Posts: 366
Threads: 26
Joined: Oct 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. yaxis).
So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.
Gottfried
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 nowleft 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
Posts: 761
Threads: 118
Joined: Aug 2007
11/16/2007, 09:09 AM
(This post was last modified: 11/16/2007, 09:10 AM by Gottfried.)
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. yaxis).
So no ambiguity is here, just use the formula of my paper, by which the alpha is determined by beta to get a real s.
Gottfried
I do not take + beta, I take + I . beta is still 0<beta<pi.
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 nowleft upper quadrant.
The question is, can we even use beta =  pi/2 as that would imply
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
Gottfried Helms, Kassel
Posts: 366
Threads: 26
Joined: Oct 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 .
And then see, what value the real part must have such that the resulting s is purely real.
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=asgrt(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 ideait 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
Posts: 761
Threads: 118
Joined: Aug 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 .
And then see, what value the real part must have such that the resulting s is purely real.
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=asgrt(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.
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 notunderstanding 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
(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.
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
Gottfried Helms, Kassel
Posts: 366
Threads: 26
Joined: Oct 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
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 notunderstanding of your concern.
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
Posts: 761
Threads: 118
Joined: Aug 2007
11/16/2007, 05:28 PM
(This post was last modified: 11/16/2007, 05:29 PM by 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.
Gottfried
Gottfried Helms, Kassel
Posts: 366
Threads: 26
Joined: Oct 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
Posts: 366
Threads: 26
Joined: Oct 2007
11/20/2007, 10:11 AM
(This post was last modified: 11/20/2007, 10:56 AM by Ivars.)
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
Posts: 366
Threads: 26
Joined: Oct 2007
11/22/2007, 01:01 AM
(This post was last modified: 11/22/2007, 07:15 AM by Ivars.)
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 nth 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 tetrationmay 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.
Posts: 761
Threads: 118
Joined: Aug 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 tetrationmay 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
Gottfried Helms, Kassel
