Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Alternate continuum sum formula?
#1
Hi.

I wonder. Is there some formula for the continuum sum of (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.
Reply
#2
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 , 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 . So what happens there? Or do you mean (i.e. multiplicative power, not compositional power, of logarithm) by "", 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 converges to zero very very quickly, so quickly (tetrationally quickly Smile ) 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.
Reply
#3
(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?
Reply
#4
(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:


Iterating formula:
.

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 is? If so, what is it?
Reply
#5
(12/20/2009, 06:31 AM)Ansus Wrote: It is possible to use this formula and substitute sexp as inverse function operator of slog.



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?
Reply
#6
(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?
Reply
#7
(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. or .
Reply
#8
(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. or .

Yes. But I do not know what inverse function expansion converges for .

Wouldn't the integral+continuum-sum thing you gave before converge for those? Because rapidly goes to zero as for such bases.
Reply
#9
(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 . 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.
Reply
#10
(12/20/2009, 06:31 AM)Ansus Wrote: It is possible to use this formula and substitute sexp as inverse function operator of slog.



Then it will be an iterating formula for slog.

But this approach is complicated.

Hmm. I noticed this when preparing for the Mueller sum:



But the last term has a singularity at 0! (note that ) Does this help "complicate" things?
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  Recursive formula generating bounded hyper-operators JmsNxn 0 1,293 01/17/2017, 05:10 AM
Last Post: JmsNxn
  Extrapolated Faá Di Bruno's Formula Xorter 1 1,823 11/19/2016, 02:37 PM
Last Post: Xorter
  2015 Continuum sum conjecture tommy1729 3 3,025 05/26/2015, 12:24 PM
Last Post: tommy1729
  Explicit formula for the tetration to base [tex]e^{1/e}[/tex]? mike3 1 2,657 02/13/2015, 02:26 PM
Last Post: Gottfried
  Another way to continuum sum! JmsNxn 6 5,942 06/06/2014, 05:09 PM
Last Post: MphLee
  Continuum sum = Continuum product tommy1729 1 2,388 08/22/2013, 04:01 PM
Last Post: JmsNxn
  applying continuum sum to interpolate any sequence. JmsNxn 1 2,618 08/18/2013, 08:55 PM
Last Post: tommy1729
  Powerful way to perform continuum sum JmsNxn 7 7,301 08/12/2013, 07:17 PM
Last Post: JmsNxn
  Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 3,668 07/17/2012, 02:12 AM
Last Post: JmsNxn
  fractional iteration by schröder and by binomial-formula Gottfried 0 2,432 11/23/2011, 04:45 PM
Last Post: Gottfried



Users browsing this thread: 1 Guest(s)