tetration base conversion, and sexp/slog limit equations sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/18/2009, 07:01 AM (This post was last modified: 02/24/2009, 09:52 PM by sheldonison.) Can someone point me to a post about base conversions for tetration involving very large numbers? For small numbers, there probably isn't a shortcut to simply taking the super-exponent of the first number (by the first base), and then taking the super-logarithm (by the second base). But for large numbers, say x larger than a googolplex or 10^(10^100), it would seem that the base conversion could be a constant. I came up with an algorithm to approximate values for base conversions for large values of x. $\text{slog}_2(x) - \text{slog}_e(x) = 1.1282$ $\text{slog}_3(x) - \text{slog}_e(x) = -0.1926$ $\text{slog}_{10}(x) - \text{slog}_e(x) = -1.1364$ Anyone know if these values are close? I am also able to use this algorithm to generate sexp estimates for any base, and have a spreachsheet with results for base 2, 3, 10, and base e. I will post more complete details when I have more time, but here's the quick version. The algorithm is to use values of b1 for the base slightly larger than e^(1/e). I used values for b1 between 1.48 and 1.6. Pick the most "linear" region of the curve, where the inflection point is, and then iterate exponentiation using base b1, until the number is very large. Then pick another more interesting base, like b2=2, e, 3, 10, and iterate taking the logarithm. The algorithm assumes that there is a unique "base conversion" constant for large values of x, for any two tetration bases large than e^(1/e). I have no idea if these results are analytic or not, or if they meet the other criteria for unique tetration results, but the graphs over the critical section of b2 look well behaved. I've looked at the function and the first derivative, as well as the even/odd deviations from linearity over the critical section. Results for critical section of base b2=10. The critical section is where the tetration curve is most linear, where the derivative is equal at both endpoints, and includes the inflection point of the curve. see http://math.eretrandre.org/tetrationforu...880#pid880 x .... 10^^x -1.45538 -0.3623 -1.40538 -0.3165 -1.35538 -0.2730 -1.30538 -0.2315 -1.25538 -0.1914 -1.20538 -0.1525 -1.15538 -0.1146 -1.10538 -0.0773 -1.05538 -0.0405 -1.00538 -0.0039 -0.95538 +0.0326 -0.90538 +0.0692 -0.85538 +0.1060 -0.80538 +0.1434 -0.75538 +0.1814 -0.70538 +0.2203 -0.65538 +0.2601 -0.60538 +0.3013 -0.55538 +0.3438 -0.50538 +0.3880 -0.45538 +0.4342 - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/19/2009, 12:10 AM (This post was last modified: 02/19/2009, 05:37 PM by sheldonison.) I did a little more work, enough to convince me that using the taylor series posted by Kouznetsov in this_post gives different results different results from what I'm getting. I think that means converting between tetration bases, even for very large numbers, is never going to give an exact constant value, but "wobbles" a little bit. The values I derived for base "e" below, are pretty smooth even out to the 4th derivative, which can be estimated by using linear difference equations. The 3rd order polynomial curve fit I'm using for base 1.485 should be accurate to more than 3 significant digits, I would guess it would be accurate to 5 or more digits. converting from base 1.485 to base "e" x my_value correct_value error term -1.00 0.00000 0.00000 +0.00000 -0.95 0.05383 0.05382 +0.00001 -0.90 0.10624 0.10626 -0.00002 -0.85 0.15747 0.15754 -0.00008 -0.80 0.20770 0.20786 -0.00016 -0.75 0.25713 0.25739 -0.00026 -0.70 0.30593 0.30630 -0.00037 -0.65 0.35426 0.35473 -0.00047 -0.60 0.40226 0.40283 -0.00056 -0.55 0.45009 0.45073 -0.00064 -0.50 0.49788 0.49856 -0.00069 -0.45 0.54575 0.54645 -0.00070 -0.40 0.59383 0.59451 -0.00068 -0.35 0.64223 0.64286 -0.00063 -0.30 0.69108 0.69163 -0.00055 -0.25 0.74049 0.74094 -0.00045 -0.20 0.79056 0.79090 -0.00034 -0.15 0.84143 0.84166 -0.00023 -0.10 0.89321 0.89333 -0.00013 -0.05 0.94602 0.94606 -0.00005 +0.00 1.00000 1.00000 +0.00000 I also tried going backwards, from the published correct tetration values for base e, to base 1.485. The wobble going in the "backward" direction was much much more pronounced, and even the first derivative curve for base 1.485 already showed defects in it when graphed. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 02/19/2009, 04:24 PM (This post was last modified: 02/19/2009, 04:27 PM by bo198214.) sheldonison Wrote:I think that means converting between tetration bases, even for very large numbers, is never going to give an exact constant value, but "wobbles" a little bit. Perhaps this is just the effect of using different slogs. If we have two slogs for one base, say $f$ and $g$, which are strictly increasing and have the same range of values $(-2,\infty)$, then $g(f^{-1}(x))$ is defined and $g(f^{-1}(x)) -x$ is 1-periodic. Or in other words $g(x)=f(x)+\theta(f(x))$ for a 1-periodic function $\theta$. Subtraction of both yields $g(x)-f(x)=\theta(f(x))$. If we consider now different bases $a$ and $b$, and looking at the difference $\delta$ of $f_a-f_b$ and $g_a-g_b$, then we see $\delta(x)=g_a(x)-f_a(x) - (g_b(x)-f_b(x)) = \theta_a(f_a(x)) - \theta_b(f_b(x))$. So even if $\lim_{x\to\infty} f_a(x) - f_b(x)$ would exist, then $\delta$ probably would be wobbly, i.e. not converge to 0. On the other hand this could be a possible uniqueness criterion. Because if $\lim_{x\to\infty} f_a(x)-f_b(x)$ exists for one slog $f$ then $\lim_{x\to\infty} g_a(x)-g_b(x)$ would not exist for another slog $g$ but wobble around $f_a(x)-f_b(x)$. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/20/2009, 10:54 AM (This post was last modified: 02/20/2009, 04:32 PM by sheldonison.) The question for sexp/slog is how to define the curve extending sexp from integers to real. One question is does there exist an sexp/slog function for which $\text{slog}_a(\text{sexp}_b(x))-x$ converge to a constant value, or does it converge to a 1-cycle periodic function? And can this be a uniqueness criterion? Consider what happens as b approaches e^(1/e) in the equation $\text{slog}_b(x)$. The curve becomes more and more linear, and there are fewer and fewer degrees of freedom for how to extend the sexp function to real numbers, and still have an increasing "well behaved" function. It must be possible to describe this rigorously in terms of limits. Here is an example. if $b=1.49208... \text{ slog}_b(e)=6$, $\text{slog}_b(2.4989)=5$, $\text{slog}_b(2.2887)=4$, $\text{slog}_b(2.0691)=3$, $\text{slog}_b(1.817)=2$, $\text{slog}_b(1.4923)=1$, $\text{slog}_b(1.0)=0$, In the limit, as b approaches $e^{1/e}$, the continuation of the sexp to real numbers will be a straight line between the second and third terms, or between $\text{log}_b(e)$ and $\text{log}_b(\text{log}_b(e))$. This segment includes the inflection point of the sexp. Then, once the curve for base b is defined, you can convert between base b and base a using $\text{slog}_a(\text{sexp}_b(x))-x$. The trick is to find a value of x, such that the $\text{sexp}_b(x)$ is large enough and equal to a defined integer value for the equation $\text{slog}_a(\text{sexp}_b(x))$. Then you have the conversion factor for large numbers, which can be used to define the sexp/slog curve for base a for all real numbers. This curve for base a assumes $\text{slog}_a(\text{sexp}_b(x))-x$ converges to a constant value as opposed to converging to a 1-cycle periodic function. Moreover, the limit as b approaches $e^{1/e}$ will eliminate any degrees of freedom in defining slog/sexp extension to real numbers for any base a. - Sheldon Levenstein bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 02/20/2009, 01:07 PM sheldonison Wrote:The question for sexp/slog is how to define the curve extending sexp from integers to real. One question is does there exist an sexp/slog function for which $\text{slog}_a(\text{sexp}_b(x))$ converge to a constant value, or does it converge to a 1-cycle periodic function? But this question is different from whether $\text{slog}_a(x)-\text{slog}_b(x)$ converges to a constant, if we assume $a,b>e^{1/e}$. If $\text{slog}_a(x)-\text{slog}_b(x)$ does converge for $x\to\infty$ then - as $\text{sexp}_b$ goes to infinity - also $\text{slog}_a(\text{sexp}_b(x))-x$ converges. But this means that $\text{slog}_a(\text{sexp}_b(x))$ can not converge to a constant and vice versa. Quote:Consider what happens as b approaches e^(1/e) in the equation $slog_b(x)$. The curve becomes more and more linear, and there are fewer and fewer degrees of freedom for how to extend the sexp function to real numbers, and still have an increasing "well behaved" function. It must be possible to describe this rigorously in terms of limits. But it never gets completely linear, doesnt it? Otherwise it would not be analytic. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/20/2009, 02:51 PM (This post was last modified: 02/20/2009, 05:15 PM by sheldonison.) bo198214 Wrote:But this question is different from whether $\text{slog}_a(x)-\text{slog}_b(x)$ converges to a constant, if we assume $a,b>e^{1/e}$. If $\text{slog}_a(x)-\text{slog}_b(x)$ does converge for $x\to\infty$ then - as $\text{sexp}_b$ goes to infinity - also $\text{slog}_a(\text{sexp}_b(x))-x$ converges. But this means that $\text{slog}_a(\text{sexp}_b(x))$ can not converge to a constant and vice versa. typo: $\text{slog}_a(\text{sexp}_b(x))-x$ is what converges to a cyclic 1-cycle function as x grows larger and may converge to a constant as x grows larger for some definitions of slog/sexp. Quote:But it never gets completely linear, doesnt it? Otherwise it would not be analytic. no, of course not. But I think the contributions of the higher order terms versus a linear estimate between the second and third terms, or between $\text{log}_b(e)$ and $\text{log}_b(\text{log}_b(e))$, becomes insignificant. I plan to try and show that the second and higher order derivatives for the two points, contribute an insignificant delta as b approaches $e^{1/e}$, and that in the limit, the linear term dominates the higher order terms by an arbitrarily large amount, and that the linear approximation suffices as a definition for the tetration for base b. I had originally intended this thread to be a search for other links about about base changes for tetration. I seem to have gotten side tracked on to defining the sexp function for real numbers for bases $>e^{1/e}$. - Sheldon Levenstein sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/21/2009, 12:18 AM (This post was last modified: 02/24/2009, 05:18 PM by sheldonison.) bo198214 Wrote:sheldonison Wrote:Consider what happens as b approaches e^(1/e) in the equation $slog_b(x)$. The curve becomes more and more linear, and there are fewer and fewer degrees of freedom for how to extend the sexp function to real numbers, and still have an increasing "well behaved" function. It must be possible to describe this rigorously in terms of limits.But it never gets completely linear, doesnt it? Otherwise it would not be analytic.again, consider the region from $\text{slog}_b(e)$ to $\text{slog}_b(\text{slog}_b(e))$. Let m=$\text{slog}_b(e)-\text{slog}_b(\text{slog}_b(e))$ and let average =$1/2(\text{slog}_b(e)+\text{slog}_b(\text{slog}_b(e)))$ If we translate the region so it is centered over -0.5 to +0.5, then the linear approximation would be: $\text{sexp}_b(x) = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 ...$ with, $a0=\text{average, and } a1=m \text{, } a2=0 \text{, } a3=0 \text{, } a4=0$ As b approaches $e^{1/e}$, m will approach zero, and average will approach e. The endpoints of this critical section of the curve have been chosen so that $\text{sexp}_b^'(-0.5) = \text{sexp}_b^'(+0.5)$ Thus a linear approximation has a continuous first derivative. However the second derivative is not continuous. It takes some algebra to show that the second derivative can be made continuous with an a3 term that is a solution of this quadratic equation. $a3*a3 + b*a3 + c = 0 \text{ with } b=(4*m-24*\text{slog}_b(e)) \text{ and }c=4*m^2$ As m gets smaller, the result for a3 approximates accurately as: $a3 \text{=~} m*m*ln(b)/6 \text{ and, } a1=m-a3/4$ The a3 maximum contribution =~ $m^2*(1/(72*e))*sqrt(1/3)\text{ at } x=sqrt(1/12)$ This 3rd order equation (with a2=0 and a4 and higher zero), approximates the critical section much better than the linear estimate. But the maximum contribution of the a3 term to the sexp equation is proportional to m squared, or proporional to m when divided by m, the delta of the critical section. This shows that the a3 contribution becomes smaller and smaller and the linear term for a1=m can be as accurate as desired as m goes to zero, as b approaches $e^{1/e}$. I have also done the algebra for a continuous 3rd derivative, which effects the a0, a2 and a4 terms, with a1 and a3 unchanged. This contribution is proportional to m cubed, so it is much smaller than the a3 contribution, and is proportional to m squared when divided by m, the delta of the critical section. So I think a linear approximation does converge as closely as one would like to the actual function, as the base approaches $e^{1/e}$. And then this approximation can be used to convert to any other arbitrary base, as described. The data posted earlier for base e and base 10 sexp results used a 3rd order polynomial to model the critical section for base b=1.485, and should be accurate to nearly six significant digits. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 02/21/2009, 12:39 PM (This post was last modified: 02/21/2009, 12:44 PM by bo198214.) I have only a rough idea about what you suggest, so please let me ask some more questions Let $\eta=e^{1/e}$, when we consider $\exp_\eta$ we realize that $\lim_{x\to\infty} \exp_\eta(x) = e$, and $\lim_{x\to\infty} {\exp_\eta}'(x)=0$. So the critical region is at infinity, where there $\lim_{x_0\to\infty}\exp_\eta(x_0-1) -\exp_\eta(x_0)=\log_\eta(\log_\eta(e))-\log_\eta(e)=0$. Vice versa $\lim_{x\uparrow e}\text{slog}_\eta (x)=\infty$ and $\lim_{x\uparrow e}{\text{slog}_\eta}' (x)=\infty$. I still dont see how you will use this for defining $\exp_\eta$. sheldonison Wrote:Consider what happens as b approaches e^(1/e) in the equation $\text{slog}_b(x)$. The curve becomes more and more linear, and there are fewer and fewer degrees of freedom for how to extend the sexp function to real numbers, and still have an increasing "well behaved" function. It must be possible to describe this rigorously in terms of limits. Here is an example. if $b=1.49208... \text{ slog}_b(e)=6$, $\text{slog}_b(2.4989)=5$, $\text{slog}_b(2.2887)=4$, $\text{slog}_b(2.0691)=3$, $\text{slog}_b(1.817)=2$, $\text{slog}_b(1.4923)=1$, $\text{slog}_b(1.0)=0$, What do you mean by "there are fewer and fewer degrees of freedom for how to extend the sexp function to the real numbers"? And otherwise of course you surely read already Jay's post about change of base, which attributes the same effect you describe here that $\text{slog}_a(x)-\text{slog}_b(x)$ converges/should converge to a constant for $x\to \infty$ if $a,b>\eta$. Moreover he gives a formula to derive $\text{sexp}_b$ from any given $\text{sexp}_a$ such that exactly this condition is satisfied. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 02/21/2009, 02:59 PM (This post was last modified: 02/21/2009, 05:25 PM by sheldonison.) bo198214 Wrote:.... And otherwise of course you surely read already Jay's post about change of base, which attributes the same effect you describe here that $\text{slog}_a(x)-\text{slog}_b(x)$ converges/should converge to a constant for $x\to \infty$ if $a,b>\eta$. Moreover he gives a formula to derive $\text{sexp}_b$ from any given $\text{sexp}_a$ such that exactly this condition is satisfied.no, I haven't read Jay's post, but I certainly will! that's exactly what I was looking for. Thank you very much for pointing me to it. It will take awhile for me to digest. Quote:I have only a rough idea about what you suggest, so please let me ask some more questions Let $\eta=e^{1/e}$, when we consider $\exp_\eta$ we realize that $\lim_{x\to\infty} \exp_\eta(x) = e$, and $\lim_{x\to\infty} {\exp_\eta}'(x)=0$. So the critical region is at infinity, where there $\lim_{x_0\to\infty}\exp_\eta(x_0-1) -\exp_\eta(x_0)=\log_\eta(\log_\eta(e))-\log_\eta(e)=0$. Vice versa $\lim_{x\uparrow e}\text{slog}_\eta (x)=\infty$ and $\lim_{x\uparrow e}{\text{slog}_\eta}' (x)=\infty$. I still dont see how you will use this for defining $\exp_\eta$. For b=e^(1/e), the critical section is at x=infinity. But for any b>e^(1/e), the critical section is at x= some finite large number, and not at infinity. I realize I'm a long way from a formal mathematical definition, especially since it would involve all of the complexities of Jay's base conversion post (which I'm reading now), extended to dealing with arbitrarily large base conversion factors as b gets closer to e^(1/e). Quote:What do you mean by "there are fewer and fewer degrees of freedom for how to extend the sexp function to the real numbers"? When the tetration base is equal to "e", the linear section only extends from sexp(-1 to 0), and after that, the function is logarithmic on the left, and exponential on the right. But for smaller bases, the transition from logarithmic to super-exponential takes much longer, as the sexp function climbs gradually, slowly making the transition from close to linear, until super-exponential behavior takes over. Because of my earlier interest in the inflection point, I took that as a good region of the curve to use to estimate the tetration curve, using bases a little bit larger than e^(1/e). This is the most linear part of the curve, and for bases a little larger than e^(1/e), the linear section lasts a long time. Then, assuming that there is a valid base conversion forumula, I extrapolated from the base a little bit larger than e^(1/e) and used that to define tetration data for other bases, like base 2, 3, e and 10. The data I got seemed a little wobbly, or 1-cyclic as you described it, for the equation slog(sexp(x)), where I'm using Kouznetsov's taylor series versus the data I generated converting from a smaller base to base e. The wobble turned out to be quite a bit larger than the error term for my estimates, so I conjectered that there is no exact base conversion formula, and that the result will be approximately some number, converging to 1-cyclic. Then you suggested that base conversion could be a uniqueness criteria, and --- a lightbulb went off! I suddenly realized that one could use any base, as long as its larger than e^(1/e) to do the conversion. And the closer you get to e^(1/e), the more linear the tetration curve would be. The contributions made by all the higher derivatives go to zero, and goes to zero faster than the slope of the critical section goes to zero. A linear approximation over the critical section is always guaranteed to have a continuous 1st derivative, no matter what base is used. As the base approaches, but is larger than e^(1/e), the tetration curve becomes more and more defined, with a linear approximation for the critical section, and with fewer and fewer degrees of freedom for alternative definitions of the critical section for how to extend sexp to real numbers. If there is a valid base conversion formula, as opposed to a 1-cyclic base conversion formula, then defining tetration for a base approacing e^(1/e) defines the tetration sexp to real numbers for all bases. Kind regards, and sorry if my posts seem confusing, - Sheldon Levenstein bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 02/21/2009, 06:36 PM sheldonison Wrote:no, I haven't read Jay's post, but I certainly will! that's exactly what I was looking for. Thank you very much for pointing me to it. It will take awhile for me to digest.Jay made really good contributions though he was a bit short in making things publically understandable. So I often had difficulties to follow him. But you are anyway closer to the topic as you made the same observation like he did. Perhaps we can help each other to get a better understanding of the topic. Quote:For b=e^(1/e), the critical section is at x=infinity. But for any b>e^(1/e), the critical section is at x= some finite large number, and not at infinity. I realize I'm a long way from a formal mathematical definition, especially since it would involve all of the complexities of Jay's base conversion post (which I'm reading now), extended to dealing with arbitrarily large base conversion factors as b gets closer to e^(1/e). ... When the tetration base is equal to "e", the linear section only extends from sexp(-1 to 0), and after that, the function is logarithmic on the left, and exponential on the right. But for smaller bases, the transition from logarithmic to super-exponential takes much longer, as the sexp function climbs gradually, slowly making the transition from close to linear, until super-exponential behavior takes over. Ah, ok, now I get it. With base going towards $\eta$ the inflection point goes to infinity and the mainly linear interval gets longer (though indeed the question is how to get this mathematically caught, but I see what you mean). The starting point of this interval is going also to infinity (?). The slope in this linear interval goes to 0 (?). How do you make use of this increased linearity to define $\text{sexp}_b$ ($b$ little bigger thean $\eta$)? Just in the usual way by $\text{sexp}(x+1)=\exp_b(\text{sexp}(x))$ and $\text{sexp}(x-1)=\log_b(\text{sexp(x))$? Quote:Then, assuming that there is a valid base conversion forumula, I extrapolated from the base a little bit larger than e^(1/e) and used that to define tetration data for other bases, like base 2, 3, e and 10. how? Quote:The wobble turned out to be quite a bit larger than the error term for my estimates, How do you obtain the error term? Quote:And the closer you get to e^(1/e), the more linear the tetration curve would be. The contributions made by all the higher derivatives go to zero, and goes to zero faster than the slope of the critical section goes to zero. A linear approximation over the critical section is always guaranteed to have a continuous 1st derivative, no matter what base is used. As the base approaches, but is larger than e^(1/e), the tetration curve becomes more and more defined, with a linear approximation for the critical section, and with fewer and fewer degrees of freedom for alternative definitions of the critical section for how to extend sexp to real numbers. I see, the proper tetrational should become more and more linear when the base approaches $\eta$ from above, and is characterized by this demand. Quote:defining tetration for a base approacing e^(1/e) defines the tetration sexp to real numbers for all bases. Here I see some difficulties to put that mathematically. You want to define tetration only for one base, but "approaching" means that you have to define tetration at least for a sequence of bases. But perhaps this is just a question of exactness. The more exact the tetration for arbitrary bases should be the closer to $\eta$ one have to choose the base of the initial tetration, something in that direction. Quote:and sorry if my posts seem confusing, Ya, mathematics is always somewhat confusing, especially if it is in the making. As long as in the end a mutual clear understanding can be achieved there is nothing bad about some fog on the way. « Next Oldest | Next Newest »

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