Solving tetration using differintegrals and super-roots - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Solving tetration using differintegrals and super-roots (/showthread.php?tid=1092) Solving tetration using differintegrals and super-roots - JmsNxn - 08/22/2016 Well about a year ago I posted my paper on the bounded analytic hyper-operators. The result was straight forward and involved expressing functions using the Mellin transform (or the exponential differintegral as I expressed it). One question that arose is the following, If we define $\Psi(n,x)$ to be the n'th super root $^n \Psi(n,x) = x$ is the function $g(u,x) = \sum_{n=0}^\infty \Psi(n+1,x)\frac{u^n}{n!}$ differintegrable? And if differintegrated, does it equal the super root for arbitrary $z$? Well I recently thought more about this question, and using an invaluable Lemma I proved I've found that in fact, for x>1 and $0 < \sigma < 1$ we do have $\int_0^\infty |g(-u,x)|u^{-\sigma}\,du < \infty$ Therein defining the function for $\Re(z) > 0$ $\Psi(z,x) = \frac{1}{\Gamma(1-z)}(\sum_{n=0}^\infty \Psi(n+1,x)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty g(-u,x)u^{-z}\,du)$ Now here's the really beautiful part, and sadly where I am stuck. If I can show that $\frac{d}{dx}\Psi(z,x) \neq 0$ then we can invert this function across $x$, to get $F$ the function $\Psi(z,F(z,x)) = x$. Taking this function observe the following $r(z) = \Psi(z,x)^{F(z,\Psi(z,x))} = \Psi(z,x)^x$ is factorizable in $z$ for x close enough to 1. Just as well $a(z) = F(z+1,\Psi(z,x))$ is factorizable in $z$ for x close enough to 1. Factorizable implies we can use the natural identity theorem. This means if $a|_{\mathbb{N}} = r|_{\mathbb{N}}$ then $a = r$ Well... $r(n) = \Psi(n,x)^x$ and $a(n) = F(n+1,\Psi(n,x)) = (^{n+1}\Psi(n,x)) = \Psi(n,x)^{^n\Psi(n,x)} = \Psi(n,x)^x$ and therefore $r(z) = a(z)$ What this means, unraveling the cryptic writing, is if $y = \Psi(z,x)$ then $y^{F(z,y)} = F(z+1,y)$ TETRATION! Therefore the requirement boils down into showing the rather painful fact that $\frac{d}{dx}\Psi(z,x) \neq 0$ for $\Re(z) > M$ for some M > 0. This will successfully construct a tetration function. I'm at a loss frankly on how to show this. The rest fell into place rather easily, nothing too exhaustive was required. It is obviously true for most x>1, but not for all. There may be exceptional values. Therefore we do have a tetration solution using the differintegral given some conditions... Numerical evidence seems to support the result too but I've only tried a little bit.