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 Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/06/2007, 01:34 PM Ivars Wrote:Im (W) = +- ipi/2 then branches k=-1 and k=+1 are accordingly -i(pi/2 - 2pi) = - 5/2pi i ; +i(pi/2+2pi) = + 5/2 pi*i and so on for all branches. 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 LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I LambertW(-1,-Pi/2)=-1.570796327*I LambertW(0,-Pi/2)=1.570796327*I LambertW(1,-Pi/2)=-1.604290913+7.647192276*I LambertW(2,-Pi/2)=-2.198342630+13.98120831*Iwhere the first argument denotes the branch. 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). Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/06/2007, 10:14 PM Hej BO, bo198214 Wrote: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 LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I LambertW(-1,-Pi/2)=-1.570796327*I LambertW(0,-Pi/2)=1.570796327*I LambertW(1,-Pi/2)=-1.604290913+7.647192276*I LambertW(2,-Pi/2)=-2.198342630+13.98120831*Iwhere the first argument denotes the branch. 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 Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 11/07/2007, 12:12 AM 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 Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 11/07/2007, 12:30 AM bo198214 Wrote:Gottfried Wrote:there is an article of L Euler, which seems to deal with it. Where did you dig out this article?! 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 Gottfried Helms, Kassel bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/07/2007, 10:12 AM Gottfried Wrote: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) So how do you finally compute the $k$th branch of $h$ or $W$ with this formula? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/07/2007, 11:59 AM Hej Gotfried, Congratulations! There are no accidents. Quote:So how do you finally compute the $k$th branch of $h$ or $W$ with this formula? Yes, how would You? And particularly, in case h(e^pi/2) and h(e^-pi/2)? Waiting impatiently even Ivars Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 11/07/2007, 12:58 PM (This post was last modified: 11/07/2007, 01:12 PM by Gottfried.) Ivars Wrote:Hej Gotfried, Congratulations! There are no accidents. Quote:So how do you finally compute the $k$th branch of $h$ or $W$ with this formula? Yes, how would You? And particularly, in case h(e^pi/2) and h(e^-pi/2)? Waiting impatiently even Ivars 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 Gottfried Helms, Kassel bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/07/2007, 01:48 PM (This post was last modified: 11/07/2007, 02:11 PM by bo198214.) It is not that difficult to compute the branches of $h$. The branches of $h(b)$ are simply the fixed points of $b^z$. In this way you can compute the branches $k=0,\dots,\infty$ via $h_k(b)=\lim_{n\to\infty}\log_{k,b}^{\circ n}(-1)$, where $\log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)}$. This is however valid only for $b\notin [e^{-e},e^{1/e}]$ as the numbering of the branches in this interval may be a bit different. For $b=e^{1/e}$ the branch 0 and -1 (which are non-real) join and for slightly smaller $b$ it splits again into two real branches. The negative branches are simply the conjugates of the positive branches:$h_{-k}(b)=\overline{h_{k-1}(b)}$. Edit: As it is so easy to compute $h$ it is perhaps more appropriate to compute $W$ from $h$ by $W(z)=z\cdot h(e^{-z})$ instead of computing $h$ from $W$. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/07/2007, 10:03 PM 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 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/07/2007, 10:26 PM (This post was last modified: 11/07/2007, 10:35 PM by bo198214.) Ivars Wrote: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. Yes the notation, mainly you have to know what $f^{\circ n}(x)$ means, it is simply the $n$ times repetition of $f$ applied to $x$, for example $f^{\circ 3}(x)=f(f(f(x)))$. In our example is then $b=e^{\frac{\pi}{2}}$. Because $k=-2$ we compute $h_{-2}(b)=\overline{h_{2-1}(b)$ which means the complex conjugate of $h_1(b)$. To compute this we consider $\log_{1,b}(z) = \frac{\log(z)+2\pi i}{\pi/2}=\frac{2}{\pi}\log(z)+4i$ and then compute $h_1(b)=\lim_{n\to\infty} \log_{1,b}(z)^{\circ n}(-1)$. This can of course only approximately computed. Have a look at some sample values: $\log_{1,b}(-1)=6*I$ $\log_{1,b}^{\circ 2}(-1)=\log_{1,b}(\log_{1,b}(-1))=1.140669505+5.000000000*I$ $\log_{1,b}^{\circ 10}(-1)=1.021323305+4.868353812*I$ $\log_{1,b}^{\circ 100}(-1)=1.021323316+4.868353806*I$ But you can see that the value converges, $h_1(b)\approx 1.021323+4.8683538*I$ hence $h_{-2}(b)\approx 1.021323-4.8683538*I$. This is indeed the same value, you get when computing $\text{LambertW}(-2,-\pi/2)/(-\pi/2)$ with Maple. « Next Oldest | Next Newest »

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