Posts: 1,924

Threads: 415

Joined: Feb 2009

01/18/2011, 01:44 PM
(This post was last modified: 01/18/2011, 01:58 PM by tommy1729.)
we have studied b^^z and z^^b alot.

they can be expressed by sexp_b and slog_b.

but what about f(z) = a , a^^a = z ?

so apart from superlog and superexp , how about a superlambert ?

what brings us to the core of the question :

is superlambert(z) always defined on C* ?

do we need " new numbers " to compute superlambert(z) ?

to give a nice equation for the superlambert , superlambert solves for 'a' in :

sexp_a(a) = z

and how do we solve that ?

maybe like this :

sexp_a(a) + a = z + a

a = sexp_a(a) + a - z

and take the iteration :

a_(n+1) = sexp_a_n(a_n) + a_n - z

to find 'a' by the limit.

however that seems dubious and/or chaotic to me.

not to mention iteration cycles.

what else can we do ?

is there an easy proof that there is always a complex 'a' ?

i think we can conclude there is always a complex 'a' because of riemann surfaces IF f(z) or z^z is locally holomorphic for each z.

this is because IF z^^z is locally holomorphic and defined for all z , then z^z has a riemann surface with range C* , so we have a mapping from C* to C*

... which can be inverted to arrive at another C* to C* mapping.

however notice the 2 IF's and the fact that we havent managed to work in all bases yet !

regards

tommy1729

Posts: 1,924

Threads: 415

Joined: Feb 2009

01/18/2011, 09:11 PM
(This post was last modified: 01/18/2011, 09:32 PM by tommy1729.)
i wanted to add that z = 0 or z = 1 are probably the only solutions such that

z^^z = 1

that means apart from those 2 branches there are no other branches for f(1) unless ALL branches intersect at one of those 2 branches of f(1).

the taylor series at that point , if existing , is thus either 1 + az + bz^2 + ... or (z-1) ( a' + b' z + c' z^2 + ...)

i find that fascinating.

also intresting is the question what is z^^z a superfunction of ?

you could define it as (f(z) + 1)^^(f(z) + 1).

and one could wonder about its fixed points.

furthermore , is z^^z actually garantueed to converge for all complex z ??

and of course it might depend alot on what kind of tetration we work with ...

tommy1729

Posts: 1,924

Threads: 415

Joined: Feb 2009

01/18/2011, 09:49 PM
(This post was last modified: 01/18/2011, 10:00 PM by tommy1729.)
oh another thing.

let * denote complex conjugate.

at least at one branch we should have f(z) = q <=> f(z*) = q* wherever the neighbourhood of f(z) and f(z*) is holomorphic and f(z) , f(z*) are on that branch.

tommy1729

Posts: 1,631

Threads: 107

Joined: Aug 2007

(01/18/2011, 01:44 PM)tommy1729 Wrote: we have studied b^^z and z^^b alot.

they can be expressed by sexp_b and slog_b.

but what about f(z) = a , a^^a = z ?

so apart from superlog and superexp , how about a superlambert ?

Superlambert?

LambertW is the inverse of \( xe^x \) not of \( x^x \).

(Though you can express the inverse of \( x^x \) by LambertW.)

Seems you suggest a misleading naming, do you?

Posts: 98

Threads: 6

Joined: Apr 2009

01/29/2011, 02:05 PM
(This post was last modified: 01/29/2011, 11:16 PM by nuninho1980.)
(01/29/2011, 10:36 AM)bo198214 Wrote: Superlambert?

I know that - this new name is inverse of x^(superE^^x) but I don't know if this new formula is correct.

superE (= super euler) ~= 3.088532...

Posts: 1,924

Threads: 415

Joined: Feb 2009

(01/29/2011, 10:36 AM)bo198214 Wrote: (01/18/2011, 01:44 PM)tommy1729 Wrote: we have studied b^^z and z^^b alot.

they can be expressed by sexp_b and slog_b.

but what about f(z) = a , a^^a = z ?

so apart from superlog and superexp , how about a superlambert ?

Superlambert?

LambertW is the inverse of \( xe^x \) not of \( x^x \).

(Though you can express the inverse of \( x^x \) by LambertW.)

Seems you suggest a misleading naming, do you?

yeah , well maybe we should rename it ... i just like the sound of superlambert , but you got a point.

im open to name suggestions.

or do you suggest solving x*e^^x instead of x^^x ??

regards

tommy1729

Posts: 98

Threads: 6

Joined: Apr 2009

I edited to change from "e" to "superE" on my post #5, sorry.

Posts: 1,924

Threads: 415

Joined: Feb 2009

(01/29/2011, 11:18 PM)nuninho1980 Wrote: I edited to change from "e" to "superE" on my post #5, sorry.

i dont know what your talking about actually.

Posts: 98

Threads: 6

Joined: Apr 2009

01/31/2011, 01:39 PM
(This post was last modified: 01/31/2011, 01:43 PM by nuninho1980.)
(01/30/2011, 06:41 PM)tommy1729 Wrote: (01/29/2011, 11:18 PM)nuninho1980 Wrote: I edited to change from "e" to "superE" on my post #5, sorry.

i dont know what your talking about actually.

superE = 3.088532... can you remember?

but this my formula is incorrect after I failed test of this formula on "kneser" code by pari/gp. I can't change formula but people may change it.

Posts: 1,631

Threads: 107

Joined: Aug 2007

02/27/2011, 12:55 PM
(This post was last modified: 02/27/2011, 01:05 PM by bo198214.)
(01/30/2011, 06:41 PM)tommy1729 Wrote: (01/29/2011, 11:18 PM)nuninho1980 Wrote: I edited to change from "e" to "superE" on my post #5, sorry.

i dont know what your talking about actually.

There is this bifurcation base 1.6353... for the tetrational:

for b<1.6353... b[4]x has two fixpoints

for b=1.6353... b[4]x has one fixpoint

for b>1.6353... b[4]x has no fixpoint

on the positive real axis.

As you see, the bifurcation base 1.6353... of the tetrational corresponds to the bifurcation base \( e^{1/e} \) of the exponential.

(Also corresponds regarding other characterizations like the point b where b[4](b[4](b[4]...)) starts to diverge or the argument where the 4-selfroot is maximal)

The normal Euler constant e is now the one fixpoint of \( e^{1/e}[3]x \).

And the Super-Euler constant is the one (positive) fixpoint of \( 1.6353...[4]x \).