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 Functional super-iteration and hierarchy of functional hyper-iterations Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/04/2009, 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. « Next Oldest | Next Newest »

 Messages In This Thread Functional super-iteration and hierarchy of functional hyper-iterations - by Base-Acid Tetration - 05/02/2009, 02:23 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/02/2009, 07:48 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by andydude - 05/02/2009, 08:32 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by andydude - 05/02/2009, 09:23 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by tommy1729 - 05/02/2009, 12:22 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/02/2009, 01:14 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by andydude - 05/02/2009, 06:38 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/02/2009, 06:14 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/02/2009, 07:27 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/03/2009, 01:08 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/03/2009, 03:25 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/03/2009, 07:21 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/03/2009, 08:13 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/03/2009, 09:45 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/03/2009, 10:27 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/04/2009, 01:06 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/04/2009, 07:56 AM RE: Functional super-iteration and hierarchy of functional composition-based operations - by Base-Acid Tetration - 05/04/2009, 08:04 PM RE: Functional super-iteration and hierarchy of functional composition-based operations - by bo198214 - 05/04/2009, 08:50 PM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by Base-Acid Tetration - 05/04/2009, 08:57 PM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by bo198214 - 05/04/2009, 09:01 PM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by Base-Acid Tetration - 05/04/2009, 09:06 PM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by bo198214 - 05/04/2009, 09:12 PM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by Base-Acid Tetration - 05/12/2009, 02:29 AM RE: Functional super-iteration and hierarchy of functional hyper-iterations - by bo198214 - 05/12/2009, 07:11 AM

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