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 Alternate continuum sum formula? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/15/2009, 12:34 PM (This post was last modified: 12/15/2009, 08:01 PM by mike3.) Hi. I wonder. Is there some formula for the continuum sum of $\frac{1}{\mathrm{tet}(x)}$ (or some other tetration-based function that decays to 0 for tetration with base e or some other non-convergent base), or as involving a continuum product whose productand(?!) decays to 1? As these could be summed using Mueller's formula and the Mueller-like product formula. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/15/2009, 08:24 PM (This post was last modified: 12/16/2009, 12:08 PM by mike3.) Hmm. How could you derive an iterating formula from this similar to the iterating formula for the original continuum sum formula? That is, one like $\mathrm{tet}_b(x) = <\mathrm{something\ involving\ tet_b(x)}>$, but that doesn't also require you to explicitly take the derivative w.r.t the base? Also, with base e we get a zero denominator since $\ln(\ln(e)) = 0$. So what happens there? Or do you mean $\ln(z)^2$ (i.e. multiplicative power, not compositional power, of logarithm) by "$\ln^2(z)$", in which case there is no problem? But this is really a cool result, since by Mueller's formula we can sum the continuum sums, and the reciprocal of tetration's derivative for real bases greater than $\eta = e^{1/e}$ converges to zero very very quickly, so quickly (tetrationally quickly ) that for any conceivable approximation level we need only a single-digit number of terms in the Mueller formula. Thus if an iterating formula can be found, it may be the fastest and most efficient method yet for the computation of the tetrational function, way better than all the other methods proposed so far (natural iteration, Carleman matrix iteration, Cauchy integral, etc.), perhaps good enough to enable the computation of tetration to extremely huge precision. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/16/2009, 08:50 PM (This post was last modified: 12/19/2009, 07:05 AM by mike3.) (12/16/2009, 02:52 PM)Ansus Wrote: Quote:Also, with base e we get a zero denominator since This is not twice-iterated logarithm, this is square of logarithm. Thanks for clearing that up. What about my other question, then, about whether there is an iterating formula for the function similar to the one from the original continuum sum formula? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/19/2009, 07:37 AM (12/16/2009, 08:50 PM)mike3 Wrote: (12/16/2009, 02:52 PM)Ansus Wrote: Quote:Also, with base e we get a zero denominator since This is not twice-iterated logarithm, this is square of logarithm. Thanks for clearing that up. What about my other question, then, about whether there is an iterating formula for the function similar to the one from the original continuum sum formula? By iterating formula, I mean this: Original continuum sum formula: $\log_z\left(\frac{\mathrm{sexp}'_z(x)}{\mathrm{sexp}'_z(0) (\log z)^x}\right) = \sum_{k=0}^{x-1} \mathrm{sexp}_z(k)$ Iterating formula: $\mathrm{sexp}_z(x) = \int_{-1}^{z} \mathrm{sexp}'_z(0) (\log z)^x \exp_z\left(\sum_{k=0}^{x-1} \mathrm{sexp}_z(k)\right)$. So is there an analogue of the second formula with these alternate formulas, that doesn't require knowing in advance (or allows one to determine) what the function of x $D_z \mathrm{sexp}_z(x)$ is? If so, what is it? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/20/2009, 08:01 AM (12/20/2009, 06:31 AM)Ansus Wrote: It is possible to use this formula and substitute sexp as inverse function operator of slog. $\operatorname{slog}_z C=-\int \left( \frac{1}{z (\ln z)^2}\sum_{q=0}^{\operatorname{slog}_z C -1}\frac{1}{D_q \operatorname{sexp}_z(q-1)}\right) dz$ Then it will be an iterating formula for slog. But this approach is complicated. Why would it be so complicated? If we represent our slog as a powerseries, we can apply the Lagrange inversion theorem. Also, what should be the bounds on the integral? And if this is too complicated, is there anything simpler? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/20/2009, 08:48 AM (12/20/2009, 08:35 AM)Ansus Wrote: Quote:Lagrange inversion theorem Not only. There are multiple methods. So then why's it so hard? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/20/2009, 09:03 AM (12/20/2009, 08:55 AM)Ansus Wrote: See for example, this: http://math.eretrandre.org/tetrationforu...hp?tid=407 And how does this relate to the integral+contiuum-sum formula you gave earlier? And it would not seem convergent for, e.g. $b > e^{1/e}$ or $0 < b < 1$. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/20/2009, 09:30 AM (12/20/2009, 09:13 AM)Ansus Wrote: (12/20/2009, 09:03 AM)mike3 Wrote: And how does this relate to the integral+contiuum-sum formula you gave earlier? And it would not seem convergent for, e.g. $b > e^{1/e}$ or $0 < b < 1$. Yes. But I do not know what inverse function expansion converges for $b > e^{1/e}$. Wouldn't the integral+continuum-sum thing you gave before converge for those? Because $\frac{1}{D_x \mathrm{tet}_b(x)}$ rapidly goes to zero as $x \rightarrow \infty$ for such bases. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/20/2009, 08:11 PM (This post was last modified: 12/20/2009, 08:12 PM by mike3.) (12/20/2009, 10:30 AM)Ansus Wrote: I do not understand what are you saying. Integral+cont. sum is not an inversion formula. Direct Taylor inversion, Lagrange inversion, polynomial inversion, rational inversion all diverge at those bases I suppose. What I'm saying is this other formula doesn't seem to work where the integral/continuum-sum would, so I wasn't sure why you mentioned it. And Andrew Robbins put together a wholly different approach, that converged to the inverse function (slog) for bases $b > e^{1/e}$. The resulting powerseries did invert under the Lagrange theorem to yield the tetrational. So inversion does work, once you have a method that can yield the right powerseries to invert. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/21/2009, 08:57 PM (12/20/2009, 06:31 AM)Ansus Wrote: It is possible to use this formula and substitute sexp as inverse function operator of slog. $\operatorname{slog}_z C=-\int \left( \frac{1}{z (\ln z)^2}\sum_{q=0}^{\operatorname{slog}_z C -1}\frac{1}{D_q \operatorname{sexp}_z(q-1)}\right) dz$ Then it will be an iterating formula for slog. But this approach is complicated. Hmm. I noticed this when preparing for the Mueller sum: $ \sum_{q=0}^{\mathrm{slog}(x)-1} \frac{1}{\mathrm{tet}'(q-1)} = \sum_{q=-1}^{\mathrm{slog}(x)-2} \frac{1}{\mathrm{tet}'(q)} = \frac{1}{\mathrm{tet}'(-1)} + \frac{1}{\mathrm{tet}'(0)} + \left(\sum_{q=1}^{\mathrm{slog}(x)} \frac{1}{\mathrm{tet}'(q)}\right) - \frac{1}{\mathrm{tet}'(\mathrm{slog}(x))} - \frac{1}{\mathrm{tet}'(\mathrm{slog}(x)-1)}.$ But the last term has a singularity at 0! (note that $\mathrm{slog}(0) - 1 = -2$) Does this help "complicate" things? « Next Oldest | Next Newest »

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