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 Tetration below 1 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/29/2007, 06:14 PM I have to admit I never considered the case $0, but I recently discussed it with Gianfranco. We know that ${}^{2n}x$ is no more injective if we allow this range. If we consider one rank below the ordinary exponentiation, we have a similar case there, that $b^x$ is no more defined for $-\infty, because $x^{2n}$ is not injective for $x<0$ so $b^{1/2}=\pm |b|^{1/2}$. This gives also trouble as then $b^n=b^{\frac{2n}{2}}=\pm b^n$. On the other hand for tetration ${}^{\frac{1}{2}} x$ is *not* the inversion of ${}^2 x$. So the rules may be a bit different here. So what do you think how tetration looks for $0? And what role plays the range $e^{-e}? Daniel Fellow Posts: 51 Threads: 20 Joined: Aug 2007 08/30/2007, 09:32 PM bo198214 Wrote:I have to admit I never considered the case $0, but I recently discussed it with Gianfranco. We know that ${}^{2n}x$ is no more injective if we allow this range. If we consider one rank below the ordinary exponentiation, we have a similar case there, that $b^x$ is no more defined for $-\infty, because $x^{2n}$ is not injective for $x<0$ so $b^{1/2}=\pm |b|^{1/2}$. This gives also trouble as then $b^n=b^{\frac{2n}{2}}=\pm b^n$. On the other hand for tetration ${}^{\frac{1}{2}} x$ is *not* the inversion of ${}^2 x$. So the rules may be a bit different here. So what do you think how tetration looks for $0? And what role plays the range $e^{-e}? My understanding is that $b = e^{-e}$ is the exceptional case here, it has a rationally neutral fixed point with a Lyapunov characteristic number of -1, see http://mathworld.wolfram.com/LyapunovCha...umber.html . It could be solved using Abel's functional equation for $f^{2 n}(x)$. The other values of $b$ have hyperbolic fixed points and could be solved using Schroeder's functional equation, Bell matrices or other equivalent methods. jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/30/2007, 11:08 PM From a practical standpoint, I've considered the solution to bases $e^{-e} < b < 1$ as being exponentially scaled sine waves of period 2 as we approach positive infinity, with iterated logarithms to get us back to the origin. This is in line with the solutions for bases between 1 and eta, which are essentially exponential near positive infinity, with iterated logarithms getting us back to the origin. For b=e^-e, we wouldn't even need an exponential scaling, just a plain old sine wave (near x=infinity), just as with eta, which becomes essentially linear near positive infinity. For solutions less than b=e^-e, we wouldn't even need an infinitesimal sine wave. A sine wave oscillating between the upper and lower point would make the most sense, with an infinitesimal perturbance which, with an infinite iteration of logarithms, would gets us back to the origin. Limits would have to be taken, of course, but it seems that a plain old sine wave would make the most sense, barring any solid evidence that the oscillations are somehow more complex than the problem would seem to dictate. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/30/2007, 11:20 PM While the sine wave idea makes sense from the standpoint of creating a simple and infinitely differentiable function, it doesn't seem to fit the idea of continuous iteration. If continuous iteration is to be considered, the sine wave idea doesn't fit. Given a value near the infinite limit, how would the function "know" which direction to go for a small fractional iteration? Is it on the upswing or the downswing? On the other hand, perhaps this is a situation where complex outputs are actually desirable? By having a spiral that hits the real plane only at the integer tetrations, we can embed information in the output that would "tell" the function where it is, and hence where to go next. Just an idea... ~ Jay Daniel Fox Daniel Fellow Posts: 51 Threads: 20 Joined: Aug 2007 08/31/2007, 12:47 AM jaydfox Wrote:While the sine wave idea makes sense from the standpoint of creating a simple and infinitely differentiable function, it doesn't seem to fit the idea of continuous iteration. If continuous iteration is to be considered, the sine wave idea doesn't fit. Given a value near the infinite limit, how would the function "know" which direction to go for a small fractional iteration? Is it on the upswing or the downswing? On the other hand, perhaps this is a situation where complex outputs are actually desirable? By having a spiral that hits the real plane only at the integer tetrations, we can embed information in the output that would "tell" the function where it is, and hence where to go next. Just an idea... For tetration $a$ converges to $h(a) = \;^{\infty}a = \frac{W(-\ln(a))}{-\ln(a)}$ $D^2 f^t(0) = f''(0) \sum_{k=0}^{t-1} f'(0)^{2t-k-2} \rightarrow D^2 \;^{n}a = 1/2 \sum_{k=0}^{n-1} {\ln(\;^{\infty}a )}^{2n-k} (1- \; \;^{\infty}a )^2$. The first few terms of the Taylor series for $\;^{n}a$ from it's fixed point are $\;^{n}a = \;^{\infty}a + \ln(\;^{\infty}a)^n \; (1- \; \;^{\infty}a) + 1/2 \sum_{k=0}^{n-1} {\ln(\;^{\infty}a )}^{2n-k} (1- \; \;^{\infty}a )^2 + \ldots$. This holds for any value of $a$ and and integer $n$. But this contains a geometrical progression that simplifies the equation based on the specific value of $a$, or more particularly the location and type of fixed point of $a^z$. As a result, $n$ will no longer be an index, but will be a complex number. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 08/31/2007, 02:10 PM (This post was last modified: 08/31/2007, 02:16 PM by bo198214.) Daniel Wrote:The other values of $b$ have hyperbolic fixed points and could be solved using Schroeder's functional equation, Bell matrices or other equivalent methods. For all values of $b$ in $(0,1)$ the function $\exp_b$ is strictly decreasing and has a fixed point $x_0$ with $\exp_b'(x_0)<0$. So Schroeder, Bell and equivalent methods do not fit. Strictly increasing and decrasing functions can be seen a bit in similarity with negative and positive numbers. For example $f^{\circ 2n}$ is strictly increasing and $f^{\circ 2n+1}$ is strictly decreasing for $f$ being strictly decreasing. Which is similar to $x^{2n} > 0$ and $x^{2n+1} < 0$ for $x<0$. Accordingly there are no strictly monotonic iterative square roots (or generally even roots) of strictly decreasing functions. (regardless whether the square root is increasing or decreasing, the iterative square must be increasing, but it is decreasing). Now you can say: Ok, then we allow non-monotonic functions (perhaps in analogy to the complex numbers, where we can extract square roots). But there is another thing: Proposition. The iterative root of a strictly monotonic continuous function (on an open interval X) is continuous if and only if it is strictly monotonic. So if we give up strict monotony of the root we also give up continuity, which then makes no more sense. Hence all the exponents $\frac{1}{2n}$ are lacking for tetration with $b<1$ and so we have no appropriate $\text{sexp}_b(\frac{1}{2n})$. If we use the analogy with real numbers then $b^x$, $b<0$, can be defined as $b^x=-|b|^x$ if $x=\frac{m}{2n+1}$ but it can not be defined for the cancelled fraction $x=\frac{m}{2n}$. (At this place I also want to correct some nonsense that I stated in my first post. It has to be: "because $x^{2n}$ is not surjective for $x<0$, $b^{1/2}$ can not be defined. This gives also trouble with $b^{n}=b^{2n/2}=?$.") So on the reals there is no way to continuously define $b^x$ for $b<0$. However if we extend our domain to the complex numbers there is the way, or rather the ways $b^x=\exp_{b,k}(x)=\exp(x(\log(b)+2\pi i k))$. Here for each $x=\frac{1}{2n+1}$ there is a $k$ such that indeed $b^x=-|b|^x$: $\log(b)=\log(-1|b|)=\log(-1)+\log|b|=\pi i +\log|b|$ $\exp_{b,k}(x)=\exp(x\cdot \log|b| + (2k+1)\pi i x)=|b|^x e^{(2k+1)\pi i x$ $\exp_{b,n}(\frac{1}{2n+1})=|b|^x e^{ \pi i }=-|b|^x$. However there is no function that $b^{1/(2n+1)}=-|b|^{1/(2n+1)}$ for all n. The suggestion of Daniel to simply iterate the strictly increasing $\exp_b^{\circ 2}$ does not really help about our problem as we still can not define the fractional iterates $\exp_b^{\circ \frac{1}{2n}}$. The only possibility I see is to extend the tetration for bases in $(0,1)$ to complex values. I think there are possibilities for (multiple) continuous iterative square roots there from which we can define a family of super exponentials $\text{sexp}_{b,k}(x)$ such that for each $x=\frac{1}{2n+1}$, we have $\text{sexp}_{b,n}(x)=1/\text{sexp}_{1/b}(x)$, where the reciprocal is surely wrong, but just there to give a rough idea. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/31/2007, 07:43 PM The interval b in (0, 1) is a very difficult region to do tetration over. The best that I have been able to do is something similar to my matrix equations, only instead of being automated, I solve each step of the way manually. From my manual solutions (i.e. using a 4-th degree polynomial, changing coefficients as necessary) I have found that tetration over the critical interval (-11 and ${}^{-2}b = +\infty$ for b<1. In order for the curve to get the "direction" in needs to go in, it needs to be more sinusoidal, than linear as with b=e. For more about the domain and the line where x=-2, see my previous post http://math.eretrandre.org/tetrationforu...287#pid287 about the domain of real-valued (and possibly real-analytic) tetraiton. Andrew Robbins jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 04:12 PM andydude Wrote:The interval b in (0, 1) is a very difficult region to do tetration over. The best that I have been able to do is something similar to my matrix equations, only instead of being automated, I solve each step of the way manually. From my manual solutions (i.e. using a 4-th degree polynomial, changing coefficients as necessary) I have found that tetration over the critical interval (-11 and ${}^{-2}b = +\infty$ for b<1. In order for the curve to get the "direction" in needs to go in, it needs to be more sinusoidal, than linear as with b=e. For more about the domain and the line where x=-2, see my previous post http://math.eretrandre.org/tetrationforu...287#pid287 about the domain of real-valued (and possibly real-analytic) tetraiton. Andrew Robbins Without going to complex values, I'd come to the same conclusion. A sinuisoidal wave would seem to make the most sense. For bases <= e^-e, we could just use a sine (cosine, same difference, I'm mostly referring to the base shape) wave. For bases between e^-e and 1, the upper and lower points converge exponentially, so we'd need an exponentially scaled sine wave. This unfortunately raises the question: do we make the integer values the peaks of the wave, or do we make them tangent to the exponential asymptotes? My personal preference is to make them tangent, but I haven't investigated. At any rate, while it seems like a potentially insightful path to pursue, I'm still leaning towards complex values. I had spent a while pondering over it yesterday, then came up with essentially what Henryk has posted (with regards to his analogy of exponentiating bases less than 0, by having a "family" of exponential functions that return complex values for non-integers), so I think he's on the right track. ~ Jay Daniel Fox GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 09/02/2007, 09:46 AM Hi everybody! Now, Enryk will certainly say that I am again late in reacting. He is right! Sorry about that. I shall improve! I still have to learn how to swim in this Forum and I really hate the TeX language, unless I am able to find an easy compiler that will help me. Coming back to the subject, I more or less agree with all the previous comments (including the expressed doubts), keeping in mind that we may say that y = b^x for x<0 does not exist, only if we decide to remain in the "reality". If we admit the "complex universe", this is no more true. Complex analytic functions would probably help us in the analysis of the entire b>0 domain, for y = b # x (b real). Unfortunately, I am not very "fluent" in that area. Therefore, as I promised to Enryk, I submit to your attention the attached pdf notes, hoping that you will not be annoyed and that find some ideas to be developed. Please see also what should, in my opinion, happen in the range 0 < b < e^(-e). It is a simulation but, just tell me what you think. In fact, I agree that that interval corresponds to the negative bases, for exponentiation. The appearence of oscillations, for b<1 must be admitted, if we accept to go out of the "reality". I think to understand a little bit better what Euler meant by saying that the "infinite towers" don't converge for b < e ^(-e). Perhaps, they just oscillate -> oo. Please also investigate, in the 1 < b < e^(1/e) interval, what could be the role of the second (upper) "unreachable" asymptote. Another fact which I should like to mention is that the superlog (slog) should be limited to b > e^(1/e) [the jayfox eta, or the self-root of e], otherwise ... we are in trouble (apart from any possible theoretical assessment). Congratulations to Andrew. Well done! But ... the war continues! Rubtsov and myself, we are investigating also in other directions. We shall keep you informed asap. Thank you G. F. Romerio Attached Files   InfTow and b.pdf (Size: 133.45 KB / Downloads: 381) bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/02/2007, 10:23 AM GFR Wrote:Now, Enryk will certainly say that I am again late in reacting. Absolutely not! Considering that you are our most experienced (not to say oldest ) member of the forum, with an age that is the smallest integer above 1 that is one less then twice its decimal mirror, that was a short time for such a comprehensive presentation. I will answer soon. « Next Oldest | Next Newest »

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