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 tetration base conversion, and sexp/slog limit equations sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 02/22/2009, 04:41 AM (This post was last modified: 02/22/2009, 04:49 AM by sheldonison.) Bo, Thanks for your reply. bo198214 Wrote: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 $\text{sexp}(x-1)=\log_b(\text{sexp(x))$? Yes, I'm using these equations. In particular, I did a lot of work on the critical section, with most of my results posted here. I rediscovered the same critical section that Jay discovered for any arbitrary base b larger than $\eta$, but I also extended the results by finding the equations for a smooth 2nd and 3rd derivative for the $\text{sexp}_b$ function for the critical unit section from $\text{slog}_b(\text{log}_b(e)) \text{ to slog}_b(\text{log}_b(\text{log}_b(e)))$. The link shows how that a linear approximation has a continuous first derivative, since the 1st derivative is equal on both endpoints of the critical section, and I also posted the equations for a continuous 2nd derivative (3rd order approximation), and a continuous 3rd derivative (4th order approximation). Quote: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? The data in this posts used b=1.485 (mostly an arbitrary number larger than $\eta \text{ sexp}_{1.485}(e) \text{ is about }6.5$). To improve convergence, I used a 3rd order polynomial estimate for the critical section (2nd order smooth derivative), and used the delta to the 4th order polynomial approximation (with a 3rd order smooth derivative) as the error term. I also tried a second base a little smaller than the first, and got similar results consistent to within about $1x10^{-5}$ in the conversion to base e. Quote:I see, the proper tetrational should become more and more linear when the base approaches $\eta$ from above, and is characterized by this demand. .... 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. Lots of difficulties. The easy part is defining a sequence that approaches $\eta$, and showing that the sequence itself has continually reducing relative error terms in the critical section, and by extrapolation, all of the sexp curve. But then for each element in the sequence, the base conversion constant must be calculated. That requires a sequence. Then one must show the base conversion factor converges to a value. I think the base conversion factor can be shown to converge to a value for each term in the sequence, but that's not the same as showing it converges to the same value as b approaches $\eta$. Then one must show that the resulting tetration curve converting from base b as it approaches $\eta$ converges when it is converted to another arbitrary tetration base curve (it may not converge). sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 02/22/2009, 04:04 PM (This post was last modified: 02/24/2009, 11:27 PM by sheldonison.) bo198214 Wrote: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. OK, here's my proposal for a rigorous mathematical approach -- going all the way back to my original post: > $\text{slog}_2(x) - \text{slog}_e(x) = 1.1282$ This is equivalent to analyzing the following equation as b approaches $\eta^+$, for increasing values of n. Subtracting the two terms slog terms from each other cancels out the fact that slog(x) increases as the base approaches $\eta^+$. There are other ways to handle this, but this is a concise way to handle it in a limit equation. In these equation, slog$_b$ refers to the slog with a linear approximation of the critical section. $\lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} (\text{slog}_b(\text{sexp}_2(n)) - \text{slog}_b(\text{sexp}_e(n)))$ First off, we can show that as n increases, the series converges for any individual value of b, (convergence as n increases is an easier problem discussed in the base conversion post by Jay; values of n>6 ought to give more or less unlimited accuracy for bases greater than 2). The harder part is to show that the limit converges as b approaches $\eta^+$. If it converges, an extension of the sexp/slog function to real numbers can be defined. As an example of how this equation could define the sexp/slog function extension to real numbers, consider the following equation, where x is a real number. If the limit above converges, then the limit below should converge to x. $\lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} (\text{slog}_b(\text{sexp}_e(n+x)) - \text{slog}_b(\text{sexp}_e(n)))=x$ With a little additional algebra, this allows defining an sexp/slog extension to real numbers, for base e, by iterating the ln function "n" times. In practice, for base e, using n=5 will give approximately a million digits of precision for positive values of x. This also works for any other arbitrary base. $\text{sexp}_e(x) = \lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} \text{ln(ln(ln(}\cdots \text{sexp}_b (x + \text{slog}_b(\text{sexp}_e(n))))))$ sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 02/24/2009, 08:24 PM (This post was last modified: 02/25/2009, 12:17 AM by sheldonison.) sheldonison Wrote:.... With a little additional algebra, this allows defining an sexp/slog extension to real numbers, for base e, by iterating the ln function "n" times. In practice, for base e, using n=5 will give approximately a million digits of precision. This also works for any other arbitrary base. $\text{sexp}_e(x) = \lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} \text{ln(ln(ln(}\cdots \text{sexp}_b (x + \text{slog}_b(\text{sexp}_e(n))))))$In this post, I will use this equation to derive the $\text{sexp}_e(-0.5)$, accurate to five significant digits, using a linear approximation of sexp/slog. I haven't made any progress in proving the limit converges, but assuming it does, earlier in this thread I derived the error term of a linear approximation as the base approaches $\eta^+$. error term =~ $m^2*(1/(72*e))*sqrt(1/3)$ relative error term =~ $m*(1/(72*e))*sqrt(1/3)$ with m=$\text{slog}_b(e)-\text{slog}_b(\text{slog}_b(e))$, as b approaches $\eta^+$, $m=e-\text{slog}_b(e)$ The error term uses the difference between a linear approximation, and a 3rd order approximation for the critical section. A linear approximation has a continuous 1st derivative, and a 3rd order approximation has a continuous 2nd derivative. For this example, use a linear approximation to calculate sexp_e(-0.5), accurate to five significant digits. The relative error term required would be $10^{-5}$, which leads to m=~1/295. Use b=$\eta$+(1/1500). As a comparison, the other results I used were $\eta$+(1/25), but using a third order approximation. The base being used here is much closer to $\eta$, and there are a lot more iterations. Starting with the limit equation, I picked a value of n=5, for sexp_e(5). My excel spreadsheet can't handle sexp_e(5), but it can handle x=slog_b(slog_b(sexp_e(5)), which is x=10355300. Then, keep iterating x=log_b(x), another 65 times, for a total of 67 log_b iterations, until the value of x is less then log_b(e). I stopped at x=2.71365. Now that x is in the linear region of the slog_b(x) curve, let x= sexp_b(slog_b(x)-0.5). For the slog_b function, I got a remainder term of 0.638016. Subtract 0.5, and use the linear approximation to begin the conversion back to the sexp_b, x=2.71194. The error term doubles to +/-0.000002 since two different linear approximations are required, one for the slog_b and another for the sexp_b. Now iterate, taking x=b^x 66 times. I got x=5.36*10^77. One last exponentiation will be too large for the spread sheet, so I pair it up with the first ln. Let x= ln(ln(ln(ln(ln(b^x))))). The result I got was sexp(-0.5)=0.497818+/-0.00002. Using the more accurate 3rd order approximation for the critical section, I got sexp(-0.5)=0.49783297. The delta between the two values was 1.5*10^-5 which is less than 2*10^-5. The next thing I plan to do is to convert the base=e sexp function posted by Kouznetsov to another base, probably b=1.45, and graph the equations and derivatives to show that Kouznetsov's sexp function requires a 1-cyclic definition of sexp/slog base conversions, as the limit cannot converge to a constant, but instead wobbles. Jay noticed the same thing for Andy's sexp/slog extension to real numbers. I will also graph the sexp_e function converting from base=1.45. Any other suggestions? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/24/2009, 09:57 PM sheldonison Wrote:$\lim_{b \to \eta^+}\text{ } \lim_{n \to \infty} \text{sexp}_e(x) = \text{ln(ln(ln(}\cdots \text{sexp}_b (x + \text{slog}_b(\text{sexp}_e(n))))))$ A very appealing formula (though the two limits belong to the right side of the equal sign)! For my taste, if I rewrite and generalize your formula a bit ad hoc (dont know whether it is true in that form): ${\exp_b}^{\circ t}(x)=\lim_{a\downarrow \eta} \lim_{n\to\infty} {\log_b}^{\circ n}(\text{sexp}_a (t+\text{slog}_a({\exp_b}^{\circ n}(x))))$ has a big similarity to Zdun's [1] formula for regular iteration (where $f$ is assumed to have fixed point at 0): $f^{\circ t}(x) = \lim_{n\to\infty} f^{\circ -n}(q^t f^{\circ n}(x))$, $q=f'(0)$. Only $q^t y$ is replaced by $\text{sexp}_a(t+\text{slog}_a(y))$. But be aware that $q^ty = \exp(\log(q)t+\log(y))$, which makes the similarity even more striking. [1] Zdun, Marek Cezary. Regular fractional iterations. Aequationes Math. 28 (1985), no. 1-2, 73--79. MR0781210 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/24/2009, 10:21 PM (This post was last modified: 02/24/2009, 10:26 PM by bo198214.) bo198214 Wrote:${\exp_b}^{\circ t}(x)=\lim_{a\downarrow \eta} \lim_{n\to\infty} {\log_b}^{\circ n}(\text{sexp}_a (t+\text{slog}_a({\exp_b}^{\circ n}(x))))$ I wonder whether this formula works also if we dont take the limit $a\downarrow \eta$ but directly put: ${\exp_b}^{\circ t}(x)=\lim_{n\to\infty} {\log_b}^{\circ n}(\text{sexp}_\eta (t+\text{slog}_\eta({\exp_b}^{\circ n}(x))))$ where of course we can again replace $\text{sexp}_\eta(t+\text{slog}_\eta(y)))={\exp_\eta}^{\circ t}(y)$ where we can compute ${\exp_\eta}^{\circ t}$ by regular iteration because it has a fixed point at $e$ (though not with Zdun's formula as his formula is only valid for $|q|<1$ while ${\exp_\eta}'(e)=1$). sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 02/24/2009, 10:54 PM bo198214 Wrote:bo198214 Wrote:${\exp_b}^{\circ t}(x)=\lim_{a\downarrow \eta} \lim_{n\to\infty} {\log_b}^{\circ n}(\text{sexp}_a (t+\text{slog}_a({\exp_b}^{\circ n}(x))))$ I wonder whether this formula works also if we dont take the limit $a\downarrow \eta$ but directly put: ${\exp_b}^{\circ t}(x)=\lim_{n\to\infty} {\log_b}^{\circ n}(\text{sexp}_\eta (t+\text{slog}_\eta({\exp_b}^{\circ n}(x))))$ where of course we can again replace $\text{sexp}_\eta(t+\text{slog}_\eta(y)))={\exp_\eta}^{\circ t}(y)$ where we can compute ${\exp_\eta}^{\circ t}$ by regular iteration because it has a fixed point at $e$ (though not with Zdun's formula as his formula is only valid for $|q|<1$ while ${\exp_\eta}'(e)=1$). In your equations, what does ${\exp_b}^{\circ n}(x)$ mean? And what does ${\log_b}^{\circ n}(x)$ mean? I assume this is the mathematical shorthand for iterating "n" times. I'm going to have to get some education in higher mathematics . What does $\text{slog}_{\eta}(x)$ mean? Is it restricted to values of x

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