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 Real and complex behaviour of the base change function (was: The "cheta" function) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/12/2009, 08:59 PM Actually I am somewhat confused about the continuation of $f_n = \log_a^{[n]}\circ \exp_b^{[n]}$ for some $n$ from the real axis to the complex plane (via the method described by Jay some posts ago, i.e. considering a/the path $\gamma$ from a point on the real axis to $z$, looking at its image under $\exp_b^{[n]}$ and determine the value of the subsequently following logarithms by their continuation along the image of that path). Has anyone a clear view to which points $f_n$ can be continued, is it even entire? sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 08/13/2009, 12:49 AM (08/12/2009, 08:59 PM)bo198214 Wrote: Actually I am somewhat confused about the continuation of $f_n = \log_a^{[n]}\circ \exp_b^{[n]}$ for some $n$ from the real axis to the complex plane (via the method described by Jay some posts ago, i.e. considering a/the path $\gamma$ from a point on the real axis to $z$, looking at its image under $\exp_b^{[n]}$ and determine the value of the subsequently following logarithms by their continuation along the image of that path). Has anyone a clear view to which points $f_n$ can be continued, is it even entire?Me too, I'm making progress on n=5,6,7 ... I think the answer will eventually come out and it will be very interesting, but I don't have enough nearly enough time to do the problem proper diligence this week. Perhaps f will even converge as n increases to infinity, just as the function converges on the real number line.... - Shel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/13/2009, 07:17 AM (This post was last modified: 08/13/2009, 07:19 AM by bo198214.) (08/13/2009, 12:49 AM)sheldonison Wrote: Perhaps f will even converge as n increases to infinity, just as the function converges on the real number line.... Walker showed a similar convergence for $f_n=\operatorname{dexp}^{[-n]}\circ \exp^{[n]}$, where dexp(x) = exp(x)-1. He showed that the limit is infinitely differentiable on the real axis. That means that he also wasnt clear about the complex behaviour otherwise he would have shown that the limit is holomorphic as a consequence of local uniform convergence. But he could prove that local uniform (or compact) convergence only on the real axis, which does not suffice to imply holomorphy (because it could be that during the convergence non-real singularities get dense towads points on the real axis). I will persue this topic in the next days and have still some unexplored ideas at my hands. sheldonison Long Time Fellow Posts: 633 Threads: 22 Joined: Oct 2008 08/13/2009, 06:48 PM (This post was last modified: 08/13/2009, 07:37 PM by sheldonison.) (08/13/2009, 07:17 AM)bo198214 Wrote: Walker showed a similar convergence for $f_n=\operatorname{dexp}^{[-n]}\circ \exp^{[n]}$, where dexp(x) = exp(x)-1. He showed that the limit is infinitely differentiable on the real axis. That means that he also wasnt clear about the complex behaviour otherwise he would have shown that the limit is holomorphic as a consequence of local uniform convergence. But he could prove that local uniform (or compact) convergence only on the real axis, which does not suffice to imply holomorphy (because it could be that during the convergence non-real singularities get dense towads points on the real axis). I will persue this topic in the next days and have still some unexplored ideas at my hands.First off, the graph below, and the earlier now incorrect results I posted to this thread correspond to the equation Henryk says Walker analyzed, except with the exponentials being "exp(x) = exp(x)-1", and the logarithms being base e. The "exp(x)=exp(x)-1" corresponds to base eta with the initial operands divided by e. Here is small slice of the complex plane, that seems to have a reasonable chance of converging as k increases. $f(x) = \lim_{k \to \infty} \log_e^{\circ k} \left( \exp_\eta^{\circ k} ( 4.7 + \Im ) \right)$ I don't have much time right now, but here is a graph for f(x), where real(x)=4.7 and imag(x) varies from 0 to i*0.5, where I show the graphs for k=5, k=6, and k=7. I haven't analyzed whether the windings work for larger values of K, but I hope they might. Also, I zoomed in on the transition near i=0.1, and the k=7 graph is off by one winding; this isn't visible at this scale. Next, I was interested in f(4.7+0.2i) and I thought it might converge to the fixed point of e. Not so! For k=7, it converges to -0.516080387 + i*0.262012723. The method of convergence is to track down those pesky windings. so as to guarantee that all of the steps in the iterated logarithms are continuous, it winds up, and eventually the imaginary portion goes negative, and it stops winding, and more or less freezes at approximately i=0.1, there is a small amount of change (not visible in the graph), when i continues growing past 0.4 For this one segment of the complex plane, F appears to converge, and for the value in question may allow for a continuous extension of Jay's base change to the complex plane, but there is much more to do. In other words, $F(4.7+0.2i)=exp(F(\eta^(4.7+0.2i))$ - Shel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/14/2009, 09:58 AM (This post was last modified: 08/14/2009, 10:27 AM by bo198214.) Now I realized that $f_n$ has a lot of singularities except for $n=1,2$ where it is a linear function. A singularity can only occur if the logarithm of 0 is taken. For $f_1$ no singularity occurs because $0$ is not in the image of $\exp$. And we can simplify it to: $f_1(z)=\log(\exp_\eta(z)) = \log(\eta) z = e^{-1}z$ Because $f_1(z)=0$ iff $z=0$ the only singularity can occur when $\exp_\eta(z)=0$ which is again not possible. So $f_2$ will also not have any singularities. It can indeed be written as a affine function. $f_2(z)=\log(e^{-1}(\exp_\eta(z)))=-1+e^{-1}z$ For $f_3$ $f_3(z)=\log(-1+e^{-1}\exp_\eta(z))$ we determine the zeros of $f_2(z)=0$: $-1+e^{-1}z=0\leftrightarrow z=e$ So whenever $\exp_\eta(z)=e$, then there is a singularity of $f_3$, this is for: $\exp_\eta (z)=\exp(e^{-1}z)=e \leftrightarrow z=e+2\pi i k e$, $k\in\mathbb{Z}$. Thatswhy $f_3$ has a singularity at each $e+2\pi i k e$. Particularly at $e$ which restricts the domain of definition of $f_3$ to $(e,\infty)$. Let us generalize this some more. Let now $f_n=\log_a^{[n]}\circ \exp_b^{[n]}$ and $f=\lim_{n\to\infty} f_n$. $\log_a^{[n]}$ has singularities exactly if $\log_a^{[m]}(x)=0$ for some $m\le n-1$. As we assume that we take the real logarithm of real numbers (otherwise singularities could be avoided by choosing non-real branches). In other words $\log_a^{[n]}$ has singularities at $A_n:=\{0,1,a,\dots,\exp_a^{[n-1]}(0)\}$. These values are reached by $\exp_b^{[n]}(z)$ at $\exp_b^{[-n]}(A_n)$ where $\exp_b^{[-1]}(X):=\{\log_b(z)+\frac{2\pi i k}{\ln(b)}: z\in X, k\in\mathbb{Z}\}$. However $f_n$ does not need to have singularities at all of these values, as some $\log_a$ may take a different branch $\frac{2\pi i k}{\ln(a)}$ instead of returning 0 (or one of $\exp_a^{[m]}$) if the argument of $f_n$ was non-real. So if we always take the primary logarithm $\log_b$ we obtain a possible set of singularities which lies inside the set $B_n:=\log_b^{[n]}(\mathbb{C})$. As the primary logarithm $\log_b$ maps $\mathbb{C}$ bijectively to $\{z:-\pi/\ln(b) < \Im(z) \le \pi/\ln(b)\}$ we conclude that $\exp_b^{[n]}$ is bijective on $B_n$. More importantly any path $\gamma$ to $z$ from $B_n$ in the upper halfplane will be mapped to $\exp_b^{[n]}(\gamma)$ in the upper halfplane, i.e. it will not wind around 0. Hence by our construction we must take the primary logarithm $\log_a$ of the point $\exp_b^{[n]}(z)$, $z\in B_n$. But the primary logarithm $\log_a$ lies again in the upper halfplane and so on, that means we must always take the primary logarithm $\log_a$ of $\log_a^{[m]}(\exp_b^{[n]}(z))$, $m\le n-1$. Particularly this is true singular choices of $z\in B_n$, they *must* yield singularities $\log_a$ can not escape to some other branch. Hence Proposition. Every element of the set $S_n := \log_b^{[n]}$$\{a,\dots,\exp_a^{[n-2]}(1)\}$$$ is a singularity of $f_n$ (which is defined via path-continuation). Where we consider $\log_b$ to be the primary branch. (0 and 1 are excluded because they have no logarithm for $n\ge 2$.) These singularities are not isolated but they are branch points. So depending how the path to a point winds around these singularites we get different results of the $f_n$. So we have to restrict ourselves to a simply connected neighborhood of the real axis where no singularities exist, there we have a unique continuation. The interesting question is now how these singularities are distributed in the limit case. Do some singularities converge to the real axis? jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/14/2009, 04:50 PM (This post was last modified: 08/14/2009, 04:52 PM by jaydfox.) (08/14/2009, 09:58 AM)bo198214 Wrote: Now I realized that $f_n$ has a lot of singularities except for $n=1,2$ where it is a linear function.But Bo, these zeroes are a trivial result of the "change of base" concept. If, starting with the number 1, we perform a change of base operation, from base e to base eta, we get a real number, approximately 6.3344. If we instead start with 0 in base e, we get approximately 5.0179. This is not surprising, as 5.0179 is log_eta(6.3344). Well, log_eta(5.0179) is approximately 4.3846. Thus, changing base from eta back to base e, we should expect 6.3344 to be 1, 5.0179 to be log_e(1), which is 0, and 4.3846 to be log(log(1)), which is (drum roll please) negative infinity. What is not as trivial to determine is where the "other" singularities are, if in fact there are any. I assume there are other singularities, but I am so far not having the best of luck in finding them. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/14/2009, 04:58 PM (This post was last modified: 08/14/2009, 06:28 PM by jaydfox.) (08/14/2009, 04:50 PM)jaydfox Wrote: (08/14/2009, 09:58 AM)bo198214 Wrote: Now I realized that $f_n$ has a lot of singularities except for $n=1,2$ where it is a linear function.But Bo, these zeroes are a trivial result of the "change of base" concept. If, starting with the number 1, we perform a change of base operation, from base e to base eta, we get a real number, approximately 6.3344. If we instead start with 0 in base e, we get approximately 5.0179. This is not surprising, as 5.0179 is log_eta(6.3344). Well, log_eta(5.0179) is approximately 4.3846. Thus, changing base from eta back to base e, we should expect 6.3344 to be 1, 5.0179 to be log_e(1), which is 0, and 4.3846 to be log(log(1)), which is (drum roll please) negative infinity. What is not as trivial to determine is where the "other" singularities are, if in fact there are any. I assume there are other singularities, but I am so far not having the best of luck in finding them.By the way, it occurs to me that for converting from base eta to base e (or any other base, I suppose), we must pick a starting point on the real line, which is either above the fixed point at e, or below it. These two approaches will give different results (for if they didn't, then regular iteration above or below would be equivalent, and we know that it is not). So bear in mind that I'm speaking from the point of view of starting above the fixed point, e.g., at about 5.0179, which would get us back to 0 in base e. Edit/Update: Sorry, Henryk, I re-read your post and realized that you had in fact specified that you were starting with a real value greater than e for n=3, and larger values as necessary for larger n. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/14/2009, 07:04 PM (This post was last modified: 08/14/2009, 08:16 PM by bo198214.) (08/14/2009, 04:50 PM)jaydfox Wrote: But Bo, these zeroes are a trivial result of the "change of base" concept. Well if you only look at the case $a=e$, $b=\eta$; it may be trivial (though singularities not zeros) because everything stays on the real axis. However I was considering general real bases $a$ and $b$ and the possibly non-real singularities in the region $B_n$. As you say yourself its difficult to determine whether the non-real possible singularities (which I completely specified) indeed exist or cancel out by appropriate choices of the logarithm. I showed that $f_3$ indeed has complex singularities that dont cancel out. And also that there are complex singularities for other bases that dont cancel out. The example of $f_3$ shows also that your suggested path-continuation does not work. If you have branch points then different paths to the same point may result in different values of $f_3$. To keep it continuous you have to specify cuts. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 12:54 AM (This post was last modified: 08/15/2009, 12:57 AM by jaydfox.) I think I have figured out conceptually how to reconcile the singularities which probably lie arbitrarily close to the real line, with the fact that we can generate a power series with a seemingly non-zero radius of convergence. To see this, consider the following contrived function: f(x) = log(log(log(log(log(x+1)+e^(e^e)))))) Note that at x=0, the value is 0. Also note that you can pick points very close to -1, and the value will barely change at all. Yet note that there is a singularity at x=-1. The function is very, very smooth at x=0, almost constant in fact, and yet there is definitely a singularity at x=-1. I think a similar effect is at work with the base change formula. For any finite n, we could in principle locate singularities by careful analysis, and as n increases, these singularities will get closer and closer to the real line. Yet they will also be less and less perceptible, unless you can manage to get really, really close to one. As such, the power series expansion would seem to have a rather large radius of convergence, because the first few thousand terms of f are determined primarily by the power series developed with f_n, with a rather small value of n. But if we could compute tens of thousands of derivatives with millions of bits of precision, we should see the radius of convergence slowly drop. Indeed, the full power series likely has 0 radius of convergence. (Or is it more accurate to say that for any finite value epsilon, the radius of convergence can be shown to be smaller than epsilon? Is that any different than saying it's 0?) It seems therefore rather miraculous that the limit is well-defined for real numbers, when it would seem to be undefined for non-real numbers. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/15/2009, 05:00 PM (This post was last modified: 08/15/2009, 05:02 PM by jaydfox.) (08/15/2009, 12:54 AM)jaydfox Wrote: I think a similar effect is at work with the base change formula. For any finite n, we could in principle locate singularities by careful analysis, and as n increases, these singularities will get closer and closer to the real line. Yet they will also be less and less perceptible, unless you can manage to get really, really close to one.In thinking about it, the singularities are trivial to find. 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. Regardless of branching, log(0) is always a singularity. And notice that, after getting this logarithmic singularity, we then proceed to take several more iterated logarithms, which would effectively diminish these singularities. As n increases, they would become very "skinny", almost undetectable when numerical precision is taken into account. Furthermore, if we increase n by 1, then this singularity disappears, because exponentiating -1 gives 1/e, and the double logarithm of the double exponentiation gives e+1. Iterated logarithms in base eta should now resolve properly, avoiding a singularity. 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. On the other hand, there aren't any "fixed" singularities, at least not of this trivial variety. The singularities themselves become more and more "insubstantial", so I'm wondering if perhaps convergence in the complex plane is indeed possible? ~ Jay Daniel Fox « Next Oldest | Next Newest »

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