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:

and lets add the initial condition:

Interpreting x as constant function, with the above lines we can derive the function

for any constant function .

And that is all! We can no derive what means for any non-constant function . Just because on the left side there are only constant functions in the second argument.

Well if we allow any function for , not only constant functions, then it is still no definition, because we have no initial condition, which stops the recursion.

For a constant function , the right side needs to evaluate in the second argument, to evaluate this it must be evaluated at and so on until one derives at m=1. This the initial condition and the recursion is finished.

If you put however any function there then on the right side in the second argument needs to be evaluated then then and so on but this recursion does never stop because 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:.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?

and

From that follows

....

[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 = 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.