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z^^z ?
#1
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
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#2
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
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#3
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
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#4
(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 not of .
(Though you can express the inverse of by LambertW.)
Seems you suggest a misleading naming, do you?
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#5
(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... Smile

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#6
(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 not of .
(Though you can express the inverse of 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
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#7
I edited to change from "e" to "superE" on my post #5, sorry. Wink
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#8
(01/29/2011, 11:18 PM)nuninho1980 Wrote: I edited to change from "e" to "superE" on my post #5, sorry. Wink

i dont know what your talking about actually.
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#9
(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. Wink

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.
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#10
(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. Wink

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 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 .
And the Super-Euler constant is the one (positive) fixpoint of .
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