02/11/2020, 04:47 PM
(This post was last modified: 02/11/2020, 11:08 PM by sheldonison.)

(02/08/2020, 05:37 PM)Ember Edison Wrote: Sorry, but please test it:

Code:`matrix_ir(1E6,,,19/20);circchart("1E6.csv")`

Code:`hexinit(1.6);[ipent(3),ihex(2)]`

hexinit(1.7);[pent(8),ihex(3)]

hexinit(22);hex(2)

pentinit(3381);pent(2)

matrix_ir(1E9,,,19/20);sexp(1)

I'm not doing updates on fatou.gp until after I finish a rigorous proof of convergence proving a simplified version of fatou.gp for base e converges to Kneser's slog.

The upper limit for tetration bases supported is somewhere around base(35000).

pentinit(1.6); ipent(3) is a non-trivial problem since there are three different real valued fixed points for tet_b=1.6. See this thread about the tetration base with an upper parabolic fixed point (b~= 1.6353). For pentation, with bases smaller than this, the result is ambiguous since there are now three real fixed points. For pentation fatou.gp arbitrarily uses the lower fixed point, which is all we have for larger bases. But it is arbitrary since there is no uniqueness criterion for pentation analogous to the uniqueness criterion for Kneser's tetration. For pent(1.6); we see that sexp_1.6(-1.62779274874299)=-1.62779274874299 which is the lower fixed point, and is where the Schroeder equation for pentation is centered. But if we start at z=3, and iterate z<=sexp_1.6(z) then we get to the 1st upper fixed point= 2.26438440348116, and if we start at z=3 and iterate z<=slog(z), then we get to the 2nd upper fixed point=4.7558078933440; neither of which is the desired lower fixed point. The tetration b=1.6 image below shows the two additional upper fixed points that make finding ipent(3) difficult when using the lower fixed point's Schroder equation for pentation.

Hopefully that explains why I wouldn't update the fatou.gp code for pentinit(1.6). This is especially true given that my current primary focus is on a tetration proof.

- Sheldon