Tetration series for integer exponent. Can you find the pattern? - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: Tetration series for integer exponent. Can you find the pattern? ( /showthread.php?tid=1053) |

Tetration series for integer exponent. Can you find the pattern? - marraco - 01/30/2016
with each integer power of n, the tetration converges, one extra coefficient (highlighted in red), to the Taylor series of , which gives the known asymptotic limit for the converged coefficients are 1, followed by the sequence http://oeis.org/A033917 (divided by n!) Some formulas for some of those tetrations are where are the Stirling numbers of the first kind. From the definition of the inverse Stirling transform, it looks like each tetration can be written as (1 plus...) a polynomial whose coefficients are inverse Stirling transform of the coefficients of some relatively simple function (coefficients not including the factorial denominator in the Taylor series) Here represents the last summation on the previous set of equations. The point of them being "relatively simple", compared to tetration, is that if we can find a general expression for , for any exponent n,it probably will be easier to generalize it to real exponents of tetration. the can be obtained by the direct Stirling transform of the coefficients shown at the start of the post. For example, the coefficients of the Taylor series of are 1 followed by [ 1, 1, 3/2, 4/3, 3/2, 53/40, 233/180, 5627/5040, 2501/2520, ...] (as shown at the start of this post). To get the , we need to remove the n! denominators of the Taylor series, and apply the Stirling transform. = [1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640, 27617832, 271481880, ...] and we get = [1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397, ...] Piece of code for calculation of Code: `tet=x+O(x^21)+1` gives this output: ^2(x+1): b_j=[1, 3, 10, 41, 196, 1057, 6322, 41393, 293608, 2237921, 18210094, 157329097] ^3(x+1): b_j=[1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397] ^4(x+1): b_j=[1, 3, 16, 125, 1176, 12847, 160504, 2261289, 35464816, 612419291, 11539360944, 235469524237] ^5(x+1): b_j=[1, 3, 16, 125, 1296, 16087, 229384, 3687609, 66025360, 1303751051, 28151798544, 659841763957] ^6(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 257104, 4480569, 87238720, 1874561291, 44057589984, 1124459440117] ^7(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4742649, 96915520, 2197675691, 54640864224, 1476693931957] ^8(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 99637120, 2323474091, 59755204224, 1676301882037] ^9(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2354318891, 61498237824, 1760945456437] ^10(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61877447424, 1786651875637] ^11(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61917364224, 1791681392437] According of OEIS, the sequence for seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725) For n=3, I made everything I could to simplify this last expression, to make it look as similar as possible to the others (for n=1, 2, and ). Can you make it simpler? Can you identify a pattern? Can you identify the exponent z in from ? (I wrote here the formulas for n=1,2,3 and ∞) RE: Tetration series for integer exponent. Can you find the pattern? - marraco - 01/31/2016
If we take for granted that the red terms on the series do not change when n grows, even for real values of n, then the difference eliminates all red terms of . So, the partial derivative has a taylor series that starts from the n+1 term, and can be taken common factor. This hints that the derivative of may have as a factor. RE: Tetration series for integer exponent. Can you find the pattern? - marraco - 02/13/2016
(01/30/2016, 09:06 PM)marraco Wrote: According of OEIS, the sequence for seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725) Following this, I had been trying to figure what a forest of fractional height is. should correspond to forests with height =. I tried various definitions for counting forests of half height, but none seems to work. Them I attempted to reverse engineer it by getting the coefficients first, and then trying to figure how to get them. My plan was to use Sheldonison's code to calculate the values of at various points, and numerically obtaining the coefficients of his Taylor series by applying finite differences. The problem is that for we need his Taylor series developed around x=1, and both knesser.gp and fatou.gp do not converge here. So, I planned to get the Taylor series around r, and with it, calculate back the Taylor around 1. I tried an r=1.01. calculated n (41) points of using knesser.gp, and from those points estimated the derivatives, and the bj values. This is the Pari/GP code I used. Code: `\r kneser.gp` Unfortunately, even using a large precision (it took more than one day running on an i7 920 processor), I only got 4 derivatives. The other have stability problems, and the ones I got are not very precise. Anyways, these are the closest numbers I got for bj of : Code: `%38 = [1.000152704474621237210741951681248733595287259524430467296, 2.861320494145353285475346429548911736678675196129994723805, -13.6475978846314412502384523327046630578, 756.92186972158813546692266102007258863]` Those should be "number of forests with height 1/2" for 1,2,3 and 4 nodes. The first number is the more accurate, but not for much. It should be 1, and the rest loses many digits of accuracy each one. So, for 2 nodes, there should be 2.86 forests of 1/2 height. But that number is surely a rough approximation. It may well be 2.5 instead of 2.86, and I find dubious that 3 nodes have a negative number of forests. I have no idea on how to improve precision. Increasing the decimals in Pari/GP does nothing. Maybe using a more accurate numerical difference equation? I used finite difference of order 40. RE: Tetration series for integer exponent. Can you find the pattern? - sheldonison - 02/14/2016
(02/13/2016, 06:07 AM)marraco Wrote: ... See http://math.eretrandre.org/tetrationforum/showthread.php?tid=729&page=2 post#15, In that post, there is a Taylor series for the 1st derivative of sexp_b(z) developed around b=2, and a Taylor series for sexp_b(-0.5), also developed around b=2. At the time I posted that, I also had developed Taylor series for the first 50 or so derivatives in the neighborhood of b=2. The key is to treat sexp_b(z) as analytic for the base b in the complex plane, in a circle around b=2. Then sample a set of points around a unit circle, and use Cauchy/Fourier which will give really good approximations to the derivatives. There is a mild singularity at b=eta which barely effects results at all, and a much much much more severe singularity at b=1. There is also reduced precision on the Shell Thron boundary. As I recall, the Taylor series posted were accurate to >25 decimal digits; using a relatively small number of sample points. I can post some pari-gp code that would work with fatou.gp to recreate that effort, if Maracco is interested. Of course, these results will be numeric, not a series. - Sheldon RE: Tetration series for integer exponent. Can you find the pattern? - Gottfried - 02/14/2016
(01/30/2016, 09:06 PM)marraco Wrote:Perhaps this is interesting for you: http://go.helms-net.de/math/tetdocs/PascalMatrixTetrated.pdf The tetration of the Pascalmatrix give that coefficients by matrix-exponentiation, and from this might possibly result smoother formulae for the expression of the single coefficients. Gottfried RE: Tetration series for integer exponent. Can you find the pattern? - marraco - 02/14/2016
(02/14/2016, 03:25 AM)sheldonison Wrote:(02/13/2016, 06:07 AM)marraco Wrote: ... As I understand, those are series for the variable in the exponent, but we need series for the variable in the base; a series for sexp_{x+r}(1.5). I mean not the series for , but for . (I write r instead of 1). That's why I used your knesser.gp to calculate points for different bases separated by n*dx steps. knesser gave me better precission than fatou. I don't know if I should had configured some parameter in fatou to get more digits. I also used a step dx=1e-15. When I tried a smaller dx=1e-20 precision was destroyed, so the error may be numerical instability inherent to the differentiation scheme. (02/14/2016, 03:43 AM)Gottfried Wrote:That's great. You found that the tetration of the pascal matrix produces the bj sequences.(01/30/2016, 09:06 PM)marraco Wrote:Perhaps this is interesting for you: http://go.helms-net.de/math/tetdocs/PascalMatrixTetrated.pdf The tetration of the Pascalmatrix give that coefficients by matrix-exponentiation, and from this might possibly result smoother formulae for the expression of the single coefficients. If we can iterate logaritms maybe we can figure what a tree of negative height is. I wonder if we can generate the tetrated pascal with a series, and if that series matches the series for some real base. That base should be important. If just we could calculate , we would get the bj for trees of 1/2 height. RE: Tetration series for integer exponent. Can you find the pattern? - sheldonison - 02/14/2016
sheldon Wrote:See http://math.eretrandre.org/tetrationforu...729&page=2 post#15, Those series are for the base not the exponent! That's why they needed to be generated with tetcomplex rather than kneser; because getting such amazingly good analytic convergence for such a series requires complex base tetration, with results for n bases, equally spaced around a circle centered on base=2. fatou.gp (with small code updates) could also generate such results. To evaluate those two referenced Taylor series, substitute x=(b-2), to evaluate the Taylor series. This is probably off topic, but bases between 1 and eta=exp(1/e) aren't the same function as Kneser tetration for real bases>eta. Also, when you develop the iterated exponential using regular iteration from the attracting fixed point for bases<eta, there is no super-exponential growth anywhere in the complex plane. There is a tetration function for bases between 1 and eta, but it is no longer real valued at the real axis. Regular iteration for bases<eta is a much simpler problem than tetration, but unfortunately it cannot be extended analytically in the complex plane to tetration for real bases>eta because it is a completely different function than Kneser tetration. RE: Tetration series for integer exponent. Can you find the pattern? - MorgothV8 - 02/14/2016
I'm not that good on math, but.... I can help as a programmer. Maybe precision problems are due to usage of "fixed" floating points (I mean fixed bit numbers). There are libraries that can work with "big number" or "arbitrary precision". For example GMP (this is a C/C++ library) gives non limited (limited by memory) number representation, so there is no problem with numbers like 10^(10^10) etc. And MPFR library does similar thing for floating point numbers ..... I've written some tetration programs using those two, and they're much more precise, but at the horrible cost of memory and time. RE: Tetration series for integer exponent. Can you find the pattern? - marraco - 02/14/2016
(02/14/2016, 07:55 AM)sheldonison Wrote: Those series are for the base not the exponent! That's why they needed to be generated with tetcomplex rather than kneser; because getting such amazingly good analytic convergence for such a series requires complex base tetration, with results for n bases, equally spaced around a circle centered on base=2. fatou.gp (with small code updates) could also generate such results. To evaluate those two referenced Taylor series, substitute x=(b-2), to evaluate the Taylor series. ...I'm royally confused now. the coefficients in the last code tag of this post are the Taylor series coefficients of the function ? Then it should be but this code gives Code: `\p 100` ... and why textcomplex.gp "generates tetration for arbitrary bases" instead of "arbitrary exponents"? Sounds like the same as knesser and fatou (the base is constant, and the exponent is the variable) RE: Tetration series for integer exponent. Can you find the pattern? - marraco - 02/14/2016
(02/14/2016, 06:25 PM)MorgothV8 Wrote: I'm not that good on math, but....I'm not even a mathematician. Just an engineer. Pari/GP uses arbitrary precision, and I used like 300 or 400 decimals. I also inverted the function for the calculation of the derivatives and verified that there is no loss of information. ...but beyond the precision used, the method of finite differences has inherent error. That may be the cause, or knesser may need more precision (with the one I choose, it took more than a day to run, but it may not be enough), or there is a bug on my posted code (ask for clarification if you want to check it), or all of those... |