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 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.

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.