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)) - bo198214 - 11/06/2007
Ivars Wrote:Im (W) = +- ipi/2 Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example Code: `LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I` The numbering of the branches is of course convention, but seems to match we the one given in your article (which seems to be a quite good article). RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/06/2007
Hej BO, bo198214 Wrote:Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example I see, thanks. I appreciate a lot the lessons I get from You. I wonder if there is analytical way to derive all branch values for W(-pi/2) as function of k and hence h(e^pi/2). What does Mathematica says about the values of h(e^pi/2) and h(e^-pi/2) along few first branches? (I have no access to such instruments...). Any kind of pattern? From paper I have I see these formulas for Quadratix of Hippias but can not yet understand how these could be used to derive values of W(-pi/2) on other branches than 0 and -1. I still think there are actually 4 values h(e^pi/2) and may be even in some more general divergent cases of real number tetration: 2 corresponding to -i, 2 to +i , but I do not know yet how to find out if that is true and what are these values. The only case when there are only 2 distinquishable values ar +-i bacause the real parts here are 0.. I also make a conjecture that as number of branch k-> infinity, h(e^pi/2) = -1 independent of way it is reached. Also, I think there must be some relation between all h(e^pi/2) and h(e^-pi/2) branch values, possibly type of Mobius transform. Again , just a conjecture, to be proved wrong to make progress Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
Ivars Wrote:But, if I understand correctly ( probably not) the picture 4. in the article http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf if nice coincidence. I found a formula in this paper, which agrees perfectly with my fixpoints-for-real-bases b>e^(1/e) formula. In that article they discuss boundaries for the lambert-w-function such that x = eta ctg(eta) + i * eta and similar (page 15, formula 4.1 - 4.5) For the complex fixpoints for real b>e^(1/e) I had the same type of formula; such that u = beta cos(beta)/sin(beta) + I * beta t = exp(u) b = exp(u/t) is real and this last expression is a branch-enabled version of b = t^(1/t) Nice... Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
bo198214 Wrote:Gottfried Wrote:there is an article of L Euler, which seems to deal with it. That was *not* complicated (as I erroneously recalled). In fact, the scan is in the Euler-archive E489 I even had the link in my July-thread in sci.math. (Where is my brain currently? ) Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
Gottfried Wrote:For the complex fixpoints for real b>e^(1/e) I had the same type of formula; such that So how do you finally compute the th branch of or with this formula? RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/07/2007
Hej Gotfried, Congratulations! There are no accidents. Quote:So how do you finally compute the th branch of or with this formula? Yes, how would You? And particularly, in case h(e^pi/2) and h(e^-pi/2)? Waiting impatiently even Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
Ivars Wrote:Hej Gotfried, No chance.... ;-) Yes, there is the coincidence; but I did not compute the W-function but the function, which gives real bases b (or, to avoid confusion with the parameter-notation in the article:bases s) for exp(u/t) where s=t^(1/t) is the principal branch, given u (or more precisely: given the imaginary part of u), and get real bases s>1 . I set imag(u) = beta = any real value -pi < beta < pi compute real(u) according to the above formula thus having u completed, then compute t = exp(u) and then compute s = exp(u/t), which is then surely real. The formula for this is in the file fixpoints.pdf In this article I've not yet included the extension of the range for beta>pi; but I've done some computations with this and found further fixpoints for the real s in the regions 2*k*pi< beta < 2*(k+1)*pi. However, there is no exact periodicity, the consecutive fixpoints for the same s approach the lower bound 2*k*pi with the index k. I can find the inverse, to compute a fixpoint t by a given s, only by numerical approximation, since s=f(u) is monotonic, using a binary iterative process. If I *had* the branch-enabled Lambert-W-function, this would be easier, but Pari/GP doesn't have it and I still have only the python-example from wikipedia for the real-valued region, and have not yet invested in programming it myself. Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
It is not that difficult to compute the branches of . The branches of are simply the fixed points of . In this way you can compute the branches via , where . This is however valid only for as the numbering of the branches in this interval may be a bit different. For the branch 0 and -1 (which are non-real) join and for slightly smaller it splits again into two real branches. The negative branches are simply the conjugates of the positive branches:. Edit: As it is so easy to compute it is perhaps more appropriate to compute from by instead of computing from . RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/07/2007
hej BO, PLEASE post at least one numerical example how to use this limit formula e.f for k=-2, h(e^(pi/2)). I am not familiar (entirely my fault) with the notations You use - what is n, how to take the limit, what is z, what is b (I assume b=e^(pi/2) in my case. Thank You in advance, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
Ivars Wrote:PLEASE post at least one numerical example how to use this limit formula e.f for k=-2, h(e^(pi/2)). Yes the notation, mainly you have to know what means, it is simply the times repetition of applied to , for example . In our example is then . Because we compute which means the complex conjugate of . To compute this we consider and then compute . This can of course only approximately computed. Have a look at some sample values: But you can see that the value converges, hence . This is indeed the same value, you get when computing with Maple. |