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.
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.
