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I have found a way to interpolate hyper operators with an entire function for natural arguments. I havent proven that they obey the recursive identity but I am trying to show that.

If you have a function $f(s):\mathbb{C}\to \mathbb{C}$ such that it decays to zero at negative infinity as well as its antiderivative (its derivative still vanishes) we say the function is kosher. (for lack of a better word.)

Define:
$E f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(-u)u^{s-1}du$

It is clear that $E\frac{df}{ds} = (Ef)(s -1)$
And Since:

$(Ef)(1) = \int_0^\infty f(-u)du = F(0) - F(-\infty) = F(0)$

By induction, and analytic continuation (since differentiation of f shifts the domain of convergence down one real s).

$(Ef)(-n) = \frac{d^nf}{ds^n} |_{s=0}$

Now we define the auxillary function defined for all $x,y \in \mathbb{N}$:

$\vartheta_{x,y} = \sum_{k=0}^{\infty} \frac{s^k}{(x\,{(k)}\,y)k!}$

If this function is kosher, as defined above, then we have the following function.

$\frac{1}{x (n-s) y} = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta_{x,y}^{(n)}(-u)u^{s-1}du$

Which agrees with natural hyper operators. and converges for all $\Re(s) > 0$

For example, lets take $\vartheta_{2,2}(s) = \sum_{k=0}^{\infty} \frac{s^k}{(2\,(k)\,2)k!} = \frac{1}{4} e^s$

Now we have:

$\frac{1}{2\,(n-s)\,2} = \frac{1}{\Gamma(s)} \int_0^{\infty} (\frac{1}{4} e^{-u})u^{s-1}du = \frac{1}{4}\frac{\Gamma(s)}{\Gamma(s)}$

This gives
$2 \,(n-s)\, 2 = 4$ and since n is arbitrary, by analytic continuation:
$2\, (s)\, 2 = 4$

If we show that as $s \to -\infty\,\,\vartheta_{x,y}(s)\to Constant$ then we have our result, since the n'th derivative will decay to zero.

On showing recursion we have the following result $\Re(s_0) > 0$:

$x \,(n-s_0)\,(y-1) = z_0 \,\in \mathbb{N}$

$x \,(n-1-s_0) \,z_0 = x\,(n-s_0)\,y$

This reduces to the following condition:
$\int_0^{\infty} [\vartheta_{x,z_0}^{(n-1)}(-u) - \vartheta_{x,y}^{(n)}(-u)]u^{s_0-1} du = 0$

It is valid when $s_0 \in \mathbb{N}$ but it remains to be shown other wise. I think the result might work itself out.

Does anyone know any hints in how to prove $\vartheta_{x,y}$ decays to a constant at negative infinity? I'm at a loss for how the function behaves. I know it is bounded above by the exponential function but I don't know what bounds it from below. I think this is a promising approach to finding hyper operators.
I've managed to do the following:

Define:

$x [0] y = y+1$ and $x [1] y = x + y$ and continue the sequence as usual.

The function $\vartheta_{x,y}(s) = \sum_{n=0}^{\infty} \frac{s^n}{(x[n]y)n!}$ as the following beautiful property:

$\frac{d^n}{ds^n} \vartheta_{x,y}(s) |_{s=0} = \frac{1}{x [n] y}$

Lets calculate it's fractional derivative at the value $\alpha \in \mathbb{C}$ then we write, using the Grunwald Letnikov derivative:

$\frac{1}{x[\alpha]y} = \frac{d^\alpha}{ds^\alpha}\vartheta_{x,y}|_{s=0} = \lim_{h\to 0^+} h^{-\alpha} \sum_{n=0}^{\infty}(-1)^n \frac{\Gamma(\alpha+1)}{\Gamma(\alpha - n + 1)n!}\vartheta_{x,y}(-nh)$

I've found an approach to showing that it satisfies the recursive identity, which is: $x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R}$

$x [s_0 + 1] (y-1) = z$
$x [s_0] z = x [s_0+1] y$

The approach is a little tricky and I don't know how to explain it so clearly. But in the mean time I have a holomorphic interpolation of the hyperoperators at natural arguments. Whoopie! Note, again, the brilliant result:

$2 [\alpha] 2 = 4$
(06/07/2013, 05:00 AM)JmsNxn Wrote: [ -> ]I've found an approach to showing that it satisfies the recursive identity, which is: $x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R}$

$x [s_0 + 1] (y-1) = z$
$x [s_0] z = x [s_0+1] y$

The approach is a little tricky and I don't know how to explain it so clearly.

Isn't this condition equivalent to you old attempt to limit the recursion to a subset of the real?
In fact your condition for the recursion

$a)\,x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R}$
$b)\,x [s_0 + 1] (y-1) = z$
$c)\,x [s_0] z = x [s_0+1] y$

translatable in

$i)\, s_0 + 1 \in \mathbb{I}_{x,y-1}$
$ii)\,x [s_0] z = x [s_0+1] y$

That bring me to ask you why you abandoned the study on the $\mathbb{I}$ sets.

Anyways, even if you limit the recursion to ALL the $\mathbb{I}$ sets you can work only on a countable subset of reals:
in other words most of the reals ($2^{\aleph_0}$ ) are out of the recursion.

PS: If was not clear, what I mean is:
if for your interpolation you can show that:

if $a)$ and $b)$ hold than $c)$ holds (recursion)

then is the same that you proved for you interpolation this:

if $i)$ holds than $ii)$ holds

as you can see, if $i)$ holds then your proof is valid only for a subset of reals.
I realized this is a more efficient way of talking about hyper operators. I will still talk about those $I$ sets, but I think it is a better approach to already have an analytic expression. It came to me in a stroke of luck when I was thinking about fractional derivatives.

The goal is to have $f(s) = \frac{1}{x[s]y}$ for all $x,y \in \mathbb{N}$ analytic and entire. The proof of recursion then only revolves around when the output is a natural number. So we only prove for a discrete set of reals.