(11/21/2011, 08:08 PM)tommy1729 Wrote: can you generate this sequence with your method ?
every intuitive solutions seems to be different , but only a few can exist.
it is unclear to me how to use carleman matrices - although i mentioned it myself - or anything else ...
i dont even have the impression anyone came close.
Oh no, I definitely cannot generate this sequence yet. I just got a little over excited with Carlemann matrices; which I think now will not work.
First of all; the law:
\( F^m \{ F^n \{ f \} \} (x) = F^{m + n} \{ f \} (x) \)
holds pretty solidly, that's really all I'm going for now. Just creating a rule by which superfunction composition follows the same way composition follows; namely
\( (f^{\circ m} \circ f^{\circ n}) (x) = f^{\circ m + n}(x) \)
Also I'm quite aware that group theory isn't limited by the number of variables. I was just referring to the fact that where your previous proof method worked, it won't now. But, I urge you to try and find a direct contradiction with the law of superfunction composition I've put forth. Even a contradiction is a step in the right direction.
Furthermore, I'm really not even working on the sequence of superfunctions themselves yet and how to generate them; just the criterion by which they need to be defined and satisfied.
(11/21/2011, 10:33 AM)Gottfried Wrote: ....
Hey Gottfried. I think I get your approach, and I'm gonna speculate that it's a coincidence the Schroeder functions for addition, multiplication and exponentiation are iterates of the exponential function.
This will not produce the hyper operations sequence when iterated; namely:
\( \exp_b^{\circ 2}(b \cdot \log_b^{\circ 2}(x)) \neq ^x b \)
However I do think that's a nifty result.
...
Returning to this superfunction sequence, I have some more tiring results
It can be proven that:
if
\( F^t \{ f \} (x) \) is a t'th superfunction of f; then \( F^t \{ f \} (x + \theta (x)) \) is also a solution where theta is a one periodic function that takes zero at \( x \in \mathbb{Z} \).
We'll also find something very similar with our superfunction sequence;
namely
if
\( F^t \{ f \} (x) \) is the t'th superfunction of f, continuous over reals, then
\( R^t \{ f \} (x) = F^{t + \theta(t)} \{ f \} (x) \) is the t'th superfunction of f, continuous over reals, satisfying the condition:
\( R \{ R^t \{ f \} \}(x) = R^{t + 1} \{ f \} (x) \) which is just like the iteration sequence we see with composition \( (f \circ f^{\circ t})(x) = f^{\circ t + 1}(x) \)
However, we would have the draw back
\( R^m \{ R^n \{ f \} \} (x) = F^{m + \theta(m)} \{ F^{n + \theta(n)} \{ f \}\} (x) = F^{m + n + \theta(m) + \theta(n)} \{ f \} (x) \neq R^{m + n} \{ f \} (x) \)
unless of course we redefine superfunction composition as; here I'll use square brackets to accentuate the differences:
\( R^m [ R^n [f] ] (x) = R^{m + n } [ f ] (x) \)
so that
\( F^{m + \theta(m)} [ F^{n + \theta(n)} [f ] ](x) = F^{m + n + \theta(m + n)} [f] (x) \)
which again, is perfectly consistent with
\( F [ F^t [f] ](x) = F^{t + 1} [ f ] (x) \)
So therefore we're going to have two layers of an infinitude of solutions for our superfunction sequence... but it continues more
we can define:
\( F \{ f \} (x) \) by Carlemann matrices, or by Kouznetsov's method, or by regular iteration of a fixpoint, or by pretty much any form that works. So which one do we choose?
This is the question I'm trying to ask; which method of making superfunctions generalizes the easiest and the best (and has the least restricted domain for the base value in the hyperoperation sequence (we all know tetration has a knack for failing b < e^(1/e); we'll probably have a similar failure for b < p involving pentation; so on and so forth) to create an infinite sequence of superfunctions, and then from that sequence, create fractional indexes.
Essentially it will be a very esoteric problem for iteration, iterating the method a superfunction is created (which most likely involves isolating a fixpoint); and then stopping halfway through that iteration to get a half superfunction.
Once we have that, we'll have to identify uniqueness faced with the dilemma \( F^t \{ f \} (x) \) is the t'th superfunction and so is \( F^t \{ f \} (x + \theta (x)) \).
And then identify another form of uniqueness given by the dillema \( F^t \{ f \} (x) \) is the t'th superfunction of f and so is \( F^{t + \theta(t)} \[ f \] (x) \).
It is all so very mind boggling!