Tetration series for integer exponent. Can you find the pattern?

[marraco]

...
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.

See http://math.eretrandre.org/tetrationforu...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

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 \( \\[15pt]

{^xb} \), but for \( \\[15pt]

{^{1.5}(x+r)} \). (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.

\( ^0(1+x) \,=\, \)\( {\color{Red} 1} \)\( + 0+ 0+0... \)

\( ^1(1+x) \,=\, \)\( {\color{Red} 1}+ x \)\( + 0+0+0... \)

\( ^2(1+x) \,=\, \)\( {\color{Red} 1}+ x+ x^2 \)\( +\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12}) \)

\( ^3(1+x) \,=\, \)\( {\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3} \)\( +\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12}) \)

Perhaps this is interesting for you: http://go.helms-net.de/math/tetdocs/Pasc...trated.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.

That's great. You found that the tetration of the pascal matrix produces the bj sequences.
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 \( \\[15pt]

{^{1.5}P} \), we would get the bj for trees of 1/2 height.