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)) - Ivars - 11/15/2007
Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis). I do not take +- beta, I take +- I . beta is still 0<beta<pi. In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant. The question is, can we even use beta = - pi/2 as that would imply h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i. Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/16/2007
Ivars Wrote:Gottfried Wrote:Well in this case, for the result s being real only, alpha is a unique function of beta. If you take +beta you need one alpha, and if you take -beta, you need another alpha adapted from this; in this case, the two alphas are even equal (the graph looks symmetrical w.r.t. y-axis). But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 . And then see, what value the real part must have such that the resulting s is purely real. And while the graph shows only u = alpha -pi*I .. u=alpha + pi*I there is no problem to extend the graph to u = alpha + (beta + 2*k*pi)*I - there is a rough periodicity with a distortion then. Just plot it in other ranges. Quote:In that case, IMAG(t) at beta=pi/2 would be -I, so graph will be on the right lower quadrant. However, at beta - pi/2, if we take +i, we get IMAG(t) = -i -> in the left lower quadrant, while at -I it will be where it is now-left upper quadrant. Yes, just look in the 3'rd quadrant; the required value for alpha is on the blue curve, and the resulting values for real and imag of t are on the magenta curves, and the resulting value for s is on the green curve. Quote:h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i. ??? hmmm.... Regards - Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/16/2007
Gottfried Wrote:But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 . What I mean is that any time You define an arbitrary complex number u you have to consider both roots of (-1) and only later get rid of one if that does not add any information. Usually it does not, so people already initially dismiss the idea of 2 roots of sqrt(-1). Euler did not, however, and that is why he never made mistakes. Which means that beta is 1, but sqrt(-1) is -i and +i simultaneously not because of beta, but because of 2 values of sqrt(-1) which can occur simultaneously for any given beta. If You take beta = - 1 , You get u=a-sgrt(-1) but sqrt (-1) again taken as usual is +-i, so You have a-+i instead of a+- i as in case of b=1. Not that it matters, but it matters in your graphs, as by using +-i instead of +- beta You will get 2 values of IMAG(t) = +i and -i in the right side of Y axis, meaning 2 symmetric curves there. Quote:And while the graph shows only u = alpha -pi*I .. u=alpha + pi*I there is no problem to extend the graph to u = alpha + (beta + 2*k*pi)*I - there is a rough periodicity with a distortion then. Just plot it in other ranges. Could You please plot few examples on the current plot just to get the idea-it is very interesting to see the character of those branches- I am really sorry, I have only excel as a tool and even that I use for business tables not for graphing functions. So please... Quote:h(e^-pi/2) diverges which it does not, and s in that case is not 4,81 = e^pi/2......... but 0,20........=i^i. I think, may be wrongly, that beta in Your case corresponds to the power of e in tetration of (e^beta). If not, please correct me. So I think if beta = pi/2 it is Ok, as h(e^pi/2) diverges and has 2 values +I , -I and ONLY then other branches, so e^pi/2 is a real value of s that corresponds to complex to real transformation as s>e^1/e. You should be able to see both I and - I at beta = pi/2. But if beta = -pi/2 something is wrong as (e^-pi/2) = 0,20..and (e^-pi/2)^(e^-pi/2)^(e^-pi/2) ....... converges, and is 0,47..... which means that is real to real case where beta should have been 0, not -pi/2- there should not have been such branch on a graph in this quadrant. So for me it looks that if You just fold the graph along Y so that left side superimposes on right side You will have the right graph+ have left side free to plot convergent values of h(e^beta) where e^beta< e^1/e so beta < 1/e (including negative beta). May be I misunderstand something still in the role of each variable, but, if You will be able to explain it to me, others will easily understand Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/16/2007
Ivars Wrote:Gottfried Wrote:But that is funny. What else is u = alpha + beta*I , so to have some U with imaginary part 1*I or -1*I, you use beta = 1 or beta = -1 . But, wait a moment. If you want I and -I in your formula for any variable x (or in my case u), what else are you doing as considering sqrt(-1)_1 = 0 + b*I and sqrt(-1)_2 = 0 - b*I when you ask for considering both complex roots of -1 ? So in my negative and positive beta I'm just doing that, what you demand; it is the negative and positive b in the above example. So since it seems to me, that I'm already doing what you ask for, this is the source of my not-understanding of your concern. Quote:I think, may be wrongly, that beta in Your case corresponds to the power of e in tetration of (e^beta). If not, please correct me.How comes that you think this? I say: h(x) has many solutions t_0, t_1, t_2, ... So what is the form of the different t? Then I say, t_0,t_1,t_2 must be an exponential of u_0,u_1,u_2,... . complex, multivalued like your example of sqrt(-1). Now: what is the complex form of all these t's? They must be the exponential of another number u_0,u_1,u_2,... multivalued again. Now what is the form of the u_0,u_1,u_2,..., which are complex numbers alpha_0 + beta_0*I, alpha_1 +beta_1*I, maybe each second beta is just the negative signed other beta, which would satisfy your request to consider positive and negative complex roots. Then I find, that if I select one imaginary part beta*I, to get a real s, the real part alpha must be of a certain value. I can insert beta_0*I, -beta_0*I and so on, compute the required alpha and always get one real s. Your demand for considering -1*I and +1*I as possible roots of -1 is perfectly modeled here - at least as far I can see. Quote:But if beta = -pi/2 something is wrong as (e^-pi/2) = 0,20..and My process goes three steps: beta*I -> u -> exp(u) -> exp(u/exp(u)) = s = real(s) and beta is not in that way related to s as you assume here. (It is not s = e^beta, for instance) Regards - Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/16/2007
Gottfried Wrote:But, wait a moment. If you want I and -I in your formula for any variable x (or in my case u), what else are you doing as considering Now I understand- and that is fine- I was just distracted by that (-pi/2) on X axis giving the same value for s=e^pi/2=4,81 (graphically it looks so ) . So that -pi/2 is just the same b=pi/2 multiplied by -1 which could be looked upon as taking the other root of (-1). OK! Quote:Your demand for considering -1*I and +1*I as possible roots of -1 is perfectly modeled here - at least as far I can see. Good! Now I think I will be finally able to place myself in the graph... On other hand, then it is just a coincidence that having b= pi/2 the result for IMAG(t) = +-i coincides with h(e^pi/2) = +- i . But please could You show a plot with few more branches of IMAG(t) on it? I am very interested what patterns You described look like now when I seem to understand where on the plot what is placed Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/16/2007
Here is a more extended plot showing a wider range for beta. The periodicity of alpha and s is not perfect; the same real s approximate the vertical axes at multiples of pi. I omitted the curves for t (= a + b*i) here. Gottfried [attachment=140] RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/17/2007
[quote=Gottfried] Here is a more extended plot showing a wider range for beta. The periodicity of alpha and s is not perfect; the same real s approximate the vertical axes at multiples of pi. I omitted the curves for t (= a + b*i) here. [\quote] Thanks. Very interesting. But can You add t=a+b*i as well? It would be easier for me to place myself on this graph, when I see the exact relation with the smaller graph I hope I have understood. Why log(log(s)) = alpha =0 at beta= pi/2+-n*pi? Is it exact relation? Would that mean that s= e there? what is IMAG( t) in these points? They seem to be roots of d^2 (log(log(s))/d(beta)^2=0 and d^2(alpha) /dbeta^2 = 0(except for n=0?) . I rememmber on the small graph, Imag (t) had inflection points at b=pi/2 while alpha=real (t) had inflection points are integer beta=1. Does it continue, the Imag (t) and real(t) to oscillate around x axis with decreasing amplitude as beta gets bigger? How does their max amplitude decay with beta- what is the functional dependance? Is there any information in a curve connecting points with the same slope in this graph on different branches? Where d/(dbeta ) of (log(log(s) ) = const? ( and in d(alpha)/dbeta=const) what is the value of alpha max in interval where beta<pi/2, so that alpha(beta=0) = ? And what is s at beta=0 ? a at beta =0 was 3, right? Exciting Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/20/2007
Ivars Wrote:Why log(log(s)) = alpha =0 at beta= pi/2+-n*pi? Is it exact relation? Would that mean that s= e there? what is IMAG( t) in these points? Ivars It seems that if ln(ln(s))=0 than s=1/e (based on Eulers idea that ln(-1) = ln(1) + imaginary periodical branches = 0 + imaginary branches) or s= e. s=e is not solvable. If s= 1/e, than s=exp(u/t) = 1/e -> u/t= -1 if u/t = -1 than u/exp(u) = -1 -> exp(u)/u = -1; epx(u) = -u u = ln (-u) or u = - Omega = - 0,5671432........ so that -Omega= ln (Omega) , which is true. but if u= - Omega = alpha+ i beta, than at points beta= npi/2 alpha= -Omega, Beta = npi/2 , i= +-i, or in this graph: slog(slog(s)=0, s= 1/e, u= - Omega+- I*n*Pi/2 than t= exp((-Omega)+-I*n*pi/2) = exp(-omega)*exp(+-I*n*pi/2) = Omega*exp (+-I*n*pi/2) = Omega*(+-(I^n)) = +-I^n*Omega. s= exp ((-Omega+-I*n*pi/2)/+-I^n*Omega) = exp(-+1/I^n+- I*n*pi/2/I^n*Omega) has values: n=0 s= exp (-+1 +-0) = 1/e, e. This by the way implies that h(1/e)= Omega h(e) = - Omega n=1 s complex s= exp ( -+1/I +- pi/(2*Omega)) = exp(+-i+-pi/(2*Omega) =exp(+-i) exp(+-pi/2Omega) = exp(+-i) *15,95.., exp(+-i) *0,062682 exp(+-i) = i^(+-2/pi) = i^0,63662 n=2 s complex s=exp(+-1+-2I*pi/(2*Omega))= exp(+-1+-I*pi/Omega) = exp(+-1)*exp(+-I* pi/Omega) = exp(+-1) *i^(+-2/Omega) n=3 s Complex s=exp((-+I+- 3pi/2*Omega)) = exp (-+I)*exp(+-3pi/(2*Omega)) n=4 s complex s= exp(+-1+-4pi I/2*Omega) = exp(+-1)*exp (+-4pi/2Omega) = exp(+-1)*i^(+-4/Omega) etc. Or do I make a mistake somewhere Best regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/22/2007
I may be messing up the thread, on other hand, this must be trivial (if it is correct) -I just wanted to check: W(2ln2)=ln2 W(3ln3)=ln3 W(4ln4)=ln4 W(nln(n))=ln n , n>=1 So h(1/4) = h((1/2)^2) = W( -ln (1/2)^2)/-ln(1/2)^2=W(2ln2)/2ln2= 1/2 h(1/27) = h((1/3)^3) = 1/3 h(1/256) = h(1/4)^4) = h((1/2)^2^2^2) = 1/4 h(1/3125)=h(1/5)^5=1/5 h(1/n^n)=1/n Which is the same as taking n-th root of n^n. so that : 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. We also have W(-ln2/2) = -ln2 So h (2^1/2) = W(-ln2/2)/-1/2ln2 = -ln2/(-1/2ln2)= 2 and simmetry ends at e^1/e Multplying h(1/x^x) * h(x^1/x) = 1 at each x if -1/e<x<e. so h(x^1/x) = 1/h(1/x^x). h(1/x^x) converges for all -<1/e x<1/e so all positive integers can be produced as n=1/(h(1/n^n))), n>=1 , fractions 1/n=h(1/n^n)) In that sense h(1/n^n) works opposite integers -when n is in denominator, h(1/n^n) is in numerator. so ln n = Integral (1/n) = integral h((1/n^n)) we can also make: m/n = h(1/n^n) / h(1/m^m) practically all integer numbers and rational fractions can be substituted by h of something, also functions like ln , Especially, h(e^pi/2) = +- i .......... but i^4=1 so in the end 1= (+h(e^pi/2))^4, -1 = (+h(e^pi/2))^2 2= (+h(e^pi/2))^4+ (h(e^pi/2))^4 So even i can be replaced by h(e^pi/2) so that i = e^+h(e^pi/2)*pi/2 My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works? You can take any infinite series and replace integer n with 1/h(1/n^n) for a starter and see what happens. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/22/2007
Ivars - very nice! Ivars Wrote:My feeling is that the whole number axis except transcendental numbers like e,pi, 2 can be constructed from infinite tetration-may be that is how nature works? I don't know whether this fits your question, but look at the postings of Gianfranco [GFR]. He stated, that every (real? complex?) number can be expressed as a powertower to a base b ( and a topexponent x ?) - don't have the link to this statement at hand. Gottfried |