Ictually I am not even sure whether iteration is not associative, but i'll assume it is not unless proven otherwise. « Next Oldest | Next Newest »

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