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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 to be the n'th super root is the function differintegrable? And if differintegrated, does it equal the super root for arbitrary ? 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 we do have Therein defining the function for Now here's the really beautiful part, and sadly where I am stuck. If I can show that then we can invert this function across , to get the function . Taking this function observe the following is factorizable in for x close enough to 1. Just as well is factorizable in for x close enough to 1. Factorizable implies we can use the natural identity theorem. This means if then Well... and and therefore What this means, unraveling the cryptic writing, is if then TETRATION! Therefore the requirement boils down into showing the rather painful fact that for 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. |