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Solution to extending tetration to real exponents


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#1 jaydfox

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Posted 07 August 2007 - 06:05 AM


Here's a discussion I started on sci.math.research:
http://groups.google...e76f3dcdd1fe33d

Hopefully that link stays valid.

Here's my latest formula, attached in MS Word format

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#2 jaydfox

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Posted 07 August 2007 - 06:09 AM

For some background, see the following links:
http://home.earthlin...tml#real-hyper4
http://tetration.itgo.com/
http://ioannis.virtu...exponents4.html

I haven't reinvented the wheel, which is to say, I haven't done all the in-depth analysis you see on some of these other pages. In some cases it's because I'm lazy, but mostly it's because they're right or mostly right, so why should I do the same work to get the same results?

As an interesting sidenote, I tried using my formula to derive a Taylor series expansion of e^^x, centered at x=0, and I met miserable failure. The first 6 or 8 terms have decreasing coefficients, but then they start to balloon out of control. Not only does the Taylor series most likely not converge on the interval [-1,1], I suspect it doesn't even converge on an interval of unit length, making it useless for constructing the entire function. I'm not quite sure how my formula can converge and be so well-behaved, and yet the Taylor series is out of control, hinting that the derivatives themselves are fairly erratic.

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#3 jaydfox

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Posted 07 August 2007 - 06:14 AM

For those unable or unwilling to download the attachment, here's the formula (screenshot from MS Word at 200% magnification). The text below fills out the rest of the document.


Below are details on a formula for calculating tetration real exponents. This solution should be valid for all bases greater than e^(1/e). Formula derived independently by Jay Daniel Fox. Unknown whether this formula has been previously derived.

[screenshot goes here]

The sample values below should be useful for getting 150-200 digits of accuracy. The value for has been determined by numerical methods.

For x=e:
a_0 =
0.014322263393121982517624482505047871744221168446203788316857822504734593323974
81731747217314400277475900613652987946087192813092324723242151039267330049105624
4
191834658508783559764260025638841454243

m = Int(3.6-y)
n = Int(30-y)

For x=2:
a_0 =
-0.604882400356630974442166216222669967821642009589460572561459555428549141575096
00784880793150315384844967796657250610556693061041848702048649907710927673394518
9
9610587120431872817670758939023610693257

m = Int(4.7-y)
n = Int(35-y)

For x=10:
a_0 =
0.545804357164790167799276018570470426204190562164585465801348831386139191659321
03193116432715149093762981816390411813262544722778125100447489778043095305428528
7
25003128802514846807262118127419590995269

m = Int(2.5-y)
n = Int(22-y)

For x=1.7632228343518967102252… (x^x = x^^2 = e):
a_0 =
-1.367538516630417502182827932933531543037417606846208800233417009461095853427291
34772928100226940660917023938541599684782863113586578945917612774908831958875300
3
0168528604828810085099567304700247508452

m = Int(5.85-y)
n = Int(38-y)

For x=1.5:
a_0 =
-7.867392707621047469441964632984898953480474513796203874950138759998527043206421
06831582476744648096542363884298138794318679011664505801124043982068818181444869
6
3887086936115832685100191979342039998622

m = Int(13.5-y)
n = Int(38-y)

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#4 jaydfox

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Posted 07 August 2007 - 06:22 AM

On the one hand, I can't realistically imagine that this hasn't been derived somewhere else. On the other hand, I've been searching on the internet for a few days, and I've found no indication that this formula is known. I found one by Peter Walker that was similar in having repeated exponentiation plus addition, but he always used 1 as the additive constant, and he needed a helper function to convert the answer back to the desired value (probably equivalent to an unknown function that calculations the constant alpha sub-zero that you see in my formula.

Ironically, I figured this formula out while trying to solve the case for 1 < x < e^(1/e). I was on the verge of making of figuring that problem out, when I had one of those Eureka moments about the case x > e^(1/e).

Anyway, if anyone can find that this formula is not original, I'd sure like to see where else it's been derived. Most of the literature I can find on tetration seems to implicitly assume that no such formula has been found yet. If indeed this formula is known, then a lot of websites need to be updated with current informaiton.

#5 caston

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Posted 07 August 2007 - 06:27 AM

Is this useful for compression and or cryptography?

#6 jaydfox

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Posted 07 August 2007 - 06:41 AM

I'm not really sure what it's useful for. It was a challenge, supposedly lacking a succinct and rapidly converging solution for real numbers, and (I think) I solved it.

#7 caston

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Posted 07 August 2007 - 06:53 AM

Nice ... :)

#8 scatha

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Posted 04 November 2007 - 05:24 PM

Hey jaydfox !... this would be a breakthrough WOW!! - generations of mathematicians have sought for it - pity it is so complicated, i wonder if inventing a new function would make it easier ?... Lambert invented one, too, to make certain equations easier to solve *G*

Maybe you will become famous :-)

#9 caston

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Posted 05 November 2007 - 12:31 AM

jay: good work :)

#10 chubtoad

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Posted 15 November 2007 - 11:59 PM

Wow how did I miss this topic before, I am also very interested in tetration and extending it to the reals (the extension to complex numbers is very nice but reals are so strange). I have not seen the formula you posted and I will definitely think about it when I have a little time. Now how about an asymptotic (In hyperpowers etc.) for the hyperfactorials 4^3^2... in the spirit of stirling approximation (a pet problem I cant figure out)? This is such an interesting area of math i dont know why more people don't think about it, but instead publish their phd thesis in such and such modular forms over the p-adic numbers of weight this and type that etc.

#11 chubtoad

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Posted 16 November 2007 - 12:07 AM

Oh and caston who asked these huge numbers come up all the time in combinatorics particularly ramsey (coloring) problems and coloring arithmetic progressions (2,5,8,11...). For example it was a big deal when the growth rate of coloring APs was shown to be wowzer (pentation iterated tetration) instead of ackerman.

#12 chubtoad

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Posted 16 November 2007 - 12:17 AM

Btw it is awesome you are able to find time to think about these kind of things, while you spend so much time in science and humanistic persuits. Nice find!

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#13 jaydfox

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Posted 27 November 2007 - 06:53 PM

Hey jaydfox !... this would be a breakthrough WOW!! - generations of mathematicians have sought for it - pity it is so complicated, i wonder if inventing a new function would make it easier ?... Lambert invented one, too, to make certain equations easier to solve *G*

Maybe you will become famous :-)

I continued my research on a newly formed forum dedicated to tetration and related research.

Alas, after further probing, I discovered that my solution wasn't all that great after all.

For those interested, extending tetration to the reals requires a couple things at the minumum:
1. Iterated exponential property
If we have a tetration function T_b(x) for a base b, then T_b(x) = b^T_b(x-1) for all real x. This is a property of a more general class of superexponentiation relations, of which tetration is a specific instance such that T_b(0)=1. It follows that T_b(-1)=0 and T_b(-2) = negative infinity

2. Infinite differentiability property
The function T_b is continuous for real x > -2, and it can be differentiated infinitely many times (with the derivatives being continuous over the same domain)

My solution satisfied these two conditions. In my haste, I found a simple and in some ways beautiful way to constuct a solution that satisfied the two conditions. However, there are infinitely many such solutions, and mine isn't all that special aside from its relative simplicity. (And yes, that might seem rather tongue-in-cheek, because it may look complicated, but it's not really that complicated once you understand how it works.)

As it turns out, we need at least a third condition to uniquely define "the" solution. I have a couple really good such conditions, and luckily both seem to favor the same solution. And no, it's not my solution. It appears that "the" solution is the inverse of the matrix solution of the Abel functional equation. For more information on the Abel function, see this post by Andrew Robbins (only hardcore math geeks need follow the link :chasses: ):
http://math.eretrand...read.php?tid=27

Andrew Robbins independently derived a matrix-based solution:
http://tetration.itgo.com/paper.html

After spending weeks studying his solution, I figured out that it's just an alternate method of calculating the basic Abel matrix solution. But seeing it solved both ways really convinced me that it's valid, and knowing that it satisfies both of my additional criteria for uniqueness just caps it off.

For the curious, my two additional criteria (which may or may not yield the same solution):
1. Odd derivatives should be non-negative (perhaps even strictly positive?) for real x > -2. This ensures that all odd derivatives are convex. Even the most insignificant deviation from "the" solution would result in negative values in some of the odd derivatives.
1b. Odd derivatives should be log-convex for real x > -2. I haven't studied this condition enough to know if it's stricter than just being convex. For a single derivative, log-convexity is stricter, but I think that if all odd derivatives are convex, then they are all log-convex. I need to prove or disprove this at some point.

2. If we extend the real solution to complex numbers (using the power series and analytic continuation), then there should be logarithmic singularities at the primary fixed points of the underlying exponential funciton. Even the most insignificant deviation from "the" solution would result in the singularities becoming non-logarithmic.




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