Hi everyone. I've been doing some exhaustive research into the Weyl differintegral and I've found a very beautiful approach to solving hyperoperators.
I'll be very frank. so I'll just state the following results:
If \( \frac{d^{-z}}{dw^{-z}}f(w) |_{w=0} = \phi(z) = \frac{1}{\Gamma(z)}\int_0^\infty f(-w)w^{z-1}\,dw \)
Is the Weyl differintegral at zero, then I've shown some equivalencies. If \( \phi \) is kosher then it is defined for at least the left half plane ending at the line \( \Re(z) = b \) and as well \( |\phi(\sigma \pm iy)|\le C_\sigma e^{\alpha |y|}\,\,\,0 \le \alpha < \pi/2 \) for all \( a < \sigma < b \) and \( |\phi(\sigma \pm iy)|\le C_{y} e^{\rho|\sigma|} \) for \( \sigma < b \) and for some \( \rho \ge 0 \)
Then:
\( \frac{d^z}{dw^z}f(w) = \sum_{n=0}^\infty \phi(-n-z)\frac{w^n}{n!} \)
As Well we obtain a new expression for \( \phi \) which is what I'm going for:
\( \phi(z)\Gamma(z) = \sum_{n=0}^\infty \phi(-n)\frac{(-1)^n}{n!(z+n)} + \int_1^\infty f(-w)w^{z-1}\,dw \)
I am currently writing a paper on this. The results are solid. I've checked them very well. I can explain how I got to them, but they take a bit to write up.
So now we say, by the following.
Define the function:
\( \frac{\Gamma(-z)}{x[z]y} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\vartheta_x(y,-w)w^{-z-1}\,dw \)
where \( \vartheta \) is an entire function of order zero defined by \( \vartheta_x(y,w) = \sum_{n=0}^\infty \frac{w^n}{(x[n]y)n!} \)
Let us assume there exists an analytic continuation of natural hyperoperators (natural z) for natural values of x (in the first argument), and complex arguments in the second argument (in y). And If \( \frac{1}{x[z]y} \) and \( \frac{1}{x[z](x[z+1]y)} \) both are kosher
then 1/x[z+1]y is kosher
and
\( \frac{\Gamma(-z)} {x[z+1](y+1)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n+1]y+1)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty \frac{(-w)^k}{(x[k+1](y+1))k!})w^{-z-1}\,dw \)
But!!! Since \( \frac{\Gamma(-z)}{x [z] (x[z+1]y)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n](x[n+1]y)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty\frac{(-w)^k}{(x[k](x[k+1]y))k!} w^{-z-1}\,dw \)
We see we are only talking about the naturals and the results are EQUIVALENT!!!!!!!
Okay, I know. This is a big if. Essentially you need to prove that the following integral on the right converges for \( \Re(z)>0 \), that \( |\omega(\sigma\pm iy)| < C_\sigma e^{\alpha|y|} \) for \( 0 \le \alpha < \pi/2 \) when \( a<\sigma < 0 \) and that \( |\omega(\sigma \pm iy)|\le C_y e^{\rho|\sigma|} \) for some \( \rho\ge 0 \). Where \( \omega \) is the following function
\( \Gamma(-z)\omega(z,x,y) = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\sum_{k=0}^\infty \frac{(-w)^k}{(x[k]y)k!}w^{-z-1}\,dw \)
If this happens and \( x[k]y \) is holomorphic in \( y \) for all natural \( x \). THEN WE HAVE AN ANALYTIC CONTINUATION OF HYPER OPERATORS!!!!!!!!!!!!!
QUESTIONS COMMENTS!
I have the evaluation that:
\( 2[z]2 = 4 \)
I'll be very frank. so I'll just state the following results:
If \( \frac{d^{-z}}{dw^{-z}}f(w) |_{w=0} = \phi(z) = \frac{1}{\Gamma(z)}\int_0^\infty f(-w)w^{z-1}\,dw \)
Is the Weyl differintegral at zero, then I've shown some equivalencies. If \( \phi \) is kosher then it is defined for at least the left half plane ending at the line \( \Re(z) = b \) and as well \( |\phi(\sigma \pm iy)|\le C_\sigma e^{\alpha |y|}\,\,\,0 \le \alpha < \pi/2 \) for all \( a < \sigma < b \) and \( |\phi(\sigma \pm iy)|\le C_{y} e^{\rho|\sigma|} \) for \( \sigma < b \) and for some \( \rho \ge 0 \)
Then:
\( \frac{d^z}{dw^z}f(w) = \sum_{n=0}^\infty \phi(-n-z)\frac{w^n}{n!} \)
As Well we obtain a new expression for \( \phi \) which is what I'm going for:
\( \phi(z)\Gamma(z) = \sum_{n=0}^\infty \phi(-n)\frac{(-1)^n}{n!(z+n)} + \int_1^\infty f(-w)w^{z-1}\,dw \)
I am currently writing a paper on this. The results are solid. I've checked them very well. I can explain how I got to them, but they take a bit to write up.
So now we say, by the following.
Define the function:
\( \frac{\Gamma(-z)}{x[z]y} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\vartheta_x(y,-w)w^{-z-1}\,dw \)
where \( \vartheta \) is an entire function of order zero defined by \( \vartheta_x(y,w) = \sum_{n=0}^\infty \frac{w^n}{(x[n]y)n!} \)
Let us assume there exists an analytic continuation of natural hyperoperators (natural z) for natural values of x (in the first argument), and complex arguments in the second argument (in y). And If \( \frac{1}{x[z]y} \) and \( \frac{1}{x[z](x[z+1]y)} \) both are kosher
then 1/x[z+1]y is kosher
and
\( \frac{\Gamma(-z)} {x[z+1](y+1)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n+1]y+1)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty \frac{(-w)^k}{(x[k+1](y+1))k!})w^{-z-1}\,dw \)
But!!! Since \( \frac{\Gamma(-z)}{x [z] (x[z+1]y)} = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n](x[n+1]y)n!(n-z)} + \int_1^\infty (\sum_{k=0}^\infty\frac{(-w)^k}{(x[k](x[k+1]y))k!} w^{-z-1}\,dw \)
We see we are only talking about the naturals and the results are EQUIVALENT!!!!!!!
Okay, I know. This is a big if. Essentially you need to prove that the following integral on the right converges for \( \Re(z)>0 \), that \( |\omega(\sigma\pm iy)| < C_\sigma e^{\alpha|y|} \) for \( 0 \le \alpha < \pi/2 \) when \( a<\sigma < 0 \) and that \( |\omega(\sigma \pm iy)|\le C_y e^{\rho|\sigma|} \) for some \( \rho\ge 0 \). Where \( \omega \) is the following function
\( \Gamma(-z)\omega(z,x,y) = \sum_{n=0}^\infty \frac{(-1)^n}{(x[n]y)n!(n-z)} + \int_1^\infty\sum_{k=0}^\infty \frac{(-w)^k}{(x[k]y)k!}w^{-z-1}\,dw \)
If this happens and \( x[k]y \) is holomorphic in \( y \) for all natural \( x \). THEN WE HAVE AN ANALYTIC CONTINUATION OF HYPER OPERATORS!!!!!!!!!!!!!
QUESTIONS COMMENTS!
I have the evaluation that:
\( 2[z]2 = 4 \)