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 Regular slog for base sqrt(2) - Using z=2 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 10:20 AM I'm still working on graphs, but it's not easy to strike a balance between using enough grid points to make a smooth graph and using few enough to keep memory usage within 1.5 GB and compute time to a fraction of an hour. By the way, these are half-sized versions of the ones I originally made. I was going to attach them (without embedding them) for those who wanted to download them, but they are over 1 MB each, and I can't attach them at that size. I suppose I can email them if anyone wants to see the larger versions, but the small versions preserve most of the interesting details. Okay, to establish (hopefully) clearly how I'm calculating the regular slog, here's a view at z=2, zoomed in pretty far. Note that each curve is almost a circle. Indeed, the further we drill down, the more they will look like circles.     An explanation is due here. Each circle is a contour of constant real part. No, they are not evenly spaced, nor even exponentially spaced. I will explain the spacing later. Essentially, I iteratively exponentiated zero, for reference 57 times. I picked a number that puts a blue ring going through the origin, based on my palette, if you're wondering why 57. There is still a fair amount of inaccuracy here, but not enough to show up on the graph. Once I had the 57th iterate of 0, I found the 58th (i.e., the 57th iterate of 1), and then subtracted each from 2. The distances from 2 are decreasing exponentially, so I take a natural logarithm of each, calling them l0 and l1 respectively (for log of "zero" and log of "one"). I then subtract, giving -0.36653913, which is very near the predicted value of ln(ln(2)). I'll call this value dl (for difference of logarithmcs). I then create a grid of values (not evenly spaced), with real parts on the interval $(l0,l0+dl)$, and imaginary part on the interval $\left(0, \text{abs}(\pi / dl))$. When I exponentiate (base e) the values in this grid, I get half a hollow disk, which I then subtract (point by point) from 2. Its inner radius passes through the 57th exponentiation of 1, the outer radius passes through the 57th exponentiation of 0. It's only a half of a hollow disk. I don't actually calculate the points above the real line; rather, I simply graph the lower half plane and mirror it to get the other half. Then, I can simply take the logarithm of this grid, point by point, to get each successive layer moving out from the fixed point. Here, I've zoomed out by a factor of 10.     Don't be fooled by the small shift in the position of the colors: we've gone more than a full cycle through the palette. Note the scale on the axes. The red circle that was clipped and only visible in the corners is now the red circle that fits comfortably between the text labels closest to the origin. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 10:37 AM (This post was last modified: 11/15/2007, 10:39 AM by jaydfox.) Zooming out another factor of 10, the non-circularity of the contours becomes obvious.     Finally, we're zoomed out enough here to see the origin. Notice that we're almost to the interesting part, where our ovals are no longer connected loops. (Technically, they're not connected now, because the imaginary parts increase counter-clockwise, due to the logarithmic singularity. However, the branches are all the same, at least when we calculate the regular slog at 2.)     This is where I can describe the peculiar spacing of the contours. When I take the logarithm of points near the origin, they will towards negative infinity. To get any kind of decent picture of what happens in the left half plane, I had to make my contours bunch up around the origin. And because of how I calculate each region as the logarithm of a region closer to the singularity, I had to effect this strategy from the outset. Therefore, I created a cyclic mapping of the real parts, $c\left(\Re(z)\right) = \Re(z) - \frac{\sin\left(2\pi \Re(z)\right)}{2\pi}$ This gives me a slope of 0 at the integers, so that I can get values very close to 0 when I get back to the origin. In the next posts, you'll see the effect I was going for. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 10:53 AM Now, here's an image zoomed out a factor of 4, and centered at the origin. This allows us to see an overview of the slog.     Note that there's a lot of detail missing to the right. So far, only the principal branch has been used for the logarithms I've been calculating. I have 64 contours for each unit interval, with about 200 points in the imaginary direction, calculated for about 80 regions. Adding even one branch begins to push the limits of my laptop. When I start showing branches, I will be reducing the number of contours per interval accordingly. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 11:06 AM Before I start showing branches, I wanted to show you zoom-ins of the singularity at z=4. I believe the term for it is essentialy singularity, as there is a line of singularities approaching it from the right, a countably infinite number of singularities that have z=4 as their limit point. Thus, we cannot remove this singularity in the traditional sense. The singularity at 2 is removable, in the sense that a Taylor series developed arbitrarily close to z=2 would have a radius of convergence limited by the singularity at z=4. I'm not sure if it's valid to develop a power series at z=2 after removing the singularity, however. Anyway, here's the first zoomin, showing both singularities, to help you appreciate the difference between them:     Notice that the red line in the middle is fairly straight. Somewhere in that vicinity, the curvature of the contours at the real points switches from left-curving to right-curving. This is a point of inflection for the rslog on the real interval between z=2 and z=4. I'm not sure what if any significance this has, but it is similar in principle to the point of inflection on the critical interval for base e, and it's worth further exploration at some point in the future when one of us can get around to it. Next, we'll zoom in properly on the singularity at z=4:     ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/15/2007, 11:39 AM (This post was last modified: 11/15/2007, 11:40 AM by jaydfox.) You might be tempted, looking at the second image in the previous post, to hope that the white space to the right of the singularity will close off, i.e., that the angle will decrease arbitrarily. In fact, we know that there will always be an angle there, because the bases of the "logarithms" at each singularity are different. A 180-degree rotation around the singularity at z=2 involves moving $\frac{\pm\pi}{\ln\ln{2}}$ units in the imaginary direction. Moving 180 degrees around the singularity at z=4, if it were an ordinary logarithmic singularity, would require moving $\frac{\pm\pi}{-\ln\ln{4}}$ units, about 12% further. If we only moved $\frac{\pm\pi}{\ln\ln{2}}$ imaginary units around the idealized singularity at z=4, we would only have gone 160.4 degrees. This remaining 19.6 degrees explains, to a first approximation, the white space in the graph (which represents the logarithms of other branches). To see this, let's zoom in a couple more times, to satisfy ourselves that this angle will indeed appear to remain. Here's a factor of 10 zoom from the previous image:     And another factor of 10:     ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 05:15 AM Apologies for the delay, but it took a while to tweak the code to get useable results. In the first few posts of this thread, I used a non-linear spacing of the contours to help bring out detail. For the graphs that follow, I've used a linear spacing, meaning contour lines with real parts at constant intervals. In this case, I've chosen 8 contours per real unit. Internally, I've calculated a "primary branch" of sorts, then calculated the first three branches up and down from there. I then took the logarithms of these six non-primary branches, which gave me the analytic continuation in the primary branch. (By the way, strictly speaking, these aren't branches of the superlogarithm, but correspond to branches of the underlying logarithm for this base. Branches imply branch cuts in a multi-valued relation. As an example, there would be a branch cut in the superlogarithm running between the singularities at z=2 and z=4, which you can't see here because the real contours wrapping around z=2 coincide.) I then moved these continuations into four non-primary "branches" (first two up and down), and took the logarithm again. I only used four branches this time, rather than 6, because I ran out of memory the first time I tried to run my script. Rather than try to squeeze more points into the same RAM, I opted simply to eliminate points. Anyway, this gave me a second layer of continuation. This process could be repeated indefinitely, had I the CPU time and memory available to do so. (Note: I also took additional logarithms in the primary branch, to fill in some detail.) First up is a zoomout, so that you can see the next branch up and down from the principal branch:     Next up, I've zoomed in a factor of 2.5, to show a little more detail but still allow you to see the entire branch (top to bottom, not left to right, obviously).     As before, I started with larger images, then shrunk them 50% for bandwidth and screen space. If you'd like to see the originals, let me know and I can email them. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/20/2007, 05:20 AM Okay, these next two images are just zoomins, a factor of 2 each, for a total zoom factor of 5 and 10, respectively, relative to the first image of the previous post. First, the factor 5 zoom:     Next, the factor 10 zoom:     ~ Jay Daniel Fox bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 11/23/2007, 09:27 AM Wow is this beautiful! By Jay D. - the artist - Fox Though it took me some while to actually understand *what* you are drawing. But I think I understand now that you draw the rslog-image of a set of vertical lines in a certain rectangle for the upper half. Where you use $n=57$ in the formula $\text{rslog}(z)=\log_{\ln(2)} \frac{2- \exp^{\circ n}(z)}{2-\exp^{\circ n}(1)$ as the approximation for the regular slog. How can one find the singularities of $f^{\circ t}(x)$ and the complex fixed points of $f(x)=\sqrt{2}^x$ in your drawing? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/25/2007, 01:15 AM Two comments. @Henryk The infinitely iterated exponential with branch indecies is probably the best way: $H_k(x) = {}^{\infty}(x)_k = \frac{W_k(-\log(x))}{-\log(x)} = e^{-W_k(-\log(x))}$ with the value substituted for x as in $H_k(\sqrt{2})$. Using the Lambert W function branches, this gives $H_{-1}(\sqrt{2}) = 4$ and $H_{0}(\sqrt{2}) = 2$ which are exactly those that you would expect. @Jay Why is it 18-periodic? I understand it would be periodic in the imaginary direction just like base-e, but why 18? Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/25/2007, 03:18 AM (This post was last modified: 11/25/2007, 03:19 AM by jaydfox.) It's not 18*I, it's $\frac{2\pi i}{\ln\left(\sqrt{2}\right)}$, which is the periodicity of the logarithm base sqrt(2). The value works out to 18.13*I or so. ~ Jay Daniel Fox « Next Oldest | Next Newest »

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