Hermite Polynomials

[mike3]

Yes, we need to find a condition for when the continuum sum will converge given a convergent \( f(z) \) Hermite series. And I suppose the bit at the beginning about the integral might need to be improved a little -- we could, e.g. show that it is always smaller than a known convergent integral to make the proof complete.

Hmm. Can we say that \( |f(x+yi)| < C_x e^{\alpha |y|} \) for \( 0 < \alpha < \pi/2 \)? for \( x \) belonging to the area we want to continuum sum. This will guarantee a converging continuum sum with a triple integral transfrom from FC. I have all this rigorously laid out. If Faulbaher's continuum sum takes \( s(s+1)(s+2)\cdots(s+n-1)\to \frac{1}{n}s(s+1)(s+2)\cdots(s+n) \) then My continuum sum may be the same as Faulbaher's. Now as for saying if the representation as continuum summed Hermite polynomials is convergent, I know some techniques from FC again that might work here. But they rely on the above and I'd have to take a closer look.

This is really quite interesting, I like this representation.

Are you going to show:

\( e^{\sum_{n=0}^{z-1} f(n)} = \frac{d}{dz} f(z) \)

Or do you have a different more convenient pattern for \( a_n \)

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

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?

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.