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 Functional super-iteration and hierarchy of functional hyper-iterations bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 03:25 PM (This post was last modified: 05/03/2009, 03:33 PM by bo198214.) (05/03/2009, 01:08 PM)Tetratophile Wrote: $\operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace,$ where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property $f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x).$ It is a bit sloppy and misunderstandable to write $x+1$ or $x$ in the second argument. As one would associate the function $x\mapsto x+1$, but actually you mean the constant function: $f \operatorname{It}_n (x\mapsto m+1) = f \operatorname{It}_{n-1} (f \operatorname{It}_n (x\mapsto m)).$ To make it a definition that makes at least on the natural numbers sense, we need to add the induction start: $f \operatorname{It}_n (x\mapsto 1) = f$ The question then however is how do we know what $\operatorname{It}_n$ for a non-constant function in the second argument is? I think to properly define the hierarchy we need an additional operator $I_n: (\mathbb{R}\to\mathbb{R})\times \mathbb{R}\to(\mathbb{R}\to\mathbb{R})$, $(f,m)\mapsto I_n^m f$ such that $(f \operatorname{It}_n g)(x) = (I_n^{f(x)} g)(x)$. with $I_0^m f = f$ $I_n^1 f =f$ and $I_{n}^{m+1} f = f \operatorname{It}_{n-1} (I_n^m f).$ My brain is flickering, I hope I got everything right. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/03/2009, 07:21 PM (This post was last modified: 05/03/2009, 07:23 PM by Base-Acid Tetration.) (05/03/2009, 03:25 PM)bo198214 Wrote: (05/03/2009, 01:08 PM)Tetratophile Wrote: $\operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace,$ where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property $f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x).$ It is a bit sloppy and misunderstandable to write $x+1$ or $x$ in the second argument. As one would associate the function $x\mapsto x+1$, but actually you mean the constant function: $f \operatorname{It}_n (x\mapsto m+1) = f \operatorname{It}_{n-1} (f \operatorname{It}_n (x\mapsto m)).$ To make it a definition that makes at least on the natural numbers sense, we need to add the induction start: $f \operatorname{It}_n (x\mapsto 1) = f$ The question then however is how do we know what $\operatorname{It}_n$ for a non-constant function in the second argument is? I think to properly define the hierarchy we need an additional operator $I_n: (\mathbb{R}\to\mathbb{R})\times \mathbb{R}\to(\mathbb{R}\to\mathbb{R})$, $(f,m)\mapsto I_n^m f$ such that $(f \operatorname{It}_n g)(x) = (I_n^{f(x)} g)(x)$. with $I_0^m f = f$ $I_n^1 f =f$ and $I_{n}^{m+1} f = f \operatorname{It}_{n-1} (I_n^m f).$ My brain is flickering, I hope I got everything right. For the function x -> x+1, you simply hyper-n-iterate the function f x0+1 times at x=x0, (eg. at x=2 you iterate 3 times, and at x=3.5 you iterate 4.5 times etc.) So I don't think that a non-constant function in the second argument would behave differently from a constant function in this respect. Also $I_0^m f = f$ would not be true, because at n=0, $I^m_0 f = f \operatorname{It}_0 m = f(m), \mbox{ not }f.$ Use of the $I$ operator would make the notation more compact, however bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 08:13 PM (This post was last modified: 05/03/2009, 08:39 PM by bo198214.) (05/03/2009, 07:21 PM)Tetratophile Wrote: For the function x -> x+1, you simply hyper-n-iterate the function f x0+1 times at x=x0, (eg. at x=2 you iterate 3 times, and at x=3.5 you iterate 4.5 times etc.) So I don't think that a non-constant function in the second argument would behave differently from a constant function in this respect. If we take the equations as recursive definitions then $\operatorname{It}$ is only defined for functions in the second argument that are constant. Thatswhy I proposed the different version. Quote:Also $I_0^m f = f$ would not be true, because at n=0, $I^m_0 f = f \operatorname{It}_0 m = f(m), \mbox{ not }f.$ Ok, I also made another mistake in the previous posting, so here again the recursive definition $(f \operatorname{It}_n g)(x) := (I_n^{g(x)} f)(x)$. $(I_0^m f)(x) := f(m)$ $I_n^1 f :=f$ and $I_{n}^{m+1} f := f \operatorname{It}_{n-1} (I_n^m f).$ From that follows $(f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x))$ $I_1^m f = f^{\circ m}$ $(f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x)$ $I_2^m f = {^{\circ m} f}$ $(f \operatorname{It}_2 g)(x) = {^{\circ g(x)} f}(x)$ .... Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/03/2009, 09:45 PM (This post was last modified: 05/03/2009, 09:51 PM by Base-Acid Tetration.) Quote:If we take the equations as recursive definitions then $\operatorname{It}$ is only defined for functions in the second argument that are constant. Thatswhy I proposed the different version. Why? If one defined the It operator for functions in the second argument the definition would not be a "recursive" definition? then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2009, 10:27 PM (This post was last modified: 05/03/2009, 10:31 PM by bo198214.) (05/03/2009, 09:45 PM)Tetratophile Wrote: If one defined the It operator for functions in the second argument the definition would not be a "recursive" definition? then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.What do you mean by "true for all functions"? That x is an arbitrary function? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/04/2009, 01:06 AM (This post was last modified: 05/04/2009, 01:51 AM by Base-Acid Tetration.) (05/03/2009, 10:27 PM)bo198214 Wrote: (05/03/2009, 09:45 PM)Tetratophile Wrote: If one defined the It operator for functions in the second argument the definition would not be a "recursive" definition? then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.What do you mean by "true for all functions"? That x is an arbitrary function? Yes. x can, in principle, be any function*; interpreting the "iterants" (the n in f [It_k] n) as constant functions rather than as simply numbers is what makes the hierarchy possible (since a function outputs numbers we can use as iterant). I should have written the iterant as g(x) to emphasize that. The order of the operations is that It_k for bigger k are performed first, and, if the same operator is nested, perform the innermost one first; top down, inside out. *as long as everything is defined: f(x), g(x), some real x, some natural n, f(x) It_(n+1) g(x), f(x) It_(n) [f(x) It_(n+1) g(x)-1] bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/04/2009, 07:56 AM (This post was last modified: 05/04/2009, 07:58 AM by bo198214.) (05/04/2009, 01:06 AM)Tetratophile Wrote: Yes. x can, in principle, be any function*; interpreting the "iterants" (the n in f [It_k] n) as constant functions rather than as simply numbers is what makes the hierarchy possible (since a function outputs numbers we can use as iterant). I should have written the iterant as g(x) to emphasize that. But anyway its no *definition*. See have your original equation: $f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x).$ and lets add the initial condition: $f \operatorname{It}_n 1 = f$ Interpreting x as constant function, with the above lines we can derive the function $f \operatorname{It}_n x$ for any constant function $x$. And that is all! We can no derive what $f \operatorname{It}_n g$ means for any non-constant function $g$. Just because on the left side there are only constant functions in the second argument. Well if we allow any function $g$ for $x$, not only constant functions, then it is still no definition, because we have no initial condition, which stops the recursion. For a constant function $m$, the right side needs to evaluate $m-1$ in the second argument, to evaluate this it must be evaluated at $m-2$ and so on until one derives at m=1. This the initial condition and the recursion is finished. If you put however any function $g$ there then on the right side in the second argument $g-1$ needs to be evaluated then $g-2$ then $g-3$ and so on but this recursion does never stop because $g-k$ never becomes a constant function for which we know how to evaluate. Think of writing a computer program to evaluate your operator. It can not guess what your intention was for non-constant second arguments. But I can guess and thatswhy I proposed the definition I gave. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/04/2009, 08:04 PM (This post was last modified: 05/04/2009, 08:56 PM by Base-Acid Tetration.) (05/04/2009, 07:56 AM)bo198214 Wrote: But anyway its no *definition*. See have your original equation: $f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x).$ and lets add the initial condition: $f \operatorname{It}_n 1 = f$ Interpreting x as constant function, with the above lines we can derive the function $f \operatorname{It}_n x$ for any constant function $x$. And that is all! We can no derive what $f \operatorname{It}_n g$ means for any non-constant function $g$. Just because on the left side there are only constant functions in the second argument. Well if we allow any function $g$ for $x$, not only constant functions, then it is still no definition, because we have no initial condition, which stops the recursion. For a constant function $m$, the right side needs to evaluate $m-1$ in the second argument, to evaluate this it must be evaluated at $m-2$ and so on until one derives at m=1. This the initial condition and the recursion is finished. If you put however any function $g$ there then on the right side in the second argument $g-1$ needs to be evaluated then $g-2$ then $g-3$ and so on but this recursion does never stop because $g-k$ never becomes a constant function for which we know how to evaluate. Think of writing a computer program to evaluate your operator. It can not guess what your intention was for non-constant second arguments. But I can guess and thatswhy I proposed the definition I gave. Your operator was defined with my operator. Here is the definition that you gave: bo198214 Wrote:$(f \operatorname{It}_n g)(x) := (I_n^{g(x)} f)(x)$. $(I_0^m f)(x) := f(m)$ $I_n^1 f :=f$ and $I_{n}^{m+1} f := f \operatorname{It}_{n-1} (I_n^m f).$ From that follows $(f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x))$ $I_1^m f = f^{\circ m}$ $(f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x)$ $I_2^m f = {^{\circ m} f}$ $(f \operatorname{It}_2 g)(x) = {^{\circ g(x)} f}(x)$ ....Your I operator is basically the same as my It operator; the only differnce is the notation. Can you tell me exactly what is different about your definition of the hyper-iteration than mine? [f It_n g](x) := [f It_n g(x)](x) - Let me clarify what it means. For any n, theoretically the procedure is 1. Evaluate g(x) at c first. 2. Hyper-n-iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. For n=1 (iteration), to evaluate this expression at any given natural x=c: 1. Evaluate g(x) at c first. 2. Iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. If you substitute g for f, it becomes: evaluate f(x) at c, then iterate f to the result, evaluate the resulting function at c. This is what (f^f)[x] f It_2 2 = $(I_n ^2) f(x)$ is. For It_2, at any natural x=c, the procedure is: 1. Evaluate g(x) at c. 2. Evaluate f(x) at c. 3. Perform the steps to calculate [f It_n f©]© g© (the OUTPUT of g(x) at c) times: 3a. Iterate f to the OUTPUT of 2. 3b. Evaluate the resulting function, at x. Make the result the input for 3a. 4. Evaluate the result of Step 2 at x. At least for integer values of g(x), your computer program could evaluate g(x), and then iterate f to the OUTPUT of the function g(x). ps How do you do © so that it comes out as ( c ) without spaces instead of a copyright symbol? This can be extended to the higher hyper-operators by repeating the steps at It_n-1 g(x) times for It n. So we both agree that it is not the FUNCTION per se, but the OUTPUT to which f is hyperiterated. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/04/2009, 08:50 PM (05/04/2009, 08:04 PM)Tetratophile Wrote: Your operator was defined with my operator. Here is the definition that you gave: I gave a recursive definition, your operator is defined with my and my is defined with yours. They are defined simulataneously by the equations (1) $(f \operatorname{It}_n g)(x) := (I_n^{g(x)} f)(x)$. (2) $(I_0^m f)(x) := f(m)$ (3) $I_n^1 f :=f$ and $I_{n}^{m+1} f := f \operatorname{It}_{n-1} (I_n^m f).$ Quote:Your I operator is basically the same as my It operator; the only differnce is the notation. Can you tell me exactly what is different about your definition of the hyper- iteration rthan mine? The difference is the declaration: Your operator It_n takes two functions and returns a function. My operator I_n takes a function and a number and returns a function. Quote:[f It_n g](x) := [f It_n g(x)](x) - Let me clarify what it means. Its absolutely clear what it means, however you see you make here the implicit distinction between It_n with a function as second argument (left side) and It_n with a number as second argument (right side). To make the difference explicit rename It_n at the right side to I_n (and change the arrangement slightly). Then this exactly (1). This equation can not be guessed! It needs to be written as part of the recursive definition! Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/04/2009, 08:57 PM (This post was last modified: 05/04/2009, 08:59 PM by Base-Acid Tetration.) bo, you needed to see my explanation. 1. Evaluate g(x) at c first. 2. Hyper-n-iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. For n=1 (iteration), to evaluate this expression at any given natural x=c: 1. Evaluate g(x) at c first. 2. Iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. The set of all ordered pairs resulting from this evaluation is {(x, [f It_1 g(x)] (x))}. « Next Oldest | Next Newest »

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