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 Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/22/2007, 10:36 PM Hej Gottfried In principle fits, as I want to get rid of real number line-I do not believe numbers are organized that way. There must be more logical and "mechanical" unified structure in mathematics which produces numbers, such where "unexpected connections" and "suprising identities" arise naturally. And so far Lambert function, and tetration related to it seems the most promising direction in finding at least some part of this structure. Also, as i and thus h(e^pi/2)= i means rotation, the processes governing math might be as well governing or even, equaling physics and flows of infinitesimals. So I think GFR is absolutely right, next question is to find which of the number expressions is the basic one. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/29/2007, 10:37 PM (This post was last modified: 12/09/2007, 02:57 PM by Ivars.) Ivars Wrote:h(1/n^n)=1/n h(1) +h(1/4) + h(1/27) + h(1/256) ...+h(1/n^n)... = harmonic series h(1)^2+h((1/4))^2+h((1/27))^2 +...+ h(1/n^n))^2...= pi^2/6 and so on for higher powers in series. Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e h((z^-y)^(z^y)) = z^-y but we know that h(e^pi/2)^h(e^pi/2) = i^i = -i^-i = e^-pi/2 on other hand, i^-i =i^(1/i) = e^( ipi/2*-i)= e^pi/2 so: h(e^pi/2)^1/h(e^pi/2) = e^pi/2 then we can generalize to z^y: h(z^y)^h(z^y) = z^-y h(z^y)^1/h(z^y) = z^y so that: h(z^y)^h(z^y)*h(z^y)^1/h(z^y) =1 or h(z^y)^ ((h(z^y))^2+1/h(z^y)) =1 which is possible always if (h(z^y))^2+1=0 -> h(z^y) = +- i; We know that the first solution is: z^y = e^pi/2 or z=e, y=pi/2 and h(e^pi/2) = +- i . But we also know that for z^y = e^pi/2 other branches of Lambert function leads to values h(e^pi/2) different from +-2pik. Question: Do there exist any other roots z and y for equations: h(z^y) = e^(i*pi/2+-I*2pi*k) ? for all k, or just for some single special k? Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/09/2007, 03:18 PM Ivars Wrote:Question: Do there exist any other roots z and y for equations: h(z^y) = e^(i*pi/2+-I*2pi*k) ? for all k, or just for some single special k? So we have in fact 3 pairs of z and y for which solutions exist: z=e, y=pi/2 , z^y= e^pi/2 h(e^pi/2)= +-i z=i, y=-i, z^y = i^-i= e^pi/2 h (i^-i)=+-i z=-i, y=i, z^y = -i^i = e^pi/2 h(i^-i) = +-i it also follows from formula h(1/(i^i)) = 1/i = -i and h(1/(-i^-i)) = 1/-i = i. so this formula h(1/z^z) =1/z must be valid not only for integer and real x, but also for imaginary and maybe complex z. I find this very intriguing. But still do not know how to extend it beyond first 2 branches of W. By the way, Gotfrieds curve which crosses imaginary axis at +- i seems to have minumal REAL value at z= 25*pi=78,5398163397448 and is equal to h(25*pi). More exactly, h(25pi) = -0,170356724523878+- I* 0,44394520769755. I could not prove it, but it seems that: h(25pi)^1/h(25pi) = 25 pi h(25pi)^(-1/h(25pi)) = 1/(25*pi) =0,012732395 I wonder if other smaller curves on Gottfrieds initial plot ( where Im(h(z) = 0) might have crossing points with imaginary axis and real minimums which have also "nice" z values. If so, the whole curve might be a reasonably nice function- or is it already well known? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/10/2007, 03:14 AM @Ivars What are you talking about? The statement "h(1/x^x) * h(x^1/x) = 1" could be interpreted in at least four ways: $h\left(\frac{1}{x^x}\right) h\left(\frac{x^1}{x}\right) = 1$ (with the usual order of operations), or $h\left(\frac{1}{x^x}\right) h\left(x^{(1/x)}\right) = 1$ $h\left((1/x)^x\right) h\left(\frac{x^1}{x}\right) = 1$ $h\left((1/x)^x\right) h\left(x^{(1/x)}\right) = 1$ (assuming the order of operations of "/" and "^" are opposite from their usual order) could you please clarify this so I can either agree or disagree with this? @Gottfried Also, it took me 3 months to understand your graphs, but I think I understand them now. It seems like you were trying to plot a graph like this: which is the contour plot of $\text{Im}(x^{1/x}) = 0$. However, I still don't understand the questions your asking, which makes it hard to get through this 8-page thread. If someone could summarize the questions and results in this thread, I would be very thankful. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/10/2007, 04:11 AM For more information about the Lambert W function, which this thread seems to be a great deal about, Wolfram's function website has a wealth of information about W and about the branches of W (you can even download PDFs of each section). To answer Ivars' question earlier in this thread, the method probably used to find the power series expansions of the Lambert W function is probably to apply general analytic continuation techniques, or to solve what coefficients satisfy the following differential equation: $x(1+W(x))W'(x) = W(x)$ as this uniquely defines the Lambert W function along with initial conditions. The general series expansion of the Lambert W function for all branches is: $W(x) = w + \frac{w}{(1+w)x_0}(x-x_0) - \frac{2w^2+w^3}{2(1+w)^3x_0^2}(x-x_0)^2 + \frac{9w^3+8w^4+2w^5}{6(1+w)^5x_0^3}(x-x_0)^3 + \cdots$ which can be obtained by repeated application of the derivative formula: $W'(x) = \frac{W(x)}{x(1+W(x))}$ from which the differential equation came. So all you need is the starting value $w = W_k(x_0)$ at the starting point $x_0$ and you can form a series expansion of W anywhere. These are just the formulas I found on Wolfram's function website. Andrew Robbins Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 12/10/2007, 07:27 AM (This post was last modified: 12/10/2007, 07:48 AM by Gottfried.) andydude Wrote:@Gottfried Also, it took me 3 months to understand your graphs, but I think I understand them now. It seems like you were trying to plot a graph like this: which is the contour plot of $\text{Im}(x^{1/x}) = 0$. However, I still don't understand the questions your asking, which makes it hard to get through this 8-page thread. If someone could summarize the questions and results in this thread, I would be very thankful. Andrew RobbinsHi Andrew - first please excuse my delay in answering your other post: I was struck in bed by sickness, and could not concentrate as I wanted to. Now, you are correct - the first attempt was just an approach to find such a contour, and since at that time I had no versatile implementation of the h()-function in Pari I had to search for an idea about this contour at all, programming the graph in Delphi-code. The second step was then, to find an analytical expression for the Im(z^(1/z))=0 in terms of one real parameter. The second type of plot is then a combined graph : u = alpha + i*beta t = exp(u) = a + i*b s = exp(u/t); IM(s)=0 where beta is the single parameter and alpha,a,b,s are dependent on beta. Everything again handwaved; the Lambert-W and h()-implementation, which was available to me at this time was still not branch-enabled (I could finally implement one using Henryk's last hint) For what this all was needed: I have these hypotheses about the composition of the eigensystem of the (infinite) Bell-matrix. This includes a relation to the fixpoints, actually the t and u-values above are needed to construct the eigen-matrices, from which then the Bell-matrix (and its fractional or general powers) can be composed. For bases s e^(1/e) I needed a possibility to compute examples to work on the verification of the hypothese, too. The found parameter-formula for s, based on the independent parameter beta, allowed then to find for an arbitrary s>e^(1/e) the according u and t (and even branches) employing regula falsi - just a poor-mans Lambert-W-replacement.... So, a sufficient (zip-) compression of this (part of the) thread would possibly be a one-liner in Maple/Mathematica & co after all ;-) Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/10/2007, 07:37 PM andydude Wrote:@Ivars What are you talking about? The statement "h(1/x^x) * h(x^1/x) = 1" could be interpreted in at least four ways: $h\left(\frac{1}{x^x}\right) h\left(\frac{x^1}{x}\right) = 1$ (with the usual order of operations), or $h\left(\frac{1}{x^x}\right) h\left(x^{(1/x)}\right) = 1$ $h\left((1/x)^x\right) h\left(\frac{x^1}{x}\right) = 1$ $h\left((1/x)^x\right) h\left(x^{(1/x)}\right) = 1$ (assuming the order of operations of "/" and "^" are opposite from their usual order) I think, second one is the right one. Best regards, Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/16/2007, 12:02 PM Gottfried Wrote:The found parameter-formula for s, based on the independent parameter beta, allowed then to find for an arbitrary s>e^(1/e) the according u and t (and even branches) employing regula falsi - just a poor-mans Lambert-W-replacement.... Gottfried And my task was to find all real s>e^(1/e) for whom h(s) would be purely imaginary, Re(h(s))=0 as for h(e^(pi/2)) = +- i - the crossing points of Gottfrieds graphs with imaginary axis. Also the real minima they have ( negative). Analytically. I have not suceeded so far. There fore I started to look into formulas involving h(z) hoping that will give answer eliminating need to compute all the steps and branches. The best and true I have found is h(1/(n^n))=1/n for all n>=1 . which leads to sum (h(1/(n^n))^2 = (pi^2)/6 etc. for other powers of h of 1/(n^n) = 1, 1/4, 1/27, 1/256, 1/3125, 1/46656, 1/823543, 1/16777216, 1/387420489, 1/10^10............. These are like infinite polynomial equations h has to satisfy. Other formulas are conjectures still. Ivars Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 12/16/2007, 02:22 PM Ivars Wrote:Gottfried Wrote:The found parameter-formula for s, based on the independent parameter beta, allowed then to find for an arbitrary s>e^(1/e) the according u and t (and even branches) employing regula falsi - just a poor-mans Lambert-W-replacement.... Gottfried And my task was to find all real s>e^(1/e) for whom h(s) would be purely imaginary, Re(h(s))=0 as for h(e^(pi/2)) = +- i - the crossing points of Gottfrieds graphs with imaginary axis. Also the real minima they have ( negative). Analytically. I have not suceeded so far. There fore I started to look into formulas involving h(z) hoping that will give answer eliminating need to compute all the steps and branches. The best and true I have found is h(1/(n^n))=1/n for all n>=1 . which leads to sum (h(1/(n^n))^2 = (pi^2)/6 etc. for other powers of h of 1/(n^n) = 1, 1/4, 1/27, 1/256, 1/3125, 1/46656, 1/823543, 1/16777216, 1/387420489, 1/10^10............. These are like infinite polynomial equations h has to satisfy. Other formulas are conjectures still. Ivars Ah, yes; that is possibly an interesting question... I'll look at it later. I think, it should be derivable by an equivalent formula as I had it for purely real bases. Let's see... Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 12/16/2007, 05:53 PM Gottfried Wrote:Ah, yes; that is possibly an interesting question... I'll look at it later. I think, it should be derivable by an equivalent formula as I had it for purely real bases. Let's see... Gottfried Which would lead to fact that those purely imaginary h(s) where s =real>e^(1/e) would have s as superroot (inverse of infinite tetration) of imaginary, as e^(pi/2) is superroot of +-i, because h(e^(pi/2)) = +-i Question: How many superroots each of these imaginary values have, and how many of them MUST be real? How are these imaginaries (crossing with imaginary axis of Gottfrieds spiderlike graphs) distributed and why? I find it very intriguing that superroot of a purely imaginary number can be real. « Next Oldest | Next Newest »

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