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 Incredible reduction for Hyper operators JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 02/13/2014, 06:20 PM (This post was last modified: 02/13/2014, 08:35 PM by JmsNxn.) 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: $2[z]2 = 4$ « Next Oldest | Next Newest »

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