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mike3 · 09/14/2009, 09:43 PM

Is there any way in which this could be done with something other than the matrix operator? For example, could a limit formula be used like that for $b = e^{1/e}$ , for this base ( $b = e^{-e}$ ) even if convergence is dog slow? If so, what would it be? One doesn't need many digits to plot graphs, and once one has enough points on an interval of towers, say from 0 to 1, it may be possible to interpolate them using a polynomial, and then one can use the functional equations to rapidly draw a graph along a wider interval of towers.

***bo198214*** · (This post was last modified: 09/17/2009, 08:02 AM by bo198214.)

(09/14/2009, 09:43 PM)mike3 Wrote: Is there any way in which this could be done with something other than the matrix operator? For example, could a limit formula be used like that for $b = e^{1/e}$ , for this base ( $b = e^{-e}$ ) even if convergence is dog slow?

Hm actually, I am not aware of such a limit formula. I only know $|\lambda|\neq 0,1$ and $\lambda=1$ but there should be also a formula somewhere for your case of $\lambda=-1$ (or generally for $\lambda^m = 1$ for some $m$ , here $m=2$ ). If I have some time I will look through some books.

The powerseries development of the regular slog/sexp probably anyway has zero convergence radius.

mike3 · 09/17/2009, 11:03 AM

(09/17/2009, 08:01 AM)bo198214 Wrote:
(09/14/2009, 09:43 PM)mike3 Wrote: Is there any way in which this could be done with something other than the matrix operator? For example, could a limit formula be used like that for $b = e^{1/e}$ , for this base ( $b = e^{-e}$ ) even if convergence is dog slow?

Hm actually, I am not aware of such a limit formula. I only know $|\lambda|\neq 0,1$ and $\lambda=1$ but there should be also a formula somewhere for your case of $\lambda=-1$ (or generally for $\lambda^m = 1$ for some $m$ , here $m=2$ ). If I have some time I will look through some books.

The powerseries development of the regular slog/sexp probably anyway has zero convergence radius.

One can get a limit formula for $b = e^{1/e}$ as the behavior at large values of the tower is somewhat like $e - \frac{2e}{x}$ (x = tower). And the bases $e^{-e} < b < e^{1/e}$ have exponential behavior (exponential-like convergence to their fixed point) at large towers. But what sort of behavior does $e^{-e}$ have at large values of the tower? It's not exponential and it's also not a simple rational function either (as it was for $b = e^{1/e}$ ).

tommy1729 · 09/18/2009, 12:16 PM

ok , i havent followed the thread , but tetration for base 0 < b < e^e seems trivial to me ?

i mean , you get a fixed point for b^x = x , so you just use that fixed point to do regular half-iterates ?

and in a similar way any positive real times of iterations !?

regards

tommy1729

***bo198214*** · 09/18/2009, 01:00 PM

(09/18/2009, 12:16 PM)tommy1729 Wrote: i mean , you get a fixed point for b^x = x , so you just use that fixed point to do regular half-iterates ?

As I already wrote there are formulas for regular iteration for $|\lambda|\neq 0,1$ and for $\lambda=1$ (where $\lambda$ is the derivative at the fixed point). In the case $b=e^{-e}$ the derivative is $\lambda=-1$ , so you have digg deeper in the literature (though the matrix power iteration at that fixed point should do also, however if $|\lambda|=1$ usually the resulting powerseries does not converge) about the case where $\lambda$ is a (in our case: second) root of unity.

For the case $0<b<e^{-e}$ we have a repelling fixed point. This gives an entire solution, which would imply that $\operatorname{sexp}(-1)\neq 0$ . This is not what mike3 wants. There should be a singularity at -2.

Its also slightly unpolite to not read the thread and then ask questions that are already answered/explained in the thread.

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