(07/06/2014, 04:14 AM)mike3 Wrote: 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 \( a_n \)