Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- Thread: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) ( /showthread.php?tid=54) |

RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - GFR - 09/21/2007
Thanks, Gottfried. I also shall read again your postings. My problem is that, as a ... bloody engineer, I am (... was?) more interested to what happens, in a more real environment, for b < eta. But, now, I think that what happens at b > eta should indeed be of the same "nature" (I mean "mathematical nature") of what hapens elsewhere. The ... "function" must be ONE. Best wishes. Gianfranco RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 09/22/2007
GFR Wrote:Thanks, Gottfried. Gianfranco, I append one post of mine of the newsgroup sci.math here. There was nearly exactly the same problem posed. Kind regards - Gottfried news://news.t-online.de:119/fd2bf9$q2e$03$1@news.t-online.com > > So as promising as this idea may seem, defining > > x^^(1/n) as a tetraroot will never give a function that will > > be continous. This is why Gottried Helms's (and others') > > functions disagree with the tetraroot -- x^^(1/n) must > > be less than the nth tetraroot for sufficiently large n in > > order for the function to be continous at zero. > > Hmm, I don't have a final answer for this yet. But with infinite series, especially in the complex domain, we have some couriosities everywhere, for instance the non-trivial zeros of the zeta-function. Also the function (1+1/x)^x x->inf, is a tricky one in this regard. Here we find, that the quantitative difference of the two approximations of 1/x->zero in the sum and x->inf in the exponent gives a surprising result. In a current thread in the tetration forum I posed the question of s^t = t for real s, complex t, and a correspondent focused this to the question of what happens, if s->inf. One can rewrite this as lim{s->inf} s = t^(1/t) then lim{s->inf} 1/s = (1/t)^(1/t) and then lim{s'->0} s' = t'^t' (1) and search for a limit in complex t', possibly sharpening by the condition, that imag(s')=0, not only in the limit. That there are solutions for finite s, thus nonzero s' seems to be obvious by numerical approximation, although I don't have a authoritative reference for this. If one resolves into t' = a + bi and log(t') = p + qi and expand (1) then we arrive at something like (a+bi)^(a+bi) -> 0 (a+bi)^a * (a+bi)^bi -> 0 (a+bi)^a * exp( bi * log(a+bi)) -> 0 (a+bi)^a * exp( bi * (p + qi)) -> 0 (a+bi)^a * exp( p*bi - q*b) -> 0 (a+bi)^a * exp(-qb)*exp( p*bi ) -> 0 where the last exp(p*bi) is just a rotation, and thus irrelevant for the convergence to zero. So it must also (a+bi)^a * exp(-qb) -> 0 and the exp()-term cannot equal zero, so the first term only must be zero, if not q or b diverge to infinity. Now factor a out to have a^a*(1 + b/a i)^a -> 0 and this reminds me to the above (1 + 1/x)^x - formula. I'm unable to proceed from here at the moment, but this seems to focus the core point of the problem. Possibly we arrive here at the famous indeterminacy, what is 0^0? It depends on the source of the term: whether the exponent or the base is -from the context of this formula- a limit-expression. On the other hand, there may be some application of L'Hospital needed here, I don't see it at the moment. Hmm. Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 10/20/2007
I've just uploaded a short discussion of the fixpoint-detail, as discussed in my previous postings. I'm trying to show, that by selection of a parameter beta I can construct a complex fixpoint, and it is made sure, that I get a purely real base b > eta. The graph and the formulae seem to indicate that the full range eta < b < inf can be constructed by a continuous function f(beta) = b for 0<= beta < pi, which would then confirm Gianfrancos and my own earlier formulae. See Fixpoints It's another sketchy manuscript of mine, but... ;-) Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 10/30/2007
Hello, I am new here- I was wandering about the imaginary unit for some time to be able to understand physics better, but I was quite positively surprised that you have confirmed by rigorous analysis what I just yesterday found out by studying the possible content of i- a definition ( or perhaps one of the definitions of i) -of imaginary unit. Perhaps there are many infinite values involved as well as this definition arises from Lambert function of a logarithm, and logarithm may take infinitely many values, but still I keep - i sign): -i = power tower (e^pi/2) = h (e^pi/2) i = - h(e^pi/2) so here z = e^pi/2 = 4,81047738097......... h(z) = - i Derivation: h(e^pi/2) = -W( -ln e^pi/2)/ ln e^pi/2 = -W(-pi/2) / pi/2 But W( -pi/2) = i*pi/2 so h(e^pi/2) = - i*pi/2 / pi/2 = - i . but - h(e^pi/2) = W(-pi/2) / pi/2 = i*pi/2/pi/2 = i There must be many more such beauties outside radius of convergence of h(z). Best regards, Ivars Fabriciuss RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/02/2007
Hey Ivars, welcome on board! Ivars Wrote:-i = power tower (e^pi/2) = h (e^pi/2) But be aware that this is mathematically wrong! Correct is: power tower (e^(pi/2)) = oo If you mean something different for example an infinite valued function or a fixed point of x^x then you have to explicitely state that. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/03/2007
hej bo198214 bo198214 Wrote:Thanks, I understood that ; I will try more carefully:Ivars Wrote:-i = power tower (e^pi/2) = h (e^pi/h2) Analytic continuation of h(z) defined as = -W(-ln(z))/ln(z) at point z=e^(pi/2) is - i. also, in this case: h(e^pi/2)^i = e^(pi/2) h(e^pi/2)^-i= e^(-pi/2) as -i^i=e^(pi/2) = 4,810477.. -i^-i = e^(-pi/2)=0,207879 i^i = e^(-pi/2)= 0,207879.. i^-i = e^(pi/2) = 4,810477 and i*ln h(e^pi/2) = pi/2 ln h(e^pi/2) = -i pi/2 = pi/2*(h(e^pi/2) (h(e^pi/2))^2 = -i^2 = e^-ipi = i^6 However, as e^-pi/2 > e^-e, h(e^-pi/2) = 0,474541.......= (2/pi)*W(pi/2) So there is no symmetry and i will not be a result of analytic continuation of infinite tetration of anything, while -i is. Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/03/2007
Ivars Wrote:However, as e^-pi/2 > e^-e, This depends on how you analytically continue h, on which path. W (and hence h) has a singularity at , so if you continue on different paths around , you possibly get different results (similar to the analytic continuation of the logarithm with singularity at 0). Surely on another path/branch . Edit: Perhaps its not at all "Surely", was just my conjecture. Edit2: Now its not a conjecture anymore but regarded as nonsense XD, instead on another branch. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/03/2007
Hej Bo, Would be interesing to see a concrete example where h(z) is analytically continued to get h(z) = i, on any branch, any z. Might contain interesting information, whatever the outcome. Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/03/2007
Ivar - there is an article of L Euler, which seems to deal with it. Its index in the Euler-archive is E489, and I found it available at some university-library with a digitizing service. I found the attached snippet interesting, and may be it inspires to have a deeper look into it. Gottfried [attachment=120] RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/03/2007
Gottfried Wrote:there is an article of L Euler, which seems to deal with it. Wow, Gottfried! Where did you dig out this article?! However it seems to be beneficial being able to read Latin ... At least I see in the text ... And there we have it already, Ivar. One branch of the LambertW function gives , hence for this branch . (Forget what I said about , that was a somewhat misled speculation). We can imagine the branches of given by the fixed points of . |