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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 11/08/2007, 02:31 PM (This post was last modified: 11/08/2007, 02:32 PM by Ivars.) hej BO, bo198214 Wrote:$\log_{1,b}(-1)=6*I$ How did You find this first value, exactly- where does b come in? Is it used as base of logarithm instead of e? so log 1,b (z) (-1) is log with base b from -1? Do I understand correctly? So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1? Which in turn is calculated in each step via normal logarithms by formula: log b (z) = (ln (z) + 2pik)/ ln (b)? Regards, Ivars bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/08/2007, 02:51 PM Ivars Wrote:bo198214 Wrote:$\log_{1,b}(-1)=6*I$ How did You find this first value, exactly- where does b come in? Is it used as base of logarithm instead of e? so log 1,b (z) (-1) is log with base b from -1? Do I understand correctly?Yes, yes, yes to the last 3 questions. We defined $\log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)}$ where $b=e^{\frac{\pi}{2}}$ hence $\log_{1,b}(z)=\frac{\log(z)+2\pi i}{\pi/2}=\frac{2}{\pi}\log(z)+4i$ and further $\log_{1,b}(-1)=\frac{2}{\pi}\log(-1)+4i=\frac{2}{\pi}\pi i +4i=2i+4i=6i$. Quote:So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1? Which in turn is calculated in each step via normal logarithms by formula: log b (z) = (ln (z) + 2pik)/ ln (b)? Yes, yes Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/09/2007, 08:30 AM bo198214 Wrote:Quote:So the problem is now turned to the properties of continuous application of logarithms with base e^pi/2 on -1? Which in turn is calculated in each step via normal logarithms by formula: log b (z) = (ln (z) + 2pik)/ ln (b)? Yes, yes So now we need analytic expressions for that infinite taking of logarithms with base e^pi/2 of (-1) as a function of k. Any ideas? I am searching as well. if we put e^(pi/2) = e^((sqrt(pi)/sqrt(2))^(sgrt(pi)/sqrt(2) ) then we can use Gamma(1/2) = sqrt(pi), so e^(pi/2) = e^(Gamma^2 (1/2)/2) and ln (e^pi/2) = Gamma^2 (1/2) /2 +- 2pik next ln would be 2 lnGamma(1/2) - ln 2 next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so in the end we will have lnln............ln (Gamma(1/2) when n-> infinity. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/09/2007, 08:43 AM (This post was last modified: 11/09/2007, 08:44 AM by bo198214.) Ivars Wrote:next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so in the end we will have lnln............ln (Gamma(1/2) when n-> infinity. Actually it does not matter with which value (instead of -1) you start the iterations of $\log_b$ as long as you dont take some hyperpowers of $b$ for example $b^{b^b}$ would lead to 0 after 4 times application of the $\log_b$ and taking again $\log_b$ would give $-\infty$, so the formula would not work with this starting value. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/09/2007, 11:51 AM (This post was last modified: 11/09/2007, 01:52 PM by Ivars.) bo198214 Wrote:Ivars Wrote:next ln2+ ln*lnGamma(1/2) - lnln2 etc. ln2 and lnln2 will cancel out, so in the end we will have lnln............ln (Gamma(1/2) when n-> infinity. Actually it does not matter with which value (instead of -1) you start the iterations of $\log_b$ as long as you dont take some hyperpowers of $b$ for example $b^{b^b}$ would lead to 0 after 4 times application of the $\log_b$ and taking again $\log_b$ would give $-\infty$, so the formula would not work with this starting value. So we have just limit log b log b log b log b ... log b (+-i*2pik) to be concerned with for branches in numerator. Right? But how do we get real part as well from this? because there is real value in denominator if we use ln to compute log b? So we have lim k, n-> infintylog b log b log b ...... log b ( +-i*2pik/(pi/2+- i*2pi/k)) So the simplest starting value is then +-i*2pik/(pi/2+-i*2pik)? Are these k in numerator and denominator the same or running independently ? Ivars bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/12/2007, 08:23 PM Ivars Wrote:So we have just limit log b log b log b log b ... log b (+-i*2pik) to be concerned with for branches in numerator. Right? ... So we have lim k, n-> infintylog b log b log b ...... log b ( +-i*2pik/(pi/2+- i*2pi/k)) So the simplest starting value is then +-i*2pik/(pi/2+-i*2pik)? Are these k in numerator and denominator the same or running independently ? I am not sure what you mean. The method is about repeated applying of $\log_{k,b}$ and not of $\log_b$. You can not pull out the $2\pi i k$. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/13/2007, 11:01 PM bo198214 Wrote:I am not sure what you mean. The method is about repeated applying of $\log_{k,b}$ and not of $\log_b$. You can not pull out the $2\pi i k$. May be I am trying to say something useless, but I am sure, that in the region where h(z) does not converge, and may be even in the region where it does converge( e.g h(sqrt(2))=2 or 4 ) e.g h(e^pi/2) You can not limit yourself to only one root of -1, but You have to use both separately, or information is lost. So the real way to get accurate results is to study both rotations anticlokwise (i) and clockwise (-i) independently. I am not sure computer programs handle it that way. I will try once more: Knoebel (1981) gave the following series for h(z) = 1+ lnz + 3^2/3!*(lnz)^2 ......... Now, if we take h(e^(pi/2) we know that answers will be +i, -i. ln(e^pi/2) = pi/2+- 2*pi*i*k -> I stress +-, k >0. If You put this in series above( I know they do not coverge, but they might give h(z) analytic continuation), you will get : h(z,k) = SUM (( (n^n-1)/n!) * ( pi/2+-2pi*I*k)^n-1 or h(z,k) = SUM (((n^n-1)/n!)* pi/2^(n-1)* (1+-4*I*k)^n-1 Therefore, if we get h(z) = i when k = 1, than we must get h(z) =-i when with k =-1. If k=0, we have a special case, as we can not get i in the series if k=0. For k=2 and -2 the values of h(z) again may be complex conjugates , but may be not. For k=3 and -3 again the imaginary values will definitely differ by sign because values for 1 and - 1 were complex conjugates. So basically for each pair of k +- You will get 2 values of h(z) which will be complex conjugates, but sometimes without imaginary part when they converge. Then there will seem to be only 1 value, a real number.But not all cases. As a conjecture, perhaps h (sqrt2) will have one real value at from ln(sqrt2) and another from ln(-sqrt2). h(cube root 2) may have 3 values, h(4th root of 2) may have 4 converging values etc. Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 11/14/2007, 05:04 AM (This post was last modified: 11/14/2007, 09:04 AM by Gottfried.) Ivars Wrote:As a conjecture, perhaps h (sqrt2) will have one real value at from ln(sqrt2) and another from ln(-sqrt2). h(cube root 2) may have 3 values, h(4th root of 2) may have 4 converging values etc. Ivars - Hmm, I've not been in your discussion with Henryk (have my mind elsewhere these days) and I don't know, perhaps I'm missing the point. What is the point of your question - besides, that we have conjugate solutions? If I recall my above graphs, for instance, then they show obviously conjugacy. And my graph for complex fixpoints for real bases b>e^(1/e)     can be continued for any number of solutions t and u, where imag(u)=some selection +- 2*k*pi (= beta in the graph). (though changing the k-factor only does not lead exactly to the same b, but this should not be seen as a problem). So there may be some problem, that I'm missing when I read your posts? Gottfried graph: alpha + I*beta = u a +I*b = t = exp(u) s = exp(u/t) = base-parameter Gottfried Helms, Kassel andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/14/2007, 05:38 AM bo198214 Wrote:So how do you finally compute the $k$th branch of $h$ or $W$ with this formula? I think the following two methods are possibly the standard branches of h: $h_k(x) = e^{-W_k(-\log(x))} = \frac{W_k(-\log(x))}{-\log(x)}$ and much easier to compute in my opinion. Andrew Robbins Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 11/14/2007, 02:10 PM (This post was last modified: 11/14/2007, 02:13 PM by Ivars.) Gottfried Wrote:Ivars - Hmm, I've not been in your discussion with Henryk (have my mind elsewhere these days) and I don't know, perhaps I'm missing the point. What is the point of your question - besides, that we have conjugate solutions? If I recall my above graphs, for instance, then they show obviously conjugacy. And my graph for complex fixpoints for real bases b>e^(1/e) can be continued for any number of solutions t and u, where imag(u)=some selection +- 2*k*pi (= beta in the graph). (though changing the k-factor only does not lead exactly to the same b, but this should not be seen as a problem). So there may be some problem, that I'm missing when I read your posts? 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 ln -sqrt2 = 1/2 ln 2 + - pi*I -> 2 different 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. « Next Oldest | Next Newest »

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