05/24/2013, 10:24 PM
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:\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:
 = \frac{1}{\Gamma(s)}\int_0^\infty f(-u)u^{s-1}du)
It is clear that(s -1))
And Since:
(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).
(-n) = \frac{d^nf}{ds^n} |_{s=0})
Now we define the auxillary function defined for all
:
}\,y)k!})
If this function is kosher, as defined above, then we have the following function.
 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 > 0)
For example, lets take = \sum_{k=0}^{\infty} \frac{s^k}{(2\,(k)\,2)k!} = \frac{1}{4} e^s)
Now we have:
\,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 = 4)
and since n is arbitrary, by analytic continuation:
\, 2 = 4)
If we show that as\to Constant)
then we have our result, since the n'th derivative will decay to zero.
On showing recursion we have the following result > 0)
:
\,(y-1) = z_0 \,\in \mathbb{N})
 \,z_0 = x\,(n-s_0)\,y)
This reduces to the following condition:
}(-u) - \vartheta_{x,y}^{(n)}(-u)]u^{s_0-1} du = 0)
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.
If you have a function
Define:
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
If we show that as
On showing recursion we have the following result
This reduces to the following condition:
It is valid when
Does anyone know any hints in how to prove
