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+--- Thread: Functional super-iteration and hierarchy of functional hyper-iterations (/showthread.php?tid=284)
RE: Functional super-iteration and hierarchy of functional composition-based operations - bo198214 - 05/03/2009
(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.
RE: Functional super-iteration and hierarchy of functional composition-based operations - Base-Acid Tetration - 05/03/2009
(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
RE: Functional super-iteration and hierarchy of functional composition-based operations - bo198214 - 05/03/2009
(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
\( (f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x)) \)
\( I_1^m f = f^{\circ m} \)
\( (f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x) \)
\( I_2^m f = {^{\circ m} f} \)
\( (f \operatorname{It}_2 g)(x) = {^{\circ g(x)} f}(x) \)
....
RE: Functional super-iteration and hierarchy of functional composition-based operations - Base-Acid Tetration - 05/03/2009
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.
RE: Functional super-iteration and hierarchy of functional composition-based operations - bo198214 - 05/03/2009
(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?What do you mean by "true for all functions"? That x is an arbitrary function?
then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.
RE: Functional super-iteration and hierarchy of functional composition-based operations - Base-Acid Tetration - 05/04/2009
(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?What do you mean by "true for all functions"? That x is an arbitrary function?
then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.
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]
RE: Functional super-iteration and hierarchy of functional composition-based operations - bo198214 - 05/04/2009
(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.
RE: Functional super-iteration and hierarchy of functional composition-based operations - Base-Acid Tetration - 05/04/2009
(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:\( (f \operatorname{It}_n g)(x) := (I_n^{g(x)} f)(x) \).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?
\( (I_0^m f)(x) := f(m) \)
\( I_n^1 f :=f \) and \( I_{n}^{m+1} f := f \operatorname{It}_{n-1} (I_n^m f). \)
From that follows
\( (f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x)) \)
\( I_1^m f = f^{\circ m} \)
\( (f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x) \)
\( I_2^m f = {^{\circ m} f} \)
\( (f \operatorname{It}_2 g)(x) = {^{\circ g(x)} f}(x) \)
....
[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.
RE: Functional super-iteration and hierarchy of functional composition-based operations - bo198214 - 05/04/2009
(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!
RE: Functional super-iteration and hierarchy of functional hyper-iterations - Base-Acid Tetration - 05/04/2009
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))}.