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...

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...