06/01/2011, 03:03 AM
(This post was last modified: 06/01/2011, 03:28 AM by sheldonison.)

(05/31/2011, 09:05 AM)bo198214 Wrote:(05/29/2011, 09:37 PM)nuninho1980 Wrote: x of max |y| may be 3.08853227...(we remember that this new number succeed "Euler")....

I think I understood now what he meant!

Like the Euler number is the argument x where b^x = x for the b where b^x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (i.e. ),

the "succeeding Euler number" is the argument x where b[4]x = x for the b where b[4]x has exactly one fixpoint - in the transition from two fixpoints to no real fixpoint (on x>0, because there is always a fixpoint between -2 and 0 for the tetrational.)

This b is around 1.635....

And it looks in the graph as if the maximum of the difference is achieved exactly at this new Euler number....

I refreshed my memory on Henryk's graph, and on the Tetra-Euler constant. Nuinho really like's that number! Last year, I posted my own tetra-Euler results and graph. But I think in this particular case, the connection between Henryk's graph and the Tetra-Euler constant is probably a coincidence.

For base= , the lower fixed point=2, and the upper fixed point=4. As I understand it, the range of Henryk's graph goes from the lower fixed point to the upper fixed point. He is graphing , where f(z) is the upper entire , and g(z) is the lower superfunction , and k is the average of the difference between the two. Going from 4=upper fixed point at -infinity, down to 2=lower fixed point at +infinity, there is a small 1-periodic wobble that defines the difference between these two functions, where .

The problem is that as the base gets closer and closer to eta, the upper fixed point can get arbitrarily close to e, so the maximum, can't occur at the Tetra-Euler number, ~3.0885322718067176544821807826411.

For example, for B=1.4409, the upper fixed point=3.07763784998, and the lower fixed point=2.42379627885. So, then the maximum can't occur at Tetra-Euler, since Tetra-Euler is outside the range of the graph, right?

- Sheldon