interpolating the hyper operators
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 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.)


It is clear that
And Since:

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

Now we define the auxillary function defined for all :

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

Which agrees with natural hyper operators. and converges for all

For example, lets take

Now we have:

This gives
and since n is arbitrary, by analytic continuation:

If we show that as then we have our result, since the n'th derivative will decay to zero.

On showing recursion we have the following result :

This reduces to the following condition:

It is valid when 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 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:


and and continue the sequence as usual.

The function as the following beautiful property:

Lets calculate it's fractional derivative at the value then we write, using the Grunwald Letnikov derivative:

I've found an approach to showing that it satisfies the recursive identity, which is:

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:

(06/07/2013, 05:00 AM)JmsNxn Wrote: I've found an approach to showing that it satisfies the recursive identity, which is:

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

translatable in

That bring me to ask you why you abandoned the study on the sets.

Anyways, even if you limit the recursion to ALL the sets you can work only on a countable subset of reals:
in other words most of the reals ( ) are out of the recursion.

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

if and hold than holds (recursion)

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

if holds than holds

as you can see, if holds then your proof is valid only for a subset of reals.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
I realized this is a more efficient way of talking about hyper operators. I will still talk about those 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 for all 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.


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