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 - 11/14/2007
Gottfried Wrote:Ivars - You can not see what I can not express properly Did You succeed in explaining why sqrt 2 tetrated might lead to both 2 and 4? Euler seems to have had an idea, and he also bothered with r=e^pi/2 in his article, but I can not read Latin good enough still. May be what I am trying to say is: We should define tetration generally with a help of some divergent series Euler style and then investigate all possible analytical developments of these series, including convergent ones in the region of convergence of h(z). I have a gut feeling I can not get rid of that somewhere the +i and - i difference= ambiquity will pop out in such case clearly. And z=e^pi/2 should be the easiest place to investigate it properly. There has to be not 2 , but 4 values at all k which sometimes overlap, but in some places adding adding pik to i and pik to - i leads to different values.E.g. ln(sqrt2) gives : ln sgrt2 +- 2 pi k =1/2ln2 +- 2 pi I k ln(-sqrt 2) gives: ln sqrt 2+-pi k= 1/2ln 2 +- pi I k but k is not 0 as negative logarithms can only be complex, never real. So totally you have : when k= 0-> 1 value ln sqrt2=1/2 ln 2 when k=1 -> 4 values ln sqrt2 = 1/2 ln 2 +- 2pi I -> 2 values, but + 2pi *I and - 2pi*I is the same ln -sqrt2 = 1/2 ln2 + - pi*I -> 2 values and +pi and - pi which does not coincide with any other values I have a feeling that tetration is the field where reducing this ambiquity early on and reaching definitions of tetration via convergent series or functions ( like W(z) ) looses important information on the way. When You see e^i*pi/2 You think about rotation of 90 degrees. When i = h(e^pi/2) then e^h(e^pi/2) * pi/2 is also a rotation by 90 degress but involves infinite operation on real numbers. NO i. It is an identity- You can replace i with h(z) and -i with another branch of h(z) . And depending on the sign of h(z) it will be either rotation anticlockwise, or clockwise. so h(e^pi/2)^2 = -1 but we clearly know that each h(e^pi/2) which is root of -1 is found via different branch, so they can not be replacable so easily . May be I am terribly wrong, as I really can not nail the place where it really matters. RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/14/2007
Ivars Wrote:You can not see what I can not express properlyHmm, surely there are more possibilities to express a value not only by convergent, but also by divergent series. May be your idea was already covered with my observation, that seemingly the same matrix Bs containing the needed coefficients for tetration can be seen as compsed of different versions of eigen-matrices Bs = W0 * D0 * W0^-1 = W1 * D1 * W1^-1 etc where W0, W1, etc are constructed by different powerseries based on the multitude of different fixpoints. So the different solutions you talk about are possibly reflected in the different versions of the eigenmatrices? I'll read your post again later and try myself to eliminate further misunderstandings, if there is still one. Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/14/2007
Hej Gottfried, Gottfried Wrote:Ivars - Could there be another symmetrical curve for b=imag(t) ? So that at point 0;0 You will have 2 curves crossing each other? Then You could take away the left 2 quadrants of the graph totally- I can not understand why they are needed as it seems to me - maybe wrongly- that there beta=-pi/2 also belongs to h(e^pi/2) which can not be true? Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/14/2007
Ivars Wrote:Hej Gottfried, Hmm; I don't know absolutely. See my derivations in the article function; it seems pretty straightforward and I don't think, there are other solutions - but... t is t=exp(u); so even if we use periodical u (including multiple 2*Pi*I*k) there is only one t according to my considerations. Don't know, whether there are other solutions possible - but if you should see another solution, I'd like to know this. Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/14/2007
Hej Gotfried, Gottfried Wrote:Hmm; I don't know absolutely. See my derivations in the article Why not: Define complex number u Euler style , as u=a+sqrt(-1)b so u= a+-ib where +i and -i must be used separately but together- they are both in sqrt(-1) always. And stick to it throughout the derivations. Then red point coordinates will be: u= 0+-ipi/2 t=0+-i s=e^pi/2 And that will correspond to h(e^pi/2) having values +i, -i in at least in first 2 branches. So on right quadrants will have 2 curves instead of 1. May be symmetric, may be not- it has to be calculated according to Your methods. On left quadrants I am not sure as I can not really understand what values in left quadrants correspond to. . and then for complex-> real case [b]0<beta<pi taken anticlockwise AND 0<beta<-pi taken clockwise[/b] Alfa, s limits i do not know-may be they change as well in this case? Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/15/2007
Hi Ivars - Ivars Wrote:Hej Gotfried,so this describes the curves in the 2'nd and third quadrant: if I use beta or -beta. So this is dealt with. Quote:t=0+-i??? t = exp(u) (whatever u) and this is unique, since exp is multiple->unique function (no two values/branches for function-result) Quote:s=e^pi/2Yes. Since you have selected one beta (+ or -) you have t and s uniquely determined. My only problem was, whether possibly there is another way/formula to force the result s being purely real - I don't know, whether my considerations are exhaustive here. Quote:And that will correspond to h(e^pi/2) having values +i, -i in at least in first 2 branches. So on right quadrants will have 2 curves instead of 1. May be symmetric, may be not- it has to be calculated according to Your methods. On left quadrants I am not sure as I can not really understand what values in left quadrants correspond to. . Best seems to me, you would compute some examples, show this/plot this and nail down the actual problem. I'm still feeling insure whether I'm getting your caveats correctly. Gottfried RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - jaydfox - 11/15/2007
Quote:ln(sqrt2) gives : ln sgrt2 +- 2 pi k =1/2ln2 +- 2 pi I kNot to nitpick, but rather than saying k is not 0, you should really specify that k is odd (and inequality with 0 is then already implied). I tend to write it as, e.g., or ln(sqrt(2)) +- (2k+1)pi I, or if you really need to factor out k and put it at the end of the expression, then ln(sqrt(2)) + pi I +- 2 pi I k RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/15/2007
Hej Gottfried, I am not trying to show You have done something wrong , I think Your graph and approach is great. I only want to make sure that reducing complex numbers to one root of sqrt(-1) = i so that a+ib work also in tetration and h(z) without loss of information. There fore I am trying to ask You since I can not do it- I understand You may feel annoyed - > if You feel any interest , of course, to carry the ambiquity of sign of sqrt(-1) through all derivation before getting rid of it. Gottfried Wrote:t = exp(u) (whatever u) and this is unique, since exp is multiple->unique function (no two values/branches for function-result) But exp(alpha+I*beta) is not equal to exp (alpha-I*beta)? so if we consistently keep u=alpha+-i beta as definition of u as complex number, than t has 2 values = e^alpha*e^(+- I*beta). If You continue from that and find that this other value t=e^alpha*e^-I*beta does not add any new information about the whole thing, than let it be. Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/15/2007
hej jaydfox, Thanks. I know how annoying such elementary things can be. jaydfox Wrote: or ln(sqrt(2)) +- (2k+1)pi I, or if you really need to factor out k and put it at the end of the expression, then ln(sqrt(2)) + pi I +- 2 pi I k But I would end up with ln(sqrt(2)) +- pi I +- 2pi I k? Ivars RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/15/2007
Ivars Wrote:Hej Gottfried, No, there is no problem wth it. But without a workout with an example I seem to have persistent problems to get your problem right. Quote:But exp(alpha+I*beta) is not equal to exp (alpha-I*beta)? 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). 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 |