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 interpolating the hyper operators JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 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 $f(s):\mathbb{C}\to \mathbb{C}$ 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: $E f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(-u)u^{s-1}du$ It is clear that $E\frac{df}{ds} = (Ef)(s -1)$ And Since: $(Ef)(1) = \int_0^\infty f(-u)du = F(0) - F(-\infty) = F(0)$ By induction, and analytic continuation (since differentiation of f shifts the domain of convergence down one real s). $(Ef)(-n) = \frac{d^nf}{ds^n} |_{s=0}$ Now we define the auxillary function defined for all $x,y \in \mathbb{N}$: $\vartheta_{x,y} = \sum_{k=0}^{\infty} \frac{s^k}{(x\,{(k)}\,y)k!}$ If this function is kosher, as defined above, then we have the following function. $\frac{1}{x (n-s) y} = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta_{x,y}^{(n)}(-u)u^{s-1}du$ Which agrees with natural hyper operators. and converges for all $\Re(s) > 0$ For example, lets take $\vartheta_{2,2}(s) = \sum_{k=0}^{\infty} \frac{s^k}{(2\,(k)\,2)k!} = \frac{1}{4} e^s$ Now we have: $\frac{1}{2\,(n-s)\,2} = \frac{1}{\Gamma(s)} \int_0^{\infty} (\frac{1}{4} e^{-u})u^{s-1}du = \frac{1}{4}\frac{\Gamma(s)}{\Gamma(s)}$ This gives $2 \,(n-s)\, 2 = 4$ and since n is arbitrary, by analytic continuation: $2\, (s)\, 2 = 4$ If we show that as $s \to -\infty\,\,\vartheta_{x,y}(s)\to Constant$ then we have our result, since the n'th derivative will decay to zero. On showing recursion we have the following result $\Re(s_0) > 0$: $x \,(n-s_0)\,(y-1) = z_0 \,\in \mathbb{N}$ $x \,(n-1-s_0) \,z_0 = x\,(n-s_0)\,y$ This reduces to the following condition: $\int_0^{\infty} [\vartheta_{x,z_0}^{(n-1)}(-u) - \vartheta_{x,y}^{(n)}(-u)]u^{s_0-1} du = 0$ It is valid when $s_0 \in \mathbb{N}$ 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 $\vartheta_{x,y}$ 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. « Next Oldest | Next Newest »

 Messages In This Thread interpolating the hyper operators - by JmsNxn - 05/24/2013, 10:24 PM RE: interpolating the hyper operators - by JmsNxn - 06/07/2013, 05:00 AM RE: interpolating the hyper operators - by MphLee - 06/07/2013, 08:29 AM RE: interpolating the hyper operators - by JmsNxn - 06/07/2013, 09:03 PM

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