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 Mizugadro, pentation, Book Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/12/2015, 12:27 AM (02/11/2015, 09:04 PM)Gottfried Wrote: http://mizugadro.mydns.jp/t/index.php/Tetration and looked at the entry for the tetration to base b=exp(1/e). I do not understand how you arrive at the polynomials, and especially how at the set of power-series in the table with the increasing m.Thank you Gottfried for your interest. The coefficients in the table are calculate using the primitive fit, quick and dirty. Quote: However, I could find the coefficients of the first row (m=0) by my standard procedure, and my method allows 12,16 or 20 correct digits for them. Can you evaluate tetration to the complex base? Quote:So if I knew, how our methods are related, and if they come out to be identical (at least for that exact base), then you could take something useful from my method, at least for the more accurate computation of the m=0 coefficients. I think, first we should check, that your tetration has the same behaviour as my one. Can you reproduce any figures from my book "Суперфункции" or from the article about "holomorphic ackermanns" with your algorithm? I mean for some definite, fixed value of base. Quote: .. Code:...:  ... 59:  1.000000000000000  0.6110954537716500   -0.2317026144766614   0.09178128765991140    -0.03756492168250494    0.01577372205052486 ...Where your coefficients were given as Code:m    c_{m,0}    c_{m,1}    c_{m,2}       c_{m,3}    c_{m,4} 0       1       0.61061   -0.23171       0.09225    -0.03757 1       0       0.69521    0.41315      -0.16027     0.07007 2       0      -0.57851    0.18323       0.49162    -0.15216 3       0       0.64730   -0.62933      -0.51128     0.51372 4       0      -0.84098    1.23261       0.42470    -0.97551 5       0       1.19090   -2.12653      -0.06895     1.57684 The data at m=0 match suspiciously well, however I've no further idea yet how to reproduce the rows at m=1,m=2 and so forth. Also, to exclude the possibility of a pure incidence/of a "false positive", it would perhaps be good if I could see more of the coefficients in that first row of your table... Why do you think that it is suspiciously well? Only 3 decimal digits. ".. precision of such evaluation is not so high (perhaps, only two or three decimal digits are significant in the estimates above).." These coefficients are estimated from the "general" fit, formula (17.1) in book "Суперфунцкии". You may extract the implementation from the generator of figure 17.1 in the same book. Then you may differentiate it so many times as you like. I doubt, if more coefficients have any sense: the precision of the fit is poor. It is barely sufficient to plot the camera-ready version of figure 17.1, but if you zoom-in the figure with scaling factor of order of 1000, you'll see the defects. For the precise evaluation of coefficients you ask, the tetration should be already implemented for the complex base, as complex tet(complex b, complex z); Do you have the efficient algorithm for the evaluation? Could you implement it in C++? Then I think, we can evaluate the coefficients of the expansion with some 14 digits. The idea is simple: You evaluate the derivatives of tetration for base $b$ at some loop around the point of interest, $t_n(b,z) = \mathrm{d}^n \mathrm{tet}_b(z) / \mathrm {d} z^n$ and $t_n(b) =t_n(b,0)$ For example, $b_k=\exp(1/\mathrm{e}) + r~ \exp(2 \pi \mathrm{i}k/N)$ for some $r\approx 0.2$ and $N=1024$ Then, you may calculate the coefficients through derivatives of these $t_n$ at zero as the Cauchi integral, approximating it as a discrete sum with respect to $k$ « Next Oldest | Next Newest »