Talk:Tetration/Archive 1

Revision
This page should have some "love" just like the hyper operator page. I have done a lot of work on tetration, and its many connections, so I think I can shed some clarity on this very confusing subject, I'm planning on adding two pages: a page about the Superlogarithm, and a page about Abel's Linearization Equation, does anyone know how to make a new page? Andrew Robbins 23:25, 13 December 2005 (UTC)


 * Its been awhile since I said this, but I'm serious about it this time. AJRobbins (talk) 02:53, 20 November 2007 (UTC)

Creating Pages

When you insert a Wikipedia link into a page, the link will either take you to the Wikipedia page if it exists or it will allow you to edit and save a new page named by the link used to create it.

Page Names

When assigning page names use might want to consider that Abel equation has 160,000 hits on google, Abel’s equation is under 10,000 hits, and Abel's Linearization Equation doesn’t appear in either the web or the book searches. This is odd because I immediately understood what you were referring to. You might what to create an Abel equation page and then set a redirect from Abel's linearization equation to Abel equation so that people could find both pages with the different search engines.

Different types of articles

A page on the superlogarithm would be interesting, but where it would belong would be a matter of its content. Ideally expositions of subjects like mathematics in Wikipedia are based on peer-reviewed material. I suspect that would constrain the article to a review of Ioannis Galidakis’ published work.

WikiProject_Mathematics is a good place to review Wikipedia standards while Wikipedia talk:WikiProject Mathematics provides good feedback and discusses proposed projects like WikiScience. PlanetMath.Org is another possibility for math web publishing.

When publishing non–peer-reviewed material on Wikipedia, mileage may vary. I watched with horror as my own wonderfully crafted pages were banished from Wikipedia while other pages I saw as being sketchy underwent serious debate and survived. You can see on own user talk page User talk:Daniel Geisler that folks were concerned about the appropriateness of my addition of a link to my own web site on tetration.

On a humorous note

Doing a search on yahoo for superlogarithm is really entertaining. I don’t know which site I like better, Free WebCam Tetration or Sex Pictures Logarithm. I couldn’t even begin to tell you what might constitute a peer-review of these sites. I can see a cocky (pardon the pun) marketing effort in newpenisenlargement.com pedaling Tetration Penis Enlargement products. Or has some kinky webcrawling AI seen the pictures and their captions from the Wikipedia Tetration entry and gone nuts? And finally, how can my tetration website make money from this? Daniel Geisler 07:19, 14 December 2005 (UTC)


 * The reason why I want to write a page about the Superlogarithm, is that I found a way to define it over all real numbers, which naturally extends tetration to all real numbers. This method, in the limit produces an infinitely differentiable function, and the method is very simple. My paper describing the method is complete, and I only need to find a place to publish it now. Where do I publish? and can I at least reference my paper in a wikipedia article on the superlogarithm? or do I need to wait until others read my paper and let someone else write the wikipedia article? Andrew 10:23, 21 December 2005 (UTC)
 * I think I remember reading somebody on the internet (was it Daniel Geisler?) who had some doubts about the superlogarithm extension to real numbers that you mentioned, which goes to show the wisdom of the "no original research" policy.  --Whiteknox 22:13, 28 November 2006 (UTC)

The applicable policy is No original research. Research that is published in a reputable place (which for mathematics means peer-reviewed journals and academic publishers, for instance) can be included. Research reported in unpublished papers and personal website will be treated with suspicion. If you are wondering in what journal to publish your paper, a good guess is to try journals that published papers you refer to.

There is no hard rule forbidding you to reference your own papers, but you need to be sensitive if you do so and expect to be asked for clarification (the relevant document is Autobiography). -- Jitse Niesen (talk) 20:59, 21 December 2005 (UTC)

Publishing works on tetration I would suggest submitting your paper to Complex Variables Theory and Appl. where Ioannis Galidakis has published his work. I wrote a paper defining tetration for complex numbers in 1990 and an much more advanced work in 2001, but I was unsucessful in getting the papers published or even obtaining any feedback on my work. I did copyright my 1990 work and believe that the work is valid with the exceptions of where the fixed points have a Lypanouv characteristic that is a root of unity. The dynamics of the real numbers 1, (1/e)^e = 0.0659880358 and e^(1/e) = 1.44466786 are different from the dynamics of all other numbers under iterated exponentiation. Even if someone correctly defines tetration for what appears to be all the real numbers, I'm still going to expect the attempt to break down at 0.0659880358 and 1.44466786. I did try and validate your value for e^^pi using the Mathematica software I've written for such tasks and concluded that your value for e^^pi is much too large. If you publish your work and I can't find a fundamental flaw with it I will gladly write an article myself. I only qualify this offer because I’ve done much work over the last twenty years trying to find ways of showing that a proposed definition for extending tetration is inconsistent. Daniel Geisler 04:32, 22 December 2005 (UTC)


 * How exactly did you define tetration in your 1990 paper? Andrew 06:11, 8 February 2006 (UTC)


 * Take Faà di Bruno's formula and substitute $$g(x)=f^{\ n-1}(p)$$ where $$p$$ is a fixed point $$f(p) = p$$. This gives the derivatives of an iterated function at a fixed point that leads to the Taylor series of an iterated function in the complex plane. All that is assumed is that f(z) has a derivative and a fixed point not at infinity. The iterated Faà di Bruno's formula works for tetratrion, pentation, hexation and so on. $$D^m f^n(p)=\sum_{k=0}^\infty {c_k f'(p)^k + D^m f^{n-1}(p)}$$ which can be solved by noting the difference equation results in a geometrical progression. In the 1990 paper I took the natural step of simplify the geometrical progressions which introduces the terms throughout the results of $$1/ 1-f' (p)^m$$. In my 1990 paper I didn’t treat the case where $$f' (p)^m = 1$$ which causes my formula for iterated functions to blow up; for tetration in the real numbers this happens at $$e^{1/e}$$ and $$e^{-e}$$. So I gave a definition for hyperbolic tetration with a fixed point. Daniel Geisler 01:08, 13 February 2006 (UTC)

Thoughts on the "Extension to low values of the second operand"
Using the relation $$n\uparrow\uparrow k = \log_n \left(n\uparrow\uparrow (k+1)\right)$$ (which follows from the definition of tetration), one can derive (or define) values for $$n\uparrow\uparrow k$$ where $$k \in {-1, 0, 1}$$.

$$ \begin{matrix} n\uparrow\uparrow 1 & = & \log_n \left(n\uparrow\uparrow 2\right) & = & \log_{n} \left(n^n\right) & = &  n \log_{n} n     & = & n \\ n\uparrow\uparrow 0 & = & \log_{n} \left(n\uparrow\uparrow 1\right) & = &  \log_{n} n    & & & = & 1 \\ n\uparrow\uparrow -1 & = & \log_{n} \left(n\uparrow\uparrow 0\right) & = & \log_{n} 1 & & & = &  0 \end{matrix} $$ Daniel Geisler

''' Tetration just like another mathematics needs to have an axiomatic basis. The trick is to properly enumerate the list of possible consistent systems which extend tetration beyond the positive integers. Negative integers, rational, real and complex numbers are possible examples. $$n\uparrow\uparrow k,$$ $$k > 0$$ is computed by iterated exponentiation is standard.''' $$n\uparrow\uparrow k,$$ $$k < 0$$ can extend k into the negative integers by using iterated logarithms instead of iterated exponentiation, but this means that $$n\uparrow\uparrow k$$ becomes multivalued.

''This confirms the intuitive definition of $$n\uparrow\uparrow 1$$ as simply being $$n$$. However, no further values can be derived by further iteration in this fashion, as $$\log_n 0$$ is undefined.''

'''Articles on arithmetic including treat $$n\uparrow\uparrow 1$$ as an axiom, not something capable of confirming an intuition. Furthermore, the dynamics of the Riemann sphere have no problem with dealing with $$\log_n 0$$ or the further logarithmic iterations from zero.'''

''Similarly, since $$\log_{1} 1$$ is also undefined ($$\log_{1} 1 = \ln 1{/}\ln 1 = 0/0$$), the derivation above does not hold when $$n = 1$$. Therefore, $$1\uparrow\uparrow{-1}$$ must remain an undefined quantity as well. (The figure $$1\uparrow\uparrow{0}$$ can safely be defined as 1, however.)''

''Again, $$0^0$$ is an undefined quantity, so values for $$0\uparrow\uparrow{k}$$ cannot be defined directly. However, $$\lim_{n\rightarrow0} n\uparrow\uparrow{k}$$ is well defined, and exists:''
 * $$\lim_{n\rightarrow0} n\uparrow\uparrow k = \begin{cases} 1, & k \mbox{ even} \\ 0, & k \mbox{ odd} \end{cases} $$

''This limit holds for negative $$n$$, as well. $$0\uparrow\uparrow{k}$$ could be defined in terms of this limit, but $$0\uparrow\uparrow2 = 0$$ would conflict with the standard undefinedness of $$0^0$$.'' '''I've seen text where $$0^0=1$$ and where $$0^0$$ is undefined. Take a look at the enlarged version of the tetration by period fractal. Zero is in the yellow circle immediately to the left of the large red area. The yellow area is period two, it doesn't converge to a single value but oscillates between two values. The following shows that defining $$0^0=1$$ is consistent with the period two behavior in the neighborhood of zero.'''

$$0=0$$

$$0^0=1$$

$$0^{0^0}=0^1=0$$

$$0^{0^{0^0}}=0^0=1$$

Tetration base $$ -{10}^{-6},$$

$$ 1.,-1.\,{10}^{-6},$$ $$ 1.00001 - 3.14164\,{10}^{-6}\,i , -9.99819\,{10}^{-7} -   8.67906\,{10}^{-11}\,i ,$$ $$ 1.00001 - 3.13987\,{10}^{-6}\,i , -9.99819\,{10}^{-7} -   8.67592\,{10}^{-11}\,i ,$$ $$ 1.00001 - 3.13987\,{10}^{-6}\,i , -9.99819\,{10}^{-7} -   8.67592\,{10}^{-11}\,i ,$$

Tetration base $$ {10}^{-6},$$ $$ 1.,1.\,{10}^{-6},0.999986, 1.00019\,{10}^{-6},0.999986, $$ Daniel Geisler 19:32, 6 May 2005 (UTC)

Notation
> How many different names and notations are there for tetration? I've seen them called super-exponents and hyper-powers; the operation tetration and hyper-4, and at least three different symbolologies (Knuth's up arrows, the related ^^ notation, and the horrid left-superscript). Should we cross-reference any of these?

> Hi I don't know if this is the place, but I've been working on tetration for years, and i've had my share of describing it to the passer-by, so I've weeded out some of the more cumbersome notations and pronunciations...

Most intersections between iterated exponentials, tetration, and hyper4 can be described with what I call "auxilliary tetration" (c`b`a) == c b a, where c, and a are superscripts. Great care should be taken to use parentheses with auxilliary notation for tetration, because (E^x)^^y != (y`E`x) so be careful! (2`b`a) == b^b^a. I also found that Ioannis' [] uses auxiliary tetration in the notation c(b, a) == (c`b`a), which I suppose I might start using, because HE's published it, i have not. Normal tetration is expressed as (c`b`1) == b^^c, and exponentiation is (1`b`a) == b^a, and multiple iterations of exp(x) n times is (n`e`x). One of the benefits of auxilliary tetration is that it has a few more axioms for moving things around: (c`b`a) == b^(c-1`b`a) == (c-1`b`(b^a)). Some interesting things you can express with auxilliary tetration are for example: googol=(2`10`2), and googolplex=(3`10`2). Oh i just remembered seeing an ASCII notation for auxilliary tetration somewhere online, i don't remember where, they had (c`b`a) == "a@b^^c" to make it look like scientific notation. The arguments of normal tetration I've heard unanimously called "base" and "order", while the final exponent in auxilliary tetration I call the "auxilliary". For the inverse functions I named them before I found anything online about them, and since I knew that it was called "tetration" i called the inverses "tetra-root" and "tetra-log" (tlog) where the tetra-root finds the base b in (c`b`a), and tetra-log finds c, the order. I saw that bit about slog, and I really don't think thats an appropriate function name, using that system, the inverse of pentation with respect to the second argument would be sslog, and that just gets silly. The tetra-root/tetra-log terminology allows for the corresponding inverses of pentation (hyper5) to also have names: penta-root and penta-log (plog). The pronunciation of the hyper function is pretty straight forward: b(d)c is pronounced "b hyper-d c", but I've had much more trouble finding a proper pronunciation for tetration and auxilliary tetration. The way its said currently without reference to tetration is "b to the b, c times", or when its about iterated exponentials, "the c-th iteration of b to the a", or a "c-th order power tower base b". The terminology I perfer the most is "the c-th tetration of b" or "b tetrated by c". When combined with an auxilliary, I've found the best way to pronounce (c`b`a) is "tetrate b by c to the a" and in a crunch, you could even leave off the tetrate if the context is clear and just say "b by c to the a", or without auxilliary: "b by c". -- Andrew Robbins and_j_rob(at)yahoo(dot)com

PS, I've also compiled a short list of the different continuous extensions of tetration I've come across, and tried myself, along with a bunch of Mathematica code for interpolation and turning the 0 This is an open problem in mathematics. I've seen attempts to define them using fractals, combinatorics, and dynamics, but there's not that much progress.

2. Is e@2 known to be irrational?? The first few decimal places are 15.154262241479.

3. How about e@3, which is around 3814279.1047602??


 * I do not think the "subject" has been sufficiently studied to give an answer to those questions. Question 1 may either be senseless or be trivial via logarithms and exponentials. Just my 2c. Pfortuny 19:06, 7 Mar 2004 (UTC)


 * In one of Rudy Rucker's books, he calls this idea "tetration". -- Tarquin 19:11, 7 Mar 2004 (UTC)

But Wikipedia doesn't have a page titled tetration.


 * I don't understand your point. I said there are other names for this concept that we should perhaps mention in the article -- Tarquin 14:13, 8 Mar 2004 (UTC)

I finally made tetration a re-direct page. But, is there a corresponding "pentation"?? If so, what is its symbol??

Who wrote the last remark? Please sign your ideas so the conversation makes sense?

Theoretically, an unlimited number of binary operations can be methodically built upon one another via iteration. Practically, there is very little compelling justification to do so, however.

On rare occasion, you will discover an equation in an area of applied math which can be expressed more concisely via tetration than involution ("exponentiation"). With successively higher binary operations, though, I do not know of any more advantageous expressions of equations that exist.

This leaves only one practical usage for higher binary operations which I can think of: a more concise expression of extremely large, combinatoric values such as Graham's number. OmegaMan

Yes, there is a pentation. The article isn't very good yet, so can someone help make it better? Hexation also exists and suffers from the same problems, although it's slightly better than the "pentation" article. --116.14.26.124 (talk) 10:26, 23 June 2009 (UTC)

Conjecture about super-exponentiation and cosines
Using Microsoft Works Spreadsheet, I found the following properties:


 * 1) 1. If n is even, the limit of x@n as x approaches 0 is 1.


 * 1) 2. If n is odd, the limit of x@n as x approaches 0 is 0.

From this, I have conjectured, but not proven, that defining a@b when b is not an integer can be done in terms of a function containing cosine in it, because these limits are the same as (cos (pi*(x/2)))^2. User 66.32.73.125

Inverse of super-exponentiation
I would like to create an article about the inverse operation of super-exponentiation, used by the same source that has the @ symbol for super-exponentiation using the & symbol. However, I don't know what to call it, and I feel afraid someone will very likely put it on vfd. Do you know?? 66.32.82.95 17:56, 2 Apr 2004 (UTC)

All inverse binary operations express what can also, of course, be expressed using ordinary, non-inverse or straightforward binary operations. So, they should only be used wherever comparatively convenient.

Consequently, subtraction is used less often than addition; evolution is used less often than involution (often awkwardly called "exponentiation"). Division is the only inverse binary operation which rivals multiplication in its commonplace usage.

In my opinion, there is no hope for an inverse binary operation of tetration (often awkwardly called "super-exponentiation") being used at all since it could only be communicated clearly in terms of tetration. Moreover, agreed standards in mathematical language and notation would become a problem. OmegaMan

> Remember that there isn't an inverse for tetration; there are two inverses. Addition and multiplication are commutative, so they have only one inverse each -- but exponentiation* has two, roots and logs. In the same way, there are 'hyperlogs' and 'hyperroots' that undo tetration.


 * I couldn't find a dictionary entry that links "involution" with exponentiation, although it does have a different mathematical meaning. I'm not sure it's wise to use such a term.

True or false??
True or false: there are plenty of Wikipedia links that change in the following category:

Originally, the link at Article A was a direct link to Article C, but later, someone modifies it and makes Article A link to Article B where Article B re-directs to Article C.

There appear to be plenty in the case of this article being C, but I want to know if there are plenty with no particular Article C. 66.245.23.108 22:56, 12 Jul 2004 (UTC)

Categorizing this article
Can anyone think of a category for this article to go into?? 66.245.77.90 00:36, 26 Aug 2004 (UTC)

Soft hyphens in long numbers
I've placed soft hyphens (&amp;shy;) into the 155-digit (206-character) value of 4&uarr;&uarr;3 because it was making the page super-wide in my browser. - dcljr 05:08, 29 Aug 2004 (UTC)

Negative "super-exponents"?
The last three entries on the page read:
 * n&uarr;&uarr;(-1) = 0 for all real numbers not equal to 1
 * n&uarr;&uarr;(-2) = negative infinity for all real numbers greater than 1
 * n&uarr;&uarr;(-2) = infinity for all real numbers between 0 and 1

Are these by definition? (Whose?) Can someone explain to me (and in the article itself) what a negative "super-exponent" (or whatever you'd call it) would even mean? - dcljr 05:22, 29 Aug 2004 (UTC)

Take the sequence negative infinity, 0, 1, 2, 4, 16, 65536. Does it make sense?? What would make more sense to you?? 66.245.127.199 21:19, 30 Aug 2004 (UTC)


 * What would make more sense to me -- and probably to dcljr -- is what I've replaced that list of identities with. Even if my TeX skills leave much to be desired. --Aponar Kestrel (talk) 20:24, 2004 Sep 15 (UTC)

But ln (x) approaches –∞ as x approaches 0. --116.14.26.124 (talk) 10:31, 23 June 2009 (UTC)

Calculation errors?
I could be wrong, but the values I get for numbers as simple as $$3\uparrow\uparrow3$$ differ from those on the page. The values I get would be:


 * $$1\uparrow\uparrow3$$ = $$\,\!1^{1^1}$$ = 1
 * $$2\uparrow\uparrow3$$ = $$\,\!2^{2^2}$$ = 16
 * $$3\uparrow\uparrow3$$ = $$\,\!3^{3^3}$$ = 19,683
 * $$4\uparrow\uparrow3$$ = $$\,\!4^{4^4}$$ = 4,294,967,296
 * $$5\uparrow\uparrow3$$ = $$\,\!5^{5^5}$$ = 298,023,223,876,953,125
 * $$6\uparrow\uparrow3$$ = $$\,\!6^{6^6}$$ = 10,314,424,798,490,535,546,171,949,056

... with similar divergences for the next rows. Have I missed something? These values are easily checkable with a pocket calculator, if anyone would care to back me up. Aydee


 * Hmm. You're doing (3^3)^3 (= 19683), the page does 3^(3^3) (= 7.62559748×1012).
 * If I go to a linear calculator (like Google Calculator) and ask it for "3^3^3", it converts it into 3^(3^3).
 * But the definition of this function on the page is $$x \uparrow\uparrow y = x \mbox{ raised to its own power }y\mbox{ times}$$. Does that mean x^x, then the result of that raised to x, etc.? Or does it mean the resolution of the symbol $$\,\!x^{x^x}$$?
 * The definition at Knuth's up-arrow notation indicates that it is the latter. But if so, then the definition of "iterated exponentiation" is not accurate. It would seem more accurate to say something like "the xth power of x, y times".
 * - KeithTyler 21:33, Sep 15, 2004 (UTC)


 * Good point. It'd probably be best to use the definition on Knuth's up-arrow notation in some form to clarify this, but I'm not sure what the best policy in these cases may be; my gut feeling is that this article needs the definition and Knuth's up-arrow notation should refer here, as the up-arrow notation is merely a method of representing tetration. On the other hand, I could be wrong. Any other ideas? Aydee 01:41, 2004 Sep 16 (UTC)


 * I just added a quick example of iterated expectation to the page. Perhaps that's sufficient to clear up any misunderstandings? - dcljr 06:37, 10 Oct 2004 (UTC)


 * There was a note in this article near the beginning: Note that when evaluating multiple-level exponentiation, the exponentiation is done at the deepest level first (in the notation, at the highest level). In other words:
 * $$\,\!2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{\left(2^4\right)} = 2^{16} = 65,\!536$$
 * $$\,\!2^{2^{2^2}}$$ is not equal to $$\,\! \left({\left(2^2\right)}^2\right)^2 = 256$$
 * It means that
 * $$4\uparrow\uparrow3$$=$$4^{\left(4^4\right)}$$ and not $$\,\! \left(4^4\right)^4$$, which is how you computed for the value. --Kevin_philippines 20:56, Sep 8, 2006 (UTC)

Page title
On July 4, 2004, User:Gdr asked this page to be renamed, but it hasn't been renamed. What happened?? 66.245.71.98 15:45, 16 Sep 2004 (UTC)


 * Your helpful wizards have finally woken up and noticed! Noel 20:01, 17 Sep 2004 (UTC)

Carets
Given that, as pointed out in the article, the up-arrow symbol (when present on a computer keyboard) is used similarly to the caret symbol (indicating superscripts), couldn't we just replace all the double up-arrows on the page (except the ones next to Knuth's name) to double carets (^^)? This would make the page readable in all browsers. Just a suggestion... - dcljr 06:56, 10 Oct 2004 (UTC)

Tetration number names
Please read very slowly and carefully:

Can anyone come up with a way to name numbers that tetration can help visualize the magnitude of?? One way that I came up with is a building with 103 floors, each of which is for numbers of various sizes; the higher the floor, the larger the numbers. The Tetrational System (the numbering system that this uses) has the following number names:

For numbers <= 10^30, whose magnitudes are easy to visualize without tetration, it uses the same number names as Rowlett. On the first 2 floors, we have:

First floor: Two, Three, Six, Nine, Twelve, Fifteen, Eighteen, Twenty-one, Twenty-four, Twenty-seven, Thirty

Second floor: Hundred, Thousand, Million, Gillion, Tetrillion, Pentillion, Hexillion, Heptillion, Oktillion, Ennillion, Dekillion

Now, let's go onto the third floor. This is where the numbers get large enough that tetration is a useful way; each term is 10 to the power of the previous term:

Third floor: Googol, Froogol; the remaining words on this floor are the same as those on the second floor only that they use -illoogol instead of -illion (that is, Milloogol to Dekilloogol are 10^(10^6) through 10^(10^30.) (Froogol is a back-formation on Froogle on the model of Googol/Google.)

Fourth throught 103rd floors: Simply take the number names on the third floor and add "-plex" for those on the fourth floor, "-duplex" for the fifth floor, "-triplex" for the sixth floor, and so on all the way to "-centuplex" for the 103rd floor. This makes the largest number in the building (dekilloogolcentuplex) 10^10^10^...10^10^10^30 with a total of 103 numbers (102 tens and a 30) between exponent signs.

Any numbers too large for this?? At this moment, I know of only one number too large for this building, Graham's number. 66.245.98.219 23:12, 16 Nov 2004 (UTC)

Jonathan Bowers created an extendtion to tetration called array notation which he explains on his Home Page. He also has many examples of Infinity Scrapers, which compared to the numbers in this building, are like Alpha Centura!--SurrealWarrior 20:23, 11 August 2005 (UTC)

Anon minor formula edit
Line 33, was: $$2 \uparrow\uparrow (n-3)$$ &minus; 3

Now is: $$2 \uparrow\uparrow (n+3)$$ &minus; 3

As I am no expert, please check if this revision is valid. -- AllyUnion (talk) 10:33, 10 Dec 2004 (UTC)

(''The following discussion was movied from the Super-exponentiation entry on Pages needing attention/Mathematics. Paul August &#9742; 20:06, Feb 7, 2005 (UTC)")


 * Super-exponentiation -- is this notation genuine? The source given is a mathforum link to where the notation is "invented" by a enthusiastic student. It seems like a dup of Knuth's up-arrow notation. Motor 19:48, 18 Mar 2004 (UTC)


 * No, Knuth's up-arrow notation is for all operations from super-exponentiation upwards, and super-exponentiation is just for that operation itself. 66.32.89.242 23:33, 2 Apr 2004 (UTC)


 * I rewrote this page using Knuth's up-arrow notation and the standard term "tetration". Really this page should be moved and redirected to Tetration. Gdr 13:23, 2004 Jul 4 (UTC)


 * Done. Noel 20:06, 17 Sep 2004 (UTC)


 * I currently put this article in Category:Mathematics, but I want to see if anyone can give a more specific category for this article. 66.245.77.90 00:44, 26 Aug 2004 (UTC)


 * http://mathworld.wolfram.com/PowerTower.html This is topic is discussed at Mathworld under the heading "Power Tower" GulDan 17:48, Sep 16, 2004 (UTC)


 * I have added information about the standard notation with supporting references. I strongly recommend using the published notation and nomenclature. MathWorld&#8217;s entry on Power Tower is great, but they do have a sub page for tetration that directs to their Power Tower entry. Technically these are not the same thing; tetration is much more comprehensive than Power Tower, but almost all published research on tetration has been confined to the Power Tower due to the profound difficulty of the subject. I have changed the category to arithmetic. User:Daniel Geisler 12:56, Dec 28, 2004 (UTC)

The other successor of exponentiation
The article seems to suggest that tetration is the logical extension of the sequence of addition, multiplication, and exponentiation. However, what about ((nn)n...) for m occurences of n? It seems to me this operation is equally an extension: Since exponentiation is non-commutative, the sequence bifurcates at this point. This "inferior super-exponentiation", if you will, has interesting properties of its own. For instance, a negative value for m corresponds to taking the nth root m times (since it can remove exponents), so m=0 yields $$\sqrt[n]n$$.

Anyway, shouldn't there be an article about this operation also? Who here knows its standard name(s) and notation(s), or does it have any? (I thought "iterated exponentiation" was reasonable, but the article's (and discussion's) usage conflicts.) --Ddawson 12:08, 20 Mar 2005 (UTC)


 * This is simply nn m $$n^{n^m}$$(NESTED SUBS/SUPS ARE BROKEN ON WIKIPEDIA >:, there is nothing especially "new" about it relative to the sense that tetration is some new operation. Dysprosia 22:25, 20 Mar 2005 (UTC)

599

Good point. (That should be $$n^{n^{m-1}}$$, though, since I count the base n.) I see what you mean: this can be defined by a fixed expression, whereas tetration must be defined recursively in a formal sense. Ddawson 15:06, 21 Mar 2005 (UTC)

Maybe there could be a quick mention of this, somewhere soon after the definition of tetration, with the explanation of why it's not nearly as interesting. My reasoning is that it's still a valid way of iterating exponentiation. Ddawson 15:25, 21 Mar 2005 (UTC)

Infinite power towers
Someone made a strange removal on the basis that "2 is greater than e" (??). I have reverted that, but then I noticed that the reason that the article stipulates r <= e, is to ensure "$$x$$ is not more than $$e^{1/e}$$". But investigation shows that r^(1/r) is never greater than e^(1/e). The article misleadingly implies otherwise when r > e. Therefore I think that that phrase should be removed or revised. Eric119 02:28, 2 August 2005 (UTC)


 * I rephrased it.--Patrick 07:40, 2 August 2005 (UTC)

I added a precise definition of infinite power towers, since I've had arguments with people before who insisted $$ x^{x^{x^{.^{.^{.}}}}} $$ is not well-defined. And I can see their point, since the result of raising x to an already existing tower depends on the value the tower already has. For instance, while the sequence $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}$$, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}}$$, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}}}}$$, ... clearly converges to 2, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^4}}$$, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^4}}}$$, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^4}}}}$$, ... just stays at 4 every step of the way, even though it would seem to eventually lead to the same infinite tower. Gmalivuk 20:45, 6 July 2007 (UTC)

New tetration and slog software
A posting on Wolfram's NKS Forum from C. A. Rubtsov and G. F. Romerio has a link to their tetration and slog software. I can't say that I support their conclusions, but then I can't really name anyone who support my own conclusions, so it's all good. They have released a tetration and slog calculator that does give a much larger value for e^^pi that I can justify, but it is more in line with Andrew Robins' estimate. I do share their interest in S. C. Woon's work.Daniel Geisler 17:50, 27 December 2005 (UTC)


 * I've checked out their calculators, and I like the interface, but the method they use for calculating superlog is the "linear" extension that uses the s(z) = (z-1) critical function, which ammounts to a 1-st degree solution using my method. The method is described in:
 * Solving for the Analytic Piecewise Extension of Tetration and the Super-logarithm,
 * which should now be available. Andrew 22:15, 2 February 2006 (UTC)

A thought about addition
$${{a + b = a + } \atop {\ }} {{\underbrace{1 + \cdots + 1}} \atop b}$$

Or, a incremented b times. Or, applying b times the Successor function to a (See Peano axioms, it's all there)

Locoluis&#778; 01:06, 31 March 2006 (UTC)
 * Just my tought. If tetration is "level 4", exponentiation is "level 3". multiplication is "level 2" and addition is "level 1" when the addition with 1 is "level 0" and "ground level". //Fabben

I have been trying to find a solution of "level 0 binary operator" for a few years. The hypothesis of increment by 1 proposed above, looks quite reasonable, however it leaves some blank spots in the whole picture.

Let's accept that a(1)b = a+b, a(2)b = ab, a(3)b = a^b etc. (just for further convenience).

Now we are discussing the problem of a(0)b. As Locoluis proposes, a(0)b = a+1. There are two logical conclusions from that:

1) The "level 0" operator is not commutative: a(0)b = a+1 != b(0)a = b+1

2) The function f(x)=a(0)x is a constant, and it does not depend on its "x" argument.

From the other hand, let's examine the following sequence:

a(1)a = a(2)2, i.e. a+a = a*2;

a(2)a = a(3)2, i.e. a*a = a^2;

a(3)a = a(4)2, i.e. a^a = a^^2 etc.

and let's try to extrapolate that sequence below "level 1". We will get:

a(0)a = a(1)2, i.e. a(0)a = a+2,

which comes into a contradiction with the assumption that a(0)b = a+1 (!)

In my opinion, the a(0)b operator definition should meet the following requirements:

1) a(0)a = a+2 (see explanation above);

2) a(0)m < a(0)n, if m<n;

3) m(0)a < n(0)a, if m<n;

4) a < a(0)b < a+b for any positive integer "a" and "b".

A very special case is when b=0. Accepting our assumption, we get

0(0)0 = 0+2 = 2, which is larger than 0(1)0 = 0+0 = 0 (!!!)

--Beloturkin 09:30, 8 September 2006 (UTC)

So far, welcome to continue discussing the Zero-Level Binary Operator... -- Unsigned

Addition can never be expressed in the same terms as further iterated operators. Note that per definition, n(2)1=n, n(3)1=n, and in general n(m+1)1=m (since you're doing n(m)n(m)n(m)...(m)n, except with only one m, leading to the degenerate case where no (m) operations actually take place). However, n(1)1 is not equal to n, even though it would be if addition were defined in terms of an iteration of a (0) operator - regardless of what that operator was. Thus, addition cannot be defined in that sense. -- Milo

Moved Comment
Isn't this page violating the third paragraph of "What Wikipedia is not"?
 * This should be added as a comment below. Apparently this is refering to the section on original research ([|What Wikipedia is not, 1.3]).  I personally think it can be deleted because nobody signed to it.  --Whiteknox 19:20, 28 December 2006 (UTC)

n ^^ 1/2
Would it help in extending tetration to the reals to extend it first to n^^(1/2) (and by extension n^^(k+1/2))? If so, what useful identities can be used to solve the equation x = n^^(1/2) for n? At first I thought about saying x = n^^(1/2) implies n = x^^2, but this really doesn't make sense, since (n^^a)^^b doesn't equal n^^(a*b) or n^^(a+b) or anything else simple. ((n^^2)^^2 = n^(n^(n+1)) = (n^^3)^n.)

Using the symbol "=/=" to denote "not equal to";

It sounds OK, the problem is though that in general n^^(1/2) =/= n^^(2/4), n^^(3/6), n^^(4/8) etc. which would be necessary for this extension to be rigorous (and mathematics is always rigorous). Other oddities crop up as well, such as the fact that 2^^(1/3) < 2^^(2/7) even though 1/3 > 2/7.

This and other problems in extending tetration to the reals is a consequence of the fact that exponentiation is neither commutative nor associative, so that in general a^b =/= b^a and a^(b^c) =/= (a^b)^c.

Meltingpot (talk) 09:19, 15 April 2008 (UTC)

It seems that there should be some identities along the lines of n^a*n^b = n^(a+b) and (n^a)^b = n^(a*b). Actually, one identity that may eventually be helpful is (n^^a)^^2 = (n^^(a+1))^(n^^(a-1)), although I don't know how helpful.

Also, in the extension to real numbers section, the slog examples seem wrong to me. It looks like they are arbitrarily equated to log. (slog10 3 = log10 3) The real inverse would be x in 10^^x = 3 which is currently undefined. (Both of the other examples make similar mistakes.) MagiMaster 07:47, 26 September 2006 (UTC)


 * Actually, does it make sense to define x^^(1/n) = y such that y^^n = x? If so, I might have a good definition for x^^y based on continued fractions.  Basically, n in the above equation doesn't have to be an integer.  Just apply that definition, for 0 < y < 1, and the definition of x^^y, for y >= 1, recursively.  For any rational n, it's a finite process.  Using those definitions, 4^^(3/5) ~= 2.2266.  Does anyone see any problems with this definition?  (Also, I have no idea how to prove monotonicity or continuity...)  Hmm... I was just reading Andrew's paper (see above) and my values don't agree with his.  (His approach looks a lot more rigorous than mine.)  MagiMaster 05:31, 28 September 2006 (UTC)

Yes, unfortunately; for consistency 4^^(3/5) would have to equal 4^^(6/10), 4^^(9/15), 4^^(12/20) as well, which it almost certainly doesn't (as I explained above).

Meltingpot (talk) 08:53, 24 April 2008 (UTC)

y=xx
Can xx be differentiated? And if not, is there proof? I think $$\frac {dx^x} {dx} = 0$$ when $$\frac {1} {e} = x$$ and it appears that $$\frac {dx^x} {dx} = 1$$ when $$1 = x$$. But I don't think I can find the derivative at first glance.--Steven Weston 12:33, 10 May 2007 (UTC)


 * $$\frac {dx^x} {dx} = \frac{de^{x \ln x}}{dx} = e^{x \ln x} \frac{d(x\ln x)}{dx} = e^{x \ln x} (1 + \ln x) = x^x (1 + \ln x). $$
 * This result is also mentioned in the table of derivatives. If you fill in $$x=1$$, you'll see that the derivative is indeed 1, as you say. Similarly, the derivative is zero at $$x=\frac1e$$. -- Jitse Niesen (talk) 13:47, 10 May 2007 (UTC)


 * Cheers. It looks so simple now...  I've looked through the table of derivatives in the past, but I can't say that it's ever caught my eye.  I've been interested in tetration for a while now.  What's the antiderivative then, or indeed is there one?  I couldn't immediately find it in the lists of integrals.--Steven Weston 19:28, 10 May 2007 (UTC)


 * I don't know how to find the antiderivative. I tried it in Maple and it did not give an answer. This probably means that the antiderivative cannot be written in terms of elementary functions like exponentials, sine, etc. Nevertheless, the antiderivative does exist; it's just that we cannot find a nice formula for it. -- Jitse Niesen (talk) 01:17, 11 May 2007 (UTC)


 * You said that it definitely exists: is there proof of this?--Steven Weston 09:28, 11 May 2007 (UTC)


 * The function $$f(x) = x^x$$ is a continuous function. There is a theorem that says that every continuous function has an antiderivative (this result is mentioned in antiderivative). -- Jitse Niesen (talk) 14:44, 11 May 2007 (UTC)

Just out of interest, can $$^x x$$ be differentiated?--Steven Weston 10:33, 15 May 2007 (UTC)


 * That depends on how you define $${}^x x$$ when x is not an integer. -- Jitse Niesen (talk) 01:39, 16 May 2007 (UTC)

Name?
Has this special case got a name? It is barely mentioned either here or on Exponentiation. —DIV (128.250.80.15 (talk) 07:43, 1 May 2008 (UTC))

Real Extension
Not my area of expertise, but the following two sentences in this section appear to be in direct conflict: Extending $$x \uparrow\uparrow b$$ to real numbers $$x > 0$$ is straightforward...   At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex numbers... Can someone clarify this? Ultraviolet777 22:58, 24 July 2007 (UTC)
 * The latter bit is talking about real or complex values of "b", the number of iterated exponents. I changed the wording a little in the article to improve this.  Doctormatt 23:18, 24 July 2007 (UTC)

real $$~b~$$
Well, Doctormatt, on the base of your update of the "real extension", could you please, name, for example, the first terms of the expansion of the tetration in the Tailor series at samall values of the number $$~b~$$ of exponentiations, ?

In other words, if

$$~{^b\rm e}=1+\sum_{n=1}^{\infty} c_n b^n $$

then, what are coefficients $$~c~$$?

dima (talk) 02:24, 21 January 2008 (UTC) P.S. What is radius of convergence of the series?dima (talk) 02:27, 21 January 2008 (UTC)


 * It appears that you haven't carefully read Doctormatt's changes or comments. $$^b a$$ is analytic in a for fixed (integer) b.  He removed (IMHO) any indication that there is a simple extension to real b of any sort, even if uxp is offered as such an extension.  &mdash; Arthur Rubin |  (talk) 07:46, 21 January 2008 (UTC)


 * Dear Artur. Sorry if my typing above is not clever enough. I understand that at fixed integer b the tetration is analytic function of a. Doctormatt typed about that case. Perhaps, his typing cannot be applied to the case of real b. The extension to the real heighs has two examples. First of them seems to be just uxp. The second one has smooth first derrivative, but it is still not analytic, and the series above in not valid. May I repeat the question: How about the analytic extension for real and for complex b? I suggest that somebody gives the reference to such an extension; overwice we should type that . dima (talk) 09:14, 21 January 2008 (UTC)


 * A paper describing a real-analytic solution for the infra log (base e) was reported to me in a USENet post in the 90s, and (I believe) here in Reference desk/Mathematics. I never looked up the paper, so I can't confirm it.  &mdash; Arthur Rubin |  (talk) 13:54, 21 January 2008 (UTC)

some small edits without author-notification
I'd like to excuse for doing edits without being logged in (usually I automatically logged in), so there is no author at my (small) series of edits just before. Don't remember, how to correct this - sorry Gottfried Helms Gotti 08:22, 2 November 2007 (UTC)

Pentation
Why not bother to add a Pentation article? Even though it's a redirect, I think it should have its own article. Pilover819 (talk) 11:37, 23 December 2007 (UTC)


 * Actually, it would be nice to have a page on pentation, as there is at least one result that has been found about it. For example Jay D. Fox notes in the Tetration Forum that the limit $$\lim_{n\rightarrow (-\infty)} x {\uparrow}{\uparrow}{\uparrow} n$$ is the lower fixed point of real-valued tetration, for example: $$e{\uparrow}{\uparrow}(-1.85) = -1.85$$ so $$e {\uparrow}{\uparrow}{\uparrow} (-\infty) = -1.85$$. Aside from this result, though, not much is known about pentation. AJRobbins (talk) 16:08, 28 December 2007 (UTC)

There now is a page on pentation. Don't know how long it will last though. --116.14.26.124 (talk) 03:45, 23 June 2009 (UTC)

Dubious
I am not satisfied with the paper. The definition of tetration is almost absent. The speculation about $$a^{a^{...^a}}_{^{~x {\rm times}}}$$ looks as property of $$~ ^xa ~$$ in the specific case of real $$~a~$$ and natural $$~x~$$. The definition should allow to see that this function is analytic with respect to each of its arguments, (at least in some range of values of arguments), reveal its singularities, and so on. Overwice we should write at the preamble that  Tetration is some mathematical function of two variables which is believed to be definable as analytical in some (still unknown) set including natural numbers. P.S. Look at the definitions of other special functions; try to do the same for tetration. Then we can keep tetration among special funcitons. dima (talk) 04:39, 13 January 2008 (UTC)

I put the subsection about uxp as special case of tetration. I tried to make it short; only definition. I suppress the doubtful statement about analiticity at a=e. Now my dissatisfaction is rather about present state of the theory of tetration, not about the article. However, the article ultra exponential function should be improved; perhaps, including the proof of the Theorem about discontinuity of derivative at integer values of argument. It seems, there is no way to define tetration as analytical function and no regular way for the precise evaluation at non-integer points. Why, for example, factorial can be generalized to analytic gamma function, but the tetration looks so ugly at the complex plane? dima (talk) 23:17, 14 January 2008 (UTC)

Discussion about "UXP"
I have moved this section from the main article, as its claims are unclear, and if I interpret them correctly, contradictory. You cannot have differentiability at one integer value (Claim C), and have non-differentiability at another integer value (Claim E) while you have a functional equation that relates these two values (Claim A). That is simply impossible. If anyone can explain to me how these claims are not in contradiction with themselves, please do so. As for you Domitori/dima and Ultra.Power, there is an excelent forum for discussing these things, called the Tetration Forum. I encourage both of you to join and learn more about tetration, and also discuss this ultra-exponential-function you are writing about. It is a great place to discuss tetration so that you can get feedback on your theories and gaps in your knowledge. AJRobbins (talk) 06:43, 19 January 2008 (UTC)


 * I have since come to an understanding that (Claim E) has a major typo in it (should be (-1)+ not 0+, which makes a big difference, as noted by Arthur Rubin). This means that the extension called "UXP" is nothing more than the method described in the "linear approximation" section of this article. This indicates that a new article ultra exponential function is unnecessary. I also agree with most of the comments in the WikiProject Mathematics discussion about "UXP", and that the "UXP" article uses non-standard terminology with non-standard notation, to introduce a standard extension of tetration in a way that makes it seem different when it really isn't. AJRobbins (talk) 02:25, 20 January 2008 (UTC)
 * This could have been avoided if Dima and Ultra had found the Tetration Forum before Wikipedia. The purpose of the forum was to have a place to talk about terminology, notation, and methods so that things like this don't happen. The Tetration Forum is also host to many discussions about series expansions, and extensions to the complex plane. AJRobbins (talk) 02:25, 20 January 2008 (UTC)

Ultra exponential function
The functional approach to the tetration leads to so-called ultra exponential function $$~{\rm f}(b)={\rm uxp}_a(b)~$$, determined with following condiitons:

(A) $$~{\rm f}(x)=a^{{\rm f}(x-1)} \mbox{for all} x>-1 ~$$

(B) $$~{\rm f}(0)=1$$

(C) $$~\rm f~$$ is differentiable on $$~(-1,0)~$$

(D) $$~{\rm f}'~$$ is non-increasing or non-decreasing at $$~ (-1,0)~$$

(E) $$\lim_{x\rightarrow 0^+}{\rm f}'(x)=\ln(a) \lim_{x\rightarrow0^-} {\rm f}'(x),$$

It happens that for real values of $$~a~$$, the conditions (A-E) unambiguously determine the funciton $$~{\rm uxp}_a~$$ , and, for integer values of the argument, the $$~ {\rm uxp}~$$ function coincides with tetration defined above . See the specific article Ultra exponential function for the details.

At $$~a \ne {\rm e}~$$ the function $$~{\rm uxp}_a~$$ is not smooth, and the alternative approaches to the extension of the tetration exist, which lead to the smooth functions.


 * I see, some users dislike the theorem above, but please, will you return to the article at least the reference? dima (talk) 03:26, 24 January 2008 (UTC)


 * I consider the theorem uninteresting, thereby making the paper uninteresting. Others (including the author) may disagree....  &mdash; Arthur Rubin |  (talk) 03:40, 24 January 2008 (UTC)

Tetration

 * $$ssrt(x)=\frac {1} {\lim_{n\to \infty}\text { }^{2n+1} \left(\frac{1}{x}\right)}$$ —Preceding unsigned comment added by Cʘʅʃʘɔ (talk • contribs) 08:15, 4 March 2008 (UTC)

What is WRONG?

y = x^x

inverse is:

x = y^y

y = x^(1/y)

y = x^(1/x^(1/y))

…

y = x^(1/x^(1/x^(…)

y = x^(1/x)^(1/x)^(…)

y = x^(1/x)^^∞

y = (1/(1/x)) ^ (1/x)^^∞)

y = 1/(1/x)^(1/x)^^∞)

y = 1/(1/x)^^(∞+1)

y = 1/(1/x)^^ ∞.

Cʘʅʃʘɔ (talk) 15:34, 3 March 2008 (UTC)


 * After careful study, the equations are correct, but convergence seems sufficiently questionable that it needs a reference that the ssrt expression in terms of the W function and the ∞z expression (also in terms of the W function) have the same or similar domains of correctness or convergence. &mdash; Arthur Rubin |  (talk) 19:26, 3 March 2008 (UTC)


 * The domain of y = lim [n -> ∞] x^^(2n+1) (not x^^n  - must be odd) is:
 * 0 < x =< e^(1/e) (s. Images: [[Image:Tetration large.png|60px]] and [[Image:InfinitePowerTower.gif|40px]]),
 * so the domain of y = lim [n -> ∞] (1/x)^^(2n+1) is:
 * (1/e)^(1/e) =< x < ∞.
 * so the domain of y = lim [n -> ∞] (1/(1/x)^^(2n+1)) is:
 * (1/e)^(1/e) =< x < ∞.


 * The minimum of x^x is x0 = 1/e; the value is y0(x0) = (1/e)^(1/e);
 * so (1/e)^(1/e) =< y < ∞.
 * Invert function has a domain:
 * (1/e)^(1/e) =< x < ∞.


 * The value of lim [n -> ∞] (1/(1/e)^(1/e) ^^(2n+1))  = 1/e,


 * The value of invert of y = x^x for (1/e)^(1/e) is y((1/e)^(1/e)) = 1/e.


 * For both functions: y(1) = 1.


 * And: lim [x -> ∞] (x^x)’ = ∞ and lim [x -> ∞] (1/(1/x)^^(2n+1))’ = 0.

Cʘʅʃʘɔ (talk) 21:57, 3 March 2008 (UTC)

Interesting pattern
I discovered this pattern to do with $$x^x$$ the other day and I thought it was quite interesting.
 * $$\frac {dx^x} {dx} = x^x (\ln x + 1) $$
 * $$\frac {d^2x^x} {dx^2} = x^x (\ln^2 x + 2\ln x + \frac {x+1} {x}) $$
 * $$\frac {d^3x^x} {dx^3} = x^x (\ln^3 x + 3\ln^2 x + 3\frac {x+1} {x}\ln x + \frac {x^2+3x-1} {x^2}) $$
 * $$\frac {d^4x^x} {dx^4} = x^x (\ln^4 x + 4\ln^3 x + 6\frac {x+1} {x}\ln^2 x + 4\frac {x^2+3x-1} {x^2}\ln x+ \frac {x^3+6x-x+2} {x^3}) $$

The nth derivative takes a similar form, with a pattern taken from Pascal's triangle and another pattern of the polynomials/(their degree) continuing down in the same positions relative to the greatest power of lnx. Lol, the only mildly tedious thing was proving the conjecture I made that this pattern was true for all natural n. If anyone has seen these polynomials before, please let me know. I'd like to know if they have a specific name. If you need to know any more of them, I wrote a program that generated the first 172 of them, before the floating point numbers stopped working.--Steven Weston (talk) 23:25, 6 July 2008 (UTC)

Yes, the coefficients of those polynomials are Lehmer-Comtet numbers Sloane's A008296. Also for more expansions of this nature, please see section 5.3.2 (page 42) of the Tetration Reference. AJRobbins (talk) 18:15, 24 September 2008 (UTC)

Ultra exponential
As proposed earlier at Talk:ultra exponential function, I'm proposing that ultra exponential function be merged into Tetration. &mdash; Arthur Rubin (talk) 02:58, 17 August 2008 (UTC)
 * Agree.--Dojarca (talk) 15:20, 31 August 2008 (UTC)
 * Agreed AJRobbins (talk) 02:32, 17 September 2008 (UTC)
 * Agree. May be, even Tetration. dima (talk) 08:41, 27 September 2008 (UTC)

Analytic extention to real heights
Since any analytic function is infinitely differentiable, it should be mentioned, that there is only one analytic extention of tetration into real heights, namely, based on infinite differentiability requirement.--Dojarca (talk) 15:55, 31 August 2008 (UTC)
 * That's not true. Well, it may be true about analytic, but I'm sure I could construct a large family of infinitely differentiable solutions.  &mdash; Arthur Rubin  (talk) 18:31, 31 August 2008 (UTC)
 * That satisfy all the integer values and functional equations? Only one.--Dojarca (talk) 09:56, 1 September 2008 (UTC)
 * cn? For a fixed base, you can add a small $$C^\infty$$ bridge (i.e., a $$C^\infty$$ function which is 0 outside the specified range) to $${}^{x}a$$ between (say) x=.2 and x=.3, and extend it by the functional equation for all real values of x.  I think it can be made $$C^\infty$$ in both x and a, but the definitions are unclear.  &mdash; Arthur Rubin  (talk) 17:06, 1 September 2008 (UTC)
 * Such addition will break analiticity: the added function is clearly not analityc if it is 0 in one range and non-zero in another range.--Dojarca (talk) 17:23, 1 September 2008 (UTC)
 * I said $$C^\infty$$ (infinitely differentiable) not $$C^\omega$$ (analytic). I don't know if there is an analytic solution, although I've been pointed to an article on fluid flows which suggests that there are at most 3 analytic solutions.  &mdash; Arthur Rubin  (talk) 18:04, 1 September 2008 (UTC)
 * Infifnitely differeintiable in all points function cannot be zero in one range and non-zero in another range.--Dojarca (talk) 19:13, 1 September 2008 (UTC)

(unindent) Umm...
 * $$f(x) = \begin{cases}e^{-1/x^2} & x>0\\

0& x\le 0\end{cases}$$ Next! siℓℓy rabbit (  talk  ) 19:24, 1 September 2008 (UTC)
 * This function is not analytic: it is not equal to its Taylor series in the neighbourhood of zero.--Dojarca (talk) 19:43, 1 September 2008 (UTC)
 * Good job! But it is "Infifnitely differeintiable [sic] in all points function" and yet is "zero in one range and non-zero in another range", contrary to your above post.  siℓℓy rabbit  (  talk  ) 19:46, 1 September 2008 (UTC)
 * Well I meant analytic function. The main point here is that there can be only one analytic extention of tetration into real heights.--Dojarca (talk) 19:53, 1 September 2008 (UTC)

Anyway, the point is moot. You can just add an (analytic) periodic function like $$\sin(\pi x)$$. This ambiguity necessarily invalidates uniqueness, unless further restrictions are imposed. siℓℓy rabbit (  talk  ) 19:51, 1 September 2008 (UTC)
 * Then it simply will not satisfy the main functional equation for tetration.--Dojarca (talk) 19:54, 1 September 2008 (UTC)
 * Ah... I was thinking that $$^{b+1}a = a^{(^ba)}$$ only needs to hold for integer values of b.  siℓℓy rabbit  (  talk  ) 19:59, 1 September 2008 (UTC)
 * Of course if you only fix the integer values, you can easily find multiple analytic solutions.--Dojarca (talk) 20:01, 1 September 2008 (UTC)
 * It's not obvious that there is any (real-)analytic solution, or that the solution is necessarily unique. Hmmm.  Actually, thinking it over, the solution cannot be unique, with the following argument:
 * Let h be a (real-)analytic function satisfying:
 * $$h(x+1) = h(x),$$
 * $$h(0) = 0,$$
 * $$h'(x) > -1,$$
 * (The third condition is required to retain the obvious monotonicity requirement for a > 1.)
 * For example, $$h(x)=\epsilon \sin(2 n \pi x)\,,$$, where n is an integer and &epsilon; is sufficiently small.
 * Let $$f^*(x) = f(x+h(x)).$$
 * Then f* also satisfies the equations, does it not? &mdash; Arthur Rubin
 * Oops. Perhaps my "3 solutions" above was an incorrect memory.  I guess some other sort of regularity requirement is necessary to obtain uniqueness.  I don't know why I didn't think of that 34 years ago when I first looked at the problem.  &mdash; Arthur Rubin  (talk) 21:24, 1 September 2008 (UTC)
 * Good one. siℓℓy rabbit  (  talk  ) 00:05, 2 September 2008 (UTC)
 * Um, how does $$f^*(x) = f(x+h(x))$$ satisfy #1? I get $$f^*(x+1) = f(x+1 + h(x+1)) = f(x+1 + h(x))$$. How does this equal $$f(x+h(x))$$ (i.e. $$f^*(x)$$)? Furthermore, how is eq. 1. equivalent to the tetrational equation: $$f(x + 1) = b^{f(x)}$$?
 * As for another stronger requirement, how about that the function have exactly one inflection point for $$x \in [-2, \infty)$$? (I'm also thinking of real bases b >= 1 only as well -- bases 0 < b < 1 seem to have a peculiar oscillating character that would necessitate a different requirement, and bases less than 0 yield complex values and all bets are off.) mike4ty4 (talk) 08:06, 17 November 2008 (UTC)
 * If $$f(x+1)=b^{f(x)}\,$$ (the relevant equation), then $$f^{*}(x+1)=b^{f^{*}(x)}$$. — Arthur Rubin  (talk) 14:53, 17 November 2008 (UTC)
 * Ah, I get it now. Thank you. mike4ty4 (talk) 06:52, 23 November 2008 (UTC)

I am surprised to admit my own ignorance, but presumably there are theorems that give sufficient conditions for the unique analytic solution of a functional equation, once appropriate boundary conditions are set. My own question is: (1) what are these theorems, and (2) which hypotheses are violated by the functional equation of the tetration. siℓℓy rabbit (  talk  ) 10:33, 2 September 2008 (UTC)
 * I think there is simple not enough functional equations. For example, you need 3 functional equations to define gamma-function.--Dojarca (talk) 10:39, 2 September 2008 (UTC)
 * I think it should be a functional equation that connects tetration with super root function. That connection would greatly simplify dealing with tetration. There is also no ormula to change the base.--Dojarca (talk) 20:43, 2 September 2008 (UTC)
 * Silly rabbit, this is a known case for functional equation:. It is known to have multiple solutions.--Dojarca (talk) 02:26, 7 September 2008 (UTC)


 * The Citizendium article seems to claim that there is a unique holomorphic solution in $$\Re(x)>-2$$. If true and sourced, it should be here.  This does not (obviously (to me, anyway)) contradict my proof above that the real-analytic solution cannot be unique, as it requires properties of the function at points of arbitrarily large imaginary part. — Arthur Rubin  (talk) 21:34, 9 November 2008 (UTC)


 * This is already mentioned, in the section "Extension to complex heights". The CZ article doesn't have any formulas either. mike4ty4 (talk) 08:14, 17 November 2008 (UTC)

Decremented tetration
I am speculating about another criteria, which may make fractional iteration unique. Assume b = sqrt(2), t2=2, t4=4 such that b = t2^(1/t2) = t4^(1/t4). Assume also Tb(x) = b^x and Tb(x,h) the h'th iterate and also the "decremented tetration" Ut(x) = t^x - 1 and Ut(x,h) the h'th iterate. Then powertowers T_b(x,h) of b can be expressed by "decremented powertowers" T_b(x,h) = (U_t2(x/t2-1,h)+1)*t2 = (U_t4(x/t4-1,h)+1)*t4 at least for integer heights. This construction relates tetrational functions of different bases with each other and introduces then new requirements for the fractional iteration as well. Although this restriction occurs yet only for "decremented" tetration, I think this idea has some importance and may be thought further... —Preceding unsigned comment added by Druseltal2005 (talk • contribs) Gotti 17:50, 13 September 2008 (UTC)
 * I don't think this is correct. Do you wish to further comment?  — Arthur Rubin  (talk) 21:34, 9 November 2008 (UTC)

?? Please, what do you mean is not correct? The formula? The idea to derive from it some requirements/restrictions?

--Gotti 19:15, 1 February 2009 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs)


 * Yes, the formula seems incorrect, as well as not being obviously related to the tetration concept(s) described here. — Arthur Rubin  (talk) 06:50, 6 February 2009 (UTC)

@Formula. Assume T-tetration T°0(x) = x, T°1(x)= b^x, T°2(x)=b^b^x,... Assume U-tetration U°0(x) = x, U°1(x)= t^x -1, U°2(x) = t^(t^x -1)-1, ... Use for an example a suitable base b = sqrt(2) = 2^(1/2). Then t = 2, u = log(2), so b=t^(1/t) = exp(u/t).

Write '-operation as x' = x/t-1 and the inverse "-operation as x"=(x+1)*t , so (x')" = x

Then T-tetration and U-tetration are connected in the following way:

T(x) = b^x = (t^(1/t))^x = t^(x/t) = t^(x/t-1)*t = (t^x' )*t = ( (t^x'-1) +1)*t = (t^x'-1)"        = U(x')"

T°2(x) = b^b^x = b^(U(x')")          = t^(U(x')"/t) = t^(U(x')+1) = t^(U(x'))*t = ((t^(U(x'))-1)+1)*t = ( U(U(x'))+1)*t = U(U(x'))"          =  U°2(x')"

Generally for any integer height

T°h(x) = U°h(x')"

where the base for U(x) is t and for T(x) is b=t^(1/t) The importance of this is, that U-tetration (or dxp or "decremented exponentiation") can be "regularly" iterated to fractional heights (powerseries has no constant term), for instance using a triangular matrix and its (well defined) fractional powers. This is then exploited as an easy interpolation-possibility for fractional iterates for T itself.

Now if b=sqrt(2), then the base for the according U-tetration can be either t= 2 or t=4. It was said, that the different shifting using t=2 and t=4 were numerically different when *fractional* iterates are computed, although for *integer* iterates the above transfer works well.

For the relation to the tetration-concept here set x=1 (keep the powerseries in terms of x!) and have actually "b^b" for "b^b^x"

--Gotti 22:01, 13 February 2009 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs) --Gotti 22:01, 13 February 2009 (UTC)


 * Seems to work for b < e1/e, where there are real fixed points available for T, and hence U — although the fact that t=2 and t=4 seem to produce different "natural" values for fractional iterates suggests that this problem is even more difficult than it appears.... — Arthur Rubin  (talk) 22:33, 14 February 2009 (UTC)

Yes, I agree. My intention was: there is also said, that there are many functions possible - for the same (fractional) iterate: just add one periodic function with half-cycle 1 with an appropriate small amplitude. This idea is used to question the uniqueness of a solution. Now: if fixpoint-shifting using 2 and fixpoint-shifting using 4 give different functions - why not try and combine both aspects (add such a periodic function) such that the resulting functions are equal for both fixpoints? (Though I've to admit, that I've not even an example...) --Gotti 07:36, 15 February 2009 (UTC) —Preceding unsigned comment added by Druseltal2005 (talk • contribs)

Decimal digits of power towers
I would like some feedback on whether adequate sources can be found to justify having a section of the article about the following interesting fact:

Given any positive integers b and d, all towers b$$\uparrow\uparrow$$n with integer n &gt; d + 1 have the same d rightmost decimal digits. The following is an extremely simple algorithm that produces these digits, assuming b is not divisible by ten (if b is divisible by ten, then the d digits are all 0):

Let x0 = b, and compute xi = bxi - 1 mod 10d (i = 1, 2, 3, ...), stopping when xi = xi - 1 mod 10d.

When the algorithm stops, xi will be the required d-digit string (in base-10, omitting any leading 0s), and the value of i will be the height of the shortest "base b" tower that has these d rightmost digits.

E.g., here are the ten rightmost digits of each sufficiently tall tower with b = 1,2,...,10, respectively &mdash; a tower is sufficiently tall if its height is equal to or greater than the indicated height h:

b  b^^x, x &ge; h    h -- -  -- 1      0000000001  1   (with leading 0s inserted) 2  ...3432948736  12 3   ...2464195387  11 4   ...0411728896  11 5   ...8408203125  4 6   ...7447238656  9 7   ...1565172343  6 8   ...6895225856  11 9   ...7392745289  10 10  ...0000000000  2

The algorithm and the "interesting fact" in italics above are described in (or, rather, I deduced them from) the discussion on this page. I'm not sure whether that's an adequate source to justify putting these conclusions into the article, though. --r.e.s. (talk) 08:02, 13 September 2008 (UTC)


 * That's certainly not an adequate source. However, if there were a published article, it could be included, I would think.  — Arthur Rubin  (talk) 21:44, 9 November 2008 (UTC)

I noted the same phenomena in a sci.math.research posting [http://groups.google.com/group/sci.math.research/browse_thread/thread/7823e4a156fa08ac Recurring digits in tetration and the Ackermann function], but soon afterward discovered that this phenomena is tied to the base one is working in. The digits reoccur in base 10 and base 6 for example but don't reoccur in base 11. Daniel Geisler (talk) 01:13, 10 February 2009 (UTC)

Derivatives
Can CX, XC, or XX be differentiated (or anti-differentiated) and if so how? srn347 —Preceding unsigned comment added by 68.7.25.121 (talk • contribs) 00:17, November 6, 2008


 * I don't see it. All I can see that you can get is by formal differentiation of the functional equation.  — Arthur Rubin  (talk) 21:40, 9 November 2008 (UTC)


 * For C=e, the evaluation of F(X)=XC and its derivatives is described in reference cited in the article, http://www.ams.org/mcom/0000-000-00/S0025-5718-09-02188-7/home.html  If can be generalized for  C>1. dima (talk) 15:37, 15 February 2009 (UTC)

Naming?
Hi.

I saw this:

"The term super-exponentiation is the most proper candidate for a name, however, Bromer published his paper Superexponentiation in 1987, and Goodstein published his paper Transfinite Ordinals in Recursive Number Theory (which coined the term tetration) in 1947, which predates Bromer. So although this is not a misnomer, the shorter and older term has gained more use."

But how is that the most proper candidate? "Super-exponential" could just refer to anything that grows faster than exponential. So it is not specific. It could be a double-exponential. It could be pentation. It could be the tetrasquare (x^x). It could even be the generalized factorial function GAMMA(x). "Tetration", and "tetrational", however, refer only to one thing. mike4ty4 (talk) 09:15, 25 January 2009 (UTC)

2^^4
It does equal 256, if not, tetration isn't hyper 4 at all. Let’s look at addition (hyper 1). 9+4=13, 9+3=12 and 9+1=10. 7+5=12, 7+2=9 and 7+3=10. 13-12=1 and 12-9=3, so (x+y)-(x+z)=y-z.

Now let’s look at multiplication (hyper 2). 8 × 6=48 right, and 8×4=32. Also, 8×2=16. I'll make up some more; say 5×7=35, 5×6=30 and 5×1=5. 48-36=16, so 8×6-8×4=8×2. 35-30=5, so 5×7-5×6=5×1. From this it's clear that (x×y)-(x×z)=x×(y-z).

Now let’s look at exponentiation‎ (hyper 3). 4^5=1024, 4^2=16 and 4^3=64. 3^6=726, 3^5=243 and 3^1=3. 1024÷16=64, so 4^5÷4^2=4^3. 726÷243=3 so 3^6÷3^5=3^1. (x^y) ÷ (x^z)=x^(y-z).

One would expect then that with tetration (hyper 3) (x^^z)√(x^^y)=x^^(y-z), and if (x^^z)√(x^^y)≠x^^(y-z) something’s wrong. If 256=2^^4 (and 4=2^^2) then the 4th root of 256 should be 4, which it is, but including to this article 256 doesn't equal 2^^4 at all, which to me makes no sense whatsoever. Wouldn't the logical thing to do be to do what every calculator in the world does already, and do tetration from left to right? Robo37 (talk) 10:09, 13 June 2009 (UTC)


 * No, you wouldn't, as exponentiation is not associative, so one has to decide whether 2^^4 is 2^(2^(2^2))) (normal convention) or (((2^2)^2)^2), which would lead to a^^b being a^(a^(b-1)), not normally considered sufficiently "hyper". — Arthur Rubin  (talk) 15:58, 13 June 2009 (UTC)


 * Can you explain in more detail? Why would it lead to a^^b being a^(a^(b-1))? You do everything else from left to right even when it does make a difference like with division and subtraction for example. Doing it from right to left would also make it easier as going from x^^y to x^^(y+1) would only involve putting x^^y to the power of x just like going from x^y to x^(y+1) only involves multiplying x^y by x and going from x×y to x×(y+1) only involves adding x to x×y.Robo37 (talk) 16:27, 13 June 2009 (UTC)


 * I'm on my way out, and may not get back to this for a day or so, but you can show by induction on b that (...((a^a)^a)...)^a) is a^(a^(b-1)), by induction on b, and using (x^y)^z = x^(y*z). See the section #Iterated powers in the article.
 * For other situations where it makes a difference, see, for example a paper by Donner(?) and Tarksi entitled Extended Operations and Relations on the Ordinal Numbers (or something like that), c. 1962. There, addition is O0, multiplication is O1, exponentiation is (except for some intial values) is O2, and something like tetration is O4. — Arthur Rubin  (talk) 16:42, 13 June 2009 (UTC)
 * — Arthur Rubin (talk) 19:34, 23 June 2009 (UTC)
 * — Arthur Rubin (talk) 19:34, 23 June 2009 (UTC)