Posts: 366

Threads: 26

Joined: Oct 2007

02/03/2008, 04:40 PM
(This post was last modified: 02/03/2008, 07:40 PM by Ivars.)
I was strugling to get in grips with infinite pentation of other base than e but failed.

So my question to experts:

Which y would satisfy the equation in the subject of the thread?

My guess is e^(-pi), so that:

e^(pi/2) [5] ( - infinity) = e^(-pi).

Ivars

Posts: 174

Threads: 4

Joined: Aug 2007

Thank you, Ivars. Let us see.

@ Andydude.

Could you please check the coordinates of the common intersection, for x < 0, and for b = e^(Pi/2), with your powerful slog and sexp machines, of:

- y = b # x = b-tetra-x;

- y = [base b]slog x, the inverse of the previous one;

- y = x, principal diagonal ?

The intersections of the three "tails" for x < 0 should correspond to b-penta(-oo).

But, I might be wrong.

GFR

Posts: 366

Threads: 26

Joined: Oct 2007

Is this very difficult or not interesting?

I still have not acquired software to be able to do it myself one day. I will proceed analytically, but that might take years

Ivars

Posts: 1,389

Threads: 90

Joined: Aug 2007

02/26/2008, 10:45 AM
(This post was last modified: 02/26/2008, 10:54 AM by bo198214.)
Ivars Wrote:Is this very difficult or not interesting?

The value must be somewhere around -2 (far from your guess of

) considering this picture showing the intersection of

with

.

I guess it is not symboblically expressable with

,

and

.

But it is not exactly -2, because

for all real

.

Posts: 366

Threads: 26

Joined: Oct 2007

Thanks!

Seems rather symmetrical, this value.

Ivars