Sum formula and "value at -1" mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 11/25/2009, 06:31 AM Hi. Is it possible that the continuum sum formula of Ansus $\frac{\mathrm{tet}'(z + z_0)}{\mathrm{tet}'(z_0)} = \exp\left(\sum_{n=0}^{z-1} \mathrm{tet}(n + z_0)\right)$ or $f(x) = K \int_{-1}^{x} (\log(b))^t \exp_b\left(\sum_{k=0}^{t-1} f(k)\right) dt$. can be iterated with only the function in a small region about x = 0 or -1... i.e. is there some way to do the integral or differentiation without knowing the values at x = 0 or -1 explicitly? Suppose a method is found that can iterate the sum operator on a Taylor series, but there is a singularity too close to 0 (within radius 1 of 0) that it cannot converge at -1. Consider base 1/4. As a Taylor series at 0, I think it has too small a convergence radius to go down to -1, due to near singularities in the complex plane, at least for the regular iteration (and I presume, as some experiments with continuum sum detailed here http://math.eretrandre.org/tetrationforu...hp?tid=370 for the simple case of tetration converging to a fixed point, which currently seems to be the only handleable case, suggest, it "should" agree with the regular iteration), so what would one do if one wanted to compute it via the continuum sum formula with a Taylor series at 0 or -1, or is this impossible (or do we need some way to sum Taylor series outside the convergence radius... Mittag-Leffler again?)? « Next Oldest | Next Newest »

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