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+--- Thread: z^^z ? (/showthread.php?tid=577)
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z^^z ? - tommy1729 - 01/18/2011
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
RE: z^^z ? - tommy1729 - 01/18/2011
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
RE: z^^z ? - tommy1729 - 01/18/2011
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
RE: z^^z ? - bo198214 - 01/29/2011
(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?
RE: z^^z ? - nuninho1980 - 01/29/2011
(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...
RE: z^^z ? - tommy1729 - 01/29/2011
(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
RE: z^^z ? - nuninho1980 - 01/29/2011
I edited to change from "e" to "superE" on my post #5, sorry.
RE: z^^z ? - tommy1729 - 01/30/2011
(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.
RE: z^^z ? - nuninho1980 - 01/31/2011
(01/30/2011, 06:41 PM)tommy1729 Wrote:superE = 3.088532... can you remember?(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.
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.
RE: z^^z ? - bo198214 - 02/27/2011
(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 \( 1.6353...[4]x \).