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 Continuous iteration from fixed points of base e jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/22/2007, 04:50 AM (This post was last modified: 11/22/2007, 04:52 AM by jaydfox.) I am somewhat embarrassed to admit that I missed a major detail of the exponential branches of the slog. I was thinking about the fixed points as the only singularities. After all, those are the "points" which repel or attract with each iteration. For base e, all the fixed points are repelling for exponentiation, and within each branch, a fixed point is attracting for logarithms (two points in the zeroeth branch). But I failed to realize that 0 is an "attracting" point of a special kind. Actually, strictly speaking, I did realize this, but not in the right context. I had realized that if you exponentiate a random complex number enough times, there is a 100% chance that you'll eventually reach a number very close to 0 (which is to say that there are points that contradict this claim, but they form a meagre set). But I was thinking of discrete exponentiations, which prevented me from seeing 0 as an attractor. You see, "infinity" is an attractor of exponentiation. As we iteratively exponentiate, we grow in magnitude towards infinity. So let's consider the line with imaginary part pi*I and real part greater than 0. If exponentiated, we'll get all real values less than -1. If exponentiated again, we'll get all values on the real line segment (0, 0.367879). But if you keep track of which values map to which, you'll see that the values from the original line with real part approaching positive infinity became those values on the second line with real part approaching negative infinity, which then became the values on the line segment which approached 0. So in the discrete case, it's not quite clear that we're approaching 0 as a limit point of infinite iterations of exponentiation, but when viewing the particular line I singled out, it's quite clear. And of course, there are an infinite number of such lines, at 2*pi*I intervals. So what does this mean for the slog? Well, if we start at the origin in the 0-th branch, and then loop around either the upper or lower singularity, we'll see a singularity at the origin, a point we can't reach by exponentiation. We can get closer and closer to it by following carefully chosen complex iterates. The singularity itself is most bizarre. It is not a logarithmic singularity, nor even a double logarithmic singularity. In fact, its very bizarreness is part of why I didn't see it initially. Let's go back to my lines with imaginary part $(2k+1)\pi i$. We don't in fact have to be right on the line. To see this, start with a disc centered at the origin, with radius $\frac{1}{{}^{100} e}$. Now logarithmicize this disc in all branches, by which I mean, find the logarithm of every point in this disc, in every branch of the natural logarithm. You should now have a half-plane, with real values less than $-\left({}^{99} e\right)$. Now logarithmicize again, in all branches. You'll get U-shaped regions bounded above by $(2k+3/2)\pi i$ and below by $(2k+1/2)\pi i$, with real part at the apex ${}^{98} e$, and stretching to positive infinity. Now we logarithmicize again, first concentrating only on the principal branch. What do we get? For the two U-shaped regions closest to the real line, the apex will go from ${}^{98} e \pm \pi i$ to approximately ${}^{97} e$. Yes, there is an imaginary part, and we could even calculate it's value very accurately: $\tan^{\small -1}\left(\frac{\pi}{{}^{98} e}\right)$, which is not zero but close enough for government work. The important part is that the U-shaped region just became incredibly thin and squished against the real line. Indeed, if we logarithmicize all of the U-shaped regions, we'll get an infinite number of thin regions compressed into a U-shaped region with apex ${}^{97} e$. And once we look at all branches, you'll see this pattern repeat. Of course, we'll need to logarithmicize 98 more times to get back to the origin. Each time we logarithmicize, we take an infinite number of U-shaped regions and contort them into a U-shaped region. By the time we get back to the origin, we have a U-shaped region with a fractal pattern of embedded regions. All the points within that fractal pattern will end up within the original disk after 101 exponentiations. All the points within the U-shaped region but not part of the fractal pattern will NOT be within the original disk. The fractal pattern itself is due to the branches, as we wind further and further around the singularity at 0. The approach to the singularity at the origin is quite unlike any function I've ever seen, though admittedly I'm inexperienced with complex singularities outside of the generic 1/z pole and the logarithmic and double-logarithmic singularities. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/22/2007, 06:10 AM (This post was last modified: 11/22/2007, 06:12 AM by jaydfox.) Actually, now that I look at this, I see a beautiful parallel to the logarithm that I had previously thought was missing. As we move away from the origin, the logarithm grows linearly as the distance from the origin grows exponentially. In parallel fashion, as we move away from the origin, the superlogarithm grows linearly as the distance from the origin grows superexponentially. Going the other way, as we approach the origin, the logarithm decreases linearly as the distance from the origin decreases exponentially. However, the logarithm decreases linearly as the distance from the origin decreases linearly. For example, log(0.049787) is about -3, but slog(0.049787) is about -0.955, give or take. Moving in further, log(0.0000454) is about -10, but the slog is about -0.99996. However, if we look in the primary exponential branch of the slog, and loop around the upper primary singularity, we'll see a singularity at the origin. As we approach this singularity, the value of the slog increases linearly as the distance from the origin decreases superexponentially! (By "decrease superexponentially", I mean the reciprocal increases superexponentially.) The branches of the superlogarithm provide many layers of complexity, something I assume would be worthy to be explored in a Ph.D. thesis. Alas, it'll be years before I could dream of embarking on such an adventure, but it's still exciting to be exploring this relatively new territory. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 11/22/2007, 09:08 PM Interestingly, the singularity is repeated at every tetrate of e (0, 1, e, e^e, etc.), and the other singularities are scaled copies of the one at the origin, plus the constant n for the n-th tetrate. The scaling factor is simple to calculate. The singularity at the n-th tetrate is ${}^{n} e$ times bigger than the scaling factor at the (n-1)-th tetrate. This can be seen by considering: $ \begin{eqnarray} {\Large e}^{\left({}^{n-1}e+\delta\right)} & = & {\Large e}^{\left({}^{n-1}e\right)} {\Large e}^{\delta} \\ & = & {}^{n}e \left(1+\delta+\mathcal{O}(\delta^2)\right) \\ & = & {}^{n}e + {}^{n}e \delta \end{eqnarray}$ ~ Jay Daniel Fox « Next Oldest | Next Newest »

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