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 Bold and Disappointing Experiment With Hermite Polynomials!!! JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 01/18/2017, 10:20 PM (This post was last modified: 01/18/2017, 10:23 PM by JmsNxn.) I only have two ideas that might make all this a bit easier. One is a simple identity I never saw you use, which looked like it could help your equations out, especially when you take the logarithm and the second derivative. $\frac{d}{dz}\sum_{n=0}^{z-1} f(n) = \sum_{n=0}^{z-1}f'(n)$ This follows very basically because $\Delta f(z) = f(z) - f(z-1)$ commutes with the derivative, and then quite obviously its inverse must equally commute with the derivative. I use this idea frequently when I'm tasked with solving indefinite sums. I work a lot on this operator, but mostly I only deal with functions of a nice exponential bound in the right half plane (something kneser's tetration definitely is not).  The second point I have, a few years ago I solved a way of representing continuum sums in a vertical strip of the complex plane. I normalized it, so that it was better written. Essentially if $\phi(z)$ is holomorphic for $0 < \Re(z) < b$ and $|\phi(z)| < Ce^{\tau|\Im(z)|}$ for some $\tau < \pi/2$ then $\sum_{j=1}^{z} \phi(j) =\frac{1}{\Gamma(z)}\int_0^\infty x^{z-1}e^{-x}\int_0^xe^{t} f(t)dtdx$ where $f(x) = \frac{1}{2\pi i}\int_{\sigma - i \infty}^{\sigma+i\infty}\Gamma(z)\phi(z)x^{-z}\,dz$ and $\phi(z) = \frac{1}{\Gamma(z)}\int_0^\infty f(x)x^{z-1}\,dx$ Now $\phi$ can be kouznetsov's iteration method because it tends to a constant as the imaginary argument grows. I'm not sure how kneser's solution behaves in the complex plane, but maybe you can represent its continuum sum this way. This leads me to wonder if we don't continuum sum the series term by term, but instead try to solve a functional equation f satisfies when $\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)tet(z)$ using the fact that $tet''(z) = tet'(z) \sum_{j=0}^{z-1}tet'(j)$ I can't continue to discuss what I mean at the moment, but I always had a rough idea on how the equations might work out using this. I'll work more out on paper so that what I say makes more sense, I'm preoccupied at the moment. I'm floored that you got so far using hermite polynomials though, I would've shied away the moment the power series attempt failed. « Next Oldest | Next Newest »

 Messages In This Thread Bold and Disappointing Experiment With Hermite Polynomials!!! - by mike3 - 01/18/2017, 09:39 AM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by JmsNxn - 01/18/2017, 10:20 PM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by mike3 - 01/19/2017, 03:14 AM RE: Bold and Disappointing Experiment With Hermite Polynomials!!! - by JmsNxn - 01/19/2017, 08:05 PM

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