I've managed to do the following:
Define:
\( x [0] y = y+1 \) and \( x [1] y = x + y \) and continue the sequence as usual.
The function \( \vartheta_{x,y}(s) = \sum_{n=0}^{\infty} \frac{s^n}{(x[n]y)n!} \) as the following beautiful property:
\( \frac{d^n}{ds^n} \vartheta_{x,y}(s) |_{s=0} = \frac{1}{x [n] y} \)
Lets calculate it's fractional derivative at the value \( \alpha \in \mathbb{C} \) then we write, using the Grunwald Letnikov derivative:
\( \frac{1}{x[\alpha]y} = \frac{d^\alpha}{ds^\alpha}\vartheta_{x,y}|_{s=0} = \lim_{h\to 0^+} h^{-\alpha} \sum_{n=0}^{\infty}(-1)^n \frac{\Gamma(\alpha+1)}{\Gamma(\alpha - n + 1)n!}\vartheta_{x,y}(-nh) \)
I've found an approach to showing that it satisfies the recursive identity, which is: \( x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R} \)
\( x [s_0 + 1] (y-1) = z \)
\( x [s_0] z = x [s_0+1] y \)
The approach is a little tricky and I don't know how to explain it so clearly. But in the mean time I have a holomorphic interpolation of the hyperoperators at natural arguments. Whoopie! Note, again, the brilliant result:
\( 2 [\alpha] 2 = 4 \)
Define:
\( x [0] y = y+1 \) and \( x [1] y = x + y \) and continue the sequence as usual.
The function \( \vartheta_{x,y}(s) = \sum_{n=0}^{\infty} \frac{s^n}{(x[n]y)n!} \) as the following beautiful property:
\( \frac{d^n}{ds^n} \vartheta_{x,y}(s) |_{s=0} = \frac{1}{x [n] y} \)
Lets calculate it's fractional derivative at the value \( \alpha \in \mathbb{C} \) then we write, using the Grunwald Letnikov derivative:
\( \frac{1}{x[\alpha]y} = \frac{d^\alpha}{ds^\alpha}\vartheta_{x,y}|_{s=0} = \lim_{h\to 0^+} h^{-\alpha} \sum_{n=0}^{\infty}(-1)^n \frac{\Gamma(\alpha+1)}{\Gamma(\alpha - n + 1)n!}\vartheta_{x,y}(-nh) \)
I've found an approach to showing that it satisfies the recursive identity, which is: \( x,y,z \in \mathbb{N}\,\,s_0 \in \mathbb{R} \)
\( x [s_0 + 1] (y-1) = z \)
\( x [s_0] z = x [s_0+1] y \)
The approach is a little tricky and I don't know how to explain it so clearly. But in the mean time I have a holomorphic interpolation of the hyperoperators at natural arguments. Whoopie! Note, again, the brilliant result:
\( 2 [\alpha] 2 = 4 \)
