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+--- Thread: Alternate continuum sum formula? (/showthread.php?tid=404)
Alternate continuum sum formula? - mike3 - 12/15/2009
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.
RE: Alternate continuum sum formula? - mike3 - 12/15/2009
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.
RE: Alternate continuum sum formula? - mike3 - 12/16/2009
(12/16/2009, 02:52 PM)Ansus Wrote:Quote:Also, with base e we get a zero denominator sinceThis 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?
RE: Alternate continuum sum formula? - mike3 - 12/19/2009
(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 sinceThis 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?
RE: Alternate continuum sum formula? - mike3 - 12/20/2009
(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?
RE: Alternate continuum sum formula? - mike3 - 12/20/2009
(12/20/2009, 08:35 AM)Ansus Wrote:Quote:Lagrange inversion theoremNot only. There are multiple methods.
So then why's it so hard?
RE: Alternate continuum sum formula? - mike3 - 12/20/2009
(12/20/2009, 08:55 AM)Ansus Wrote: See for example, this:
http://math.eretrandre.org/tetrationforum/showthread.php?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 \( 0 < b < 1 \).
RE: Alternate continuum sum formula? - mike3 - 12/20/2009
(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.
RE: Alternate continuum sum formula? - mike3 - 12/20/2009
(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.
RE: Alternate continuum sum formula? - mike3 - 12/21/2009
(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?