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interpolating the hyper operators
#1
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.)

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
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.
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Messages In This Thread
interpolating the hyper operators - by JmsNxn - 05/24/2013, 10:24 PM
RE: interpolating the hyper operators - by JmsNxn - 06/07/2013, 05:00 AM
RE: interpolating the hyper operators - by MphLee - 06/07/2013, 08:29 AM
RE: interpolating the hyper operators - by JmsNxn - 06/07/2013, 09:03 PM

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