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Well this theory is structured as a plausible solution to complex hyper operators. This is a proof sketch and it requires a bit more work here and there; requires an evaluation of a limit; but there is an over all sense of plausibility.

We start by defining a sequence of analytic functions that obey the following rules:

$\vartheta_n : \mathbb{C} \to \mathbb{C}$

$\vartheta_n(n) = 1$ and $k \in \mathbb{N}_0\,\,\,\vartheta_n(k) = 0 \,\,\Leftrightarrow\,\, k \neq n$

Necessarily; these functions can be arbitrary and they pose the true uniqueness criterion. Nonetheless we suggest these functions and give a plausible solution.

${\bf I}:\,\,\,\,\vartheta_n(s) = \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\,\,\,;\,\,\,n \ge 1$

$\vartheta_0(s) = e^{e^s-1} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{-\frac{s}{k}}$
Reason for this is it has fast convergence and satisfies the required conditions.

Now we perform a trick. Considering hyper operators; which are written by the following:

$x\,\,\bigtriangleup_{s}\,\,(x \,\,\bigtriangleup_{s+1}\,\,y) = x \,\,\bigtriangleup_{s+1}\,\,(y+1)$

$x \,\,\bigtriangleup_0\,\,y = x + y$

and the identity of $\bigtriangleup_n$ for any natural $n$ is $1$.

We get the usual sequence where zero is addition, one is multiplication, two is exponentiation, etc... We then write that complex operators are products of natural operators with complex exponents. Here; $\vartheta_n$ is as before.

${\bf II}:\,\,\,\,x \,\,\bigtriangleup_s\,\, y = \prod_{n=0}^{\infty}(x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} = (x + y)^{\vartheta_0(s)} \cdot (x\cdot y)^{\vartheta_1(s)} \cdot (x^y) ^{\vartheta_2(s)} \cdot (^y x)^{\vartheta_3(s)} \cdot ...$

It's easy to identify that at natural arguments the product converges to the natural hyperoperator it indexes.
This product; however; converges if and only if the series:

$\sum_{n=0}^{\infty} \ln(x\,\,\bigtriangleup_n\,\,y)\vartheta_n(s)$ converges.

With this; we perform the ratio test.

$L = \lim_{n \to \infty} |\frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) \vartheta_{n+1}(s)}{\ln(x \,\,\bigtriangleup_{n}\,\, y) \vartheta_{n}(s)}|$

It is clear that
$\lim_{n\to\infty} |\frac{\vartheta_{n+1}(s)}{\vartheta_{n}(s)}| = 0$

Without proof; we can safely assume when x and y aren't 2 or one of them 1:

$\lim_{n \to \infty}| \frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) }| = \infty$

The question is if they cancel each other out. Or if one takes over the other. I'm hoping it's zero and the result is entire--the $s$ disappears in $L$; so the equation either diverges for all $s$ or converges for all $s$ or inconclusive. I do not believe that $\frac{\ln(x \,\,\bigtriangleup_{n+1}\,\, y) }{\ln(x \,\,\bigtriangleup_{n}\,\, y) }$ grows too too fast as n varies. So maybe ${\bf I}$ will induce convergence. If it doesn't; we'll use a function for $\vartheta$ that decreases to zero faster than the definition as it is now. Essentially, the true difficulty is finding a solution for $\vartheta$ that allows for the convergence of ${\bf II}$.

The recursive pattern becomes much simpler to find a solution for because we only have to account for instances where $x \,\,\bigtriangleup_s\,\,y = z\,\,\in \mathbb{N}_0$. Where of course we must have $x\,\,\bigtriangleup_{s-1}\,\,z = x\,\,\bigtriangleup_s\,\,y+1$. However. Most of the returned values will not be natural and for these most of our hyper operators are undefined. So the question becomes much simpler. It's almost as if we can extend the hyper operators beyond natural numbers. If you cannot get what I mean; consider:

$x\,\,\bigtriangleup_{1.5}\,\, 2 = z \,\, \not \in \mathbb{N}_0$

however:

$x\,\,\bigtriangleup_{0.5}\,\,z = x\,\,\bigtriangleup_{1.5}\,\, 3$

This allows us to extend operators to some real or complex numbers. Most definitely probably not all.

A brief note about the operators. They are definitely not commutative for almost all values. I haven't proved this yet however; so in the fractal chaos there may emerge a commutative operator. They are not associative for almost all values. Again; haven't proved this yet; but I think this one is pretty obvious. There may or may not exist identities for the operators. This one may be difficult to prove.

I'm mostly working on trying to prove convergence at the moment. One step at a time. Once that is proven we'll have an entire function that passes through every hyper operator at every two natural arguments. Then we tweak $\vartheta$ to be recursive. Maybe; if possible; tweak it to have a right hand identity.
If we want. We can try to treat this as a uniqueness claim for extending hyperoperators to complex numbers. That if we find a formula for $\vartheta$ that satisfies recursion for natural numbers it should also satisfy recursion for complex numbers. Ergo; uniqueness for how hyper operators behave for complex numbers. It's not very lucid yet. Again; one step at a time.

EDIT:

evaluation of limit

$\begin{eqnarray*}
\vartheta_n(s) &=& \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\\ &=& \frac{s e^{e^s} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{\frac{-s}{k}}}{n e^{e^n} \prod_{k=1}^{\infty}(1-\frac{n}{k})e^{\frac{-n}{k}}} = \frac{\psi(s)}{\psi(n)}\\
\end{eqnarray*}$

$\begin{eqnarray*}
f(t) &=& \ln(x\,\, \bigtriangleup_{t}\,\, y)\\
&=& \ln(x+y) \frac{\psi(t)}{t} + \sum_{k=1}^{\infty} \frac{\psi(t)}{\psi(k)} \ln(x\,\, \bigtriangleup_{k}\,\, y)\\
&=& \psi(t) \cdot (\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{\ln(x\,\, \bigtriangleup_{k}\,\,y)}{\psi(k)} )
\end{eqnarray*}$

Therefore we can rephrase the limit as; keeping in mind the first term $\ln(x+y)/t$ disappears:

$\begin{eqnarray*}
L &=& \lim_{t \to \infty} | \frac{f(t+1)}{f(t)} \cdot \frac{\vartheta_{t+1}(s)}{\vartheta_{t}(s)}|\\
&=& \lim_{t \to \infty} | \frac{\psi(t+1) \vartheta_{t+1}(s) } {\psi(t) \vartheta_{t}(s)} \cdot \frac{ \frac{\ln(x+y)}{t+1} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)}}{\frac{\ln(x+y)}{t} + \sum_{k=1}^{\infty} \frac{f(k)}{\psi(k)} } |\\
&=&1
\end{eqnarray*}$

Which means convergence is inconclusive. Using this method generally we will have this result; so long as $\vartheta$ remains a quotient. Anyone know any other convergence tests? I'm pretty sure this outlaws the root test as inconclusive as well. The integral test? I'll keep looking... Maybe I made a mistake. I tried using the integral test and I got divergence. The methods were a little questionable however. I'll hold out on posting that just yet.
(07/19/2012, 04:44 PM)JmsNxn Wrote: [ -> ]We start by defining a sequence of analytic functions that obey the following rules:

$\vartheta_n : \mathbb{C} \to \mathbb{C}$

$\vartheta_n(n) = 1$ and $k \in \mathbb{N}_0\,\,\,\vartheta_n(k) = 0 \,\,\Leftrightarrow\,\, k \neq n$

Necessarily; these functions can be arbitrary and they pose the true uniqueness criterion. Nonetheless we suggest these functions and give a plausible solution.

${\bf I}:\,\,\,\,\vartheta_n(s) = \frac{s}{n} e^{e^s-e^n} \prod_{k=1}^{\infty}\frac{1-\frac{s}{k}}{1-\frac{n}{k}}e^{ \frac{n - s}{k}}\,\,\,;\,\,\,n \ge 1$

$\vartheta_0(s) = e^{e^s-1} \prod_{k=1}^{\infty}(1-\frac{s}{k})e^{-\frac{s}{k}}$
Reason for this is it has fast convergence and satisfies the required conditions.

Now we perform a trick. Considering hyper operators; which are written by the following:

$x\,\,\bigtriangleup_{s}\,\,(x \,\,\bigtriangleup_{s+1}\,\,y) = x \,\,\bigtriangleup_{s+1}\,\,(y+1)$

$x \,\,\bigtriangleup_0\,\,y = x + y$

and the identity of $\bigtriangleup_n$ for any natural $n$ is $1$.

We get the usual sequence where zero is addition, one is multiplication, two is exponentiation, etc... We then write that complex operators are products of natural operators with complex exponents. Here; $\vartheta_n$ is as before.

${\bf II}:\,\,\,\,x \,\,\bigtriangleup_s\,\, y = \prod_{n=0}^{\infty}(x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)} = (x + y)^{\vartheta_0(s)} \cdot (x\cdot y)^{\vartheta_1(s)} \cdot (x^y) ^{\vartheta_2(s)} \cdot (^y x)^{\vartheta_3(s)} \cdot ...$

Dont you need to prove the identity of the infinite product ?

e.g. does the infinite product formula hold for x + y and x*y ??

regards

tommy1729
Yes. By definition. $\vartheta_n(n) = 1$ and $\vartheta_k(n) = 0 \,\,\Leftrightarrow\,\, k \neq n\,\,k \in \mathbb{N}_0$ therefore the rest of the product disappears at natural values. That's the whole trick . I'd take the limit as $\lim_{h \to n}\vartheta_n (h) = \vartheta_n(n) = 1$ just for technicality sake.

$x\,\,\bigtriangleup_0\,\, y = (x + y)^1 \cdot (x \cdot y)^0 \cdot (x^y)^0 \cdot ...$

$x \,\,\bigtriangleup_1\,\, y = (x + y)^0 \cdot (x \cdot y)^1 \cdot (x^y)^0 \cdot ...$

It's real for real $s$ and complex for complex $s$ and natural for natural $s$. Very nice.
i think you need to modify to get "hyperoperator speed" towards zero such that we have convergence.

i think it is possible.

thats good.

but are the hyperoperator equations still valid for s not an integer ??

intresting idea.

regards

tommy1729
Exactly what I was thinking. I was trying to think of functions that grow fast; extremely fast.

The first problem is getting convergence. I do think it's possible as well. The main problem was that my function could be broken into a quotient: $\vartheta_n(s) = \frac{\psi(s)}{\psi(n)}$ This was a technique I used to ensure $\vartheta_n(n) = 1$ but I think it backfires now.

I'm going to think about possible functions for $\vartheta$. I think it only needs double exponential: $e^{-e^t}$ since the series is the logarithm of the hyper operators.

The second problem is seeing if we can recover recursion at natural x and y. I think I have a technique at accessing this. I'll have to think about it. The beauty is that we are dealing with natural numbers so I'm thinking we'll be able to do some kind of proof by induction. This is difficult to vocalize without strictly going through the process of what I mean but I'll get to it probably tomorrow. I'm still trying to think hard about convergence.

The third problem is getting a right hand identity. Proving non commutativity and non associativity. Then I'll say I have a solution.

I have induced convergence!

We use Ackermann numbers!

$\vartheta_n(s) = \frac{\sin( \pi (s- n))}{ \pi(s-n)}(n \,\,\bigtriangleup_n\,\,n)^{s-n}$

This is a much more sophisticated function.

These grow faster than anything in the planet. Certainly faster than $x \,\,\bigtriangleup_n\,\,y$. Unfortunately; no clue how to prove this. I know that it decreases fast enough to zero to dampen the growth of hyper operators. Because it grows faster than any hyper operator. Sort of how n! grows faster than any natural iterated multiplication of x; $x^n$.

$x\,\,\bigtriangleup_s\,\,y = \prod_{n=0}^{\infty} (x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)}$

Now to tackle recursion. I will need a third Aha! to conquer this one. Anyone have any tips on how to prove this converges? I think the ratio test on this one is pretty useless.
I think the following will work to show convergeance

assume x , y and s to be reals > eta

1) the value of the product = 0 => done

2) the value is not zero => take the log => we get a sum

2b) show that the tail of the sequence goes to 0 faster than constant / n^2 hence we have convergeance. ( because of the famous Basel problem " zeta(2) = Pi^2 / 6 " )

As for the recursion , i dont know if it helps but remember

to go from s+1 to s do :

( we take the inverse with respect to y , notation : ^-1 )

f(x,y,s) = f(x,f^-1(x,y,s)+1,s)

which should be valid for all sufficiently large real non-integer s too.

regards

tommy1729
I think that is a clever way of writing recursion. You made a typographical mistake though.

If we write: $\mathcal{L}_s(x, x \,\,\bigtriangleup_s\,\,y) = y$

then we have recursion as:

$x \,\,\bigtriangleup_{s-1}\,\,y = x\,\,\bigtriangleup_s\,\,\mathcal{L}_s(x,y) + 1$

However. I am only concerned with when $\mathcal{L}_s(x, y) \in \mathbb{N}$. Since going to reals requires an universal extension of hyper operators to reals.

$\mathcal{L}_s(x,y) = \prod_{n=0}^{\infty} \mathcal{L}_n(x,y)^{\zeta_n(s)} = (y-x)^{\zeta_0(s)} \cdot (\frac{y}{x})^{\zeta_1(s)} \cdot (\log_x(y))^{\zeta_2(s)} \cdot (\text{slog}_x(y))^{\zeta_3(s)}\cdot...$

$\zeta_n(s) = \frac{\sin(\pi (s-n))}{\pi (s-n)} \psi_n(s)$

for some undetermined $\psi$

This would actually allow us to extend to some reals. Thanks for this formation of recursion. I think it's a more efficient formula.
I've been experimenting looking at these operators and I've become much more familiar with its structure. I can prove convergence now.

Let's suppose by contradiction that; for sufficiently large n:

$\frac{x\,\,\bigtriangleup_n\,\,y}{(n\,\,\bigtriangleup_n\,\,n)^{n-s}}\,\, > \,\frac{1}{n^2}$

However. It is clear that for some $n\,\, >\,\, x,y$

1: $n\,\,\bigtriangleup_n\,\,n\,\, >\,\, x\,\,\bigtriangleup_n\,\,y$

Therefore we write:

$(n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, n^2(x\,\,\bigtriangleup_n\,\,y)$

Now we know that $n^2\,\, <\,\, x\,\,\bigtriangleup_n\,\,y$

Therefore:

$(n\,\,\bigtriangleup_n\,\,n)^{n-s}\,\, <\,\, (x\,\,\bigtriangleup_n\,\,y)^2$

However this is a contradiction because for sufficiently large $n$ the left equation becomes much larger than the right. This is easy to deduce by the relation 1 above. Therefore to prove convergence we just need add the claim:

$\ln(x\,\,\bigtriangleup_n\,\,y)\,\, < \,\,x\,\,\bigtriangleup_n\,\,y$

$|\vartheta_n(s)| \,\,\le \,\,|(n\,\,\bigtriangleup_n\,\,n)^{s-n}|$

YES! We have convergence for all s.

$x\,\,\bigtriangleup_s\,\,y = \prod_{n=0}^{\infty} (x\,\,\bigtriangleup_n\,\,y)^{\vartheta_n(s)}$

$\vartheta_n(s) = \frac{\sin(\pi(s-n))}{\pi(s-n)}(n\,\,\bigtriangleup_n\,\,n)^{s-n} \psi_n(s)$

An important theorem I have to prove is the following, I consider it a stern requirement of hyperoperators. For all $\Re(s) \ge 0$ and $\epsilon \,\,>\,\, 0$ and $x,y\,\,> 1$

$|x\,\,\bigtriangleup_{s+\epsilon}\,\,y| > |x\,\,\bigtriangleup_{s}\,\,y|$

I'll mull over that for awhile.

also; hopefully:

$\frac{d^n}{ds^n} x \,\,\bigtriangleup_s\,\,y\,\,\,\,> 0$ for at least $\Re(s) > 0$
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