Posts: 368

Threads: 44

Joined: Sep 2009

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

, for this base (

) 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.

Posts: 1,386

Threads: 90

Joined: Aug 2007

09/17/2009, 08:01 AM
(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 , for this base () even if convergence is dog slow?

Hm actually, I am not aware of such a limit formula. I only know

and

but there should be also a formula somewhere for your case of

(or generally for

for some

, here

). 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.

Posts: 368

Threads: 44

Joined: Sep 2009

Posts: 1,358

Threads: 330

Joined: Feb 2009

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

Posts: 1,386

Threads: 90

Joined: Aug 2007

(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

and for

(where

is the derivative at the fixed point). In the case

the derivative is

, so you have digg deeper in the literature (though the matrix power iteration at that fixed point should do also, however if

usually the resulting powerseries does not converge) about the case where

is a (in our case: second) root of unity.

For the case

we have a repelling fixed point. This gives an entire solution, which would imply that

. 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.