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 Real and complex behaviour of the base change function (was: The "cheta" function) bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/15/2009, 05:36 PM (This post was last modified: 08/15/2009, 05:37 PM by bo198214.) (08/15/2009, 05:00 PM)jaydfox Wrote: In thinking about it, the singularities are trivial to find.You mean *some* singularities?! Quote:For a=eta, b=e, anywhere that the exp^[n-2](x) is equal to -1, we will have a singularity. The double logarithm of the double exponentiation, in the respective bases, will be 0. These are the singularities induced by $f_3$. Quote:This makes me wonder, then: for any given n, there are singularities near the real line, and as n increases, these singularities get arbitrarily close. Can you make a picture for those that dont currently sit down with a computer algebra system computing exactly this? Imho the $\log^{[n-2]}(-1)$ converges to the upper primary fixed point of $\exp$. So why should they come arbitrarily close to the real axis? jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 06:40 PM (This post was last modified: 08/15/2009, 06:49 PM by jaydfox.) (08/15/2009, 05:36 PM)bo198214 Wrote: Can you make a picture for those that dont currently sit down with a computer algebra system computing exactly this?A picture I can do. Quote:Imho the $\log^{[n-2]}(-1)$ converges to the upper primary fixed point of $\exp$. So why should they come arbitrarily close to the real axis? But Henryk, remember that, due to branching of logarithms, we must start at a point on the real line and then follow a path. Well, if we start at $\exp_{e}^{[n-2]}(x=0)$, you arrive at a rather large number. If you then try to create a simple path to $\exp_{e}^{[n-2]}(x)=-1$, and then perform n-2 logarithms, you will not get a simple path. It will loop wildly around the upper fixed point. However, let's consider the point $\exp_{e}^{[n-3]}(x)=0+\left(2\,\mathrm{floor}(\exp_{e}^{[n-3]}(0)/2\pi)+1\right)\pi i$. This point is roughly the same distance from the origin as the [n-3] exponentiation of 0, but it has real part 0, and imaginary part equal to (2k+1)*pi, integer k. Thus, on the next exponentiation, it's -1. Note that we can use a rather large quarter circle (or approximately so) to connect this point to the real line. If we take the logarithm of this arc, in the primary branch, we get a line segment from the real line to a complex value with roughly the same real part, but imaginary part equal to pi/2. Note that this line segment now gives us a very simple image as we iteratively perform logarithms, such that it ends up very, very close to 0, with a very small imaginary part (the higher the value of n, the closer to the real line it gets). Now this approach only got us the "closest" singularity to whatever point on the real line we started at (0 in my example). Instead of a quarter circle, we can do a 3/4 circle, 5/4 circle, etc., and we can pick different k values in the (2k+1)*pi*i formula, to find arbitrarily many singularities, and depending on the exact path being used, we can find these singularities in arbitrarily many different branches. I will draw up some pictures do demonstrate. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 07:13 PM (This post was last modified: 08/15/2009, 07:17 PM by jaydfox.) (08/15/2009, 06:40 PM)jaydfox Wrote: Now this approach only got us the "closest" singularity to whatever point on the real line we started at (0 in my example). Instead of a quarter circle, we can do a 3/4 circle, 5/4 circle, etc., and we can pick different k values in the (2k+1)*pi*i formula, to find arbitrarily many singularities, and depending on the exact path being used, we can find these singularities in arbitrarily many different branches. I will draw up some pictures do demonstrate.Alas, I'm at home and don't have access to a math library. When I'm at home, I typically go to http://www.sagenb.org. By the way, I recommend it to anyone who wants quick access to a powerful library! However, sagenb.org appears to be having problems, as the SAGE libraries aren't loading. Pictures might have to wait until Monday. Anyway, it occurs to me that for n>=4, we can create a very nice "grid" of singularities, in the exp^[n-4](x) image. Simply pick all the points whose real part is equal to log((2k+1)*pi), k a non-negative integer, and whose imaginary part is equal to (2m+1)/2*pi, for m an integer. Exponentiating once will get you to +/- (2k+1)*pi*i, which exponentiating again will get you to -1. Note that this grid of points covers the entire right half of the complex plane, so that when we iteratively perform logarithms, we can always find points close to the real line. As n goes to 5, 6, etc., with each logarithm, we get a strip of points between -pi*i and pi*i, and if we use other branches, we can always keep the complex plane filled with points that will be singularities. These are what I would call the "trivial" singularities, because they are singularities, no matter what branch of the logarithm of eta we use. The non-trivial singularities are the ones Henryk pointed out, where we get eta, eta^eta, etc., but only in the primary branch of the logarithm of eta! This is not so easy a feat to accomplish as n gets very large, because we have to wind in and around singularities to make sure we are in the primary branch of the logarithm of base eta. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/15/2009, 09:44 PM (This post was last modified: 08/15/2009, 10:06 PM by bo198214.) (08/15/2009, 07:13 PM)jaydfox Wrote: Simply pick all the points whose real part is equal to log((2k+1)*pi), k a non-negative integer, and whose imaginary part is equal to (2m+1)/2*pi, for m an integer. Exponentiating once will get you to +/- (2k+1)*pi*i, which exponentiating again will get you to -1. Thats good for visualization. Quote:Note that this grid of points covers the entire right half of the complex plane, so that when we iteratively perform logarithms, we can always find points close to the real line. But still I dont get this. Iterated (branches of) logarithms take any point to one of the fixed points of exp. Why do they come arbitrary close to the real axis with increasing n? I hope the picture will clarify. Even if this is the case, and I trust you enough to believe it, the bigger question is whether the singularities come arbitrarily close to *any* point of the real axis where the base change is defined. Or are there points with a certain neighborhood that does not contain a singularity of any $f_n$? If so, we would just use such a point for the powerseries development. Quote:because we have to wind in and around singularities to make sure we are in the primary branch of the logarithm of base eta. I think we anyway have to specify a cut system, as we have (branching) singularities. The functions $f_n$ are multivalued (if we continue passing a cut). If I understand that right, you set the cuts of $f_n$ to be the points that would be mapped to the negative real axis of any involved $\log_\eta$. See, if you define the value via a path, then you will usually get different values for non-homotopic paths (i.e. any deformation of one into the other would cross a singularity). So this is the same as a multivalued function. (The situation gets even worse if singularities appear on certain but not all branches.) The proper path that leads to using primary branch of the logarithm would then be a path that does not cross any cut of $f_n$. I think the cut system is like a labyrinth and hope you can provide a picture of it. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 10:40 PM (08/15/2009, 09:44 PM)bo198214 Wrote: (08/15/2009, 07:13 PM)jaydfox Wrote: Note that this grid of points covers the entire right half of the complex plane, so that when we iteratively perform logarithms, we can always find points close to the real line. But still I dont get this. Iterated (branches of) logarithms take any point to one of the fixed points of exp. Why do they come arbitrary close to the real axis with increasing n? I hope the picture will clarify. I hope so. The reason, of course, has to do with our choice of starting point. We want log^[n](-1), which as you point out, would seem to go to a fixed point. But consider the first of the n iterated logarithms. If you choose the log(-1) at pi*i, then yes, you are going to descend into a fixed point as you continue taking iterated logarithms. But notice that my first choice of log(-1) is way out on the imaginary line, maybe at something like (10^100 + 1)*pi*i. Notice that, even in the primary branch of the logarithm, it will take quite a while for us to get near the fixed point, and indeed, the iterated logarithms will get very close to the real line. Well, without further ado, I will start showing some pictures (sagenb.org came back online):     This first picture is to show the case of n=4. I find log(log(-1)), in the principal branch, and I get 1.14472988584940 + 1.57079632679490*I. I create a straight line from 0 to this point, which is the blue line in the picture. The exponentiation of this line gives me a curve with 1 and pi*i as endpoints, which is the green curve in the picture. Finally, the second exponentiation gives me a curve with endpoints of e and -1. Note that with two iterations remaining, I will get a singularity when I get to the third iterated logarithm in base eta. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 10:46 PM Now I will show the case of n=5. I find log(log(log(-1))), and for simplicity, I will keep using the principal branch. This gives me a value of 0.664571922248822 + 0.941029487312600*I. I will show two graphs.     The first shows the straight line with endpoints 0 and log(log(log(-1))), in the color of blue. The image of the first exponentiation of this line is shown in a sort of cyan color, and it has the same endpoint as the straight line from the case of n=4, incidentally. The second exponentiation gets to pi*i, and is show in a sort of yellow-green hue. Finally, the third exponentiation gets to -1, and is shown in red.     This second graph is simply zoomed out, so you can see the full curve of the third exponentiation. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 11:02 PM To demonstrate that there are singularities all along the positive real line, I will show the first three, with lines connecting them to x=1. I will show several more, without exponentiating them, just to give you an idea of where they are. The color scheme is the same as before: blue is the original image, cyan is the first exponentiation, yellow-green is the second, red is the third. Note that I did the first 100 such singularities, yet they get so close together that the dots just overlap in this graph:     And now a zoomout. Note that, due to the large real part of the yellow line (up to about 16 in places), the red line will be an arc with a radius of several million in places. I could zoom out to show this, but I trust everyone can imagine it.     ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 11:20 PM Now for n=6. As before, I plotted multiple lines, and just to show the density of the singularities, I showed the first 1000, which do not even reach as far as having real part 1. The color scheme: blue is the original image, cyan is the first exponentiation, yellow is the second, green the third, and red the fourth.             ~ Jay Daniel Fox bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/16/2009, 11:15 PM Ok, my explanation for the phenomenon is roughly: The orbit of a point $z_0$ from the upper halfplane under $\log$ is a right winding spiral around the primary fixed point in the upper halfplane. This spiral gets bigger the farther $z_0$ is located from the origin. The spiral getting bigger means the lower part of the spiral gets closer to the real axis. Indeed already the second iteration of $z$ (though it is usually not the minimum of the spiral) converges to the real axis when the imaginary part of $z_0$ goes to infinity. $z_0 = x_0 + iy_0$ $x_1 = \ln(|z|)$ $y_1 = \arctan(y_0/x_0)$ $y_2 = \arctan(\arctan(y_0/x_0)/\ln(\sqrt{x_0^2+y_0^2})$ for big arguments the logarithm grows faster than the arctan, thatswhy if we increase a big $y_0$ the quotient $\arctan(y_0/x_0)/\ln(|z|)$ decreases and hence $y_2$ decreases. As long as the radius of a number $z_n$ with $0<\Im(z_n)<\pi$ is too big, the imaginary part of the logarithm $\Im(z_{n+1})=\Im(\log(z_n))$ (which is the angle of $z_n$) will be smaller than the imaginary part of $z_n$. But doesnt this mean that the singularities that come close to the real axis somehow accumulate around 0? At least the minimum of the spiral will be around 0. I think I could even calculate bounds on the real axis around 0 in which the minimum of every such spiral has to live. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/16/2009, 11:38 PM (08/16/2009, 11:15 PM)bo198214 Wrote: But doesnt this mean that the singularities that come close to the real axis somehow accumulate around 0? At least the minimum of the spiral will be around 0. I think I could even calculate bounds on the real axis around 0 in which the minimum of every such spiral has to live.Don't take this wrong, as I mean no offense, but I'm not really sure what you're getting at. I'm not sure what you mean by singularities accumulating around 0. Indeed, for any given n, they accumulate around the real line, even for n as low as 5, and the accumulation gets denser as we move up the real line, to larger and larger real part. But perhaps we are speaking of two different things? ~ Jay Daniel Fox « Next Oldest | Next Newest »

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