Hermite Polynomials

[JmsNxn]

I would find that last identity dubious -- that would make the continuum sum just the log of the derivative!

That's an identity for tetration. If \( f \) is tetration its derivative is the continuum product. (at least im pretty certain of that, tommy's always saying it and a little algebra shows it)

So you're suggesting to guarantee the convergence, you should limit \( \alpha \) to be less than \( \pi/2 \)? (Which is not a problem for tetration since it's actually bounded on the strip) is this to try and prove equivalence between this and your fractional calculus continuum sum?

Well I meant if you have alpha less than pi/2 I can give you a continuum sum. It may not be faulbahers but it probably is (since continuum sums are unique if they are exponentially bounded on the real line and of exponential type less than pi/2 on the imaginary line.) It's a triple integral transform that's a little ugly but it works.

The coefficients \( a_n \) depend on whatever function you're trying to continuum sum -- I don't get what you mean by a "more convenient pattern". It'll depend on the function, just as whether or not there's a "convenient pattern" for the Taylor series coefficients depends on the function.

I meant, the following.

if \( f \) is tetration base e, then \( e^{\sum_{j=0}^{z-1} f(j)} = \frac{d}{dz}f(z) \)

With this relationship you can solve for \( a_n \) using what I presume is a lot of summation identities.

Is this what you plan to do? Or do you plan to find another way to solve for \( a_n \). I.e: "a more convenient pattern."