• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Breaking New Ground In The Quest For The "Analytical" Formula For Tetration. mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 03/20/2011, 03:45 AM (This post was last modified: 03/20/2011, 03:47 AM by mike3.) Hi. I just managed to put together the hard proof of the explicit, non-recursive formula for the solution of general recurrences of the form that I mentioned here: http://math.eretrandre.org/tetrationforu...42#pid5542 The note with the proof is attached to this post. Comments, critique, etc. would be welcomed. This proves the existence of the explicit non-recursive expression for the coefficients of the regular Schroder function of , and indeed, of the coefficients of regular Schroder functions in general. So the question now becomes: is there a more "elegant" expression? I'll give some attempts at trying to take a whack at it here. 1. The regular Schroder function (at the natural fixed point 0) coefficients are given by the above recurrence formula, or the general explicit formula, with . (where is a Stirling number of the 2nd kind.) Let the resulting coefficients "" be denoted . Then the Schroder function is . 2. Now, we go and expand these out. The first few coefficients look like: . ... Factoring the denominators and pattern recognition suggests where are the so-called "Schroder function magic numbers". 3. We want to turn this into a recurrence on the numerators. Let denote the numerator of the nth term . Then we have . Now insert the recurrent expression for : . But beyond here we run out of luck. The next step would be to set the sum for in this and try to solve for a recurrence for the and then try to come up with an explicit non-recursive formula for that, but it gets hairy and we don't have an expression for the expanded-out product there. So the first order of business here would be to find the explicit, non-recursive formula for the coefficients of . Any ideas? The degree of the resulting polynomial is . Also, . We don't set it to 1, since if you look at the denominator formula, you'll notice it is equal to -1 there and we have . Attached Files ExplicitFormulaProofq.pdf (Size: 68.76 KB / Downloads: 525) mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 03/28/2011, 11:51 AM (03/20/2011, 03:45 AM)mike3 Wrote: But beyond here we run out of luck. The next step would be to set the sum for in this and try to solve for a recurrence for the and then try to come up with an explicit non-recursive formula for that, but it gets hairy and we don't have an expression for the expanded-out product there. So the first order of business here would be to find the explicit, non-recursive formula for the coefficients of . Any ideas? The degree of the resulting polynomial is . Also, . We don't set it to 1, since if you look at the denominator formula, you'll notice it is equal to -1 there and we have . Add: For this product, we can write where is another sequence of magic numbers defined for . To find the equations for , we use then . Now, letting , we have The last formula allows for equating coefficients, yielding a three-piece recurrence: , for , for , for with the appropriate initial conditions, e.g. , , for , . Not sure if this is of any use or helps in getting explicit coefficient formulas, though. mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 03/28/2011, 11:04 PM (This post was last modified: 03/29/2011, 03:37 AM by mike3.) Going further, we can get recurrence formulas for . We have . . Now consider all occurrences of for a given j. That will be the recurrence for generating the . Let's see where that goes. We multiply the two sums together to get . Then we have . To get the coefficient for a given j, we note that the sum over j will only include that j when , which means . So we get as the atrocious recurrence for the magic numbers. What to do with that? (P.S. when should be 1) mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 04/10/2011, 08:33 PM Breaking news: On this thread on sci.math: http://groups.google.com/group/sci.math/...e91f?hl=en I discussed what amounted to a special case of finding the coefficients for the finite product in the formulas. Apparently, this does not have a known solution, and the solution for a related problem (partition numbers) looked to be extremely non trivial, involving sophisticated mathematics. So it seems that for now, the binary-counting based formula will just have to do... tommy1729 Ultimate Fellow     Posts: 1,480 Threads: 354 Joined: Feb 2009 04/10/2011, 09:46 PM (04/10/2011, 08:33 PM)mike3 Wrote: Breaking news: On this thread on sci.math: http://groups.google.com/group/sci.math/...e91f?hl=en I discussed what amounted to a special case of finding the coefficients for the finite product in the formulas. Apparently, this does not have a known solution, and the solution for a related problem (partition numbers) looked to be extremely non trivial, involving sophisticated mathematics. So it seems that for now, the binary-counting based formula will just have to do... what about contour integrals ? euler expressed it as an n'th derivative and those can be given as contour integrals. its not a sum though nor simple. ( btw as for p(n) being a fractal , i said that before. ) tommy1729 mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 05/09/2011, 05:08 AM (This post was last modified: 05/10/2011, 01:06 AM by mike3.) Meh, it looks like the whole approach of trying to directly find the "magic" numbers is insoluble. So it seems we're left with trying to simplify/"elegantize" the "ugly" binary formula in some other way. Anyway, I noticed that that "binary" formula can be written simply as a sum over subsets of an interval of naturals (note that binary = indicator functions), giving a neater, more compact form: . Then, . I didn't try a form like this before since I didn't consider it "explicit" enough for my liking (I'm a fan of sums over straight indices, and I was hoping the straight-index form would help, but alas, no dice.) But this lets us write the Schroder coefficients as . We can eliminate the factorial quotient from the product since the product multiplies it to , giving . A "simpler" formula for this would then mean one that sums over fewer terms and also has straight indices and no binary/subset/etc. stuff. This sums over terms. In theory, it should be possible to get it to . How can we do this? The above is very general -- we can express Bernoulli numbers and what not in a similar form. But in that case, and in a lot of other cases, there exists a form with fewer terms, straight indices and no binary/subset/etc. stuff -- just some nested sums and products. Is that possible here as well? Is this formula already known somewhere? « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post There is a non recursive formula for T(x,k)? marraco 5 2,523 12/26/2020, 11:05 AM Last Post: Gottfried Recursive formula generating bounded hyper-operators JmsNxn 0 3,095 01/17/2017, 05:10 AM Last Post: JmsNxn Extrapolated Faá Di Bruno's Formula Xorter 1 4,215 11/19/2016, 02:37 PM Last Post: Xorter Explicit formula for the tetration to base $$e^{1/e}$$? mike3 1 5,224 02/13/2015, 02:26 PM Last Post: Gottfried Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 6,546 07/17/2012, 02:12 AM Last Post: JmsNxn fractional iteration by schröder and by binomial-formula Gottfried 0 3,883 11/23/2011, 04:45 PM Last Post: Gottfried simple base conversion formula for tetration JmsNxn 0 4,563 09/22/2011, 07:41 PM Last Post: JmsNxn Change of base formula using logarithmic semi operators JmsNxn 4 11,848 07/08/2011, 08:28 PM Last Post: JmsNxn Constructing the "analytical" formula for tetration. mike3 13 27,564 02/10/2011, 07:35 AM Last Post: mike3 One very important formula Ansus 7 16,038 11/03/2010, 04:21 AM Last Post: Ansus

Users browsing this thread: 1 Guest(s) 