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 Regular "pentation"? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/01/2010, 11:56 AM What happens if you use the same regular method as to do regular tetration, and apply it again, to get regular "pentation" for bases $1 < b < e^{1/e}$? Like what does $\sqrt{2} \uparrow \uparrow \uparrow x$ look like on the real line/complex plane? Using the regular tetration, it seems that $\sqrt{2} \uparrow \uparrow \uparrow 1 \approx 1.41421356$ $\sqrt{2} \uparrow \uparrow \uparrow 2 \approx 1.51994643$ $\sqrt{2} \uparrow \uparrow \uparrow 3 \approx 1.54305998$ $\sqrt{2} \uparrow \uparrow \uparrow 4 \approx 1.54793050$ $\sqrt{2} \uparrow \uparrow \uparrow 5 \approx 1.54894876$ ... which suggests that $\sqrt{2} \uparrow \uparrow \uparrow n\ <\ {}^{n}(\sqrt{2})$, at least for $n \ge 1$, and also that it approaches a limiting value ~1.54921732 as $n \rightarrow \infty$, a fixed point of the tetrational $^{x} (\sqrt{2})$. It would be interesting to determine what the upper bound of the regular interval for the bases of pentation would be. We know that for tetration it is $e^{1/e}$, but what about pentation? (I suppose this would require tetration for b greater than $e^{1/e}$ to investigate, though, so we'd need other methods like the Abel iteration or the continuum sum (I'm a big fan of continuum sums, by the way )) Using the superlog to descend into negative values of $n$, $\sqrt{2} \uparrow \uparrow \uparrow 0 = 1$ $\sqrt{2} \uparrow \uparrow \uparrow -1 = 0$ $\sqrt{2} \uparrow \uparrow \uparrow -2 = -1$ $\sqrt{2} \uparrow \uparrow \uparrow -3 \approx -1.41264443$ $\sqrt{2} \uparrow \uparrow \uparrow -4 \approx -1.51722115$ $\sqrt{2} \uparrow \uparrow \uparrow -5 \approx -1.53991429$ ... which suggests the approach to another fixed point of the tetrational, here $-1.54590582$. (this suggests some curve vaguely like arctan) This means the pentational approaches fixed points at both positive and negative infinity, which means there are 2 possible options for the regular iteration, not as clearly differentiated as is in the case with tetration. Though I suppose for consistency, one would use the attracting fixed point. What does the graph of the two regular pentationals at this base look like? (I suspect the real-line graphs will be too close to discern the difference, but the same may not be so on the complex plane.) And especially at the complex plane... I'm not sure what those branch cuts in $\mathrm{tet}$ and $\mathrm{slog}$ are gonna do... tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/01/2010, 03:21 PM dear mike the regular interval that you seek has been computed by andy. its somewhere here on this forum. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/01/2010, 08:54 PM (This post was last modified: 05/01/2010, 08:58 PM by bo198214.) (05/01/2010, 03:21 PM)tommy1729 Wrote: dear mike the regular interval that you seek has been computed by andy. its somewhere here on this forum. Lazy boy! Andrew gave a rough overview of the behavior of the higher operations here. Due to the fact that $\operatorname{sexp}_b(x)\to -\infty$ (below identity) for $x\to -2$ and $\operatorname{sexp}_b(0)=1$ (above identity), tetration always (for every base) has a lower fixed point (I guess this is true for every even rank), which can be used for regular iteration giving a pentation (or the odd rank above). Interesting question would indeed be for which bases pentation has a lower fixed point (but indeed I think this was computed already on the forum, perhaps do a search for pentation). mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/02/2010, 12:07 AM (This post was last modified: 05/02/2010, 12:09 AM by mike3.) Actually, I already saw some of those posts. I'm not sure if the use of regular iteration at the lower (repelling) fixed point would be the best choice for pentation of bases greater than the constant where the pentation goes off toward infinity, just as we don't use the lower fixed point of exponential to build the tetration at $b > e^{1/e}$. The reason for this is that I think the result of the regular iteration at the upper and lower fixed point differs (doesn't it?) for $1 < b < K$, where $K$ is the constant beyond which $^{x} b$ no longer has an attracting fixed point and so $b \uparrow \uparrow \uparrow x \rightarrow \infty$ as $x \rightarrow \infty$. Instead, I'd think it'd make more sense to use the upper fixed point for $1 < b < K$, same as the case for the tetration, and then one needs more sophisticated methods like the Abel-matrix, matrix-power, or continuum-sum for bases beyond that constant $K$ (according to one post it's somewhat over 1.6). andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2010, 09:12 AM Might I also point your attention to this post, in which I actually give some coefficients of pentation. Also, I'd like to point your attention to these pictures of tetration's fixedpoints in the complex plane (which is listed as exp^z(1) - z). As you can see, there are 3 fixed points for base-e, 2 are complex conjugates of each other as in tetration, and 1 is "lower" as you put it. The coefficients I calculated above are calculated with natural/intuitive iteration, so there is not explicit fixedpoint in the calculation. I forgot what I was going to say next. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/02/2010, 11:29 PM (This post was last modified: 05/02/2010, 11:58 PM by Base-Acid Tetration.) (05/02/2010, 09:12 AM)andydude Wrote: As you can see, there are 3 fixed points for base-e, 2 are complex conjugates of each other as in tetration, and 1 is "lower" as you put it. The coefficients I calculated above are calculated with natural/intuitive iteration, so there is not explicit fixedpoint in the calculation. I think you are mistaken. there's LOTs more     You see these "color wheels that go counterclockwise"? they have to have a zero (a fixed point) in the middle. In fact, for every one of those red bands in the color graph (of infinite length) there are infinity fixed points in the corresponding "whitish region" (= some region including red band + 1*)'s boundary. *the region itself and the GRAPH of the region has to be disginguished carefully... Are these gonna matter? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/03/2010, 02:54 AM (05/02/2010, 11:29 PM)Base-Acid Tetration Wrote: I think you are mistaken. there's LOTs more Yes, I should have said "at least 3". (05/02/2010, 11:29 PM)Base-Acid Tetration Wrote: Are these gonna matter? From what we've learned about tetration, each fixedpoint will produce a different pentation using regular iteration. However, since a fixedpoint is not required for intuitive/natural iteration, it will be more consistent, and not dependant on a fixedpoint. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/03/2010, 03:27 AM I'm curious: how did you get so much detail in that graph? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/07/2010, 03:10 AM (05/03/2010, 03:27 AM)mike3 Wrote: I'm curious: how did you get so much detail in that graph? Well, that is because I used Dmitrii's coefficients from here. They're much more precise (or "accurate") than any coefficients I've calculated. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/07/2010, 03:17 AM (05/01/2010, 11:56 AM)mike3 Wrote: It would be interesting to determine what the upper bound of the regular interval for the bases of pentation would be. We know that for tetration it is $e^{1/e}$, but what about pentation? (I suppose this would require tetration for b greater than $e^{1/e}$ to investigate, though, so we'd need other methods like the Abel iteration or the continuum sum (I'm a big fan of continuum sums, by the way )) I think I finally understand what you're talking about here. There is an interval that plays the same role in pentation as $(e^{-e}, e^{1/e})$ plays in tetration. I talked about the upper bound of this interval here, but whether or not there is a lower bound to this interval is unknown. Just as the point $(x=e^{1/e}, y=e)$ is the "highest" point on the graph of $y = x {\uparrow}{\uparrow} \infty$, so is $(x=1.6353, y=3.0885)$ the "highest" point on the graph of $y = x {\uparrow}{\uparrow}{\uparrow} \infty$. « Next Oldest | Next Newest »

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