• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Road testing Ansus' continuum product formula mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/23/2009, 09:10 AM (This post was last modified: 09/23/2009, 09:15 AM by mike3.) Where's that? The only thing I could find looked to have the Taylor coeffs. (or some other coeffs) hardwired. That's not what I want: I want one that will compute more coeffs. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/23/2009, 09:40 AM (This post was last modified: 09/23/2009, 09:56 AM by mike3.) Never mind, I realized I could experiment using the regular iteration at bases like b = sqrt(2), plus a cheap numerical differentiation procedure, to see how the coeffs. behave. It seems the terms in $b_k = \sum_{n=1}^{\infty} \frac{f^{(n-1)}(0)}{n!} {n \choose k} B_{n - k}$ for k = 2 (i.e. 2nd coeff of continuum sum of sqrt(2) tetration expanded about 0), f = regular iteration of b = sqrt(2), grow up in magnitude mildly hypergeometrically (but alternate in sign). I get (rounded) term n = 38: -143.931066425221 term n = 40: 1281.50456722963 term n = 42: -12659.7473479491 term n = 44: 138050.651934470 term n = 46: -1654016.61498652 term n = 48: 21681557.9712844 term n = 50: -309748957.739627 The ratio of magnitudes looks like (mn = magnitude of term n) m40/m38 = 8.90359947 m42/m40 = 9.87881563 m44/m42 = 10.9046925 m46/m44 = 11.9812300 m48/m46 = 13.1084282 The differences of these are (rn = ratio of term n to term n-2) r42 - r40 = 0.97521616 r44 - r42 = 1.02587687 r46 - r44 = 1.07653752 r48 - r46 = 1.12719814 In other words, the ratio of the magnitudes of successive coefficients grows slightly faster than linear, but not too much so. I'd venture it is between linear and quadratic. I heard that the Borel summation is not applicable to sums that grow hypergeometrically like this, but I heard here: (http://mathworld.wolfram.com/Borel-RegularizedSum.html) that it originates wth the summing of divergent hypergeometric functions, whose term magnitude ratios may grow up even faster than the ones here do. So could the Borel summation be applicable here? It would seem not to have so many fussy parameters, but here: http://en.wikipedia.org/wiki/Borel_summation I hear it needs an analytic continuation of a certain function to the whole positive real line, but how can you analytically continue a more-or-less arbitrary series like that? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/23/2009, 07:54 PM (09/23/2009, 09:52 AM)Ansus Wrote: Did you try this formula btw? $f_a(x)=\log_a\left(f_a(x)+\sum_{n=0}^\infty\frac{f_a^{(n+1)}(0)}{n!}B_n(x) \right)$ What function is $f_a$? Is that just some arbitrary function, or is it Tetration? If it's tetration, then it would seem to be similar to the Faulhaber formula. Addendum: Just tested it assuming f is tetration w/base sqrt(2) and numerical differentiation values for the derivatives. Yepper, conventional sum diverges just like the other. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/23/2009, 08:26 PM Interestingly, though, it seems Borel summation might work with the Faulhaber (and probably also this) formula, due to the not-too-bad divergence of the terms. I tried sum numerical tests using 1 integral and 60 derivatives, but that only seemed to get the Borel-function (the thing inside the Borel-summation integral) to converge in a radius of around 13, which isn't much, yet integrating up to that limit (from 0 to 13) got the coefficient of x^2 in the Tayor expansion of the sqrt(2) regular tetration's continuum sum at 0 as 0.439, and the coefficient of x as 0.666, both of which seem to agree with differentiation of the left hand side of the sum formula (the log with the quotient of the derivative of tetration, etc. inside it bit), though as I mentioned, I can't try for more accuracy than this (as I'll need to have to actually try generating series expansions of tetration for other bases, esp. the fabled b = 0.04, which I think the Borel summation may be more suited to as it doesn't require all those wacky parameters the other did) due to limited convergence radius. Yet the theory mentions about analytically continuing. There's also the possibility of trying a higher-order one (adds another factorial in the denominator) with double integrals, yet that would be even nastier to compute due to the whole 2d grid thing. However, I'm not sure how you'd analytically continue the sum in the Borel integrals past its convergence radius on the positive real axis to get more accuracy. Is there any good code for doing Borel summation that I could test with? « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post fixed point formula sheldonison 6 12,561 05/23/2015, 04:32 AM Last Post: mike3 Numerical algorithm for Fourier continuum sum tetration theory mike3 12 22,245 09/18/2010, 04:12 AM Last Post: mike3

Users browsing this thread: 1 Guest(s)