z^^z ? tommy1729 Ultimate Fellow Posts: 1,455 Threads: 350 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 tommy1729 Ultimate Fellow Posts: 1,455 Threads: 350 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 tommy1729 Ultimate Fellow Posts: 1,455 Threads: 350 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 bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 01/29/2011, 10:36 AM (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? nuninho1980 Fellow Posts: 96 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... tommy1729 Ultimate Fellow Posts: 1,455 Threads: 350 Joined: Feb 2009 01/29/2011, 03:32 PM (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 nuninho1980 Fellow Posts: 96 Threads: 6 Joined: Apr 2009 01/29/2011, 11:18 PM I edited to change from "e" to "superE" on my post #5, sorry. tommy1729 Ultimate Fellow Posts: 1,455 Threads: 350 Joined: Feb 2009 01/30/2011, 06:41 PM (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. nuninho1980 Fellow Posts: 96 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. bo198214 Administrator Posts: 1,391 Threads: 90 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 $1.6353...[4]x$. « Next Oldest | Next Newest »