Real Tetration Solution

Hi,

Before getting to my solution of the continuous tetration problem, let me tell you about how I got into the subject. I was one of those people who re-invented the wheel when school just "stoped" at exponentiation, and I kept on going. I've been working on tetration on and off for about 7 years now, and I've been to almost every place online I can find. So far I have not seen any real solutions. My favorite by far, although not smooth, is the:

x^^y ~= y+1 where y in (-1, 0)

approximation. Its the simplest, and will suffice, when you're trying to find orders of magnitude, or one or two decimal places. In my research I've found a lot of really insightful discussions -- like those of Ioannis Galidakis, Robert Munafo, and Daniel Geisler. Good Job Guys! you all inspired me to continue searching when I thought no one else in the world cared about tetration.

Most of my research recently has been numerical, I've tried to extend tetration algebreicaly, as have many others, without much success. When I focused purely on numerical solutions, I started trying to find different ways of stating the problem of tetration. I started by assuming that the solution be infinitely differentiable, rather than simply continuous, and also felt that monotonic, or always positive derivatives were neccessary. So I tried plugging the conditions into Solve[] because DSolve[] wacks out, and still no solution. I tried using series expansions around several different expansion points, and still no solution. Some points do give a solution, but its oscillating, and changes as soon as you add one term to the series expansion, so the sequence of series terms doesn't converge.

Recently I did something different, and it worked! I did it! I solved the problem of continuous tetration! But in a very roundabout way. I did it by providing the conditions that slog_b(x) be infinitely differentiable (actually only n-times differentiable) into Solve[], and got numbers that would CONVERGE for higher values of n. So we can now find the true value of e^^pi == x because we now know the true value of slog_e(x), and can solve that for when slog_e(x) == pi.

I'm thinking of writing a paper about this, but I have no idea where I should submit it, any suggestions would be nice. But before I do, anyone interested can ask me at and_j_rob(at)yahoo(dot)com and I will try and send something to anyone interested in this numerical solution.

To sum up, here is the true value of e^^pi (one of the questions at tetration.org) where the (@n) indicates using an n-th degree polynomial which means solving for a n-times differentiable function:

e^^pi = 3 715 229 069.085 273 (@30)
e^^pi = 3 715 084 943.035 024 (@40)
e^^pi = 3 715 033 138.003 963 (@50)
e^^pi = 3 715 011 255.457 622 (@60)
e^^pi = 3 714 998 605.150 005 (@70)
e^^pi = 3 714 991 271.249 438 (@80)
e^^pi = 3 714 987 492.820 818 (@90)
e^^pi = 3 714 985 215.721 806 (@100)

and I can do this for any degree solution, which isn't so different from having a formula, a formula is just a way to compute the true value, and I have a way to compute the true value! but for now:

e^^pi ~= 3,714,985,000 (± 5000)

I'm thinking that the uncertainty should be related to the difference between an approximation (@n) and (@n-1), but I'm not sure. Anyways, here are the first 5 derivatives of slog_e(x) at zero, which you can use to build a series expansion of slog, the tetra-logarithm:

slog(0) == -1

slog'(0) = 0.9164429566214118 (@8)
slog'(0) = 0.9161324503564959 (@10)
slog'(0) = 0.9160223347014526 (@12)
slog'(0) = 0.9159779530658628 (@14)
slog'(0) = 0.9159586198921122 (@16)
slog'(0) = 0.9159498538999188 (@18)
slog'(0) = 0.9159458665561289 (@20)
slog'(0) = 0.9159452632662496 (@50)
slog'(0) = 0.9159455362746406 (@60)
slog'(0) = 0.9159459089141265 (@90)
slog'(0) = 0.9159459510955973 (@100)
slog'(0) ~= 0.9159459
slog'(0) ~= 3389/3700

slog''(0) = 0.49700991788382987 (@8)
slog''(0) = 0.49772820110974303 (@10)
slog''(0) = 0.4981019910013889 (@12)
slog''(0) = 0.49831176561867124 (@14)
slog''(0) = 0.498437127162984 (@16)
slog''(0) = 0.4985160725571452 (@18)
slog''(0) = 0.4985680382605074 (@20)
slog''(0) = 0.4987044656658535 (@50)
slog''(0) = 0.4987070538863207 (@60)
slog''(0) = 0.4987089862517204 (@90)
slog''(0) = 0.4987090968049525 (@100)
slog''(0) ~= 0.49870
slog''(0) ~= 8311/16665

slog'''(0) = -0.6837515467534577 (@8)
slog'''(0) = -0.6710185343558521 (@10)
slog'''(0) = -0.6664583394500544 (@12)
slog'''(0) = -0.6645499371435517 (@14)
slog'''(0) = -0.6636688645176111 (@16)
slog'''(0) = -0.6632347770975023 (@18)
slog'''(0) = -0.6630115768664259 (@20)
slog'''(0) = -0.6627687321227633 (@50)
slog'''(0) = -0.6627739897538185 (@60)
slog'''(0) = -0.6627837773978416 (@90)
slog'''(0) = -0.6627850619414579 (@100)
slog'''(0) ~= -0.66278

slog''''(0) = -2.2367370315668955 (@8)
slog''''(0) = -2.235806360084065 (@10)
slog''''(0) = -2.2396379014371224 (@12)
slog''''(0) = -2.243227212846931 (@14)
slog''''(0) = -2.245972485470915 (@16)
slog''''(0) = -2.2479905166908463 (@18)
slog''''(0) = -2.249472420086772 (@20)
slog''''(0) = -2.254231314353492 (@50)
slog''''(0) = -2.2543472192602274 (@60)
slog''''(0) = -2.2544491989906366 (@90)
slog''''(0) = -2.254456942999395 (@100)
slog''''(0) ~= -2.2544

slog'''''(0) = 2.4184938232589905 (@8)
slog'''''(0) = 1.674064907059518 (@10)
slog'''''(0) = 1.4270431214503754 (@12)
slog'''''(0) = 1.3218432461262983 (@14)
slog'''''(0) = 1.2705786937372705 (@16)
slog'''''(0) = 1.2433020424136674 (@18)
slog'''''(0) = 1.2278296498331995 (@20)
slog'''''(0) = 1.2002880657516535 (@50)
slog'''''(0) = 1.200134054342864 (@60)
slog'''''(0) = 1.200237964111215 (@90)
slog'''''(0) = 1.2002709742029045 (@100)
slog'''''(0) ~= 1.2002

So as you can see, using a 100-degree polynomial, and solving such that the function is 100-times differentiable, we get at least 5 significant figures for the first 5 derivatives. So far 100 is the highest degree solution i've tried to compute. I think it took about an hour to compute. I left it alone for about 3 hours and came back and it was done, maybe next time I should time it.

Andrew Robbins
.

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