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RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/03/2007
bo198214 Wrote:Wow, Gottfried! Where did you dig out this article?!;-) Well, I'd like to be able to recall where I finally found it. It was either a polish, czech, russian or rumaenian library - I even only vaguely remember that it was a university. If I ever find that adress in my "history"-file I'll let you know here. Hint: I started at the Euler-archives, clicked through some links and got ideas for keywords for specific google-search. Gottfried Here is the title of the article: [attachment=121] RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/04/2007
Gotfried, Thanks for the article,I tried to avoid Latin texts of Euler, as I do know 0 Latin. However: I looked at article; Euler never specifies the sign of sgrt(-1). We do, when we use i. When he sets angle to pi/2, he gets r=e^pi/2. Power tower e^pi/2 = sgrt(-1) = e^sgrt(-1) pi/2 May be I miss something, but as long as he shows no signs , it is OK,if we always take the right sign = the same in both cases. Though he did put a constraint on angle pi, but I could not understand what that means. However, when we calculate h(e^pi/2) = -W(-ln(e^pi/2)/ln(e^pi/2) we specify that we will use +i via W(-pi/2) = i*pi/2; and we get h(e^pi/2)=-i; Euler would have written W(-pi/2)= sgrt(-1) *pi/2, and got h(e^pi/2)=sgrt(-1).No problems. If we put W(-pi/2) = -i*pi/2, we would get h(e^pi/2) = +i. If we look at power series for W(-pi/2): i*pi/2= - pi/2-p^2/4-pi^3/2*2^3 - 8/3*pi^4/2^4 - 125/24*pi^5/2^5 ..... i= -1-pi/2 - pi^2/2*2^2 - 8/3*pi^3/2^3 - 125/24*pi^4/2^4......= e^(-pi/2)*sum ( -1^(n)*(n^(n-2)) -i = 1+pi/2+pi^2/2*2^2+ 8/3*pi^3/2^3 +125/24*pi^4/2^4.... = (e^pi/2)*sum(n^(n-2))? These both should be divergent, but definitely not to the same value, Ivars Fabriciuss RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/04/2007
Ivars, have a look at the imaginary part of the Lambert W function (picture from Mathworld) You see there two branches for the negative real values . For one branch is and for the other branch is . A power series development is always attached to the point of the development and the convergence radius is determined by the nearest singularity. The standard powerseries development is developed at 0 and the singularity of W is at so the convergence radius is . If you want to analytically continue the function along a certain path you pave the path with overlapping discs and develop the (coefficients of the) power series in their centers consecutively. In this way you can compute the value of the Lambert W function if you approach by a northern bow around (which yields ) or if you approach it by a southern bow, which yields . I explained the method (though in another context) already some posts ago even with a picture attached. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/04/2007
Hello, Bo bo198214 Wrote:Ivars, have a look at the imaginary part of the Lambert W function (picture from Mathworld) This is clear now, very good explanation. Especially the picures are now clear. Thanks a lot. I hope You are not tired yet Next Item how to find power series expansion along any branch and the formulas You show in previous post I am afraid I am not ready yet to even comment, but I will take notice, and return to the series expressions of +- i. So, depending on branch, h(e^pi/2) will be either -i if W(-pi/2) = +ipi/2 or h(e^pi/2) = +i W(-pi/2) = - ipi/2. That means that even if h^(e^pi/2) can have values i+pi*k, but they can not be reachable simultaneously because as soon as we fix branch (if we are moving along it, figuratively speeking), we are doomed to change of sign for i: so h(e^-pi/2) = -i-2pik or h(e^-pi/2 = +i+2pik Euler would have written h(e^pi/2)=sgrt(-1)+- 2pik and W(-pi/2) = sgrt(-1) *pi/2 +-2piK (which he did without mentioning +-2pik specifically because he limited himself to angle<=pi in the work Gotfried mentioned). Leaving question open. In Eulers works, sgrt(-1) has argument: pi/2+- 2pi; +- 4pi; +- 6pi etc. Or do I make a mistake somewhere? Regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/04/2007
Hello, Bo bo198214 Wrote:Ivars, have a look at the imaginary part of the Lambert W function (picture from Mathworld) This is clear now, very good explanation. Especially the picures are now clear. Thanks a lot. I hope You are not tired yet Next Item how to find power series expansion along any branch and the formulas You show in previous post I am afraid I am not ready yet to even comment, but I will take notice, and return to the series expressions of +- i. So, depending on branch, h(e^pi/2) will be either -i if W(-pi/2) = +ipi/2 or h(e^pi/2) = +i W(-pi/2) = - ipi/2. That means that even if h^(e^pi/2) can have values i+pi*k, but they can not be reachable simultaneously because as soon as we fix branch (if we are moving along it, figuratively speeking), we are doomed to change of sign for i: so h(e^pi/2) = -i-2pik or h(e^pi/2 = +i+2pik Euler would have written h(e^pi/2)=sgrt(-1)+- 2pik and W(-pi/2) = sgrt(-1) *pi/2 +-2piK (which he did without mentioning +-2pik specifically because he limited himself to angle<=pi in the work Gotfried mentioned). Leaving question open. In Eulers works, sgrt(-1) has argument: pi/2+- 2pi; +- 4pi; +- 6pi etc. Or do I make a mistake somewhere? Regards, Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/04/2007
Ivars Wrote:so h(e^-pi/2) = -i-2pik Yes Different branches are not necessarily apart. Though this is true for the logarithm, it is for example not true for or . To compute (a closed form of) the power series coefficients of at say -1 depending on the branch, seems a rather laborious task. I am not sure how Mathematica and Maple compute the Lambert W function outside the convergence radius of its standard power series. Perhaps they dont use a power series expansion but an iteration formula. For example the different branches of can be computed by infinitely applying the logarithm: (corresponding with the branches of the logarithm) or as we know already by infinitely applying the exponential for the range . In this post I showed the dependency of the fixed points of from . The fixed points of are the different branches of . RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/04/2007
Though it is not mentioned on Mathworld one can probably use the quite easy iteration formula: with directly derived from iff by making the iteration formula out of it. The branches of the logarithm then again correspond to the branches of . For example for I get by this iteration formula: when using the main branch of the logarithm. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/04/2007
hej Bo, Thanks again. But: if you calculate W (-pi/2) via iteration of logarithm, it will have the same 2pik difference between different branches at point -pi/2. Logarithm itself has 2 branches such that log (-1) = +i *pi or - i*pi. So we are in the same position again- if we choose one branch, we end up with W(-pi/2) = +ipi/2, and h(e^pi/2) = - i ; if another, W(-pi/2) = - ipi/2 and h(e^pi/2) = + i , but it does not mean h(e^pi/2) = +- i at the same time. Which means that value of h(e^pi/2) depends on choice we make about usage of +i or -i on the way. Now e^pi/2 is real, positive and has one value. The principal value of h(e^pi/2) is -i or i depending on our choice of +i or - i earlier in the process of calculations, so it is dependant on the direction we choose to go around singularity -1/e in W or negative numbers in x= ln (y/x). Now if the value of power tower is dependant on the choice of +i or - i along the way, we have to keep for further calculations values: h(e^pi/2) = +-i +- 2pik. This look very similar to situation with ln(-1) = +-i*pi+- 2pik but the order is different.If we take ln(-1)=+ip, we get h(e^pi/2) = - i So ln(-1) = -(pi*h(e^pi/2) -+2pik). I do not see any other way to differentiate between sgrt(-1) and h(e^pi/2) except that they seem to have opposite order of signs -if we choose +i somewhere, we get - h(e^pi/2) and if we choose - i we get +h(e^pi/2), or one could write h(e^pi/2)= -(sgrt(-1) +2pik) A mistake again? RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/04/2007
Ivars Wrote:Logarithm itself has 2 branches such that log (-1) = +i *pi or - i*pi. The logarithm has a branch for each integer , that is for you get . This is because the logarithm is the inverse of the exponential and . So its not really about chosing the sign of but about chosing the . Quote:h(e^pi/2) = +-i +- 2pik. Again, this is not true. If we have a fixed point of then (or ) is usually not again a fixed point (and hence not a branch of ). For example then but also . PS: the name is sqrt and not sgrt RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/05/2007
Thanks again, So we choose one branch of logarithm with k=-1-> ln(-1) = i*pi and by that got to value of W(-pi/2) on the branch which gives = ip/2 and by that got value of h(e^pi/2) = - i 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 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. but h(e^pi/2) = - W(-pi/2)/ ln(e^pi/2) so as ln(e^pi/2) = pi/2 +- 2k*pi *sqrt(-1) h(e^pi/2) = - (+- i(p/2+-2pi*k))/ (pi/2 +-2 k*pi * i)= = (-+i *pi/2-+i *2*k*pi) / pi/2+- i* 2 k*pi) = pi/2 (-+i-+i*4k)/ pi/2*(1+-i*4K) = (-+i-+i*4* k)/ (1+-i*4*k) So every time we take -i - 4*k*i in upper part, we get 1+4*k*i in denominator, and vice versa. Now what would be the period of such ratio? h(e^pi/2) = -i-4k*i/(1+4ki) if k=0, h(e^pi/2) = -i if k=1, h(e^pi/2) = (-i - 4i)/(1+4i) = -5i/(1+4i) = 1/ (i/5-4/5) if k=2 h((e^pi/2)= (-i - 8i)/(1+8i) = (-9i)/(1+8i) = 1/(i/9-8/9) if k=3 h(e^pi/2) = (-i-12i)/(1+12i) = (-13i)/(1+12i)= 1/(i/13-12/13) if k=-1 h=(e^pi/2) = (-i+4i)/ (1-4i) = 3i/(1-4i)= 1/(1/3i-4/3) if k=-2 h(e^pi/2) = (-i+8i)/(1-8i) = 7i/(1-8i) = 1/(1/7i-8/7) if k=-3 h(e^pi/2) = (-i+12i)/(1-12i) = 11i/(1-12i) = 1/(1/11i-12/11) if k=-4 h(e^pi/2) = (-i+16i)/(1-16i) = 15i /(1-16i)= 1/(1/15i)-16/15) we can do the same calculations starting with i+4ki in upper part, 1-4ki in lower: h(e^pi/2) = i+i 4k / (1-4ki) if k=0, h(e^pi/2) = i if k=1 h(e^pi/2) = (i+4i)/(1-4i) = 5i/(1-4i) = 1/(-i/5-4/5) if k=-1 h(e^pi/2) = (i-4i)/(1+4i) = -3i/(1+4i) = 1/(-1/3i-4/3) seems like a complex conjugates to first 2 sets of values. I do not know about periodicity, there seems to be none , but these might be other fixed points for e^pi/2? Or have I made a mistake again Best regards, And thank You once more for time spent advising me. Ivars |