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 Infinite tetration giving I*Omega costant=I*0.567143... Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/27/2008, 10:09 PM (This post was last modified: 02/28/2008, 10:33 AM by Ivars.) Please could You check analytically or numerically if this derivation is valid? h(((e^(pi/2))^(1/Omega))*(e^I)) =-W((-pi/(2*Omega)-I))/(pi/(2*Omega)+I)=(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329.. The other branch of W should give -I*Omega? The argument for h is complex, a product of: e^(pi/(2*Omega))= e^(pi/2*0.567143..) = e^2,769663952 =15,95327204.. And e^I = cos1+I*sin1= 0,540302306+I*0,841470985 So numerically h( 8,619589667+I*13,42421553)= I*0,56714329 Real counterpart of this is simpler: h(Omega^(1/Omega))=h(1/e) = -W(-ln(1/e)/(ln(1/e))= W(1)/1=Omega=0,56714329 Square superroot of (Omega^1/Omega) : ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= -1/W(-1)= -1/(-0.318131505204764 + 1.337235701430689*I) = 0.16837688705553+0.707755195958823*I. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/05/2008, 03:04 PM (This post was last modified: 03/05/2008, 03:54 PM by Ivars.) According to definition of infinite tetration, this I*Omega has to a result of infinite tetration of a self root of type: (I*Omega)^(1/(I*Omega)) = 8,619589667+I*13,42421553= e^(pi/2))^(1/Omega))*e^I And that is true. The angle of self root of I*Omega is (obvious from previous expression): atan (13.42421553/8.619589667) = atan (1,557407725.) = 1 rad= 57,29577951 degrees. And module is : 15,9532720360977 So the polar form of self root of I*Omega is: (I*Omega)^(1/(I*Omega)) = 15,9532720360977*e^I = e^((pi/(2*Om))+I) Taking a self root of a real number, in polar coordinates looks like as t increases, graph is tying a 1 loop knot, at same angle , whose tan= approx. 10,6 angle= 1,476 approx 84,6 degrees always. All self roots cross point where t=1 . What does a selfroot of complex number looks like I do not know. I do not have Complex function plotting software. Would be interesting to see z^(1/z) in 3D. The point I calculated has to be there as well, also i^(1/i)=e^(pi/2). Is it? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/06/2008, 04:00 PM Ivars Wrote:Please could You check analytically or numerically if this derivation is valid? h(((e^(pi/2))^(1/Omega))*(e^I)) =-W((-pi/(2*Omega)-I))/(pi/(2*Omega)+I)=(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329.. Well, if we dont consider branches we just need to verify that $(I\Omega)^{\frac{1}{I\Omega}}=e^{\frac{\pi}{2\Omega}+I}$ And this is straight forward computing: $(I\Omega)^{\frac{1}{I\Omega}}=e^{\frac{\ln(I\Omega)}{I\Omega}}$ $\frac{\ln(I\Omega)}{I\Omega}=\frac{\ln(I)}{I\Omega}+\frac{\ln(\Omega)}{I \Omega}=\frac{\pi}{2 I \Omega}-I\frac{\ln(\Omega)}{\Omega}$ Now you defined $\Omega$ to satisfy $\Omega e^\Omega = 1$ so $\Omega=e^{-\Omega}$, and $\ln(\Omega)=-\Omega$. Putting this into the above equation: $\frac{\ln(I\Omega)}{I\Omega}=\frac{\pi}{2I\Omega}+I$ q.e.d. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/06/2008, 04:05 PM (This post was last modified: 03/06/2008, 04:05 PM by bo198214.) Ivars Wrote:What does a selfroot of complex number looks like I do not know. I do not have Complex function plotting software. Would be interesting to see z^(1/z) in 3D. The point I calculated has to be there as well, also i^(1/i)=e^(pi/2). Is it? The problem is that 3d is not enough for plotting a complex function. There are 2 dimensions for the argument and 2 dimensions for the value, which makes together 4. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/06/2008, 09:51 PM (This post was last modified: 03/07/2008, 12:28 PM by Ivars.) Thanks for proving my little formula. So nice. Of course, its 4D, that is needed. So we either need to know where to look for projection to 3D, or have some mapping that reduces dimensionality ( if there is one). I need to add few more formulae and check before we can explain(? ) oscillations related to Omega and W(1),I mentioned in another thread : Omega^(1/(I*Omega) = e^I Omega^(-1/(I*Omega)=e^-I sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))) cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega))) (I*Omega)^(1/Omega) =-0.342726848178+I*0.13369214926.. Module ((I*Omega)^(1/Omega)) = (1/e) = Omega^(-1/Omega) Arg ((I*Omega)^(1/Omega)) = atan(-2,5632)=-1,198826..rad = -68,6876759..grad An Interesting complex number with module 1/e. The angle between these 2 formula values is 2,1988261.. rad =125,983.. degrees. Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/07/2008, 09:03 AM Ivars Wrote:Of course, its 4D, that is needed. So we either need to know where to look for projection to 3D, or have some mapping that reduces dimensionality ( if there is one). Yes, if you look at wikipedia or mathworld for example for the Lambert W function you get plots of the real and imaginary part and of the absolute value of the function in dependence on the complex argument (in the horizontal plane). Quote:I need to add few more formulae and check before we can explain(? ) oscillations related to Omega and W(1),I mentioned in another thread : Omega^(1/(I*Omega) = e^I Omega^(-1/(I*Omega)=e^-I sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))) cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega))) But Ivars you can show that on your own, its quite similar to the derivation in my previous post (if you use the reply button you can even see the tex source and learn how to write things nicer with tex). Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/07/2008, 10:57 AM bo198214 Wrote:But Ivars you can show that on your own, its quite similar to the derivation in my previous post (if you use the reply button you can even see the tex source and learn how to write things nicer with tex). Yes, of, course, these are quite obvious. I just posted them as intermediary result. Would be interesting to place them on Complex plane, all these roots involving Omega. I was thinking about learning tex as well from Your posts, I just want to have a more complicated and more important formula to show with tex. I have one in mind, still checking. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/07/2008, 11:02 AM (This post was last modified: 03/07/2008, 11:25 AM by Ivars.) Interestingly, if we take z=I*ln(phi)= I* ln(1.6180399..) and z =I*logomega (phi)= I* (ln(1.6180399..)/ln(Omega))= -0.8484829...., using: sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))), cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega))) sin(I*logomega (phi))= (-I/2) cos (I*logomega (phi)) = (1/2) *(sqrt(5))= phi-1/2=1.6180399-0.5=1.1180399 but (I/2)=sin(I*ln(phi), so sin(I*ln(phi)*sin(I*logomega (phi)) = 1/4 sin(I*ln(phi)+sin(I*logomega (phi)) =0 sin(I*ln(phi)/sin(I*logomega (phi)) =-1 sin(I*ln(phi)-sin(I*logomega (phi)) =-I Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/07/2008, 09:19 PM (This post was last modified: 03/08/2008, 10:52 AM by Ivars.) I just came up with a numerical result I can not verify yet analytically, numerically my software is also poor, but: Consider circle map (Arnold map) : $\theta'= \theta+\Omega-{\frac{K}{2*\pi}}*sin(2\pi\theta)$ Let $\Omega=0.567143..$ And $K= {\frac{\pi}{2*\Omega}..$ then map becomes: $\theta'= \theta+\Omega-{\frac{1}{4*\Omega}}*sin(2\pi\theta)$ I did 1800 iterations with 50 digit accuracy ( This was my first try) and the resulting conjecture is: $lim (n->infinity) {\frac{\theta n}{n} = 1$ monotonically from below, no oscillations. So the resulting angle is 1 rad again. I was expecting it,as $\Omega$ seems to be kind of a self frequency of a flow on unit circle? But than I do not know much about circle maps. Perhaps that is true for any $\Omega$? Not only Omega constant? Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/07/2008, 09:30 PM (This post was last modified: 03/07/2008, 09:31 PM by bo198214.) Ivars Wrote:$lim (n->infinity) {\frac{\theta n}{n} = 1$ What do you mean by $\theta n$? Btw. $\lim_{n\to\infty}$ has tex code Code:\lim_{n\to\infty}. « Next Oldest | Next Newest »

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