Talk:Exponentiation

10.1 Exponentiation in Abstract Algebra

From my abstract algebra knowledge of a group, there is no operation called division. Why is it shown here? The proper way to write the inverse is just x^-1 never 1/x, which is not meaningful in an algebraic group. Perhaps this should better be explained in the page, and remove references to 1/x. In general math, yes, 1/x is certainly a valid way to write x^-1, but not in abstract algebra. Conceptually we all handle it the same way, convert 1/x to x^-1 but 1/x is not the operation 1 divided by x. It is the multiplicative inverse of x, that's all. In the group Z3, under multiplication, there is no element 1/2, since there are only elements 0,1,2. However, 1/2 means 2^-1 which is defined, and ultimately 2^-1 = 2. This is easier to explain than the nonsense that results by using the operation division, which would be 1/2 = 2... 24.34.198.111 00:27, 27 April 2007 (UTC)

log (a^b) = b log(a) valid or not valid?

Sorry, I had to correct my question, I have made several mistakes when I wrote it before.

What I meant was:

In the computation of power for two complex numbers a and b as:
a^b
it's used that:
a^b=e^{b log(a)}

But then, in the section Failure of power and logarithm identities, it is said that the identity log(a^b)=b log(a) do not hold in general, and it's prooven with an example.

How is it possible then to equal a^b=e^{b log(a)} , since you need to use the property log(a^b)=b log(a) to write it in that way? What makes that true in this case if a and b are any complex numbers?

Kaexar 01:50, 19 July 2007 (UTC)

What the article says is that $\log(a^{b})$ may be different than $b\log(a)$. For example if a is -i and b is 2, we have $(-i)^{2}=-1$ and $e^{2\log(-i)}=e^{2(-i\pi /2)}=e^{-i\pi }=-1$, as expected. The identity that fails is when you try to take logarithms using principal values, because -iπ is not a legal principal value. — Carl (CBM · talk) 18:32, 18 July 2007 (UTC)

I thank you for your answer but unfortunatelly a parsing error appeared and made half of your post unreadable by the browser :(

But you made me notice that I wrote my question wrong. Thanks again.

Kaexar 01:50, 19 July 2007 (UTC)

Sorry, I have fixed my error. You are right that if you already knew what a^b was then you would need something to rewrite it as e^{b log(a)}. But in the context of complex numbers, this identity is used as the definition of a^b, after the exponential function is defined (via a power series). — Carl (CBM · talk) 01:59, 19 July 2007 (UTC)

power series of e^x

Makholm and CBM. Note that the power series of e^x is not used in this elementary article at all. Nor is the differential equation or the continued fraction. All that stuff is taken care of in the article exponential function. This article on exponentiation is aimed at readers who do not know about exponentiation already. Superfluous material is likely to confuse the reader and make him stop reading. Holes in the argumentation likewise. The power series is a hindrance for reading and understanding. The limit without explanation is a hindrance for reading and understanding. Not everything that is true should be written everywhere. It should serve a purpose. Bo Jacoby 17:16, 8 August 2007 (UTC).

The limit definition of e^x is not used either, except to define e^x. I agree the continued fraction definition isn't needed, but the power series definition is quite important. You will need to justify why you think that the mere inclusion of a formula hinders understanding. Moreover, the "derivation" seemed to add very little to the article. Checking that the derivation is correct would be beyond the ability of the average reader of this article (especially since it used the uncommon notation $\lim _{|n|\to \infty }$ without defining it). We can just give the most important formulas for e^x without "proving" that they are correct. — Carl (CBM · talk) 12:13, 9 August 2007 (UTC)

The limit definition of e^x is used to show definition compatibility. The power series is important, but not here. To many readers a formula is a hindrance, and an unnecessary formula is an unnecessary hindrance. If e^x is defined by the power series the reader will think:

"The most honorable wikipedia editor first defined

e²_{old definition} = e · e

where

e = 1+1+1/2+1/6+1/24+1/120+1/740+...,

and later

e²_{new definition} = 1+2+2+4/3+2/3+4/45+8/315+...

I don't see whether these two definitions provide the same result. I must commit seppuku".

If, on the other hand, e^x is defined by the limit the reader will think:

"The most honorable wikipedia editor first defined

e²_{old definition} = e · e

where

e = lim(1+1/n)ⁿ,

and later

e²_{new definition} = lim(1+2/n)ⁿ = lim(1+2/(2·m))^(2·m) = (lim(1+1/m)^m)² = (lim(1+1/n)ⁿ)² = e² _{old definition} .

I see that the two definitions provide exactly the same result. I'm a genius".

To the service of the reader we should stick to the limit definition. You are welcome to explan |n|→ ∞.

Bo Jacoby 14:15, 9 August 2007 (UTC).

The limit definition is not that much simpler than the power series definition, and the limit calculations are not as elementary as you claim. Readers who don't want to see formulas at all should avoid reading mathematics - we don't need to pander to people who stop reading the first time they come to a formula. Now that the derivation involving the limit is gone, we can replace |n| with the more standard n. — Carl (CBM · talk) 19:05, 10 August 2007 (UTC)

The limit definition is not simpler than the power series definition, but the definition compatibility is much easier to see using the limit definition than using the power series definition. Do you understand the argument above? Of course people reading mathematics should read formulas, but the formulas should serve a purpose, and the power series formula serves no purpose in this context. Why do you want to make these changes, Carl? Bo Jacoby 22:43, 10 August 2007 (UTC).

There is no need to demonstrate the definition's compatibility in this context - we can just assert it. The justification that used to be here for the limit definition was far from accessible to the average person - it required a good understanding of real analysis to transform it into a complete argument. The purpose for including the formulas in this article is to have them accessible if a reader comes to this article to look them up. — Carl (CBM · talk) 22:51, 11 August 2007 (UTC)

By now the article does not even assert compatibility between the two definitions. The main article on exponential function does not explain the compatibility either. The reader is left clueless. The number e is not explicitely defined, and the two formulas for e^x are neither shown to be equal to one another nor to be equal to the old definition of e^x for integer x. So the subsection is useless to the beginner and comprehensible only to the reader to whom it is superfluous. The important formula e^x = lim_n(1−x/n)⁻ⁿ, which before was included as e^x = lim_|n|(1+x/n)ⁿ, has now been omitted, and it is not found in exponential function. The proof for definition compatibility does not really require advanced real (or complex) analysis, because if convergence fails no alternative definition is offered and so no incompatibility issue occurs. You are welcome to include convergence arguments in exponential function, but exponentiation should be kept elementary. Your present edit is a backwards step. I do not question your good faith, but you should take more care. Bo Jacoby 07:59, 13 August 2007 (UTC).

Why do you think that this article "should be kept elementary"? There are no policies and guidelines that imply so. On the contrary, our article about "exponentiation" should say everything that an encyclopedia could be expected to tell about "exponentiation", even if some of it will not be understood by everybody who understands some of the article. We group content by subject, not by levels of sophistication, and there is no requirement that every reader must understand either all of an article or nothing in the article. To keep demanding that is just a wrong editing strategy. –Henning Makholm 09:14, 13 August 2007 (UTC)

See Wikipedia:Make_technical_articles_accessible. Bo Jacoby 06:38, 15 August 2007 (UTC).

That manual of style page reminds us "Do not "dumb-down" the article in order to make it more accessible. " The issue here is not whether to make the article accessible - I'm sure we all have that goal. The issue is more subtle than that. — Carl (CBM · talk) 11:41, 15 August 2007 (UTC)

The name of the article exponentiation is attractive to beginners, (unlike for example Quantum gravity which is more scary), and the article links to the main article, exponential function for readers who want details. The purpose of introducing the exponential function e^x in this article is to prepare the way for the exponentiation a^x. Bo Jacoby 11:30, 13 August 2007 (UTC).

Sorry, it is simply not valid to reason from "the article's title is attractive to beginners" to "the article must not say aynthing except what beginners can understand". That is not how an encyclopedia ought to be organized. The article about exponentiation should tell everything relevant about its subject, not be cut off at some arbitrary level of sophistication. –Henning Makholm 00:14, 16 August 2007 (UTC)

I agree. But the power series for e^x is not relevant for the subject exponentiation. We have exponential function for that. The subsection on integer powers of e should define e and show that the two definitions of e^x for integer x produce the same results. Bo Jacoby 07:53, 16 August 2007 (UTC).

There is a a second reason for defining the exponential function, and that is for defining complex exponentiation. In the context of complex analysis, the power series definition of the exponential function is the important one. — Carl (CBM · talk) 12:52, 13 August 2007 (UTC)

In the context of complex analysis the limit definition e^z = lim_n(1+z/n)ⁿ, is equally important, and the geometrical interpretation of e^i·x = lim_n(1+i·x/n)ⁿ is more straightforward than the geometrical interpretation of e^i·x = Σ_n(i·x)ⁿ/n! Bo Jacoby 19:24, 13 August 2007 (UTC). Note also that the subsection you changed, Exponentiation#Powers_of_e, is part of the section Exponentiation#Exponentiation_with_integer_exponents, where it used to belong but no longer belongs. Please clean it up. Bo Jacoby 06:17, 15 August 2007 (UTC).

I remember that that material was placed with complex exponentiation originally, which I thought made more sense, but you moved it to its current location. The original layout of the article had the first section just on integer exponents of integers. — Carl (CBM · talk) 11:41, 15 August 2007 (UTC)

Do you disagree on the present general layout: integer exponents - powers of positive numbers - powers of complex numbers ? Bo Jacoby 17:06, 15 August 2007 (UTC). By the way, I did not move the power series to the current location, but I removed it, as it is found in many places elsewhere, in e (mathematical constant) and in exponential function and in Taylor's series, and as it is not used in this article. But I didn't find the proof, that the two definitions of eⁿ are compatible, elsewhere in WP, so I included it, and you removed it. I still believe that readers are puzzled by a new definition, without a word of explanation, of a concept that has just been defined. Don't you? Bo Jacoby 07:44, 17 August 2007 (UTC).

If we give two definitions and just say "they're equivalent", readers will believe us. If they wonder why they are equivalent, they can look at Characterizations of the exponential function, linked from exponential function. — Carl (CBM · talk) 01:55, 20 August 2007 (UTC)

Exponentiation has defined a^k for integer k and now defines e^k for integer k without saying (or showing): "they're equivalent" . It's not about lim_n(1+x/n)ⁿ = Σ_nxⁿ/n! for real or complex x, but about lim_n(1+k/n)ⁿ = (lim_n(1+1/n)ⁿ)^k = e^k for integer k . Characterizations of the exponential function does not show it. Nor does e (mathematical constant). Bo Jacoby 06:31, 20 August 2007 (UTC).

You have your priorities reversed. In the context of complex exponentiation, one first defines e^x, and then defines e as e^1. One does not first define e and then define e^x from it. The point is that e^x is very straightforward to define and prove analytic, unlike 2^x or π^x. — Carl (CBM · talk) 14:04, 20 August 2007 (UTC)

One can define a function exp(x) . Without saying (or showing) that exp(1)^k = exp(k) for integer k, the notation e^x is unjustified, and it makes no sense to the reader to include the exponential function in exponentiation . Bo Jacoby 09:09, 21 August 2007 (UTC).

Principal n^th root

In " If a is a real number, and n is a positive integer, then the unique real solution with the same sign as a to the equation \ x^n = a is called the principal n^th root of a, and is denoted \sqrt[n]{a}."

When you say "the unique solution" is it not implicit that there always is one? (not the case!!) And why use the "the same sign"? (what about a=0??) Wouldn't it be more clear/didactic to explain what happens dividing in cases, the n's into even or odd and the a's into positive, 0, and negative?

Why use the term "principal n^th root"? It only makes real sense after someone studied complex numbers, which still takes a few years for most students. Ricardo sandoval 04:28, 24 August 2007 (UTC)

The principal nth root is unique. The nth root is not always unique, even when operand and results are real numbers. The sentence would not be correct without "principal". Principal does not refer to complex numbers. See n-th root or square root. About "with the same sign as a", in the case that a=0, n=0, and the convention 0⁰ = 1 is adopted, you are right... In all the other cases, the expression "with the same sign as a" is correct. The special case can be briefly described later as an exception (but this should be done also in the article about n-th root).

It is indispensable to explain in this article the principal nth root (because it introduces the rational powers). It is not indispensable to explain the nth root. Before my 21 August edit, the authors only described the simplest case of principal square root, with positive a, n and x, and they did neither call this "pricipal n-th root" nor "one of the two possible n-th roots", but simply and incorrectly "the n-th root".

However, there's a separate article about n-th root, and the internal link is now clearly specified on top of the section... Paolo.dL 09:53, 24 August 2007 (UTC)

I was pointing to the absence of principal even root of negative numbers as in n-th root. So the term "the unique" doesn't apply. Someone could define the principal nth root for these number but they would not be real anyway.

I see the text is an improvement over the previous one (to which you referred) and it is even better now. But there is still the problem above. When remarking about the nth root of zero, I was pointing to the more trivial fact that zero has no sign, in the usual sense ( I am not considering the sgn function as a definition of sign). Although someone should expect the principal nth root of someone "without sign" (0) should also be "without sign" (0), this is not stated explicitly. Ricardo sandoval 03:43, 25 August 2007 (UTC)

Well, the sentence about the n-th root of negative numbers (copied from the lead of the article n-th root), was already included at the end of the section. About the n-th root of zero, it's a very special case, and it is easily understandable that "zero has the same sign as zero", even if zero has no sign. This for sure will not create misunderstandings or ambiguity. I don't see the need to say it explicitly here, in a section that is a summary of a clearly specified "main article", which the reader can easily check in case of doubt. Paolo.dL 10:17, 25 August 2007 (UTC)

In the section Rational powers of positive real numbers it is implicit that the result is also positive, i.e that the negative root should not be taken, or the principal value should be used. I think that should be made explicit. —Preceding unsigned comment added by TerryM--re (talk • contribs) 00:11, 4 November 2007 (UTC)

GA on hold

This article has been reviewed as part of Wikipedia:WikiProject Good articles/Project quality task force. In reviewing the article against the Good article criteria, I have found there are some issues that need to be addressed. The article is under-referenced. Also, it's not specific enough for stating references in general. I am giving seven days for improvements to be made. If issues are addressed, the article will remain listed as a Good article. Otherwise, it will be delisted. If improved after it has been delisted, it may be nominated at WP:GAC. Feel free to drop a message on my talk page if you have any questions. Regards, OhanaUnited^{Talk page} 21:32, 8 September 2007 (UTC)

Whoever can take care of the images that display equations, check out the first "Justification" under the heading "Zero to the zero power." It claims X^X -> 1, where it should be X^0 NinjaSkitch 06:06, 1 October 2007 (UTC)

"a to the power n"?

I realize that "a to the power n" is correct terminology, but there is another correct notation that could be included there, "a to the exponent n". My math teacher tells us that that is the correct form, and that "a to the power n" is incorrect. However, I don't believe her, and I think both are correct. Is my math teacher wrong, or is she right? ZtObOr 01:42, 9 October 2007 (UTC)

EDIT: I forgot to include this bit. My math teacher's reasoning is that since the "n" in "a to the power n" is called the exponent, that it should really be called "a to the exponent n", since the word "exponent" refers to the exponent "n" in the first place, even if the word "power" is in its place.. —Preceding unsigned comment added by Ztobor (talk • contribs) 01:45, 9 October 2007 (UTC) (Edited 4/10/2008)

It's better not to argue with your math teacher. — Carl (CBM · talk) 02:34, 9 October 2007 (UTC)

Of course, if my math teacher is indeed wrong, then it should be okay to argue with him. (I'm in a different grade now, and we have a male math teacher now.) ZtObOr 02:04, 5 October 2008 (UTC)

Perhaps it is too pedantic to say that "a to the power n" is not a notation; it is a verbal description, or terminology. The notation is $a^{n}$. As to the answer to your question: $n$ is the exponent, as your teacher rightly says. It is also the power. The exponent is the symbol that we use to designate a power. Therefore I prefer to say "a to the power n" as it doesn't confuse the thing with the symbol that we use to describe it. This is rather like the distinction between a number and a numeral. Of course you may choose to finesse the problem by saying "a to the n".TerryM--re 00:01, 4 November 2007 (UTC) I should add that the term index is also almost synonymous with exponent. In fact now that I think about it, I'm not sure that the terms are well defined; they are certainly used almost interchangeably by many people.TerryM--re 00:31, 4 November 2007 (UTC)

I don't get your saying n is both the "exponent" and the "power". If n is the exponent, doesn't that preclude it from being the entire expression?

But you do have a point. I say "a to the power n" because the word "power" refers to the entire expression, not just "n". If I were to say "a to the exponent n", that would mean that aⁿ is the exponent, and not just n. ZtObOr 01:15, 6 October 2008 (UTC)

In most mathematical contexts, "index" means the number that identifies one of several variables that have been given names from a numbered sequence. An index is often notated as a small lowered number following the variable letter, whereas an exponent is notated as a small raised numbers. For example, $a_{3}x^{5}$ means a-with-the-index-3 multiplied by x to the fifth power. In tensor calculus some indices are written as raised numbers, rather than lowered ones, but that is a specialty usage. –Henning Makholm 01:58, 4 November 2007 (UTC)

The index can also refer to the "index" of a radical, or the exponent by which you reverse the operation by. ZtObOr 01:56, 5 October 2008 (UTC)

I have never seen the phrase "a to the n-th" ever used in any literature, nor have I every heard this phrase being used. If one says "a to the n-th" without "power" following, what does this mean? For example, if I say, "a to the fifth," do I mean a⁵ or a^1/5? European and American usage differ here, and "a to the n-th" is, to my mind, ambiguous. Xantharius (talk) 16:03, 6 October 2008 (UTC)

Seems I was wrong then. I didn't realize that "a to the nth" could also mean a^1/n. But usually you'd say "a to the 1-nth" if you were referring to that, no? ZtObOr 02:29, 19 October 2008 (UTC)

I'm British, where saying "a to the n-th" would really have no meaning, or at least would not be very clear. But I teach in a US college, where students would probably know what I meant if I said "a to the third," for example, to signify a³. I continually point out to my students that this is ambiguous: did you mean a³ or a^1/3? So, perhaps, one might say "a to the 1-nth" to mean the fractional version, but it actually took me a while to realize what you meant in the abstract case of 1/n. I think omitting this phrasing, for the sake of clarity, and readers beyond the US, is probably best, with the more straightforward "a to the n-th power" meaning aⁿ and not leaving out the word "power". Xantharius (talk) 04:46, 19 October 2008 (UTC)

For what it's worth, I think there are a lot of people (including myself) who pronounce x⁵ as "x to the fifth", and I don't remember hearing anyone say this when they mean x^1/5. I don't see why "x to the fifth" would be any more ambiguous than "x to the fifth power". On the other hand, when the exponent is a variable, I pronounce xⁿ as "x to the n". --FactSpewer (talk) 05:39, 19 October 2008 (UTC)

Notation: cis (x) = cos (x) + i * sin(x) = e^(ix)

Is there any article to link to, to mention the cis function? This function can make working with Euler and De Moivre's formulas a lot easier. It is clearer to write on paper and leads to less mistakes, not to mention faster. An online example of cis, although not specific: http://oakroadsystems.com/twt/sumdiff.htm

Also, a link to an article describing implementing complex exponention in software: http://www.efg2.com/Lab/Mathematics/Complex/Numbers.htm The code is in Pascal/Delphi. JWhiteheadcc 06:05, 2 December 2007 (UTC)

I do not think the name "cis" is standard or widespread, except perhaps as a didactic device used when teaching the complex exponential to beginning students. The function itself is often needed in many applications (quantum physics, AC electronics, higher algebra, to name a few), but there it is invariably written e^ix which is shorter and eliminates the need to remember a different notation for the special case of x being real. Euler's formula is already in the article, but perhaps deserves to be displayed. –Henning Makholm 12:25, 5 December 2007 (UTC)

I have the impression that 'cis' is more common in engineering. — Carl (CBM · talk) 13:43, 5 December 2007 (UTC)

OK; I don't have many engineering texts to compare with. –Henning Makholm 17:41, 5 December 2007 (UTC)

Powers with infinity

This section, when I came upon it, claimed that Cantor's theorem implies that $\infty <2^{\infty }$ in the context of calculus. This is nonsense. If $2^{\infty }$ is interpreted as the limit of 2^g(x) as x goes to infinity (where g(x) is a real valued function which goes to infinity), then this $2^{\infty }$ is just a limit of real numbers and not some kind of cardinal which is "larger" than another infinite limit of real numbers. Not larger in the sense of Cantor's theorem, anyway. This $\infty$ is not a cardinal.

So I fixed up the section so that it at least says correct things instead of nonsense, but I don't think it's a great section. I wouldn't mind if someone just deleted it, but then someone else would probably repost the same misconceptions later on.

By the way, I don't know where on earth I could find citations for this kind of thing. It's clear to any mathematician familiar with the relevant definitions. —Preceding unsigned comment added by 70.245.113.97 (talk) 01:43, 11 December 2007 (UTC)

Dismissing Cantor's theorem as nonsense is itself nonsense. The text as it was, was a necessary counterbalance to the way people are flinging around infinity in the section. If you want to say it makes no sense from the POV of calculus then say it, don't delete the fact that it makes complete sense from a set POV. --Michael C. Price ^talk 06:43, 28 August 2008 (UTC)

Michael, really, that text doesn't make sense, and I don't think 70.245 was claiming there was anything nonsensical about the theorem, just about its application here, which genuinely is out of place. The symbol ∞ is not used for transfinite cardinals. I'm going to assert that flatly; it is simply not used that way, at all, in set-theoretic publications. (It is used, occasionally, for the notional "cardinality" of the collection of all ordinals, but that's so far from being a reasonable reading of the symbol in the current context that it doesn't even come up here.)

The text cannot stand as it is. My only doubt is whether to remove that paragraph alone, or to cut deeper, maybe deferring the whole section to extended real number line, which is where this discussion really belongs. --Trovatore (talk) 08:11, 28 August 2008 (UTC)

Then perhaps there should be a section about exponentiating transfinites or alephs or whatever. Eiether way the subjects should be mentioned somewhere in this article. --Michael C. Price ^talk 08:43, 28 August 2008 (UTC)

There's already such a section: Exponentiation of cardinal and ordinal numbers. --Zundark (talk) 11:54, 28 August 2008 (UTC)

Which nowhere mentions Cantor's theorem, nor transfinites, aleph etc. If the article is really going to have misleading statements such as 7^∞ = ∞ in it, then let's see some citations.... --Michael C. Price ^talk 12:14, 28 August 2008 (UTC)

The section on exponentiation of cardinal and ordinal numbers doesn't mention Cantor's Theorem, but it's where Cantor's Theorem should be mentioned if this article is going to mention it at all. The paragraph on cardinal exponentiation contains little more than a definition, and the definition doesn't depend on whether the cardinals are finite or infinite (= transfinite = an aleph). More details are in the Cardinal exponentiation section of the cardinal number article. As for 7^∞ = ∞ (which is apparently intended as a shorthand for 7ⁿ → ∞ as n → ∞), I wouldn't object if someone simply removed the entire "Powers with infinity" section. --Zundark (talk) 12:47, 28 August 2008 (UTC)

I'm inclined to agree.--Michael C. Price ^talk 14:26, 28 August 2008 (UTC)

I agree too; things like 7^∞ are just poor notation, as is most of the rest in that section. Shreevatsa (talk) 15:47, 28 August 2008 (UTC)

Am I right in thinking that $n^{\infty }$ tends to $\infty$ when n>1, remains at 1 when n=1, tends to 0 when abs(n)<1, and tends to $\pm \infty$ when n<-1? Note: This being my first contribution, I don't want to make changes directly, in case of misinterpretation of the intent of the article! MessyBlob (talk) 18:48, 5 April 2008 (UTC)

What would be correct to say is that n^k tends to ∞ as an integer k tends to ∞ when n > 1; tends to 1 when n = 1; and tends to 0 when abs(n) < 1. The expression n^k cannot converge to ±∞ if n ≤ −1, because the notion of convergence means that it gets close to one value, not more than one. In this case it would therefore probably be okay to write n^∞ = ∞ if n > 1, that n^∞ = 0 if abs(n) < 1, that n^∞ = 1 if n = 1; but note that n^x might not even be a real number if n is negative, and I do not immediately see any reasonable way to define n^∞ if n ≤ −1, because the value of n^k oscillates between 1 and −1 if n = −1, and between increasingly divergent positive and negative numbers if n < −1. Xantharius (talk) 19:50, 5 April 2008 (UTC)

Agree that convergent and divergent series is the way to go on this. It might be useful to express the case of n = −1 in terms of a wave in the imaginary axis, using Euler's equations as basis. For n < −1, the same would apply, but tending to infinite amiplitude. You could probably reach intermediate values for n =−1 using a cosine function of phase, like $-1^{\phi }=\cos {\pi \phi }$. —Preceding unsigned comment added by MessyBlob (talk • contribs) 17:12, 6 April 2008 (UTC)

0 to power of 0

In this section, it is heavily implied that the Microsoft Calculator is a programming language, which is obviously wrong. Maybe it should be kept for simplicity or perhaps a rewording of the statement is needed. —Preceding unsigned comment added by 88.108.69.114 (talk) 19:37, 17 May 2008 (UTC)

I edited the text to fix this, and added reference to Microsoft Excel. Paolo.dL (talk) 16:06, 18 May 2008 (UTC)

the only way i see this = 1 is the ^0 is subtracting 1 less so 0^0 = 0-(-1)= 1 —Preceding unsigned comment added by 208.91.185.12 (talk) 21:31, 2 October 2008 (UTC)

0 to the power 0: Knuth

I'm surprised the 0⁰ section does not mention all that Knuth has said in his paper Two Notes on Notation. Can someone add it? The paper: TeX source, doi:10.2307/2325085, arXiv:math/9205211.

Just to follow up on this: The article now cites it (more precisely, it cites the published version instead of the preprint). --FactSpewer (talk) 01:58, 18 October 2008 (UTC)

Possible error in section 1.5

I'm not a math person, but I think I see an error in section 1.5 Identities and properties and would like someone help me check this out.

In the last line of the section the statement is a^b^c = a^(b^c), but the example preceding this seems to show just the opposite.

Then there is an inequality a^b^c ≠ (a^b)^c. Should this be an equality not an inequality?

And, finally, the article states "Without parentheses to modify the order of calculation, the order is usually understood to be from right to left". Should this be "from left to right"?

```` —Preceding unsigned comment added by Lou27182 (talk • contribs) 16:14, 27 July 2008 (UTC)

a<sup>bc</sup> means a<sup>(bc)</sup>, and (a<sup>b</sup>)<sup>c</sup> is written a<sup>bc</sup>. Bo Jacoby (talk) 16:32, 27 July 2008 (UTC).

0^0 = 1

The arguments for defining $0^{0}$ to be 1 are overwhelming even in analysis, and I feel that the article should be rewritten to reflect this. For instance, in addition to the analysis reasons already given (the power series for $e^{x}$ and the derivative of $x^{n}$), there is the $x=0$ case in the geometric series formula

${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{for }}|x|<1,$

itself a special case of the binomial formula for general exponents.

I feel that the arguments for leaving $0^{0}$ undefined stem from a misguided insistence that functions be continuous or differentiable on their domains. Here are rebuttals to the arguments currently in the article:

• The first reason given was essentially that the two-variable function $x^{y}$ on the region defined by $x,y\geq 0$ and $(x,y)\neq (0,0)$ would cease to be continuous on its domain if $0^{0}$ were defined. Would one similarly argue that the absolute value of 0 should be left undefined since defining it renders the absolute value function non-differentiable on its domain?
• The second reason given was that for complex numbers $z$, one defines $z^{z}$ by choosing a branch of $\log \,z$ and setting $z^{z}=e^{z\log \,z}$, but no branch of $\log \,z$ is defined at 0; it is not possible to define $z^{z}$ as a holomorphic function on an open neighborhood of 0. Would one similarly argue that since $z^{1/2}$ for complex $z$ is defined as $e^{{\frac {1}{2}}\log \,z}$, one should not define ${\sqrt {0}}$?

The best solution is to define $0^{0}=1$, and to accept that the function $x^{y}$ will be discontinuous at $(0,0)$ and that no holomorphic branch of $z^{z}$ will exist in a neighborhood of 0. Of course, defining $0^{0}=1$ is not the same as saying that $\lim f(x)^{g(x)}=1$ whenever $f(x)$ and $g(x)$ are tending to 0; the article should remain clear on this point.

BjornPoonen (talk) 05:31, 11 October 2008 (UTC)

0^0 has been the subject of much prior discussion see archive. Its not wikipedia to job to make the definition, mearly report on which sources define it or not. --Salix (talk): 06:42, 11 October 2008 (UTC)

I agree with BjornPoonen. Your objections are to the point. The quote: "it may be best to treat 0<sup>0</sup> as an ill-defined quantity" is unjustified and without references. Bo Jacoby (talk) 10:52, 11 October 2008 (UTC).

I fully agree with Salix here. The article deals with the problem of 0^0 fairly, it cites an authority saying the value taken depends on the circumstances, and that certainly has been my experience. It lists a number of times when 1 is a reasonable value and some others where it isn't. To blanket the lot saying '0^0 is 1' would simply be wrong. It isn't up to wikipedia to make up the rules. Dmcq (talk) 11:43, 11 October 2008 (UTC)

Thank you, Salix and Dmcq, for your comments, and in particular for pointing out the archive (even if it does not contain any arguments resembling my analogies with $|0|$ and ${\sqrt {0}}$). Your arguments that the article should continue to mention that some authors leave $0^{0}$ undefined are well taken, and I agree with you on this point. What I feel is unjustified is the implication that the definition $0^{0}=1$ is useful only in the discrete side of mathematics. I would suggest removing "There are two principal treatments in practice, one from discrete mathematics and the other from analysis." I would also suggest removing the blanket statement "In general, mathematical analysis treats $0^{0}$ as undefined...". Citing some textbooks that leave it undefined does not mean that this is the general approach in analysis, when other textbooks define $0^{0}=1$. BjornPoonen (talk) 17:26, 11 October 2008 (UTC)

I think the article is fine as it stands. It describes several contexts in which it is usual to define 0^0 to be 1 in order to avoid special cases; it describes other contexts in which it is normally left undefined. This is a balanced NPOV treatment. Gandalf61 (talk) 17:37, 11 October 2008 (UTC)

How about replacing the second section of "Zero to the zero power" which starts "In many settings..." by 'In most settings not involving continuity the value of 0^0 is defined to be 1, this simplifies the theorems and removes special cases. In analysis the form 0^0 is treated as an indeterminate form and the value of each such expression is defined if possible as a limit value so as to preserve continuity." Then just list where 0^0 is commonly defined as 1 and where it isn't without saying they are justifications for anything in general. Then there could be a subsection about controversy saying that some people want to define 0^0 to be always 1 even in analysis and cite Knuth or whoever for this point of view. I think this would make the whole business a bit simpler and clearer for beginners coming along. Dmcq (talk) 10:41, 12 October 2008 (UTC)

I agree; something along these lines would be not only simpler and clearer, but also more neutral than the current version, which seems to intersperse editorial opinions among the statements about the settings in which 0^0 is defined or not. Many of these statements themselves are non-controversial, even though different people draw different conclusions from them. BjornPoonen (talk) 15:10, 12 October 2008 (UTC)

I'm going to attempt an implementation of Dmcq's suggestion. FactSpewer (talk) 03:02, 13 October 2008 (UTC)

I think the suggestion is not too bad, though I would look for a way to make it clearer that "most settings not involving continuity" is not the same as "most settings" period. My worry is that it's too easy to skip the words "not involving continuity".

On another note, I misread the bit FactSpewer added about ordinal exponentiation; on rereading I agree it's correct. I still don't think it should be added at that spot, though; it's sort of too clever and insufficiently enlightening. Also I don't recall seeing it in the literature. --Trovatore (talk) 05:32, 13 October 2008 (UTC)

Actually I was not the one who added the definition of ordinal exponentiation! It predates my editing; I haven't bothered to go through the history to find out who added it! In fact, I agree with you that it is too clever. FactSpewer (talk) 17:46, 13 October 2008 (UTC)

Everyone please slow down!

Two new editors, FactSpewer and WardenWalk, have been making large revisions to the article without discussing them in detail. The volume of changes in a short time makes it difficult to check them all for correctness and appropriateness. Please, if you want to make a large revision to an established article, it is best to write it up somewhere else (say, a subpage of your user page) and point people there for discussion. --Trovatore (talk) 16:47, 13 October 2008 (UTC)

Your point is well taken; I should try harder not to live up to my username. I have followed your suggestion and put a link to a proposed version at User:FactSpewer, taking care to include not only my own edits, but to restore also your own minor edit, along with those of Warden Walk, and to include the subsequent edit of Xantharius, which I also agree with. I hope that you and the rest of the community can come to a consensus about these changes soon, before other edits are made to the Exponentiation page, so that these newer edits are not accidentally overwritten or forgotten. Suggestions for improvement are welcome. --FactSpewer (talk) 20:04, 13 October 2008 (UTC)

I took a few minutes to review this version from just before the revert. It looks somewhat OK to me; the tone is still adequately neutral to my taste. It might use a little trimming. I would move the introductory text back up to the top of the section; as it is the section starts too abruptly to my taste, without establishing the context for the bullets to come. — Carl (CBM · talk) 00:11, 14 October 2008 (UTC)

Trovatore also recommended changing the opening sentence to place in context what would be following it. His suggestion was incorporated, so the current proposal at the link at User:FactSpewer opens the section in a slightly different way than the version you referred to. You might look to see if you prefer that version, although it may still not go far enough in the direction you suggest. By "the introductory text" do you mean the first sentence of the "History of differing points of view" subsection together with the two bullets beneath it? Moving that to the top of the 0<sup>0</sup> section would not be hard to do, if this is what you are suggesting. But this would partially defeat the purpose of separating out the controversial statements, so I'm not sure I would favor it. --FactSpewer (talk) 04:33, 14 October 2008 (UTC)

Sorry, I wound up working really late and still had to work out afterwards. I'll trust Carl if he likes it. --Trovatore (talk) 07:38, 14 October 2008 (UTC)

It could still do with a bit of polishing but it looks much better than what's there at the moment. The second section seems to imply that the value of 0^0 is never set to a specific value other than 1. Some individual points still sound like justifications rather than anything enlightening. I noticed a few things elsewhere but not to do with the edits - I wonder who put in the bit about not needing to study trigonometry! Overall I'd be happy for it to go in as is, the structure looks good so little edits can do what's left Dmcq (talk) 08:21, 14 October 2008 (UTC)

As for "The second section seems to imply that the value of 0^0 is never set to a specific value other than 1": yes, this reflects the outside literature; all the texts I know that set it to a specific value set it to 1; the others leave it undefined and/or say that the value should be selected based on context. If someone can provide a scholarly reference with a serious case for setting it to a different particular value, this should be added, of course. --FactSpewer (talk) 12:47, 14 October 2008 (UTC)

I remember looking through several sources to work on the section before, and never found one where the value was anything other than 1 (except as a spurious result using some computer algebra systems, as someone else wrote in the lower section). — Carl (CBM · talk) 13:26, 14 October 2008 (UTC)

By the way, I noticed FactSpewer wrote "I distinguish the statement that 0^0 is an indeterminate form (in the context of limits) from the statement that 0^0 itself is undefined." on his/her talk page. That is a another viewpoint which I did not find in the literature on indeterminate forms (which is surprisingly limited). I did find two authors who directly state that 0^0 is not or may not be defined, however, and who post-date Knuth's 1992 paper. — Carl (CBM · talk) 13:32, 14 October 2008 (UTC)

The authors post-dating Knuth's paper are interesting if they refer to Knuth and argue against him. Otherwise they are merely repeating outdated points of view. Bo Jacoby (talk) 14:01, 14 October 2008 (UTC).

Regardless whether you personally feel the views are outdated, publications from the 1990s are equally contemporary with Knuth's paper. Another possibility is that these other authors didn't mention Knuth's argument because they simply didn't find that argument persuasive enough to spend ink on a reply. Who knows! The point here is to describe the situation with some degree of detachment. — Carl (CBM · talk) 15:23, 14 October 2008 (UTC)

I agree that this should not be a question about who published last. In any case, distinguishing "the statement 0^0 is an indeterminate form (in the context of limits) from the statement that 0^0 itself is undefined" is not something I invented; it is in the literature; namely, it is a paraphrase of Knuth (1992). Since there is controversy in the outside literature about whether "0^0 is indeterminate in the context of limits" implies that "0^0 is undefined", I refrained from stating this implication as fact in the draft, and instead state only the (hopefully) noncontroversial claim that "0^0 is an indeterminate form in the context of limits" in the main section. Only later in the "History of differing points of view" section do I write that some authors interpret this as saying that 0^0 should be left undefined. --FactSpewer (talk) 15:35, 14 October 2008 (UTC)

I think it's clear from the arguments in Knuth 1992 (p. 6) that he is concerned with having notation that lets him state things like the binomial theorem elegantly. So when he says "It has to be 1", I read this as "It has to be 1 in order to avoid inelegant statements of theorems I am interested in". For, someone who didn't like the convention that 0^0 = 1 can still accept the binomial theorem written in alternate notation. Similarly, the number of mappings from the empty set to itself has to be 1, but we don't have to denote it 0^0 (of course this is a standard notation for it). We could call it |Hom(0,0)| or something like that.

The next-to-last paragraph on p.6 seems to make a distinction between different meanings of 0^0; Knuth seems to agree that 0^0 is undefined as a "limiting form". The idea that some things are "less defined" than others seems muddy to me. — Carl (CBM · talk) 16:54, 14 October 2008 (UTC)

The article indeterminate form has some simple examples where 0^0 is 2, 1, or 0. Dmcq (talk) 18:24, 14 October 2008 (UTC)

In fact one can remove the lim+ bit by replacing x with x<sup>2</sup>. Dmcq (talk) 18:26, 14 October 2008 (UTC)

Every book and paper that defines 0<sup>0</sup> makes the definition that 0<sup>0</sup>=1, so there is no controversy as to how 0<sup>0</sup> is defined. Many books and papers do not define 0<sup>0</sup>. For example the bible and the qur'an do not define 0<sup>0</sup>. This does not imply that wikipedia must state the fact that some authors do not define 0<sup>0</sup>. Bo Jacoby (talk) 21:25, 14 October 2008 (UTC).

The Bible does not comment on the situation either way. Whereas in this situation, we have respectable sources that explicitly state that 0^0 is indeterminate. The article should remain as it stands; FactSpewer's version is not an improvement. —Steven G. Johnson (talk) 21:41, 14 October 2008 (UTC)

I really just am not going to start saying that (2<sup>-1/x2</sup>)<sup>x2</sup> is 1 when x is zero. That really would be quite abominable. Dmcq (talk) 22:26, 14 October 2008 (UTC)

I think it's only Bo Jacoby who is arguing for that sort of thing; the proposed version by FactSpewer doesn't make any sort of claim like that. Stevenj, could you comment in more detail? I was planning to copy over Factspewer's version and then edit it to be a little closer to the existing text. But I'll wait for your thoughts. — Carl (CBM · talk) 23:36, 14 October 2008 (UTC)

To Steven: The statement that " 0<sup>0</sup> is indeterminate " reflects the fact that the function x<sup>y</sup> cannot extended from {(x,y):x>0,y>0} to {(x,y):x≥0,y≥0} by continuity, but it does not imply that 0<sup>0</sup> is, or must remain, undefined. Nor does it define it of course. To Dmcq: the function (2<sup>−1/x2</sup>)<sup>x2</sup> is undefined for x=0 because 1/0 is undefined. The limit lim<sub>x→0</sub>((2<sup>−1/x2</sup>)<sup>x2</sup>) =2<sup>−1</sup> is defined for real x, but not for complex x, while (lim<sub>x→0</sub>(2<sup>−1/x2</sup>))<sup>lim<sub>x→0</sub>(x2)</sup> =1 for real x, but not for complex x. A discontinuity may be abominable, but it cannot always be helped. To Carl: I think you argue ad hominem. Bo Jacoby (talk) 13:10, 15 October 2008 (UTC).

Bo, you've been presented with several published references in which the authors state 0^0 is undefined. The article needs to reflect their opinion as well as Knuth's opinion that 0^0 "has to be 1". I think you are the only person here who has not expressed general agreement with that sentiment. — Carl (CBM · talk) 13:31, 15 October 2008 (UTC)

Carl, if anybody defines 0<sup>0</sup>, then it is defined, no matter how many books do not define it. Bo Jacoby (talk) 13:46, 15 October 2008 (UTC).

In case this is not already clear, I must say that I agree with Carl here: We need to acknowledge that respectable references state that 0^0 is undefined. By the same token, I think we (including Steven G. Johnson) need to acknowledge that other respectable references state that 0^0 = 1: not just Knuth, but also Bourbaki. Mentioning both possibilities as points of view that have been published is the point of the version posted at User:FactSpewer. (By the way, if Knuth and Bourbaki are not respectable, I don't know who is!) --FactSpewer (talk) 14:10, 15 October 2008 (UTC)

I have a few criticisms of the version of User:FactSpewer:

• It is strange to include the limit of x^x among the "settings not involving continuity". Also, it is a rather unconvincing justification for 0^0 = 1, and the situation is already much better explained in the second paragraph in the context of limits along different curves. So why not just remove this x^x item?
• Some trimming could be done (as Carl said). For instance, in the item about e^x, one could delete the second sentence. It is more or less just repeating what is in the first sentence (or at least it is an obvious consequence of the first sentence).

On the whole, however, the version is an improvement over what came before. I personally have more sympathy for the argument that 0^0 = 1, so if it were up to me I would write it more with this in mind, but after reading everyone's comments I understand that it is not so much an issue of what is "more correct" as it is an issue of accurately reflecting what published references say.BjornPoonen (talk) 15:14, 15 October 2008 (UTC)

2<sup>−1/x2</sup> with the value 0 at x = 0 is a perfectly well behaved continuous and differentiable function on the reals. What it does for complex numbers is irrelevant unless one is considering such functions. When dealing with Fourier series the step function always has a value halfway between the values on either side of the discontinuity. That doesn't mean that in all circumstances everywhere in mathematics that must hold otherwise something weird would have to happen when one multiplied a step function by itself. Dmcq (talk) 16:47, 15 October 2008 (UTC)

Yes, and? Bo Jacoby (talk) 16:59, 15 October 2008 (UTC).

"Undefined" means that it has never been defined. Definitions are not "undone". If an author treats x<sup>y</sup>, and does not define 0<sup>0</sup>, and is unaware that 0<sup>0</sup> has been defined elsewhere, then he will in good faith publish that 0<sup>0</sup> is undefined. But in this article, where 0<sup>0</sup> is defined, it is evidently untrue that it is undefined. The reader will become confused. Bo Jacoby (talk) 16:59, 15 October 2008 (UTC).

Er no. In mathematics, defined and undefined are used to explain whether or not expressions have meaningful, sensible, and unambiguous values. This is different to whether anyone has ever defined something. which 0^0=1 may be meaningful and sensible it is not unambiguous hence its status. Its a good thing we don't take the first definition of a mathematical entity else we might be stuck with π = 3. --Salix (talk): 17:14, 15 October 2008 (UTC)

I'll explain further about :2<sup>−1/x2</sup> with the value 0 at x = 0 raised to the power of x<sup>2</sup>. When x = 0 then this is of the form 0<sup>0</sup>, no ifs or buts about it. This is an indeterminate form and in the domain of continuous differentiable functions the value is defined by l'Hôpital's rule. And the value comes out as 1/2. This is not undefined. It is defined. And it is not 1. Any other handling of this would cause immense trouble to a whole swathe of mathematics never mind a lot of practical applications. The business about Fourier step function I was saying is the same. In its domain there is a clearly defined value which always works and is nice and clean just like the 0<sup>0</sup> = 1. And one can even define a step function that always acts like that anywhere. But one can't then go outside that domain and expect that everything that looks like a step function will act like the step function that was defined to be useful with Fourier series. Sometimes discontinuities have other values defined at the discontinuity. With exponentiation we'd have to use a different sign for exponentiation in analysis if the current sign and usage was changed to say that 0<sup>0</sup> was always 1. It simply would no longer be a very useful operation. Dmcq (talk)

To Salix. No expression can be defined and undefined at the same time. The approximation π = 3 was not abandoned to leave π undefined, but merely refined to make π known with better precision. To Dmcq. The function f<sub>1</sub> defined by f<sub>1</sub>(x)=2<sup>−1/x2</sup> has f<sub>1</sub>(0) = 2<sup>−1/02</sup> = 2<sup>−1/0</sup> which is undefined because you cannot divide by zero. The definition may be supplemented by defining the function f<sub>2</sub> by f<sub>2</sub>(x)=f<sub>1</sub>(x) for x≠0 and f<sub>2</sub>(0)=0. This function f<sub>2</sub> is continuous for x=0 because lim<sub>x→0</sub>(f<sub>2</sub>(x)) = 0 and f<sub>2</sub>(lim<sub>x→0</sub>x) = 0 so lim<sub>x→0</sub>(f<sub>2</sub>(x)) = f<sub>2</sub>(lim<sub>x→0</sub>x). Consider the function f<sub>3</sub> defined by f<sub>3</sub>(x)=(f<sub>2</sub>(x))<sup>x2</sup>. This function f<sub>3</sub> is discontinuous for x=0 because lim<sub>x→0</sub>(f<sub>3</sub>(x)) = lim<sub>x→0</sub>(2<sup>−1</sup>) = 2<sup>−1</sup> while f<sub>3</sub>(lim<sub>x→0</sub>x) = 0<sup>0</sup> = 1, so lim<sub>x→0</sub>(f<sub>3</sub>(x)) ≠ f<sub>3</sub>(lim<sub>x→0</sub>x). So consider the function f<sub>4</sub> defined by f<sub>4</sub>(x) = f<sub>3</sub>(x) for x≠0 and f<sub>4</sub>(0) = 2<sup>−1</sup>. This function f<sub>4</sub> is continuous for x=0. Care must be taken when handling discontinuous functions, and it is not solved by either redefining or undefining 0<sup>0</sup>. Bo Jacoby (talk) 00:02, 16 October 2008 (UTC).

Plenty of notations in mathematics are defined in different ways in different contexts, or are meaningful in some contexts but not in others. Whatever you or I may think of 0^0 conventions, we have plenty of reputable modern sources that explicitly call 0^0 undefined or indeterminate. The article reports the situation accurately, without passing judgement, as it should. This has been argued ad nauseam here, and the conclusion is always the same. —Steven G. Johnson (talk) 06:15, 16 October 2008 (UTC)

An expression being meaningless in some context does not make it undefined. I don't suppose you have reputable modern sources that explicitely call any expression defined and undefined at the same time. The discussions here have disclosed misunderstandings amongst the editors and contributed to improvements of the article. If they cause you nausea you should seek medical advise. While the matter is argued it is not concluded. The quote "In some contexts where the exponent varies continuously, it may be best to treat 0<sup>0</sup> as an ill-defined quantity" is without reference and simply not true. The discontinuity is not removed by re-defining or ill-defining or un-defining 0<sup>0</sup>. Bo Jacoby (talk) 09:48, 16 October 2008 (UTC).

An expression is only relevant to the theory it is part of. It is perfectly okay for 0<sup>0</sup> to be always 1 when talking about integers. Expressions on their own are just symbols and have no meaning. The expression I was talking about was in analysis about real functions. The value there was 1/2. For functions of complex numbers there is no definite value. The exponentiation is a different function in each but they are all expressed using the same notation because most people have no problem with it. In a computer system they are all expressed differently internally but that's not what people do, and wikipedia just reports on things as they are. Wiki also reports on what some people want when there are enough references to make it notable, but what Bo goes on and on about is not current general practice, and I doubt it ever will be, because it would cause unnecessary problems. Dmcq (talk) 12:18, 16 October 2008 (UTC)

Exactly which problems are avoided by leaving 0<sup>0</sup> ill-defined or un-defined? I don't think you sincerely suggest redefining 0<sup>0</sup> to be 1/2 based on you calculation above, because another calculation, (a<sup>1/x2</sup>)<sup>x2</sup>=a for 0<a<1, will provide 0<sup>0</sup>=a when x=0. It is an error to assume continuity of x<sup>y</sup>. It is not an error to define a discontinuous function. If it "is perfectly okay for 0<sup>0</sup> to be always 1 when talking about integers", then 0<sup>0</sup>=1 always, because 0 is an integer. Bo Jacoby (talk) 15:23, 16 October 2008 (UTC).

You're not talking about what is but what you'd like. And to take the argument the other way I just had a look at my copy of 'An Introduction to Combinatorial Analysis' by John Riordan, a person who should agree with 0<sup>0</sup>=1 if anyone would. In his introduction to Blissard calclus, what's more commonly called Umbral calculus he says "A sequence a<sub>0</sub>, a<sub>2</sub>, a<sub>2</sub>, ... may be replaced by a<sup>0</sup>, a<sup>1</sup>, a<sup>2</sup>, ... with the exponents treated as powers during all formal operations, and only restored as indexes when operations are completed; note that a<sup>0</sup> is not necessarily equal to 1." Here we have a power of 0 which is not 1 by a combinatorist. And the expression is treated as an exponential. Personally I would treat integers even with 0<sup>0</sup> as not 1 and consider it only some relation to Umbral calculus which sticks n the 1's - so in fact it is the bit Riordan says isn't 1 that I think mainly gives rise to the 1 people want. Dmcq (talk) 16:58, 16 October 2008 (UTC)

I don't understand your answer. You say that Riordan "should agree with 0<sup>0</sup>=1 if anyone would" and then you quote him for writing that "a<sup>0</sup> is not necessarily equal to 1". Don't you see the contradiction? Did you reply to my question "Exactly which problems are avoided by leaving 0<sup>0</sup> ill-defined or un-defined?"? Bo Jacoby (talk) 18:27, 16 October 2008 (UTC).

No I don't see any contradiction in what I said. As to the problems caused by saying 0<sup>0</sup> is always 1 for real numbers it means that simple algebraic manipulations of formulae may lead to wrong results instead of values which are indeterminate at some points. For continuous functions which is what most applied mathematicians want the indeterminate points can be filled in again using something like l'Hopitals rule. If the value becomes 1 then simple transformations lead to errors instead where the function is discontinuous. It just makes things horrible. Anyway what you are saying is not general practice and that is what is really important to Wikipedia. Dmcq (talk) 09:01, 17 October 2008 (UTC)

The problem, that wrong arguments may give wrong results, cannot be solved. If some calculation involving limits gives 0<sup>0</sup> as a result, then you should be alarmed that you may wrongly have assumed continuity. Your fine example above can be simplified a little: Let a be a number, 0<a<1. Then (a<sup>n</sup>)<sup>1/n</sup>=a for all integers n>0, and so the limit for n→∞ is lim((a<sup>n</sup>)<sup>1/n</sup>)=a, while (lim(a<sup>n</sup>))<sup>lim(1/n)</sup>=0<sup>0</sup> . The wrong assumption that lim((a<sup>n</sup>)<sup>1/n</sup>)=(lim(a<sup>n</sup>))<sup>lim(1/n)</sup> leads to the wrong result that a=0<sup>0</sup>. The convention that 0<sup>0</sup> is undefined leads to the wrong result that a is undefined. The convention that 0<sup>0</sup>=1 leads to the wrong result that a=1. The error is not in the definition of 0<sup>0</sup>, but in assuming that x<sup>y</sup> is continuous. Conclusion: undefining 0<sup>0</sup> solves nothing. Bo Jacoby (talk) 11:10, 17 October 2008 (UTC).

You are mixing up indeterminate and undefined. The idea of an indeterminate form allows people to do straightforward algebraic manipulations on continuous functions without having to worry all the time. It may stick in points where an indeterminate form arises but the value is still defined via limits so the function is not changed or made undefined at some points or anything like that. Saying 0<sup>0</sup> is 1 would break that. In your example splitting the limits leads to an indeterminate form the value of which needs to be determined otherwise if the function is continuous. It doesn't lead to the result that a is undefined. And it doesn't lead to a wrong result unless you assume that 0<sup>0</sup>=1. Dmcq (talk) 17:27, 17 October 2008 (UTC)

I am not the one mixing up. The section Exponentiation#Zero to the zero power calls 0<sup>0</sup> undefined rather than indeterminate. The words are: "justifications for treating 0<sup>0</sup> as an undefined quantity include". (My italics). The editors should not assume that the reader understands the difference between undefined and indeterminate. However it might be more acceptable to the reader that 0<sup>0</sup> is defined and indeterminate form at the same time, than that 0<sup>0</sup> is defined and undefined at the same time. Why not change undefined quantity to indeterminate form ? To me it would be a step in the right direction. Bo Jacoby (talk) 18:17, 17 October 2008 (UTC).

I think you are still ignoring the sources provided:

1. "... Let's start at x=0. Here x^x is undefined." Mark D. Meyerson "The Xx Spindle." Mathematics Magazine, v. 69 n. 3, Jun 1996, pp. 198-206.
2. According to Benson (1999), "The choice whether to define 0^0 is based on convenience, not on correctness."

Note that neither of these authors said anything about indeterminate forms; they directly referred to the question whether 0^0 is defined. On the other hand, there is no source that directly comes out and says "0^0 is both defined and indeterminate", and the only source that even hints this might be possible is Knuth (1999), but he goes on to say that in a sense he has described, "... the value of 0^0 is less defined than, say, the value of 0 + 0." So Knuth seems to be playing both sides of the question there. — Carl (CBM · talk) 18:34, 17 October 2008 (UTC)

Bo is quoting the current article, the draft replacement does exactly what he says which is replace undefined with indeterminate form. I'm happy we can both agree with that at least. Can we go ahead and stick in the replacement then? I'd probably want to expand the second section to say what indeterminate form means and try for a better example - perhaps the 2^etc. one I was using above. However I agree with what is there and it seems a big improvement to me. Dmcq (talk) 00:25, 18 October 2008 (UTC)

OK, the proposal for the 0^0 section has been available for comment for several days now, and I think I've incorporated all the specific suggestions for improvement that all of you had, so I've implemented the proposal. Hopefully small subsequent edits can fix whatever problems remain. Of course, discussion is welcome. I hope that if anyone thinks it should be reverted in its entirety, that person can initiate a discussion on the Talk:Exponentiation page and wait for a reply before taking action. --FactSpewer (talk) 01:54, 18 October 2008 (UTC)

By L'Hôpital's rule it can be shown that the limit of x<sup>x</sup> as x approaches 0 is 1. So it sure makes sense to define 0<sup>0</sup> as 1. Also, by repeated application of the abovementioned rule it can be shown that x<sup>xx</sup> has a limit of 0 as x approaches 0 and that x<sup>xxx</sup> has a limit of 1 as x approaches 0 and that in general, when there are an even number of xs in this power tower that the limit is 1 and when there are an odd number of xs in this power tower that the limit is 0. And don't mention the many hours I wasted on proving this. --116.14.26.124 (talk) 02:42, 23 June 2009 (UTC) Shit the formatting, but I got kinda stumped with the coding. Where's that <math> template?

You have shown only that the limit for x<sup>x</sup> is 1, for other functions f(x) and g(x) both going to zero f(x)<sup>g(x)</sup> can be anything one wants. The article gives examples of functions that do that. One can't say 0/0 is 1 since the limit of x/x is 1. The limit of 2x/x is 2. Dmcq (talk) 09:40, 23 June 2009 (UTC)

"Rule of calculus" for lim f(x)<sup>g(x)</sup>

The changes made have improved the article and I wish to thank the editors involved. I dislike, however, the following sentence: "The rule in calculus that lim<sub>x→a</sub> f(x)<sup>g(x)</sup> = (lim<sub>x→a</sub> f(x))<sup>lim<sub>x→a</sub> g(x)</sup> whenever both sides of the equation are defined would fail if 0<sup>0</sup> were defined", (my italics), because the italiced assertion is not a general rule of calculus, and because it refers to 0<sup>0</sup> being defined which we have just allowed, and because an equation (defined number)=(indeterminate form) is as false as (defined number)=(another defined number) and (defined number)=(undefined number). I hope we can agree to omit the sentence. Bo Jacoby (talk) 12:30, 18 October 2008 (UTC).

0<sup>0</sup> doesn't have a defined value in calculus, it is an indeterminate form. If 0<sup>0</sup> was defined as always 1 it wouldn't be an indeterminate form. Also the limit round an indeterminate form doesn't have to be indeterminate, it can be defined. The sentence seems perfectly okay to me. 13:10, 18 October 2008 (UTC)

After looking it over, I too think that the "rule of calculus" sentence should be removed or at least qualified, but for a different reason than Bo Jacoby: according to the definitions in the article,

• $\lim _{n\to \infty }(-2)^{(2n+1)/(2n-1)}=-2$
• $\lim _{n\to \infty }(-2)^{2n/(2n-1)}=\lim _{n\to \infty }2^{2n/(2n-1)}=2$ (because the numerator of the exponent $2n/(2n-1)$ is even).
• One can easily turn these limits of sequences into limits of functions. For instance, let f(x) = -2 (constant function), let g(1)=1, for x > 1 let g(x) be the number of the form 2n/(2n-1) that is closest to x, breaking ties arbitrarily, and for x < 1, let g(x) = g(2-x); then $\lim _{x\to 1}f(x)^{g(x)}=2$.
• So the "rule of calculus" fails not only for 0<sup>0</sup>, but also for (-2)<sup>1</sup>.

Should we interpret this as saying that (-2)<sup>1</sup> should be left undefined? (I hope not! That would be a disaster!) --FactSpewer (talk) 15:55, 18 October 2008 (UTC)

Good point. The rule only holds for areas where exponentiation is continuous. If the -2 was a complex number and not a real then one could have (-2)<sup>x</sup> being continuous but you wouldn't get the same values as the rule for negative reals. And silly me I said the complex exponentiation of a complex number with zero imaginary part would give the same result as for reals when one used the principal value but I should have said this for positive reals only. It does lead to the language being rather a bit more cumbersome but better safe than sorry. Dmcq (talk) 17:12, 18 October 2008 (UTC)

If we restrict the rule to where the 2-variable function is continuous, then it no longer says anything about what is happening at (0,0). In particular, it is then compatible with 0<sup>0</sup> = 1, say, so there is no point in mentioning it as a justification for leaving 0<sup>0</sup> undefined. On the other hand, if we don't restrict the rule, then it forces (-2)<sup>1</sup> to be undefined, as mentioned above, which would be terrible. So I recommend simply removing the rule from this section. As for the example that was added, this should go too, in my opinion; there is already a long footnote about the different limits of x<sup>y</sup> that can be obtained as (x,y) approaches (0,0). Or maybe it could be mentioned in the next subsection; i.e., instead of saying that "S" gave a counterexample, we could say that "S" gave the counterexample (e<sup>-1/x</sup>)<sup>x</sup>, which according to Knuth is what actually happened. --FactSpewer (talk) 17:57, 19 October 2008 (UTC)

A limit point does not have to be included in a limit process, the article on indeterminate form mentions this point explicitly. As it says at the start of limit point "In mathematics, informally speaking, a limit point of a set S in a topological space X is a point x in X that can be "approximated" by points of S other than x itself." Notice the "other". Dmcq (talk) 20:09, 19 October 2008 (UTC)

If "S" used e instead of 2 then that is a good reason to put it in, it can then have a reference. I only used 2 because it was the version in indeterminate form, possibly because it is less mystical than e. Dmcq (talk) 20:36, 19 October 2008 (UTC)

I agree that the limit of a function f(x) as x approaches a can be defined without f(a) itself being defined (I gather that this is what you meant by your remark about limit points), but the point I was trying to make was this:

• The statement in the current version of the article says "The rule in calculus that lim<sub>x →a</sub> f(x)<sup>g(x)</sup> = (lim<sub>x→a</sub> f(x))<sup>lim<sub>x→a</sub> g(x)</sup> whenever both sides of the equation are defined and exponentiation is continuous would fail if 0<sup>0</sup> were defined."
• What I think you mean by this rule is, for instance, that the rule of calculus does not apply to limits where f(x) and g(x) are approaching -2 and 1, respectively, because the function x^y is not continuous at (-2,1), and similarly that the rule of calculus does not apply to limits where f(x) and g(x) are both approaching 0, because the function x^y is not continuous at (0,0).
• So far, so good. But if I now interpret the statement literally, then even if 0^0 is defined, the rule still does not apply because x^y is not continuous there, so there is no failure.
• And if you do see a failure here, one that implies that 0^0 must be undefined, why doesn't the same logic force (-2)^1 to be undefined?

In summary, I think this sentence about the "rule of calculus" sentence really needs to be removed. --FactSpewer (talk) 06:44, 20 October 2008 (UTC)

You're saying the right thing about a limit and then you miss the point. It isn't that the exponentiation function is discontinuous at (-2,1) that makes the limit rule invalid there, it is that it is discontinuous approaching there. If exponentiation was continuous for the reals around (-2,1) but was defined with a discontinuity at that point then the rule would lead to a contradiction but that isn't true. And anyway its better to say an indeterminate form as undefined implies there is no defined value for an expression. Dmcq (talk) 07:16, 20 October 2008 (UTC)

Dear Dmcq, I appreciate your effort to explain the "rule of calculus" sentence to me, but I am still unclear as to what it is trying to say. I have no objection to keeping it if it can be made precise and a reference for it (in the form you are stating it) is given. Let me suggest a couple possible readings of the sentence (expanding it in more detail, just to make sure we are on the same page), and ask you which one you mean. I believe it is trying to say the following:

There is a rule of calculus that says that if

• f(x) and g(x) are real-valued functions defined in a neighborhood of a real number a (but maybe not at a itself),
• lim<sub>x →a</sub> f(x) exists, and equals b, say,
• lim<sub>x →a</sub> g(x) exists, and equals c, say,
• b^c is defined,
• lim<sub>x →a</sub> f(x)<sup>g(x)</sup> exists and equals d, say, and
• "exponentiation is continuous",

THEN d = b^c.

Moreover, if 0^0 were defined, then this rule would fail.

My question is what is meant by "exponentiation is continuous". Which of the following is intended as a hypothesis for the application of the rule?

1) The 2-variable function x^y is continuous at (b,c).

2) The 2-variable function x^y is defined on a neighborhood of (b,c), and is continuous at (b,c).

3) There is a sequence of points approaching (b,c) such that the 2-variable function x^y is continuous at each of these points.

4) The 2-variable function x^y is continuous at each point of the form (f(t),g(t)) for t within the domains of both f and g.

5) The 2-variable function x^y is continuous in an open neighborhood of the set of points of the form (f(t),g(t)) for t within the domains of both f and g.

6) The 2-variable function x^y is continuous at every point it is defined.

7) None of the above (if this is the case, what did you mean?)

(Admittedly, some of these would render the "rule" false or useless.) Of course, I'm not suggesting that we write all this in the article.

Finally, in what textbook do you find this "rule of calculus"? For the time being, I'll flag the sentence as needing a citation.

Thank you, --FactSpewer (talk) 02:47, 22 October 2008 (UTC)

I have to say that this once Bo is right. Dump the sentence. Let's avoid excessive attempts at explanation, especially unsourced explanation. "Just the facts, ma'am", and the facts are the different definitions used in the literature, not so much our interpretation of the reasons behind them. --Trovatore (talk) 04:15, 22 October 2008 (UTC)

OK, that's fine with me. --FactSpewer (talk) 05:03, 22 October 2008 (UTC)

I know it isn't a primary source but perhaps you an agree the statement at the start of Indeterminate form is correct:

In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.

It then goes on to say 0<sup>0</sup> is an indeterminate form and if you look in any textbook they'll say the same thing. Now if the algebraic operation after replacing the subexpression gives a defined number then the substitution does give enough information to determine the original limit. This is quite different from the case of Pow(x,y)=x<sup>y</sup> where (x,y)≠0 and Pow(0,0)=1. This does not lead to an indeterminate form. One cannot say that limit Pow(x,y) = Pow(limit x, limit y) if both sides are defined. Pow is not an algebraic operation in the terms of the definition.

Of course it would be possible to say the usual rules for indeterminate forms only applies for addition, subtraction, multiplication and division but don't apply for exponentiation. But that is not what is done. What is done is that 0<sup>0</sup> is treated as an indeterminate form.

As to the question about continuity the answer is number 3. Think of the same question as applied to the indeterminate form 0/0. If it had a defined value then it wouldn't be an indeterminate form. And division isn't continuous at 0,0. If divide(x,y)=x/y except divide(0,0)=1 was defined then it couldn't be used as an indeterminate form where one could move the limits inside the brackets. Without the defined value divide(0,0)=1 we could use it as an indeterminate form with a bit extra about moving limits in this particular function. Indeterminate form is just a special case of a function with a discontinuity that has no defined value at the discontinuity and is used because people didn't want to have a slew of special rules for dealing with straightforward expressions. Dmcq (talk) 09:35, 22 October 2008 (UTC)

More exact about continuity: Exponentiation in the domain of the x used in the limit must be a continuous function. That's it. As it says in the limit article it need not be defined at the limit point. This just excludes zero and negative reals from being used in the domain of the limit. Dmcq (talk) 10:16, 22 October 2008 (UTC)

Dear Dmcq, I agree that your interpretation of "exponentiation is continuous" as my number 3 makes the "rule of calculus" sentence a mathematically correct statement. We could adjust the sentence so that it says this. On the other hand, it is a little convoluted, and I doubt that there exists a textbook containing a rule stated in this way, so calling it a "rule" is perhaps a stretch. Probably Bo Jacoby and Trovatore are right, that we should just remove the sentence (along with the following sentences illustrating it with an example). What do you think? --FactSpewer (talk) 16:08, 22 October 2008 (UTC)

Not much. That example is what "S" used as described in the history section. I am not sure what exactly is going through your mind when you say that that definition of continuous makes it a mathematically correct statement if you then want to get rid of that example. I'll look at getting a good statement of that rule, it wasn't me wrote the equation in but I did add the continuous bit, but it is just a particular version of the general indeterminate form rule. I'm frankly a bit surprised here has been so much debate over what a limit or continuous means, never mind this idea that somehow a limit might be something other than a value and a divide or exponentiation would know its operands are limits. Dmcq (talk) 17:08, 22 October 2008 (UTC)

Feedback on proposed draft (only!)

To keep the discussion constructive, I'd like this new section to be devoted exclusively to feedback on the proposed draft. Absolute statements that 0^0 is 1 or that 0^0 is undefined can go elsewhere.

No one over the past few days has had any objection to the proposed corrections outside the 0^0 section, so I will now restore those corrections, while leaving the 0^0 section untouched for the time being.

As for the proposed 0^0 section, a few people on the Talk:Exponentiation page have come out generally in support of it while also saying that it will need some minor fixing that can be done later. I believe the only person so far who has explicitly said that it is not an improvement is Steven G. Johnson. Steven: I would like to understand better what specifically in the new draft you are objecting to. I think we all agree with you that several references state explicitly that 0^0 is undefined, but I do not understand why this is incompatible with the proposed draft. Indeed, the proposed draft cites some such references, while also citing references for the opposite point of view, that 0^0 is 1. --FactSpewer (talk) 18:10, 16 October 2008 (UTC)

I just updated the proposed draft. Now it contains only the 0^0 section (the corrections in other sections having been implemented already). I also attempted to incorporate the suggestions made by those of you who commented on it. In particular, the opening sentences were rewritten so that it starts less abruptly, and so that it begins the discussion in a balanced way. Feedback on this draft is welcome, and should go here, in this section of Talk:Exponentiation. --FactSpewer (talk) 15:53, 17 October 2008 (UTC)

Identities for complex logarithm and complex exponentiation

The sentence

The identities do hold if the complex logarithm is defined as a multivalued function, or better as a function whose domain is a Riemann surface, but no ordinary single valued function on the complex plane can satisfy the identities.

looks mathematically incorrect to me.

Consider the identity log a<sup>b</sup> = b log a, for instance, for complex numbers a and b with a ≠ 0.

If we are treating log as a multivalued function, and A denotes one possible value of log a, then log a is the set of numbers of the form A + 2πin, where n ranges over integers. So b log a is the set of numbers of the form b(A + 2πin).

On the other hand, a<sup>b</sup> is defined as exp(b log a), which is the set of numbers of the form exp(b(A + 2πin)), so log a<sup>b</sup> is the set of numbers of the form b(A + 2πin) + 2πim, where m and n range over integers.

So the set log a<sup>b</sup> contains the set b log a, and for most values of b, it is strictly larger, meaning that the identity fails. --FactSpewer (talk) 01:30, 17 October 2008 (UTC)

Yes, the section needs improvement. Bo Jacoby (talk) 04:35, 17 October 2008 (UTC).

Yes sorry I was thinking it needed tightening up when I put it in. I'll remove it until I can think of something more satisfactory. Dmcq (talk) 08:50, 17 October 2008 (UTC)

Thank you, Dmcq. Your other recent edits to Exponentiation look good, by the way. --FactSpewer (talk) 14:41, 17 October 2008 (UTC)

Recent edits by Dmcq

I am unhappy with some of the very recent edits by Dmcq:

• In general, it is better to avoid talking about multivalued functions, because people tend to forget that their outputs are sets and make errors by writing equalities between their values as if the values were numbers. There are (at least) two ways to resolve the multivaluedness for the complex logarithm (and hence also z^b where z is a nonzero complex number:
  • One, as you wrote to me, is to reconsider the multivalued function as a function on the associated Riemann surface. This yields a single-valued function, but it then can no longer be evaluated at numbers, because its domain is a set of points of a complex manifold, not a set of complex numbers.
  • Another is to preface each equality involving log(z) or z^b with a choice of branch of the complex logarithm, which can be either the principal branch, or another one. If one varies the branch, one recovers all the information contained in the function of the Riemann surface, but the big advantage is that one can now write down equalities that make sense!
• The statement that "powers of negative numbers are...discontinuous even where defined" whereas "powers of complex numbers are continuous" is confusing, first because it is not entirely clear whether the argument of the functions you are considering is the base or the exponent or both jointly, and second because negative numbers are special cases of complex numbers, and third (a matter of wording) because it is not powers that are continuous or not, but functions.
• Although there are two numbers whose square is 1, it is standard that 1<sup>1/2</sup> means the positive square root, so it is strange to say that 1 = (-1.-1)<sup>1/2</sup> is wrong.
• It is strange to say that the first line of the Clausen argument is wrong. What is wrong is going from the second line to the third, which assumes the identity. I understand that Dmcq wants to treat e as a complex number and define e^z by using a nonprincipal value of log(e) (i.e., a value of log(e) other than 1), but this is not the standard way of interpreting e^z.
• The claim that the real number e and the complex number e+0i are different is (at best) a nonstandard point of view. The real numbers form a subset of the complex numbers, so e is a complex number whether or not it is written as e+0i.

--FactSpewer (talk) 06:19, 20 October 2008 (UTC)

I'll put here what I said when FactSpewer put these points on my talk page with a couple of small edits. I haven't tried to deal with the additions above yet:

To take your points in order.

The modern viewpoint of the logarithm in mathematics is as far as I'm aware that it is defined on a Riemann surface, this is referred to as the log function and the principal value version is called Log. Talking about a multivalued function is certainly old-fashioned but it is closer to current practice than just dealing with the single valued form. Factspewer has added some suggestions above about dealing with thisbut putting in 2, I'll have to think a bit about that but putting in the branch by adding 2πik where k is some k rather than all k as for the multiple value form could be made to work and be less confusing I believe. Dmcq (talk) 07:54, 20 October 2008 (UTC)

Powers of negative reals are discontinuous, they are only defined for some rational and even then keep changing sign. By contrast the principal value of (-r+0i)^(p+i0) is defined as e^(p log(r) + p π i) which is continuous. I agree the text should be reworded to make it clear exactly what's discontinuous but reals are not a special case of imaginary numbers - and the rules for exponentiation are radically different for negative reals.

The original also wanted to remove multiple values of i<sup>i</sup>, Ill add some references. Here's two articles on the web dealing with the multiple values of i<sup>i</sup> and the second has references to a couple of books and magazines.

What is i to the Power of i?

Complex number to a complex power may be real

Plus the wiki Imaginary unit article mentions the multiple values of i^i.

There are two possible square roots of 1 and that's what the multi-valued view says, and I had explicitly qualified the statement saying that view was being considered.

If you have a look above at the business of the powers of negative number being discontinuous you can see one of the problem with treating real numbers and complex numbers as the same. Exponentiation of e+i0 uses the rules for complex powers whereas exponentiating e as a real uses the rule for real numbers. Logically it is just wrong and when it is done one needs to be quite careful as the paradox shows.

Thinking about it though the equation doesn't actually give a wrong result yet from the principal value point of view only the multiple value point of view. The first error from the principal value point of view does come in the next line as FactSpewer says.

(original point about limits has a separate section here). Dmcq (talk) 07:54, 20 October 2008 (UTC)

Clausen's paradox

I've had a go at the multi-valued bits at the start of the complex exponentiation section and for Clausen's paradox and tried to make them a bit less off-putting. Dmcq (talk) 08:00, 21 October 2008 (UTC)

Regarding the e + 0i bit, I think it can also be explained as follows: in the first line, that e<sup>...</sup> is really intended to be exp(...), but the second line replaces the exp() with an actual complex exponent. I don't know that this is any easier to source, though. — Carl (CBM · talk) 13:48, 22 October 2008 (UTC)

I'll put in braces for multiple values as they are sets. Some people do that and it should make things a bit more reasonable to read. Dmcq (talk) 13:51, 22 October 2008 (UTC)

powers of positive real numbers.

The present structure of the article is that powers of real numbers are treated before powers of complex numbers. I suggest that powers of positive real numbers are finalized before powers of negative real numbers and powers of complex numbers, because the discontinuous rational powers of negative numbers are relatively unimportant and confusing. Comments, please. Bo Jacoby (talk) 11:46, 20 October 2008 (UTC).

I think that would be reasonable okay. Dmcq (talk)

Looks much better with them separated. It's possible to notice now that we didn't say anything about negative powers of positive real numbers. I think I'd move out all the stuff about imaginary and complex powers too to a separate section 'Complex powers of real numbers'. It all clutters up the basic stuff unnecessarily. I'm not sure if it should be before or after the negative reals bit, perhaps after it and just before all the complex powers of complex numbers. Dmcq (talk) 17:09, 20 October 2008 (UTC)

Plus I think powers of e should be just before Real powers so the contents list would be:

1 Exponentiation with integer exponents

1.1 Positive integer exponents

1.2 Exponents one and zero

1.3 Combinatorial interpretation

1.4 Negative integer exponents

1.5 Identities and properties

1.6 Powers of ten

1.7 Powers of two

1.8 Powers of one

1.9 Powers of zero

1.10 Powers of minus one

2 Powers of positive real numbers

2.1 Principal n-th root

2.2 Rational powers

2.3 Powers of e

2.4 Real powers

3 Powers of negative real numbers

4 Complex powers of real numbers

4.1 Imaginary powers of e

4.2 Trigonometric functions

4.3 Complex powers of e

4.4 Complex powers of positive real numbers

5 Powers of complex numbers

5.1 Powers of the imaginary unit

5.2 Complex logarithms

...

Dmcq (talk) 17:29, 20 October 2008 (UTC)

zero to the zero'th power

QUOTE:

$\lim _{x\to 0^{+}}(2^{-1/x})^{x}=1/2,\!$

$\left(\lim _{x\to 0^{+}}{2^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=0^{0}.\!$

If 0<sup>0</sup> was defined as 1 then both sides of the rule would be defined giving the false conclusion that 1/2 = 1.

UNQUOTE

In the chain of equalities

$1/2=\lim _{x\to 0^{+}}(2^{-1/x})^{x}$ $=\left(\lim _{x\to 0^{+}}{2^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=0^{0}$ $=\left(\lim _{x\to 0^{+}}{3^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=\lim _{x\to 0^{+}}(3^{-1/x})^{x}$$=\left(\lim _{x\to 0^{+}}{3^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=1/3.\!$

the weak links are equality sign number two and five, and not number three and four, no matter how $0^{0}$ is defined.

Bo Jacoby (talk) 12:24, 20 October 2008 (UTC).

The limit isn't defining 0<sup>0</sup>. If 0<sup>0</sup> was defined and this rule of calculus isn't thrown out then both expressions would be equal but in this type of application 0<sup>0</sup> is treated as an indeterminate form. What you've written is like saying since the limit of 2x/x as x tends to 0 is 2 but also 0/0 and the limit of 3x/x s 3 but also 0/0 then 2 = 3. Dmcq (talk) 12:37, 20 October 2008 (UTC)

ps you're quite right about the weak link being in equality signs number two and five. They only hold if both sides are defined. Dmcq (talk) 13:14, 20 October 2008 (UTC)

In the chain

$2={2x \over x}=\lim _{x\to 0^{+}}\left({2x \over x}\right)={\lim _{x\to 0^{+}}(2x) \over \lim _{x\to 0^{+}}(x)}={0 \over 0}$ $={\lim _{x\to 0^{+}}(3x) \over \lim _{x\to 0^{+}}(x)}=\lim _{x\to 0^{+}}\left({3x \over x}\right)=\lim _{x\to 0^{+}}(3)=3$

the weak links are equality signs number three and six, and not number four and five, no matter how or if ${\tfrac {0}{0}}$ is defined. The point is whether ${\tfrac {x}{y}}$ is continuous or not.

If you are allowed to write ${\lim _{x\to 0^{+}}(2x) \over \lim _{x\to 0^{+}}(x)},$ then you are also allowed to write ${0 \over 0},\,$ and if you are allowed to write $\left(\lim _{x\to 0^{+}}{2^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}},$ then you are also allowed to write $0^{0}.\,$ Bo Jacoby (talk) 15:37, 20 October 2008 (UTC).

Yes exactly. The equalities don't hold because 0/0 is an indeterminate form. If 0/0 was defined as 1 say then the limit rule wouldn't be valid as it would say 2=1=3. Dmcq (talk) 16:40, 20 October 2008 (UTC)

The "limit rule" is invalid independently on whether 0/0 is defined or not. A definition of 0/0 does not invalidate any legitimite calculation. The weak link is "lim(2x/x)=(lim 2x)/(lim x)", not "(lim 2x)/(lim x)=0/0" which is true because lim 2x=0 and lim x=0. (Her "lim" means "lim<sub>x→0</sub>"). Similarily, defining 0<sup>0</sup> is not harmful in any context, and 'undefining' 0<sup>0</sup> solves no problem at all, and the 'justifications' for leaving 0<sup>0</sup> undefined are logically invalid. Bo Jacoby (talk) 10:40, 21 October 2008 (UTC).

I'm just going by the generally accepted convention which is what wiki should principally report on. Of course we could define 0/0=1, many people have wanted to do exactly that, and all this would really affect in analysis is that division would have to be treated as a general function rather than following the simpler rules that the concept of an indeterminate form allows. So one could have pow(0,0)=1 or divide(0,0)=1 no problem just like one has sign(0)=0. It would cause lots of problems though because people move round exponents easily just like the association rules and suchlike for addition. It's much easier to track problems with discontinuities with named functions. Saying 0<sup>0</sup> is defined as 1 would just cause too many problems in analysis. Much as I respect Knuth his opinion on this matter is just that and no more, he can be quoted as an authority on one side but he doesn't set the standard. Dmcq (talk) 12:33, 21 October 2008 (UTC)

Defining 0/0 solves no problem, but defining 0<sup>0</sup> does solve a problem, and creates no problem. It is OK to report that some authors do not define 0<sup>0</sup>, but it is not OK to report logically invalid justifications which seem to be original research. See WP:OR. Bo Jacoby (talk) 14:21, 21 October 2008 (UTC).

I fully agree that defining 0^0 as 1 solves some problems in some circumstances. That doesn't mean it is an accepted way of proceeding in analysis. If it is defined as 1 then exponentiation can't be treated like the other straightforward functions like division by zero. What original research are you talking about? Everything I've said has been the way things have been done for at least the last 150 years or more. I'm not inventing anything. Dmcq (talk) 14:53, 21 October 2008 (UTC)

The rule in calculus that lim x→a f(x)^g(x) = (lim x→a f(x))^(lim x→a g(x)) whenever both sides of the equation are defined and exponentiation is continuous is not a general rule of calculus but was coined for the purpose. "If 0^0 was defined as 1 then both sides of the rule would be defined giving the false conclusion that 1/e = 1" is not true, because exponentiation is not continuous if 0^0 is defined. Bo Jacoby (talk) 16:08, 21 October 2008 (UTC).

Please see indeterminate form or the definition on the Mathworld site Weisstein, Eric W. "Exponentiation". MathWorld.. If 0/0 or 0^0 or whatever was 1 then the limit would have the defined value 1 and would not be indeterminate. Think of it as how a computer would evaluate it. Stick in both limits into divide and it either returns indeterminate or 1, returning indeterminate and 1 at the same time is a possibility but I haven't seen anybody doing that. In computers Not a Number is used for indeterminate. Having degrees of definedness like Knuth with "In this much stronger sense, the value of 0<sup>0</sup>is less defined than, say, the value of 0+0", really doesn't cut it as a mathematically precise way of going round things. Dmcq (talk) 17:46, 21 October 2008 (UTC)

The MathWorld site says "Certain forms of limits are said to be indeterminate when merely knowing the limiting behavior of individual parts of the expression is not sufficient to actually determine the overall limit" while indeterminate form says "an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form." So 0<sup>0</sup> is an indeterminate "form" only means that in a limit of the form f(x)<sup>g(x)</sup> with f(x)→0 and g(x)→0, you need more information to determine the limit. It does not have any bearing on whether 0<sup>0</sup> itself is defined or not; the indeterminate form article gives examples of expressions that are undefined but are not indeterminate forms, and MathWorld has separate articles for Indeterminate and Undefined, making it clear that they are different concepts. Shreevatsa (talk) 18:57, 21 October 2008 (UTC)

First let me remind everyone that we are not here to discuss the merits of the question. I know it's very tempting, but it's not going to get us anywhere.

When it comes to interpreting and incorporating the sources, there are two points I'd like to make:

1. MathWorld is not a reliable source. There's good information there, but there's also a lot of very idiosyncratic definitions and terminology. See for example Wikipedia:Articles for deletion/Radical integer and Wikipedia:Articles for deletion/Regular number. Apart from MathWorld's merits (of which, obviously, I have a low opinion) WP:RS warns against heavy reliance on tertiary sources.
2. This modern use of indeterminate form to focus exclusively on the functional form of the expression, leaving open the possibility that the expression might be well-defined when you plug in values, is, I think, a bit of a novelty, and not everyone has gotten the memo. It is clear from context that when many of these authors say 0<sup>0</sup> is an indeterminate form, they do indeed mean that it is undefined. This redefinition of indeterminate form is not unreasonable on its own, but for us it unfortunately complicates the task of understanding what the sources are saying.

--Trovatore (talk) 19:17, 21 October 2008 (UTC)

I was only really pointing to those to show it isn't original research, not as original sources. I think that pushing back 0/0 etc to being 'algebraic forms' which actually might have defined values, i.e. the indeterminate and 1 at the same time option, is just pushing back the problem of limits in analysis so it causes a lot more problems elsewhere. Are people really trying to propagate the problem into algebra? In algebra at the moment if you have two numbers and apply an operation you get another number or an indeterminate form or an undefined value. With this change I really hate to think how basic operations are going to be explained. As far as I know this is not the general consensus opinion and I haven't seen a mathematical way of saying it rather than some hand waving like Knuth. Treating exponentiation of integers differently from that of reals is fine, messing up algebra sounds a distinctly iffy proposition. Am I really that out of date and the consensus now is that mathematical operations in algebra should sometimes actually give their defined value and at other times should be treated as indeterminate? Why is that better than the business as described? And what exactly are the rules to govern this? By all means stick it in if that is the general consensus but I'd like to see some evidence of a theory of how it is all supposed to work. Dmcq (talk) 22:06, 21 October 2008 (UTC)

Algebraically it does make sense to take 0<sup>0</sup>=1 — there is no algebraic notion of raising a value to a real power, so implicitly this must be a something-to-integer or something-to-natural-number type exponentiation. For those the argument for the value of 1 is solid (though the argument for saying so in the article is much less so, given the state of the sources). Where it might (in my opinion, does) make sense to leave 0<sup>0</sup> undefined is in the real-to-real or complex-to-complex case, where there's no algebraic argument to make. But at that point we've left algebra behind. --Trovatore (talk) 22:29, 21 October 2008 (UTC)

I think you're going a little far here saying algebra doesn't deal with real numbers. All those children finding how long it takes 5 men to dig 3 ditches would disagree. ;-) Dmcq (talk) 22:54, 21 October 2008 (UTC)

I didn't say algebra doesn't deal with real numbers. I said there's no algebraic notion of raising a value to a real power. --Trovatore (talk) 22:57, 21 October 2008 (UTC)

Oh sorry I see what you mean, the definition of integer or rational powers as opposed to the other ones which depend on continuity. Dmcq (talk) 23:02, 21 October 2008 (UTC)

Yes I'd like to write 0.0<sup>0.0</sup>or something like that for the indeterminate case. I'm not sure how that would go down, someone here already complained about me distinguishing between the real number e and the complex number e+i0. Dmcq (talk) 13:24, 22 October 2008 (UTC)

Although I agree with your sentiment, I don't think there are any sources for it, because there's no reason for a complex analysis book to spend too long distinguishing between the natural number 0 and the complex number 0. Rather than making that distinction, we could say that exponentiation is often defined as a<sup>b</sup> = exp(b log a) in continuous settings, and that definition leaves 0^0 undefined. Certainly that definition can be found in numerous complex variables texts, and no complex variables text attempts to define log(0). — Carl (CBM · talk) 13:34, 22 October 2008 (UTC)

Branches of the complex logarithm

Isn't the section on Exponentiation#Branches of the complex logarithm a bit over the top? I don't think this article needs anything quite like that if it can be dealt with better in the branch cut or complex logarithm articles. I certainly was quite reluctant to put in even the bit about Riemann surface but I though a quick reference and a picture were called for. Dmcq (talk)

I agree it should be moved elsewhere. — Carl (CBM · talk) 13:42, 22 October 2008 (UTC)

Dear Dmcq and Carl, I agree with your suggestion that it would make sense to move the stuff about branches of the complex logarithm to the complex logarithm page. That makes a lot of sense. I would be happy to attempt this move, but give me a little time, since the complex logarithm page is in need of serious revision and streamlining in my opinion. (Do you agree?) My plan is to put a draft of a new complex logarithm page on my user page for discussion. Sound good? --FactSpewer (talk) 15:55, 22 October 2008 (UTC)

Sounds fine by me. I'm going to move the sections here mentioning complex numbers so they are together like I said in the proposed contents list somewhere above. I'll not change any of the actual text of the sections. Dmcq (talk) 17:44, 22 October 2008 (UTC)

OK, a proposed draft of the complex logarithm page can now be found at my user page User:FactSpewer. If anyone has suggestions for improvement, probably it's best to write them on the Talk:Complex logarithm page. Discussion is welcome! --FactSpewer (talk) 17:43, 25 October 2008 (UTC)

Redundant equations?

Dear Dmcq, Thank you for the new sentence in the 0^0 section. It's better than what was there before. We might still consider removing the sentences

For example:

$\lim _{x\to 0^{+}}(e^{-1/x})^{x}=1/e,$

$\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=0^{0}.$

If 0^0 were defined as 1 then replacing the subexpressions in the first expression by their limits would give the false conclusion that 1/e = 1.

since they are redundant with what came before and with $(e^{-1/x})^{x}$ being a counterexample to Moebius's statement. Anyway, I leave it to others to decide this.

--FactSpewer (talk) 00:01, 25 October 2008 (UTC)

The 'false conclusion' comes from the discontinuity of x<sup>y</sup>and not from defining 0<sup>0</sup>. I support the omission of the sentence. Bo Jacoby (talk) 07:11, 25 October 2008 (UTC).

I'm not sure what your problem is with this. There are lots of examples of where 0<sup>0</sup>=1 is a good decision, why remove a good example of what a indeterminate form is? This is the original example that explained things 190 years ago. If you wish to remove this then please raise this issue to arbitration. Dmcq (talk) 08:20, 25 October 2008 (UTC)

Looking at the wiki process I believe the next stage if you really want to go on with removing traces of 0^0 being considered anything but 1 to a history section would be to raise the matter for mediation in the talk page at Wikipedia talk:WikiProject Mathematics. Dmcq (talk) 10:17, 25 October 2008 (UTC)

I wasn't sure if you were responding to Bo Jacoby or to me, but in any case I think everyone can agree that there are much more than traces of 0^0 being undefined in the current article, and no one (except maybe Bo Jacoby) is advocating eliminating this point of view. I believe that (e^{-1/x})^x should be kept; all I was pointing out is that it is now in the article twice, and it seems to fit better with being a counterexample to Moebius's claim (as you say, it is 190 years old) than here where it was supposed to be an example of a "rule of calculus" that has since been removed. Anyway, I'm not going to do anything; as I said, I leave this to others. One other minor nitpick about the second equation: it is strange to claim that the two sides are equal if one is arguing that both sides are undefined. (It would not be standard to say that 0/0 = 0/0, for instance.) --FactSpewer (talk) 16:00, 25 October 2008 (UTC)

I consider you and Bo Jacoby as one on this matter and that you will go on and on. Remove the equation and you are in an edit war and the matter goes to mediation. Dmcq (talk) 16:33, 25 October 2008 (UTC)

I think the equation should stay (the later occurrence is only a mention), but it could stand some clarification. Something like:

For example:

$\lim _{x\to 0^{+}}(e^{-1/x})^{x}=1/e,\!$

while if one tried to evaluate the limit by replacing subexpressions by their limits, one would get

$\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}=0^{0},\!$

which shows that it is not valid to do so.

As it stands, it does not make clear what exactly being an indeterminate form means. Shreevatsa (talk) 17:20, 25 October 2008 (UTC)

There is a wikilink to indeterminate form for more information and a reference note just after the quick explanation which also describes it. I'm not sure how your change would improve matters it adds a bit and subtracts a bit and others would complain about other small differences. It might be worth you looking at the indeterminate form article and seeing if that explains what it is about properly though - a new eye can always spot something that would put off an naive user looking things up. Dmcq (talk) 17:39, 25 October 2008 (UTC)

Why do you think the edit was worse than what is currently there? (I only meant that the LaTeX markup needs to be cleaned up...) The problem I (and others) have with the current version is the notion that $\lim _{x\to 0^{+}}(e^{-1/x})^{x}=\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}$ (it is not, exactly because 0<sup>0</sup> is an indeterminate form) based on which it says that "if 0<sup>0</sup> was defined as 1, it would give the false conclusion...". What it should say instead is that 0<sup>0</sup> is an indeterminate form, so one cannot evaluate $\lim _{x\to 0^{+}}(e^{-1/x})^{x}$ as $\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}$, irrespective of whether 0<sup>0</sup> is defined as 1 or left undefined or whatever. Shreevatsa (talk) 23:28, 26 October 2008 (UTC)

If you read the citation it gives a quote. The operative words are that 0/0 is meaningless and he says the same is true of the other indeterminate forms. That's what that authority thinks. There may be authorities on this that think otherwise. I'd be interested to see how they interpret the situation. Please provide a citation if you find one. Otherwise one has to treat the exponentiation as annotated saying it came from a limit so in this case one mustn't get the value or one mustn't actually substitute the limits of the subexpressions only potentially do that as part of the original limit operation or some such thing. In this case one can substitute the value of the subexpressions. The value of each subexpression is 0. What is the value of 0<sup>0</sup>? The author thinks meaningless so it is an indeterminate form and doesn't determine the original value. It is pretty clear it is not regarded as some magic value which really is 1 but somehow because of some maths one can't really specify doesn't determine the original limit. Dmcq (talk) 08:43, 27 October 2008 (UTC)

Would you write the following?

$\lim _{x\to 0^{+}}{\frac {2x}{x}}=2,$

${\frac {\lim _{x\to 0^{+}}{2x}}{\lim _{x\to 0^{+}}{x}}}={\frac {0}{0}}.$

No. You would say that replacing subexpressions the second equation is not valid, because 0/0 is an indeterminate form. The quote you have provided says that "In general the limit of φ(x)/ψ(x) when x=a in case the limits of both the functions exist is equal to the limit of the numerator divided by the denominator. But what happens when both limits are zero?" so the "authority" (BTW rather obscure book and authors...) also thinks that in the case of an indeterminate form one cannot replace subexpressions by their limits. The definition of "indeterminate form" in the Wikipedia article, and in that section in this article, also agree. So after saying (thrice!) that it is not valid to replace subexpressions, we should not go ahead and do it ourselves! Shreevatsa (talk) 12:52, 27 October 2008 (UTC)

Yes, it is perfectly valid to replace the subexpressions in that limit. The fact that you get an indeterminate form means that some other method (in this case, simplifying the fraction) is needed to compute the limit. This article currently says, "Limits involving algebraic operations can often be evaluated by replacing subexpressions by their limits; if the resulting expression does not determine the original limit, the expression is known as an indeterminate form." If you get an indeterminate form, this does not mean the original limit doesn't exist, it only means that some other method has to be used to compute it.

That is exactly what happens when you replace the subexpressions in your limit - the resulting fraction 0/0 no longer helps in computing the original limit. This happens more often with derivatives - if you take any continuous function, write the limit for the derivative, and replace the numerator and denominator with their limits, then you always get the indeterminate form 0/0, regardless whether the derivative exists or not. This does not mean that the substitution is invalid in any way. — Carl (CBM · talk) 13:24, 27 October 2008 (UTC)

Amen to that. The author says it is an indeterminate form because it is meaningless. What is your less obscure source on the subject? Dmcq (talk) 13:29, 27 October 2008 (UTC)

That book is also the primary reference for the Classification of discontinuities article. Dmcq (talk) 14:30, 27 October 2008 (UTC)

Sorry, I didn't mean to make an attack on the book; was just remarking that the authors don't seem to be analysts and an "authority" in the mathematical world, although of course, for us they indeed are, being a published book :) Shreevatsa (talk) 14:39, 28 October 2008 (UTC)

Both equations $2=\lim _{x\to 0^{+}}{\frac {2x}{x}}$ and ${\frac {\lim _{x\to 0^{+}}{2x}}{\lim _{x\to 0^{+}}{x}}}={\frac {0}{0}}$ are valid. The equality sign means that the left hand side and the right hand side are either both undefined, or both defined having the same value. If ${\frac {0}{0}}$ is indeterminate or undefined, then so is ${\frac {\lim _{x\to 0^{+}}{2x}}{\lim _{x\to 0^{+}}{x}}}.$ The substitution $\lim _{x\to 0^{+}}{\frac {2x}{x}}={\frac {\lim _{x\to 0^{+}}{2x}}{\lim _{x\to 0^{+}}{x}}}$ is unjustified because the fraction ${\frac {x}{y}}$ is either not continuous or not defined for y=0. The simplification $\lim _{x\to 0^{+}}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\to 0^{+}}{f(x)}}{\lim _{x\to 0^{+}}{g(x)}}}$ is justified only when $\lim _{x\to 0^{+}}{g(x)}\neq 0.$ The situation that the error was detected late does not mean that it occurred late. The error may not be detected until the alarming expression ${\frac {0}{0}}$ emerged. I read Carl's contribution as if you consider $\lim _{x\to 0^{+}}{\frac {2x}{x}}={\frac {0}{0}}$ to be valid. Am I wrong? Bo Jacoby (talk) 15:49, 27 October 2008 (UTC).

What is the purpose of all that stuff you wrote? Are you disputing the interpretation of a source? Or are you wanting a lesson in analysis? Dmcq (talk) 15:58, 27 October 2008 (UTC)

I was asking the meaning of Carl's contribution above. Bo Jacoby (talk) 23:18, 27 October 2008 (UTC).

Yes, I consider $\lim _{x\to 0^{+}}{\frac {2x}{x}}={\frac {0}{0}}$ to be a valid computation, but not at all helpful in computing the limit. It is certainly true that standard practice uses the equals sign there to mean something other than literal equality (random example on the web). But just because you get an indeterminate form when computing a limit does not mean that the limit itself is undefined. It can mean you need to simplify the limit expression, or need to apply L'Hopital's rule, or need to find the right estimate to compute the limit. — Carl (CBM · talk) 23:46, 27 October 2008 (UTC)

Thank you, Carl. Perhaps we are approaching an understanding of the sources of our disagreement. My position is that 2=0/0=3 is invalid because both equations 2=0/0 and 0/0=3 are unjustified, while your position is that 2=0/0=3 is invalid, even if both equations 2=0/0 and 0/0=3 are valid, because 0/0 is indeterminate. Am I right? That would explain why any definition of 0/0 or 0<sup>0</sup> is so distasteful to you, but harmless to me. We do agree that 2=0/0 is abuse of the equality sign to mean something other than equality. Do you think that this wikipedia article on exponentation should abuse the equality sign without a word of warning to the reader, just because such abuse is done elsewhere on the web? Bo Jacoby (talk) 09:53, 28 October 2008 (UTC).

Yes, you have my opinion correct. Of course a definition of 0^0 in algebraic contexts, or in the notation for infinite series, is fine by me, and I use it all the time. In the article, we can avoid using expressions like

$\lim _{x\to 0^{+}}{\frac {2x}{x}}={\frac {0}{0}}$

by saying something such as

If the numerator and denominator in the limit $\lim _{x\to 0^{+}}2x/x$ are replaced by their respective limits, the indeterminate form 0/0 is obtained.

That avoids the issue of the equals sign entirely, and is also more clear about how the 0/0 was obtained. — Carl (CBM · talk) 13:08, 28 October 2008 (UTC)

Yes, or something such as "If the numerator and denominator in the limit $\lim _{x\to 0^{+}}2x/x$ are replaced by their respective limits, the expression 0/0 is obtained, but that replacement is not correct because the fraction x/y is not continuous for y=0, no matter if or how x/0 is defined". We should not trap the reader into using the equality sign and then pretend innocent just because we did not explicitely use the equality sign ourselves. Bo Jacoby (talk) 13:31, 28 October 2008 (UTC).

I wouldn't say it's "not correct", I would say it doesn't help in computing the actual value of the limit, because an indeterminate form is obtained. A specific reason not to say it's incorrect is that many students are explicitly taught to make that replacement as the first step in evaluating a limit... — Carl (CBM · talk) 13:36, 28 October 2008 (UTC)

I too agree that something like "If the numerator and denominator in the limit $\lim _{x\to 0^{+}}2x/x$ are replaced by their respective limits, the indeterminate form 0/0 is obtained" is a good idea. Shreevatsa (talk) 14:39, 28 October 2008 (UTC)

I'll put down a bit more explicitly how I interpret what's written. I interpret it that they can follow a program something like this:

Subroutine Evaluate_Limit(expression expr, variable x, limit_value a)

If constant(expr) return value(expr)

if variable(expr,x) return a

if operation(expr) then

try_value = Apply(operator(expr),

Evaluate_Limit(operand1(expr),x,a),

Evaluate_Limit(operand2(expr),x,a))

if regular_value(try_value) then return try_value

if infinite(try_value) then return try_value

If indeterminate(try_value) then

Say I couldn't find the limit by pushing the limit to subexpressions

Comment try l'Hopitals rule for instance

....

if function_call(expr) then

etc etc

End of Evaluate_Limit subroutine

Dmcq (talk) 15:05, 28 October 2008 (UTC)

In that programmed way of evaluating limits I would have

function Apply(operation oper, expression expr1, expression expr2)

if oper = Exponentiation then

if is_number(expr2) and is_number(expr2) then

if expr1 <> 0 then return expr1<sup>expr2</sup>

if expr2 < 0 then return projective infinity, may be + or -

if expr2 > 0 then return 0

return Indeterminate_form(Exponentiation, 0, 0)

...

To deal with 0<sup>0</sup> being always defined as 1 I would have to return something like {value 1, indeterminate form} or else have special duplicate versions of the Apply function say 'Apply_within_limit' which returns something different. Personally I think it is easier to accept 0<sup>0</sup> as an indeterminate form and that the actual value is 1 in most circumstances but that can't be determined just by looking at 0<sup>0</sup> on its own. In particular no contradiction will be reached by setting it to 1 if continuous limits are not involved and this is the obvious useful value to define for it when the arguments are integers. Dmcq (talk) 14:40, 14 December 2008 (UTC)

You have all missed the point, I believe

See the example above in "redundant equations", by Factspewer. This example does not show that 0^0 is undefined in the algebra of limits for real numbers, it is much more basic than that! Suitably altered, it shows that exponentiation cannot be, in general, defined on the real numbers. In particular, there is no sensible way to define the binary function of exponentiation for the input pair of real numbers (0,0).

This is where it helps to know the (a) construction of integers in terms of numbers (0,1,2,3,4,5,... for which there can be no conceptual reduction - you can only draw a tally in the sand and wave your arms around until people understand what you mean), rationals in terms of integers, and reals in terms of rationals. More precisely for example, we have a definition of integers in terms of equivalence classes of number-number pairs, rationals in terms of equivalence classes of integer-integer pairs, reals in (for example) terms of equivalence classes of Cauchy sequences of rationals. Knowing about this sort of thing can clarify things. Some may perjoratively call this stuff "logic", but it isn't really, it's just very nit-picking mathematics. But it is this sort of contemplation that led to the theory of p-adic numbers, so it is not entirely stupid.

Anyway, the confusion around "0^0=1" (see next paragraph for why the inverted commas), present even in professional mathematicians, results from a conflation of integers and reals and their power functions. These should technically not be conflated, it technically makes no sense (not wrong, nonsensical) to say the real number 5 equals the number 5. Of course we identify numbers with the corresponding real numbers via the embedding of N in R. Addition, multiplication and order are preserved, but exponentiation is not quite. This embedding does not allow one to talk about x^y, where x and y are real number variables ranging over R. So for the purposes of resolving talk about 0^0 where 0 signifies the real number 0, one has to abandon this habitual (usually harmless) identification of 0 and 0', 1 and 1', 2 and 2', etc.

Note that any phrase "0^0=..." makes no literal sense until you have clarified that you mean the real number 0. The binary power (to stop confusion with exp) function on numbers (0,1,2,3,4,5,...) can very well be defined at the number pair (0,0), by defining 0^0=1. This causes no problems and makes perfect sense. Likewise for integers. For reals it is not so simple, in fact there is no way to define exponentiation at (0,0).

Let us use the Cauchy sequences definition of reals. Then in the above example write 2 instead of e and write 1/n instead of x, to get a convergent (and hence Cauchy) sequence. Then insist that power:RxR->R is such that 0^0=1. Bang! You have destroyed the Cauchy definition of the real number system.

Okay, then we reject power:RxR->R being such that 0^0=1. What about the stuff raised in the article as a reason why 0^0=1 might be good to have, for example infinite series for e^x at x=0, geometric series at x=0, binomial theorem at x=0, power rule for derivatives at n=1 and x=0, etc. We want to get that sorted. And we can, because nothing I said above indicates that we cannot have a function power:RxN->R with all the usual properties. R is a field, and this is just the usual field exponentiation applied to R. Naturally we can have power:RxZ->R also.

In power:RxZ->R, power(0,0) is defined and the usual interpretation of power(0,0)=1 is true. In power:RxR->R, power(0,0) is undefined. We cannot have a binary operation on R with the properties that we want a power function to satisfy, that is defined everywhere. Se la vi, there isn't always a function that does what you want. Bozo9 19:59, 25 October 2008

The dispute here is not really about any essential truth, more about what referenced authorities say. At least that's what it should be about for wiki.

As to your points: It would be better to distinguish reals and natural numbers like you say, it would reduce acrimony, but it isn't essential. Continuous limits don't say anything about things which are only defined for integer values so one can't use the reasoning of indeterminate forms to say that any of the cases where 0<sup>0</sup>=1 is used are wrong. And even if one did extend an integer formula to a continuous real version in any of the cases where having 0<sup>0</sup>=1 matters you'd be practically bound to get 1 by the reasoning from indeterminate forms anyway. From the other point of view of course one could redefine indeterminate forms to leave out the 0<sup>0</sup> case or do something about having extra information attached to the exponential limit case like people writing atan<sup>−1</sup>(y/x) do when they say it produces the Arg function. Now that's a source of bugs! By the way there is a push to have the IEEE floating point standard changed to have two versions of power like you say, pown for the integer exponent case which returns 1 and powr with a real exponent which return NaN - their version of indeterminate. Dmcq (talk) 19:24, 25 October 2008 (UTC)

Dear Bozo9: I like your idea of letting there be two exponentation functions, one defined on R x N, and the other defined on (a subset of) R x R. If you know a reference for this point of view, I would suggest adding it as a third possibility in the "differing points of view" section. (But if not, probably it is best to leave things as they stand, for the reason just cited by Dmcq.) By the way, Dmcq, that IEEE proposal you mention is interesting; I didn't know about that; if you have a reference, perhaps that would be worth mentioning in the "treatment in computer languages" section. --FactSpewer (talk) 21:26, 25 October 2008 (UTC)

For pown powr see for [1] and it is in some other comments as desirable but I don't think it has been marked as a blocker issue by anyone. William Kahan stuck in the original pow(0,0)=1 into IEEE as he believed it was the most useful thing to do, I'm a bit surprised about that as he is usually very careful but I guess it is like the atan2 case, who'd in their right mind would return NaN for the (0,0) case of that in computing even though arg(0) is formally undefined. I wouldn't stick pown, powr in the main article without something far more definite. Dmcq (talk) 22:05, 25 October 2008 (UTC)

The integers are embedded in the reals. It is original research (and makes no sense) to distinguish between 0 and 0.0, between e and e+i0, or between 0<sup>0</sup> and 0<sup>0.0</sup> . Any function defined for integers is defined for reals having integer values. "The dispute here is not really about any essential truth, more about what referenced authorities say. At least that's what it should be about for wiki". Yes, but bringing referenced authoritites together in the same article may lead to logical inconsistencies and nonsense, as is the case now. The meaning of the statement: "0<sup>0</sup> is undefined", is context dependent. 0<sup>0</sup> is undefined in the local context of a book in which 0<sup>0</sup> is undefined, but 0<sup>0</sup> is not undefined in the wider context of an encyclopedic article in which 0<sup>0</sup> is defined. Bo Jacoby (talk) 08:25, 26 October 2008 (UTC).

You talk the talk but you don't walk the walk. No original research. Leave alone what you perceive as contradictions. Write a book about it and you can become an authority yourself. Dmcq (talk) 09:14, 26 October 2008 (UTC)

Re Bo: You're right that the literature is inconsistent about the status of 0^0. This article does a reasonable job of explaining how and (to an appropriately small degree) why the literature is inconsistent. That's all we can expect it to do, in the end. The article itself does not explain the distinction between 0^0 for integer values and 0^0 for real values; that distinction was just discussed on the talk page as a way of figuring out what is going on in the literature. — Carl (CBM · talk) 12:16, 26 October 2008 (UTC)

The assertion $\lim _{x\to 0^{+}}(e^{-1/x})^{x}=\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}\!$ is false even if 0<sup>0</sup> is undefined. That is just an elementary observation made by a reader who need not write a book about it. It is not a reasonable job to write a false statement in the article. Bo Jacoby (talk) 14:25, 26 October 2008 (UTC).

I understand you feel it is wrong. That's just not the point. Think of youself like those one on the IEEE floating point discussion talking about max(x,NaN)=x. For a person steeped in the idea of GIGO this is nonsense. But it's going in. And if it is reported on wiki it will be reported factually perceived contradiction and nonsense or not. Just like sqrt(-0)=-0 is defined in IEEE now. Make of that what you will. That's what collaboration and no original research is about. Dmcq (talk) 15:08, 26 October 2008 (UTC)

IEEE evaluates the boolean expression (e^(-1))=NaN to false, and (e^(-1))=1 evaluates to false, so, also according to IEEE, $\lim _{x\to 0^{+}}(e^{-1/x})^{x}=\left(\lim _{x\to 0^{+}}{e^{-1/x}}\right)^{\lim _{x\to 0^{+}}{x}}\!$ is false irrespectively on whether 0<sup>0</sup> is NaN or 1. Thank you for this referenced authority in support of my point of view. What I feel has nothing to do with it. Bo Jacoby (talk) 16:04, 26 October 2008 (UTC).

250 kilobytes

That's how much discussion has been spent on the 0^0 question in this Talk page to date. Realize that you're not going to convince someone who has been arguing with essentially all other editors on this subject for two years now.

To quote CMummert from last year:

All of these concerns have already been addressed, as you must be aware. There is no requirement that others must continue to respond to your comments when you raise no new issues. CMummert · talk 13:14, 28 March 2007 (UTC)

Need we continue now? —Steven G. Johnson (talk) 19:17, 28 October 2008 (UTC)

Oh I've written docments much longer that that! ;-) Dmcq (talk) 19:28, 28 October 2008 (UTC)

e^x

The derivation that e^x = e^k for integer x=k was removed by Dmcq. Why? Bo Jacoby (talk) 00:05, 30 October 2008 (UTC).

Because although the result was correct the derivation had a step in it which was implicitly wrong and would have required quite a bit of explanation to fix. I didn't seem necessary. If we're going to have proofs in the article they should at least be good ones. Dmcq (talk) 00:14, 30 October 2008 (UTC)

What was the problematic step? Bo Jacoby (talk) 00:17, 30 October 2008 (UTC).

I think that, correct or incorrect, the derivation isn't needed. In an article at this level, the typical reader isn't going to be interested in or aided by proofs. Moreover, the notation in the derivation that was there,

$e^{k}=\left(\lim _{|n|\rightarrow \infty }\left(1+{\frac {1}{n}}\right)^{n}\right)^{k}=\lim _{|n|\rightarrow \infty }\left(\left(1+{\frac {1}{n}}\right)^{n}\right)^{k}=\lim _{|n|\rightarrow \infty }\left(1+{\frac {k}{n\cdot k}}\right)^{n\cdot k}=\lim _{|m|\rightarrow \infty }\left(1+{\frac {k}{m}}\right)^{m}$

involves things like "|n| → ∞" that are not typically encountered in a first calculus course. Also, e itself is defined as

$e=\lim _{n\rightarrow \infty }\left(1+{\frac {1}{n}}\right)$

higher in the article (without the absolute value). I don't see that the appearance of the absolute value would be at all clear to a naive reader. But I also don't think it would be worth writing a long explanation for this derivation, which is already a tangential point in the article. — Carl (CBM · talk) 01:03, 30 October 2008 (UTC)

Sorry for wasting your time, the original was fine. It was just me I have a bad cold and I was going to bed. In fact I'm only here now as I'm down again to get myself a lemon drink and thought I better write an apology while I was up. Dmcq (talk) 01:57, 30 October 2008 (UTC)

I'll have a go at trying to fix that section into something reasonable. At the moment it is just stuck there and a naive reader wouldn't have the foggiest why it was important. That proof probably should be in as part of the reasoning. Dmcq (talk) 08:35, 30 October 2008 (UTC)

To Dmcq. Thank you for the apology. I hope you get better. To Carl. When k<0 and n>0 then m=nk<0 , and one needs the limit $\lim _{m\rightarrow -\infty }\left(1+{\frac {1}{m}}\right)=e.$ The exponential function is central to the generalization from integer to noninteger exponents. Without a hint about the new definition of e^x being a generalization of the old definition, the notation e^x is unmotivated. The text can be improved but I don't think it should be omitted. Bo Jacoby (talk) 08:54, 30 October 2008 (UTC).

The one thing I tried to do to this section was put a little motivation at the beginning so it isn't just a funny constant stuck in for no obvious reason. That has been reverted. What on earth is wrong with trying to make the mathematics seem as though it has any meaning or motivation? Less important is the business about sticking a proof inline. I believe even for important proofs it is better for something like Wikipedia to just state the result and results first and leave the proof for those who want to work through them. Wikipedia is not a textbook, and especially not the sort like those professors who write a whole lot of formula on the board without saying anything and with their back to the students. Dmcq (talk) 12:48, 31 October 2008 (UTC)

Bo Jacoby, I am reverting your change to this section on the grounds that it is clearly against the style guidelines for Wikipedia in general and Wikipedia:WikiProject Mathematics in particular. Could you please have a good look at Wikipedia:Make technical articles accessible and also Wikipedia:WikiProject General Audience about the problems caused by aiming too much at the subject rather than the audience. Also see the Wikipedia:Manual of Style (mathematics)#Proofs for guidance why inline proof is deprecated. Dmcq (talk) 13:19, 31 October 2008 (UTC)

There is an article on exponential function, and an article on e, so I think that this subsection of the article on exponentiation should not contain more detail than what is used in the later subsections. That's why I removed a sentence. I do not think that my change made the text neither inaccessible nor aimed at a special audience. (Remember the rule: improve rather than revert). I made the change for a reason. It is important to note that not only must the exponent be large, but also the deviation from unity be small in order that Euler's indeterminate form $\left(1+{\tfrac {1}{\infty }}\right)^{\infty }$ be finite. (Happily we are not discussing indeterminate forms here). So the words "while the number goes to one" were included, as I still think they ought to be. An explanation why (e)<sup>k</sup>=e<sup>k</sup> for negative integer k is still missing, but an explanation why e<sup>x+y</sup> = e<sup>x</sup>·e<sup>y</sup> would suffice, I think. Then we could also omit the proof that (e)<sup>k</sup> = e<sup>k</sup> for positive integer k. I don't consider the number 1.001<sup>1000</sup> to be an 'inline proof'. Bo Jacoby (talk) 14:34, 31 October 2008 (UTC).

The inline proof was that e raised to the k<sup>th</sup> power is the same as e<sup>k</sup>. The bit you removed on 1.0001<sup>1000</sup> was to show the form was the same as (1+1/n)<sup>n</sup> otherwise more text is needed to say the same thing. You removed the justification for having a separate section on e<sup>x</sup> when you removed the first sentence. It isn't about removing everything possible and using as few words as possible, it is about removing what is irrelevant to the subject of the article. There is nothing inherently wrong with words if they make the subject relevant to the reader, in fact that is a good thing. It already said about numbers close to 1 and about going to the limit. Dmcq (talk) 14:44, 31 October 2008 (UTC)

Let's take one step at a time. Do you object against "and is defined as the limit as the power goes to infinity" being replaced with "and is defined as the limit as the power goes to infinity while the deviation from one goes to zero" ? Step two. Do you think that the reader cannot identify 1.001<sup>1000</sup> and (1+1/1000)<sup>1000</sup> ? 15:58, 31 October 2008 (UTC).

Yes, it is better to show exactly how the two are related. And to the second part yes at this level it may not be quite obvious. Dmcq (talk) 17:05, 31 October 2008 (UTC)

exp(x)

The function exp(x) is important because it equals its own derivative, and hence arises in any application where something grows or decays in proportion to its current value. The number e is important not because it is the limit of (1+1/n)^n, but because it is the number such that e^x = exp(x) and hence gives a convenient notation for the exponential function that reminds the user of its multiplicative property.

With this in mind, one possibility might be to define exp(x) first (as the function equaling its derivative with exp(0)=1, and/or by the power series, and/or by the limit - probably it is best to mention all three, with the first one being primary), and then to state that there is a number e such that e^x (as defined earlier in the article) equals exp(x). Then there is no confusion between the notations, and as a bonus the formula that exp(k) equals e.e....e is an automatic consequence. --FactSpewer (talk) 01:44, 1 November 2008 (UTC)

The exponential function has got an article of its own. The article on exponentiation did not assume that the reader knows about derivatives, but integer exponentiation has just been explained. Therefore the definition e = lim(1+1/n)<sup>n</sup> is more appropriate here. The challenge here is to explain that the new definition of e<sup>k</sup> matches the old definition for integer values of k, positive, negative and zero, in short: (e)<sup>k</sup> = e<sup>k</sup>, so that the chain of logic is not broken. A proof that e<sup>x+y</sup> = e<sup>x</sup>·e<sup>y</sup> would be helpful for showing (e)<sup>k</sup> = e<sup>k</sup>, but e<sup>x+y</sup> = e<sup>x</sup>·e<sup>y</sup> is not necessarily true for square matrices. Bo Jacoby (talk) 07:20, 1 November 2008 (UTC).

Defining exp(x) and then saying it is equal to e<sup>x</sup> sounds way over the top here, but I can see you don't like having the x in the place of a power until there is some justification for that. I'll have a go at moving the power law identity up as Bo Jacoby had it before, we don't need an immediate proof, and then add a little bit on the proof that is in there saying it also showed e<sup>x</sup> satisfied the power law identity for positive integers. We can just state that this can also be proved for negative and rationals numbers too, and in the limit for real and complex numbers, though it fails for instance for matrices which do not commute. Dmcq (talk) 08:53, 1 November 2008 (UTC)

A couple of thoughts and questions:

• I agree with Bo Jacoby that the stuff on the exponential function would fit better in the exponential function article.
• There is also already an article on e. What is the purpose of this section on e^x? (I don't mean for this to be a rhetorical question; I really want to know what your vision of this section is, in relation to the rest of the exponentiation article.) Given what already exists on these other pages, might it suffice just to say something like "There is a real number e about 2.718... such that the function e^x plays an important role, as explained in exponential function?"
• Are there readers with the mathematical sophistication to understand the proofs you are hoping to include in this section and who have not heard of derivatives?
• Are you sure you want to include all these proofs? If so, do you also want to justify the limit definition, i.e., that the limit exists? (If you are trying to keep this elementary, I'm guessing no.) Without justification for the definition, the subsequent proofs are conditional.
• If we really want to include some proofs, it might be easier to understand the power series definition of the exponential function instead of the limit definition. (I know that the series is defined as a limit, but people who never seen a limit before can still pretend to themselves that they understand what an infinite series is.) The formula exp(x+y) = exp(x) exp(y) is really easy from this definition (assuming that you know the binomial formula, and are willing to accept rearranging the infinite series).--FactSpewer (talk) 03:39, 2 November 2008 (UTC)

I think that the article on exponentiation should explain a<sup>x</sup> where a is a positive real number and x is an arbitrary complex number. The exponential function e<sup>x</sup> is a convenient intermediate step. (This is to answer FactSpewer's question of the purpose of the section on e<sup>x</sup>). There are two common definitions of e<sup>x</sup> , namely the limit lim(1+1/n)<sup>n</sup> and the Taylor series Σx<sup>n</sup>/n! . The power a<sup>n</sup> where n is an integer has just been explained, but some readers may at this stage not be comfortable with the theory of derivatives, which is more advanced than the theory of exponentiation. That's why I preferred the limit definition to the Taylor series definition in this context. Perhaps we should have a very short big exponent subsection , telling that lim<sub>n→∞</sub>a<sup>n</sup> is zero for |a| < 1 and infinite for |a| > 1 and equal to unity for |a| < 1 and undefined elsewhere, e.g. for a = −1 . This motivates investigating a big power of a number close to one. It is interesting that (1+x/n)<sup>n</sup> is virtually independent on n for big values of n. (1% of interest in 10 years almost equals 0.5% of interest in 20 years). I don't know an equally elementary motivation for studying the Taylor series. I think that the equally easy formulas lim(1+(x+y)/n)<sup>n</sup> = lim(1+x/n)<sup>n</sup>·lim(1+y/n)<sup>n</sup> and Σ(x+y)<sup>n</sup>/n! = (Σx<sup>n</sup>/n!)·(Σy<sup>n</sup>/n!) should be found in the article on exponential function. Bo Jacoby (talk) 06:30, 2 November 2008 (UTC).

Agree with Bo Jacoby mainly though I think we can leave out anything extra about limits.

There are two reasons I believe the section in included:

The exponential function is needed to define real and complex exponentiation

The exponential function is a very notable exponentiation. It is an exponentiation with its own special name.

The first is mentioned explicitly at the start and the second is implicit. I do not believe the section would deserve inclusion here for a reason like that it is equal to its derivative. And π<sup>x</sup> is not included because it is not notable enough.

Limits need no definition here. They have been illustrated in going from the particular case of n=1000. There is a link to the limits article (though it will probably need updating when some argument about articles on limits is resolved). Limit isn't what is important to the article. Giving some justification that e<sup>x</sup> is really an exponentiation is the point of the proof, any extra properties of the exponential function are not relevant.

I don't see that Taylor series would be any advantage - it would mean having two limits instead, and anyway the definition here is older and more intuitive for its connection to exponentiation. It's not needed so why change the definition?

Polished up proofs can go into wiki but they need to be stuck into an article pitched at the right level and the exponentiation article is just not that, especially not a section like this one just discussing real exponentiation. PlanetMath or the Mizar system is the place to go if the proof is all the and the reader is unimportant. It quite possibly is notable enough to be worth having a separate article for proving the property in general with the full rigour and then both this and the exponential function could link to it.

Personally as to the exponentiation article the main thing I feel is currently missing from this section is a reference pointing to an external source for a full proof. Dmcq (talk) 08:37, 2 November 2008 (UTC)

Just been looking at he exponential function and it never shows that it satisfies exp(x+y)=exp(x).exp(y) and I haven't found it elsewhere in wiki. Using differential calculus it is easy and wouldn't be out of place there, it could then be referred to from this article for the full generality. Dmcq (talk) 08:54, 2 November 2008 (UTC)

The two reasons that Dmcq gave are the same as the ones I would have given, except that it is not needed for real exponentiation (which has already been defined via the limit process!) Given that these are the only two goals, and that most of the material here fits better in either the e article or the exponential function article, I would suggest eliminating this section and placing the following sentence in the section where exponentiation of positive real numbers is defined.

There is a real number e, about 2.718, such that e<sup>x</sup> equals the exponential function, which plays an important role in calculus.

And then of course, the function e^z would be mentioned later on when defining complex exponentiation. Is there anything else that needs to be said here that is not redundant with what is (or should be) in the e and exponential function articles?--FactSpewer (talk) 08:07, 3 November 2008 (UTC)

e<sup>x</sup> is not needed for real exponentiation, but it is needed for complex exponentiation, of which real exponentiation is a special case. So the definition of real exponentiation as a limit of rational exponentiation is the one that is not needed. Bo Jacoby (talk) 08:39, 3 November 2008 (UTC).

Notability is the first priority for wiki and e<sup>x</sup> is very notable. And whatever about not being needed in a mathematical sense to define the powers of real numbers that is how they are normally calculated. Everytime I look at a mathematical article someone is trying to remove any sense or meaning and polish it into ivory tower purity and aloofness according to their idea of the one true way to wisdom. For instance the bit on large powers was good but it relegated the reasons for e<sup>x</sup> being included to a footnote so the next section was again reduced to being like a professor writing formulae on the board with his back to the room and leaving the room without ever saying a word. Dmcq (talk) 11:37, 3 November 2008 (UTC)

it seems as if Dmcq and I agree that e<sup>x</sup> should be defined here, as it is, in order to prepare for defining a<sup>x</sup>, but that FactSpewer disagrees. Bo Jacoby (talk) 22:03, 3 November 2008 (UTC).

Yes, I guess so. I suppose it's just a matter of taste as to how much to put in here, and how much to refer to other articles that contain the same material.

In any case, I agree with Dmcq that the limit definition of a^x for a>0 should stay here. It is a definition that is common in many textbooks, and if it doesn't go into this article, I worry that it might not be covered anywhere else in Wikipedia. Exponentiation is the natural place for it.--FactSpewer (talk) 02:44, 5 November 2008 (UTC)

Actually I didn't think I said much about the limit definition of a^x, in fact now we have two definitions of that one from limits under rational powers and one from e^x and ln. I guess the one under rational power should be moved down so they were in the same section. I' have a go at doing that now. Dmcq (talk) 11:03, 5 November 2008 (UTC)

I think I'll also move the definition of e down from the end of the powers with big exponents section to the start of the e^x section again. In its defintiion both the number and the power are being varied at the same time and it gives motivation for the e^x section Dmcq (talk) 11:28, 5 November 2008 (UTC)

The latest reordering of the subsections is not satisfactory. The value of (1+n<sup>−1</sup>)<sup>n</sup> for big integer values of n belongs naturally in the subsection big exponents of the section on integer exponents. The definition of e<sup>x</sup> belongs naturally under the heading powers of positive real numbers, and equally naturally immediately after the definition of e, which was last in the previous section. The characterizations of a<sup>1/n</sup> = e<sup>(ln a)/n</sup> as the positive solution of the algebraic equation x<sup>n</sup> = a should be included, but not confuse the chain of logic. Bo Jacoby (talk) 22:53, 5 November 2008 (UTC).

The value of e doesn't really fall under big exponents. The number that is being raised to a power is being changed rather than just a number being raised to a big exponent. And there isn't really any point in sticking e in the integer powers section, it only starts becoming relevant when you get to real powers. Once it is in the exp section there is no rationale for moving exp up. It also puts the rationale back into the exp section.

I put both of the definitions of raising to a real power together as it seemed strange separating them and the one using a limit was in the wrong place under rational powers. nth root and rational power needs to be done before using rational powers in the limit definition of a real power. Dmcq (talk) 23:19, 5 November 2008 (UTC)

For big values of n, (a+bn<sup>−1</sup>+cn<sup>−2</sup>+···)<sup>n</sup> goes to infinity if a > 1 and towards zero if |a| < 1 , and towards a positive value, (e<sup>b</sup>), if a = 1. These facts belong to big exponents. Bo Jacoby (talk) 00:00, 6 November 2008 (UTC).

That's just not notable enough to go into the integer section. It has very little to do with big exponents. You might as well say it belongs with powers of one wich i doesn't either. It is related to exp and what is said about exp is quite enough on the matter. And it is hardly a good reason to mangle the rest of the article. Dmcq (talk) 00:11, 6 November 2008 (UTC)

How then make the logical connection between the integer exponents and the exponential function? Bo Jacoby (talk) 07:56, 6 November 2008 (UTC).

In the section about the exponential function. I don't see that splitting it into two different sections makes for a better logical connection.

I notice you also tend to move any overall summaries to the bottoms of the sections and just dive into the maths at the top. I guess this is also part of the business about logical building up. However I think for this medium you're much better off following the dictum 'tell them what you're going to say, say, it, tell them what you've said'. Dmcq (talk) 08:35, 6 November 2008 (UTC)

If a section is called powers of e, then e should be defined or explained already. Otherwise the reader cannot understand the header. The reader should be able to read exponentiation without having to study exponential function. The dictum of repetition is fine for oral presentations, but not necessarily for an encyclopedia, where the reader is free to reread several times. I would hate to sacrifice logic on the altar of smalltalk. Bo Jacoby (talk) 12:13, 6 November 2008 (UTC).

With that logic you can't ever say what you are about to describe which explains why you remove explanations at the beginning. It is a wonder you can abide even having titles to articles. Even the Mizar system describes things before delving into them. It isn't a viable position for an encyclopaedia. Dmcq (talk) 13:15, 6 November 2008 (UTC)

I don't think we disagree as much as it seems, nor that I remove explanations in the beginning. I am the one who want to define e before it is used. :-) Bo Jacoby (talk) 13:30, 6 November 2008 (UTC).

We do differ. You want to move e to a section where it is not mentioned in the head and seems unrelated and then start the exp section without saying anything about its purpose. You already moved out the heading which gave purpose and moved the purpose down to the bottom of the big exponents section from where I moved it back. You also moved what was originally at the start of the real powers section to the very bottom after everything else to do with real powers and I went and restored that also to ts original position. And you put exp at the start of the real powers section without any text at all except the powers of e heading between the real powers heading and a definition of the exponential function. That sounds like removing explanations at the beginning to me. Dmcq (talk) 19:02, 6 November 2008 (UTC)

I know what I did and I know what you did: you introduced a big gap between big exponents and powers of e which are closely connected logically. The algebraic definition of fractional exponents does not lead to complex exponents, but you gave it priority in favour of the powers of e section that is the road towards generalization of exponentiation. Yes, here we differ. But we do not differ in our attempts to write a good encyclopedia. You are not right in assuming that I want to remove explanations. Bo Jacoby (talk) 23:30, 6 November 2008 (UTC).

The logical connection is small unless you stick e into the big exponents section where it doesn't belong. It isn't a big exponent expression. it is simply e. The removing of all explanatory text about real powers between the two was especially unnecessary, even if they were related there was no need for them to be together. And even if e was given in the big exponents section the explanatory bit at the top of the exp section should have been left there rather than moved under big exponents where it wouldn't be found. You seem to have some idea that people will read the stuff serially like a book whereas most of the readers will have the attention span of gnats and just want to drill down to the bit they are interested in using the contents list. As to the other business you moved rational powers after real powers when limits of rational powers are used in a definition of real powers. Dmcq (talk) 01:03, 7 November 2008 (UTC)

I'd like to have

1. the big exponent subsection of the integer exponent section to include the definition e = lim(1+1/n)<sup>n</sup>, because that is how e can be explained at this level of sophistication, and n is a big, integer exponent.
2. the exponential function e<sup>x</sup> = lim(1+x/n)<sup>n</sup> to be defined in the beginning of the section on real powers of positive real numbers, because that is the road leading all the way to a<sup>z</sup> for positive a and complex z.
3. the proof or explanation why e<sup>x</sup> = (e)<sup>x</sup> for integer k, positive, negative and zero. It goes without saying that e<sup>1</sup> = e, so one need not have read the big exponent subsection definition of e in order to understand this definition of e<sup>x</sup> .
4. the real exponent subsection defining a<sup>x</sup> = e<sup>x ln(a)</sup> for positive a and real x.
5. the rational exponent subsection explaining that x=a<sup>1/n</sup> solves a=x<sup>n</sup>. A comment that this has historically been an alternative definition, but it is a blind alley not leading to defining complex exponents. The radical notation of n'th root.
6. then follows the complex powers of positive real numbers section.

Both serial and random access reading of the article should be possible. Bo Jacoby (talk) 09:08, 8 November 2008 (UTC).

I have already explained why I consider what you are doing to be against the idea of an encyclopaedia and you are repeating that you want to go ahead with it. Firstly there should be explanations of what sections are about at the top of the sections. Secondly the terms in the headers do not need to be explained before a section, they can be explained at the start of a section - otherwise articles couldn't have titles. Thirdly there is no one true way to a power of a real number via exp and log, using limits is an alternate notable route and more obvious too.

We have discussed this and I see no prospect of coming to a consensus agreement. Therefore I am raising the matter to Wikipedia_talk:WikiProject_Mathematics in the hope someone there can mediate something. Dmcq (talk) 12:37, 8 November 2008 (UTC)

I have raised mention of this at Wikipedia_talk:WikiProject_Mathematics#Exponentiation wars Dmcq (talk) 13:36, 8 November 2008 (UTC)

1. there should be explanations of what sections are about at the top of the sections. I generally agree.
2. the terms in the headers do not need to be explained before a section. No, but it is an advantage.
3. there is no one true way to a power of a real number. No, but a presentation has got to use one of the ways, and then explain that the other way also exists. One cannot walk two ways simultaneously.
4. I have already explained. Perhaps you haven't explained it well enough.
5. what you are doing. Perhaps you did not understand what I was doing.
6. Exponentiation wars. This is a discussion, not a war. I am listening to you and I am answering. And it takes place on the discussion page, not in the article.

Bo Jacoby (talk) 23:11, 8 November 2008 (UTC).

Name for basic identity

It occurs to me that I don't have a standard name for the basic identity a<sup>b+c</sup>=a<sup>b</sup>a<sup>c</sup>. Is there one? Dmcq (talk) 09:55, 1 November 2008 (UTC)

There may be names for it, but I don't think they are universally used. When a is e, I think some people call it the addition formula for the exponential function, or the functional equation for the exponential function. I'd suggest leaving it unnamed.--FactSpewer (talk) 03:43, 2 November 2008 (UTC)

Notation in Big exponents

The notation in big exponents is fairly obvious, but not the usual notation used elsewhere that I know of. Is this a standard way of writing limits anywhere? I found I had problems translating it into English, the 'for' in the statement I translated as 'as'. Dmcq (talk) 16:07, 5 November 2008 (UTC)

That's a fairly standard notation, in my experience. The alternative is using $\lim$. I would normally say "as" rather than "for". --Tango (talk) 14:57, 8 November 2008 (UTC)

Okay I've stuck in 'for' so it says 'for n going to infinity' instead of 'as n goes to infinity'. It doesn't quite read right to me but can't always be happy Dmcq (talk) 15:22, 8 November 2008 (UTC)

Ulp, misread - I'll have to rethink Dmcq (talk)

I would say "f(n) tends to L as n tends to a" written "f(n) -> L as n -> a". --Tango (talk) 15:36, 8 November 2008 (UTC)

Non

About Loadmaster's edit: Does Wikipedia have a standard for hyphenation in words like nonzero? If not, is there a notable reference that recommends inserting a hyphen? Table 6.1 of the Chicago Manual of Style lists non as a prefix forming closed compounds --- nonzero, nonnegative, etc.WardenWalk (talk) 21:36, 8 November 2008 (UTC)

I believe nonzero has a definite edge in maths. Put a question on Loadmaster's talk page if you want to see why he did it, that's the best way of getting a response from an editor. And if you really don't like non-zero revert it, doing that once when there's been no discussion is reasonable. Dmcq (talk) 22:06, 8 November 2008 (UTC)

I made the changes so that all the non-xxx words were consistent in the article (e.g., non-integer, non-negative, non-rational, etc.). There were also instances of both nonzero and non-zero. It seems stange that nonzero is preferred over the hyphenated form while the opposite is true for all other terms. Personally, I find the hyphenated form clearer. But no, I do not have a specific grammatical guide as a reference. — Loadmaster (talk) 19:00, 11 November 2008 (UTC)

Yes, I understand that you were trying to make it consistent. But on what basis do you claim that the hyphenated form is preferred for all terms other than nonzero? Dictionary.com recognizes nonnegative as a word, but not non-negative (even though apparently the latter does come up elsewhere on Wikipedia). The Mac OS X Dashboard dictionary recognizes nonzero and nonnegative but not non-zero or non-negative. In fact, I would say that most other words involving non are typically spelled without a hyphen: nonsense, nonprofit, etc. There is a longer list in the Chicago Manual of Style, and in the Mac OS X Dashboard dictionary (enter "non"). The grammar references I have checked suggest that the hyphen is used only when omitting it would lead to confusion or would suggest an incorrect pronunciation, or when the rest of the word is capitalized (e.g., non-English), though some say that British English has a tendency to hyphenate more than American English. Do you have any reason, beyond your personal preference, for keeping the hyphens?WardenWalk (talk) 16:40, 14 November 2008 (UTC)

No. By all means, if you have authoritative references for such a style, use them. It would also be useful if the use of non prefixes was mentioned in more detail in WP's style guides. The only thing I've found is in the Wikipedia:Manual of Style. It's interesting to note that even that page itself uses hyphenated/non-hyphenated non prefixes inconsistently. Just remember that the goal is to make the articles self-consistent. | Loadmaster (talk) 20:09, 14 November 2008 (UTC)

Thank you for pointing out the hyphenation information in Wikipedia:Manual of Style. I should have thought of looking there, even though it is not so clear what it is advocating, and you're right that it is inconsistent itself! For now, I've removed the hyphens in this article.WardenWalk (talk) 00:29, 16 November 2008 (UTC)

Negative odd powers of negative reals

I'm wondering if the rule for negative reals with odd integer powers holds when a non-integer power is split into a product having an integer exponent and a fractional exponent. Given a non-integer a and non-zero x, then for some integer n such that a = n+f,

$x^{a}=x^{n+f}=x^{n}\cdot x^{f}$,

Does it then follow that if x is a negative real and n is an odd integer, that the result has a negative real component? For example, is it true that

$(-7)^{-2.6}=(-7)^{-3}\cdot (-7)^{0.4}=-(7^{-3})\cdot (-7)^{1.4}=-\left({\frac {1}{7^{3}}}\right)\cdot (-7)^{0.4}\quad (?)$

Compare this to an equivalent equation that uses an even integer n:

$(-7)^{-2.6}=(-7)^{-4}\cdot (-7)^{1.4}=+(7^{-4})\cdot (-7)^{1.4}=+\left({\frac {1}{7^{4}}}\right)\cdot (-7)^{1.4}\quad (?)$

It appears that the two results differ in sign, which seems to be inconsistent. Another example:

$(-7)^{1.2}=(-7)^{-3}\cdot (-7)^{4.2}=-(7^{-3})\cdot (-7)^{4.2}=-\left({\frac {1}{7^{3}}}\right)\cdot (-7)^{4.2}\quad (?)$

$(-7)^{1.2}=(-7)^{-2}\cdot (-7)^{3.2}=+(7^{-2})\cdot (-7)^{3.2}=+\left({\frac {1}{7^{2}}}\right)\cdot (-7)^{3.2}\quad (?)$

Is this correct, or did I miss something? — Loadmaster (talk) 20:27, 11 November 2008 (UTC)

You seem to be assuming that (-7)<sup>0.4</sup> and (-7)<sup>1.4</sup> have the same sign (for the real part), but that's clearly not the case since (-7)<sup>1.4</sup>=-7*(-7)<sup>0.4</sup>. Multiplying by a negative number is going to change the sign of the real part. --Tango (talk) 20:35, 11 November 2008 (UTC)

Doh. I thought it might be something more complicated, like what is dicussed about odd/even rational exponents in the "Powers of negative real numbers" section. — Loadmaster (talk) 21:10, 11 November 2008 (UTC)

Section on zero to the power of zero

I like this section very much. However, in the discussion of the limit of f(x)<sup>g(x)</sup> when f(x) and g(x) tend to 0, these functions are required to be "positive-valued". To me, this makes perfect sense for f. But it seems pointless for g, unless we want to allow f to take the value 0, which is of little benefit.

I would suggest we require f to be strictly positive, and remove any restriction on the sign of g. That way, the possible limits can be anything in [0,+∞], instead of [0,1]. The part about [0,1] gives one the misleading impression that 0<sup>0</sup> is really trying to be a number between 0 and 1, if one doesn't pay close attention to the restriction imposed on g. 67.150.253.241 (talk) 14:39, 15 November 2008 (UTC)

Sounds reasonable, especially since that is what indeterminate form uses! Might as well be consistent. Why not have a go and see if you can do it better? It was only done this way to fit in with the example where you'd get e<sup>+∞</sup> as x went to 0 from below. It would be nice to keep that example as it is historically important. Dmcq (talk) 15:02, 15 November 2008 (UTC)

Actually, I just noticed that a footnote mentions the limit of x<sup>y</sup> along the x axis. Certainly, there are other paths to the origin which give the limit 0, but this is the simplest. So what example to use will take some thought once we've disallowed the value 0 for f. I'll come back to this sometime soon.

It might be worth mentioning that the function x<sup>y</sup> defined on R<sub>+</sub><sup>*</sup> × R, viewed as a subspace of $\mathbf {\overline {R}} ^{2}$, has a limit at all accumulation points of the subspace, except for (0,0), (+∞,0), (1,+∞) and (1,-∞). 67.150.254.179 (talk) 17:44, 15 November 2008 (UTC)

Limits of powers

I looked back and there used to be a section 'Powers with infinity'. I propose bringing something similar back to be a real or complex number equivalent to Large exponents the main purpose of which would be to talk about real and complex limits, and I'd put it just after the 0^0 section. I note a bit has just been added to the 0^0 section which would fit nicely as a basis for the new section. The new section would duplicate a bit in the current Large exponent section but that would be unsuitable for adding to as it is right at the top dealing with integers. Dmcq (talk) 12:28, 19 November 2008 (UTC)

Trimming complex logarithm material

About a month ago, I tried rewriting the complex logarithm material in exponentiation, but the consensus here was that the material belonged in the complex logarithm article instead. An expanded version of this material has since been put there, so I will now trim the complex logarithm material in exponentiation, replacing it with links where appropriate.

The section on powers of the imaginary unit now seems out of place. The main dichotomy in the definitions of exponentiation is between integer exponents and real/complex exponents. Integer powers of i belong in the Exponentiation with Integer Exponents section (it matters very little whether the base is real or not). --FactSpewer (talk) 05:37, 26 November 2008 (UTC)

It was there a while back and I moved it down to the complex section for a couple of reasons.

Exponentiation is a function of both its parameters not just one. There are some general principles in the integer section as well as integers and reals to an integer power but there's no general principle shown by i to an integer power any more than a matrix to an integer power.

Article should be aimed at the level of the readers. There are a couple of levels with exponentiation and I was trying to put the simplest first. Just because i can be raised to an integer power doesn't make it simple. It has more context in the complex number to a power section.

I didn't think it necessary to separate the integer to an integer power and real to an integer power sections at the start, nor the various different types of power of a complex number, but I wonder now with the way some people think. Dmcq (talk) 10:04, 26 November 2008 (UTC)

I think just changing the section to 'Integer powers of complex numbers' and having the bit about powers of i as a sentence within it would be best. Dmcq (talk) 10:23, 26 November 2008 (UTC)

Just struck me reading it again that powers of negative real numbers should reference forwards to complex numbers as a domain where there are solutions even if there aren't within the reals. Also I rather liked the picture of the spiral for logs illustrating the point about the ambiguity of values. Different people learn in different ways and one should put n illustrations of ideas as well as words and symbols. Pity we can't put in sound and feelies too. Dmcq (talk) 10:32, 26 November 2008 (UTC)

By the way I was intrigued to see that besides the sectio removal and moving bits of text elsewhere, you removed as well as the images the following sentence: "The computation of complex powers is facilitated by converting the base a to polar form, as described in detail below." Why remove both images and a helpful pointer? It doesn't harm to make things easy for the user. Dmcq (talk) 11:17, 26 November 2008 (UTC)

I don't feel strongly about the positioning of the "powers of i" section, so if you do, then let's just leave it where it is. By the way, could you explain what you meant by "general principles"? Also, why is the definition of i^n more complicated than the definition of r^s for real numbers r and s? I would have said that the opposite is true (look at how much space is devoted to the definitions). I'm not sure I understood the points you were making here.

I agree that having the "powers of negative real numbers" section refer to the complex number section is reasonable.

I like the spiral picture for logs too! It hasn't been deleted from Wikipedia, but just moved to the complex logarithm page. Same for the color map; these items are primarily relevant to the complex logarithm.

The removal of the sentence "The computation of complex powers is facilitated..." was unintentional; sorry about that; I'll try to put it back. On the other hand, probably that section that it refers to could use some trimming too. Do we think that exponentiation formulas like

$\left(r^{c}e^{-d\theta }\right)\left[\cos(d\log r+c\theta )+i\sin(d\log r+c\theta )\right]$

are worth remembering? --FactSpewer (talk) 01:01, 27 November 2008 (UTC)

Thanks the changes you've made seem good.

Can't say that expression above looks pretty!

One can put the same picture in different articles if they fit the context.

Euler believed in putting complex numbers at the start of his book but it isn't common practice nowadays so to cater for readers it's probably better to have the non complex number bits first. I see the complex number section starting off at about the powers of negative real numbers section. It could be rearranged a little I think - for instance I was wondering would moving Complex power of a complex number after Roots of arbitrary complex numbers be better.

By general principles I just meant the exponential identities and properties which are described informally in the integer section. This is similar to the stuff in the Exponentiation in abstract algebra section but more accessible.

In fact there's four topics in the integer section that I can see, the identities and properties, the combinatorial interpretation, the special integers, and the limits. I suppose one could forward reference complex numbers from the powers of -1 if i wasn't in the integer powers section but just a special number. Dmcq (talk) 01:25, 28 November 2008 (UTC)

Regarding "same picture in different articles": agreed. I just didn't feel that these pictures, though, fit the context very well here. So I feel that it's better not to include those pictures here.

Moving Complex power of a complex number after Roots of arbitrary complex numbers is an interesting idea, but I think it is better to keep it as it is, given that this is an article about exponentiation. My feeling is that the section on roots is there just to explain how complex exponentiation (after it is defined) relates to the concept of roots.

I'll try adding your idea to have the -1 section refer forward to the i section. --FactSpewer (talk) 20:04, 29 November 2008 (UTC)

Speed

What about the speed of calculation is it polynomial timme? --Melab±1 ☎ 01:12, 30 December 2008 (UTC)

0<sup>0</sup> (yet again)

I don't mean to start a whole 'nuther discussion on 0<sup>0</sup>, but I do want to make the observation that if 0<sup>0</sup> = 1 while 0<sup>x</sup> = 0 for all x > 0, doesn't this define a discontinuous function? | Loadmaster (talk) 19:58, 30 December 2008 (UTC)

Yes. The function x<sup>y</sup> is necessarily discontinuous at x=y=0, no matter how one defines 0<sup>0</sup>. You get different limits for 0<sup>t</sup> and t<sup>0</sup> as t goes towards 0, so there is no hope of continuity from all directions. –Henning Makholm (talk) 06:05, 14 April 2009 (UTC)

Limit argument for x^0=1

User:Anonymous Dissident inserted the following argument at the end of the "exponents one and two" section:

The conclusion that $n^{0}=1$ where $n\neq 0$ is further supported by systematically comparing values derived from taking a set number to powers below 1.

$3^{1}$ = 3

$3^{0.5}\approx$ 1.7320508075689

$3^{0.25}\approx$ 1.3160740129525

$3^{0.1}\approx$ 1.1161231740339

$3^{0.01}\approx$ 1.0110466919379

$3^{0.0001}\approx$ 1.0001098672638

Thus a limit can be derived to express how $3^{x}$ behaves as x approaches 0.

$\lim _{x\to 0}3^{x}=1$

I removed it with the summary: This argument is logically out of place; non-integral exponents are only defined much later in the progression -- with arguments that assume all integers done already. AD now writes this on my talk page:

Could you explain why you thought this argument was out of place? The section was discussing the 0 exponent. —Anonymous Dissident<sup>Talk</sup> 05:48, 14 April 2009 (UTC)

Yes, the conclusion is about exponent 0, but the arguments are about rational exponents that converge towards exponent 0. In the ordinary development of exponentiations, rational exponents come after integral exponents, so properties of rational exponents are not convincing arguments for how to define things for exponent 0. If the only thing that justfied x^0=1 were that it would allow a nice limiting property later in the development, it would make sense to give it here, but given that there is plenty of independent justification, this forward reference just introduces needless complexity. –Henning Makholm (talk) 05:58, 14 April 2009 (UTC)

So, essentially, you contend that my argument is too complex for that place in the article? I think I'd have to disagree; the section is titled "exponents one and zero", so therefore all information pertinent to that should be incorporated there. —Anonymous Dissident<sup>Talk</sup> 06:04, 14 April 2009 (UTC)

No, I contend that your argument is circular. It presupposes definitions about rational exponents which themselves are informed by how things work out for integral exponents. It is not a compelling argument because it can only be made after the decision it argues for has been made. (I might be convinced otherwise if you can cite a respected textbook that includes your argument when it tries to justify the defintion of x^0). –Henning Makholm (talk) 06:09, 14 April 2009 (UTC)

I see what you're saying now, but I don't think I understand what you refer to when you call the argument circular. In any case, this is not a mathematical lemma and the intent was not to argue or prove anything. I just thought it was an easy way to explain why x^0 = 1, as it presents a visible demonstration of values of 3^x getting closer to 1 as x gets closer to 0. —Anonymous Dissident<sup>Talk</sup> 06:19, 14 April 2009 (UTC)

Okay, perhaps I should have said ... part of a circular argument.

Mathematically speaking, the only correct answer to why x^0=1 is: "Because that is how it is defined". If we dig deeper and ask why it was defined that way, we necessarily get into "fuzzy" arguing about utility and convenience, and you say that was not your intention. You won't get red marks for digging that deep immediately, but then what you dig up had better be a compelling argument for choosing that definition in the first place, rather than calculations that assume as a given the mathematics that follow from that definition.

Furthermore, even if the argument is only to be taken as an instructive example, we really ought not to suggest to our readers that such limiting arguments are trustworthy when it comes to x^0. In particular, the limit does not work with x=0! You can compute 0<sup>ε</sup> for as small a positive ε as you please, and the result will always come out 0. But at the moment the exponent reaches 0 -- and only then -- the result (according to the usual definitions) jumps to 1! –Henning Makholm (talk) 06:54, 14 April 2009 (UTC)

(outdent) Well, it was explicated in the first sentence how n must be nonzero; for the limit, perhaps that requirement need be restated. As regards 0^0 being equal to 1, that's a matter that requires extensive discussion (which is given). Also, it's immaterial if 0^0 suddenly becomes 1 when 0^0.0000001 = 0; a limit describes the behaviour of a function as its input approaches a given value a, but what happens at f(a) is irrelevant information. —Anonymous Dissident<sup>Talk</sup> 07:04, 14 April 2009 (UTC)

Nonzero is not enough; in fact it works only for positive n. The second part of your reply I don't understand at all. The very point of your argument is to use the limit of f(x) as x goes towards 0 as an explanation of why f(0) has the value it has. I should say that it is very material as to whether the argument is a good one that it fails to work for f(x)=0^x -- why, then, should one trust it for f(x)=3^x? It happens to point to the right result for 3^x, but there is nothing about the argument itself that indicates that it would work better for 3^x than for 0^x. This makes the argument dangerous -- you need to know by external means when it will work and when it won't, and by far the easiest way to know that is to already know the correct answer. What does the argument help, then? Who does it help? –Henning Makholm (talk) 08:20, 14 April 2009 (UTC)

It should be trusted because it is demonstrated, and can be replicated at any level you like. It's seen that as x approaches 0 for 3^x, your output gets closer and closer to 1. That list there corroborates it. Thus the limit is derived; but, as you say, perhaps more specification for n is required. The case of 0^0 is unique, and it has a unique treatment as such in the article. At the end of the day, though, this debate's going nowhere, and I am only an amateur. I thought that little factoid might have been useful, but if it's flawed, I have no problem with its exclusion. —Anonymous Dissident<sup>Talk</sup> 08:33, 14 April 2009 (UTC)

Oh, the fact that n^x converges to 1 for positive n is true enough. If I gave the impression that I was disputing the truth of that statement, I apologize. I merely contend that it would not be helpful for a reader to have it pointed out in this context; on the contrary it would create confusion.

The mathematics articles differ from the rest of Wikipedia in that we have an infinity of indisputably true facts to choose from that we might add to the article. Therefore, the tough editorial choices is not so much about whether the content of a claim can be verified (which, of course it must, but that does not exclude enough to get an article down to a readable size) as whether it is an interesting fact, or one that helps understand another fact that, in turn, is interesting. We do reject a lot of clearly true facts as original research because, while true, they do not further the encyclopedic goals of the article. I'm sorry if I didn't make it clear enough that this, rather than truth, was the level I was working at. –Henning Makholm (talk) 10:31, 14 April 2009 (UTC)

Okay, now I see where we're at. Of course, when you say "interesting", I think, perhaps, you refer to how useful a particular piece of text or statement is in effectively conveying information; after all, what's "interesting" is extremely subjective (I know many sorry people who would consider none of the content of this article remotely interesting, unfortunate as that is). —Anonymous Dissident<sup>Talk</sup> 10:37, 14 April 2009 (UTC)

"Interesting" may not even be about how effectively information is conveyed, but whether that particular information ought to be conveyed at all. Yes, that is subjective. Contrary to public perception, mathematics is full of subjective disputes. About what is worth proving and what is not, about what is worth teaching, explaining, remembering and what is not, about which proof is the nicest one, or about whether among two equivalent basic definitions one should choose that which makes this generalization or that one look more straightforward. We resolve those disputes like WP in general does: first try for consensus, and if that fails, apply WP:NOR and WP:V. If we cannot agree whether a fact is worth writing down, we look for reliable sources that support the viewpoint that it is (implicitly, by having it written down). Conversely, if no sources can be found to show that publishing mathematicians have found some fact worthy of presenting, Wikipedia probably doesn't need to present it either. –Henning Makholm (talk) 11:28, 14 April 2009 (UTC)

I agree what you've said above; but it's somewhat contrary to your previous statement. —Anonymous Dissident<sup>Talk</sup> 13:06, 14 April 2009 (UTC)

I think that the text in question is out of place in that section, because that section is only about integer exponents, and rational exponents are covered later in the article. There are also other issues that Henning Makholm has pointed out (such as continuity being assumed), but even if those could be fixed through longer exposition, the argument would still be out of place. — Carl (CBM · talk) 11:32, 14 April 2009 (UTC)

The title of the section would suggest otherwise, as aforementioned. If the section header is to be "one and zero exponents", it should treat that subject completely, irrespective of other content. That's how a comprehensive article is formed. So while I'll concede (if more experienced mathematicians advise me) that my argument is not required, I'll continue to argue that it was in a fine location based on principles covered in the WP:MOS. I placed it there because of the section header – and when the article continues to be built, such section headers and the comprehensiveness they demand and they afford is of important consideration. —Anonymous Dissident<sup>Talk</sup> 13:09, 14 April 2009 (UTC)

Perhaps that section should be split, and the information on 0 exponents moved lower in the article. I think I read section headings hierarchically: that subsection header refers to "Exponents one and zero" only in the context of "Exponentiation with integer exponents"; in the latter context one cannot raise a number to a non-integer exponent.

The larger issue, which I think Henning alluded to, is that there two related but distinct ideas about exponentiation. The first is that it refers to repeated multiplication. In this context one can start with positive integer powers, then add negative integer powers via the formula $a^{-n}=1/(a^{n})$, then add zero powers via the rule $a^{0}=1$. One can also define rational powers, for non-negative numbers at least, by the rule $a^{m/n}=(a^{m})/(a^{n})$. However, there is no concept of limit in all this; it is merely algebraic. And so there is no reason to expect that it will be continuous; continuity cannot be used to justify anything yet.

The second way of viewing exponentiation is analytic. For example, one can prove that exponentiation, as defined in the algebraic way, is continuous on a certain subset of the rational plane, and thus extends continuously to an operation on a subset of the real plane. But this requires one to first prove that the algebraic definition does give a continuous function on an appropriate domain. So in any case, the continuity is something that must be proved, not something that can be used to justify values of the exponential. — Carl (CBM · talk) 13:45, 14 April 2009 (UTC)

You do have a point about the section hierarchy, but a problem exists for the article in that a treatment of the 0 exponent is not given in a non-integer context – and that's quite needed. I think I agree that the section needs to be split off, and I'd not think that a whole level 2 section concerning the zero exponent and a) how it works in relation to positive and negative indices b) how all numbers raised to 0 give 1 and why and c) the discussion of 0^0 would be a bad idea. Cheers, —Anonymous Dissident<sup>Talk</sup> 13:54, 14 April 2009 (UTC)

I think by treatment of the power of 0 in a non-integer context you must be meaning a limit of 0 since 0 is an integer. There's three sections which treat powers of 0, the main one Exponentiation#Powers of zero which deals with all the cases except of 0<sup>0</sup>, the section Exponentiation#Zero to the zero power which deals with the explicit case of 0<sup>0</sup>, and the section Exponentiation#Limits of powers which deals with the business the OP was on about limits of zero. They are split and ordered fairly logically at the moment I think and sticking them together would mix up things of quite varying difficulty in a bit of mess. Arguing from the last to try and justify the first would be entirely the wrong way to go about things I thnk. Dmcq (talk) 14:35, 14 April 2009 (UTC)

Isn't 0^0 exactly the same as 0/0?

If we look at multiplication we can say that:

X×5=X+X+X+X+X

X×4=X+X+X+X

X×3=X+X+X

X×2=X+X

X×1=X

X×0=X-X

X×-1=X-X-X

X×-2=X-X-X-X

X×-3=X-X-X-X-X

X×-4=X-X-X-X-X-X

X×-5=X-X-X-X-X-X-X

Correct? There’s nothing wrong with that, so wouldn't it make sense to apply the same rules to exponentiation?.........

X^5=X×X×X×X×X

X^4=X×X×X×X

X^3=X×X×X

X^2=X×X

X^1=X

X^0=X÷X

X^-1=X÷X÷X

X^-2=X÷X÷X÷X

X^-3=X÷X÷X÷X÷X

X^-4=X÷X÷X÷X÷X÷X

X^-5=X÷X÷X÷X÷X÷X÷X

That’s right, isn't it? Why have 0/0 an acceptation to this rule when it looks pretty clear that it doesn’t need to be? Robo37 (talk) 20:20, 14 May 2009 (UTC)

Not sure what the question is. They are both the same in that both are indeterminate forms and you need some other information to determine the value if any. Dmcq (talk) 21:59, 14 May 2009 (UTC)

Dear Robo37: I too am not sure I understand the question, but let me try. Your argument seems to be that to go from X^n to X^{n-1}, one divides by X. So for instance, X^4 = X^5/X. If one follows your reasoning, one is led to believe that 0^4 is the same thing as 0^5/0 = 0/0. Of course this is nonsense, and should not be used as justification for leaving 0^4 undefined. For the same reason, such an argument should not be used to justify leaving 0^0 undefined. (On the other hand, as Dmcq said, 0^0 is an indeterminate form, which means that limits involving exponentation of functions tending to 0 can have different values, depending on the particular functions involved.) --FactSpewer (talk) 03:59, 15 May 2009 (UTC)

If you're trying to equate 0<sup>0</sup> = 0⁄0, you could use similar logic to equate 0<sup>0</sup> = 0×0. But as Dmcq says, simplistic logic like that is not enough to derive a meaningful answer. — Loadmaster (talk) 23:22, 2 June 2009 (UTC)

Well how about this; 0^0 = 0^1/0^1 = 0/0 Robo37 (talk) 16:14, 23 June 2009 (UTC)

x<sup>x</sup>

I have been looking for an article on the function x<sup>x</sup>, but cannot find one. Does anybody know where to look? --72.197.202.36 (talk) 22:35, 22 May 2009 (UTC)

It will usually be written e<sup>x·log(x)</sup>. It deserves no article of its own. It is not that important or useful. Bo Jacoby (talk) 16:16, 24 May 2009 (UTC).

Just a guess, but it might be related to the Lambert W function. — Loadmaster (talk) 23:29, 2 June 2009 (UTC)

I think it could be worthy of an article. Some topics to cover: extreme points / limits, lack of elementary antiderivative, generalization to tetration, connection with the Lambert W function, connection with factorials (Stirling's formula). Although it doesn't have a name (as far as I know), it's a nontrivial function that no doubt very many people have considered. Fredrik Johansson 19:41, 23 June 2009 (UTC)

History of differing points of view has error

The claim that "erroneously claiming that $\lim _{t\to 0^{+}}f(t)^{g(t)}=1$ whenever $\lim _{t\to 0^{+}}f(t)=\lim _{t\to 0^{+}}g(t)=0$. A commentator who signed his name simply as "S" provided the counterexample of $(e^{-1/t})^{t}$." is error.

Because, $\lim _{t\to 0^{+}}F[f(t),g(t)]\neq F[A,B]$ for $\lim _{t\to 0^{+}}f(t)=A,\lim _{t\to 0^{+}}g(t)=B$ if A and\or B $\in \{0,\infty \}$. This is common rule of limits theory. Therefore,"S" is mistaken because $\lim _{t\to 0^{+}}(e^{-1/t})^{t}\neq (\lim _{t\to 0^{+}}e^{-1/t})^{\lim _{t\to 0^{+}}t}$. Left formula is not indeterminacy of type $0^{0}$.

There is conflict like in sequence of calculation: Rule 1 "x/x = 1" and Rule 2 "0/0 is undefined". $\lim _{t\to 0^{+}}{\frac {t}{t}}\neq {\frac {\lim _{t\to 0^{+}}t}{\lim _{t\to 0^{+}}t}}$!!!

Therefore, true text is "correctly claiming that $\lim _{t\to 0^{+}}f(t)^{g(t)}=1$ whenever$\lim _{t\to 0^{+}}f(t)=\lim _{t\to 0^{+}}g(t)=0$, where f(t) and g(t) are not exponential functions. Therefore, they cannot have degree with divisor t. Otherwise, there can there be conflict of commutation of operations between ${\frac {x}{x}}=1$ and ${\frac {0}{0}}=$ Indeterminate and $\infty \cdot 0=$ Indeterminate with different results of calculation ways. Therefore, f(t) and g(t) should be are simplified that contained only indeterminacy of true type 0." —Preceding unsigned comment added by 62.181.43.110 (talk • contribs) 2 June 2009

I think what the OP is saying is that it isn't a case of the indeterminate form 0<sup>0</sup> because the result isn't 1 and he is sure the result must be 1. Which is basically what all the argument before has been about, if you define 0<sup>0</sup> to be 1 then something with that form as a limit must not be treated as an indeterminate form but using the general handling of limits. For instance it is quite possible to define 0/0 to be 1 but that would make handling of such limit forms much more error prone. Dmcq (talk) 18:07, 2 June 2009 (UTC)

Dear 62.181.43.110: The published paper by Knuth was correct in saying that "S" was right. This is a mathematical truth independent of whether you believe that 0<sup>0</sup> should be defined. --FactSpewer (talk) 02:21, 3 June 2009 (UTC)

Brackets does not influence on computation process. This typical mistake of dilettantes. $(a^{b})^{c}=a^{(bc)}$ is law. Therefore, $(e^{-1/t})^{t}=e^{(-1/t\cdot t)}=e^{-1}$. Left expression is not type 0<sup>0</sup>, because middle expression is type indeterminacy 1<sup>0/0</sup>, that gives determinacy. It is result of limits theory.

Functional "limit" there is not distribution law $L[F[f(x),g(x)]]\neq F[L[f(x)],L[g(x)]]$ at critical points.

This is typical error at dilettantish in limits theory. U should use $a^{bc}=e^{bc\cdot \ln a}$ always to calculate 0<sup>0</sup> indeterminacy. This expression shows $(a^{b})^{c}==a^{(bc)}$ law for always real expressions.

Therefore, Mobius was right if f(t) and g(t) are not exponential functions.

Using exponential f(t) form is not correctly, because then will be indeterminacy not 0 at f(t), and Exponentiation form like a<sup>b/0</sup>.

U erroneously propagate dilettantish theory. Or true text, or neutral text must be without grade.

The ultimate touchstone in wikipedia is WP:NOTE, whether there are citations supporting what is written. The text cites a book saying what is in the article. Can you point out a book saying that "S" was wrong? If not then there's no way what you say can be put in however much you believe you are right. I'm sorry if you think that is dilletantish but this is an encyclopaedia not a maths work and it can't engage in original research, it just reflects what others have done. Dmcq (talk) 11:59, 3 June 2009 (UTC)

It is impossible. Brackets cannot influence on results of calculation because there is calculated not function, and Functional! U can use L'Hopital rule 0/0 only to indeterminacy resolve! 0<sup>0</sup> and other indeterminacies are resolved reduction to L'Hopital rule 0/0 only and other ways are not exists! This way is one - using of Derivative. This is standard of limit theory and it is not "original research" that differents of the error text! Now this text is original research to give error! Therefore, the wikipedia has incorrect unauthority claims now. 62.181.43.110 (talk) 12:50, 3 June 2009 (UTC)

It is possible for the same limit to be transformed so that it has several different indeterminate forms; each individual indeterminate form is obtained by taking the limits of the main subexpressions naively. So, for example $\lim _{x\to 0}\left(x\cdot {\frac {1}{x}}\right)$ has indeterminate form $0\cdot \infty$ and $\lim _{x\to 0}(x/x)$ has indeterminate form 0/0. Similarly, the limit $\lim _{x\to 0}(e^{-1/x})^{x}$ has indeterminate form $0^{0}$ and $\lim _{x\to 0}e^{-x/x}$ has indeterminate form $e^{0/0}$. So brackets do indeed influence the computation of the indeterminate form of an expression.

Brackets do not change results of functional. In limits theory functionals are used, but not functions. Therefore, limit (x/x) = (x)/(x) independently of brackets exists. Functional independents from decomposition. Functional is not composition of function. Value of functional calculated only as a whole. Remember of integrals' rules!!!62.181.43.110 (talk) 13:44, 3 June 2009 (UTC)

You are right that there is no "distribution law" for limits at indeterminate points; this is why it is useful in some settings for $0^{0}$ to be indeterminate.

Therefore, limits calculates as a whole by using L'Hopital rule 0/0. These calculations give to 0<sup>0</sup>=1 always.62.181.43.110 (talk) 13:44, 3 June 2009 (UTC)

Finally, the statement "if f(t) and g(t) are not exponential functions" is not clear at all to me. It would be easy to make f and g discontinuous everywhere except 0, with the same indeterminate behavior of $\lim _{x\to 0}f(x)^{g(x)}$. — Carl (CBM · talk) 12:05, 3 June 2009 (UTC)

By the way I'd be a bit careful of $(a^{b})^{c}=a^{(bc)}$. It's normally not going to cause any trouble and doesn't really affect the discussion here, but see the third example in Exponentiation#Failure of power and logarithm identities where it goes wrong even with a=e. Dmcq (talk) 12:43, 3 June 2009 (UTC)

You can use L'Hopital rule 0/0 only to reduce any indeterminacy. Therefore, if f(t) has exponential functions then you convert its into two function ff(x)<sup>fg(x)</sup>. Using L'Hopital rule does it automatically. We using real functions here with real limits. 62.181.43.110 (talk) 12:50, 3 June 2009 (UTC)

You're certainly right that L'Hopital's rule is usually stated for the indeterminate form 0/0, but that does not change the fact that the indeterminate form of $\lim _{x\to 0}\left(x\cdot {\frac {1}{x}}\right)$ is $0\cdot \infty$ if you view the multiplication as the final operation. Syntactic manipulations of expressions inside limits can change the indeterminate form. — Carl (CBM · talk) 13:00, 3 June 2009 (UTC)

It is incorrect claim. "Syntactic manipulations" cannot change of results of any limit that there is not divergent series. In limits theory functionals are used, but not functions. From manipulations the results cannot change!!! —Preceding unsigned comment added by 62.181.43.110 (talk) 13:10, 3 June 2009 (UTC)

I have no idea what you mean by "functionals are used, but not functions". The definition of a limit refers to a function, not to a functional. I have never seen a calculus textbook talk about "functionals". — Carl (CBM · talk) 16:00, 3 June 2009 (UTC)

Please stop changing the article without a citation and removing cited material. That is disruptive editing. You may believe you are right but cited material is what you need if you're going to edit cited material. Find a book that agrees with you and give its name and the page where it says "S" was wrong or that "S" didn't do what the text says he did. Dmcq (talk) 14:10, 3 June 2009 (UTC)

Dear Dmcq, Using of terms "unconvincing" and "erroneously" to authority scientists of Libri and Möbius it is impossible. We don't known who is "S" - it is unauthority unknown dilettante. Its view is wrong. He is not know base of limits' theory! By consensus, you should delete those words in the text! 62.181.43.110 (talk) 14:17, 3 June 2009 (UTC)

Provide a book supporting your point of view. The argument from "S" appeared in a reputable journal, it wasn't contested. It stopped the argument. People have quoted it including Knuth who is cited in the article. As I have said before it does not matter if it is right or wrong, this is an encyclopaedia not a maths text. If you do not cite a book or journal your reasoning and changes simply cannot stand against a quote from Knuth. Dmcq (talk) 14:37, 3 June 2009 (UTC)

Quot from Knuth Concrete Mathematics p.162

"Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant."

See also http://www.cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html from U have rewrited the text. Terms "unconvincing" and "erroneously" there is not exists. These terms in wikipedia were added without consensus, by author's "political" viewing. 62.181.43.110 (talk) 14:59, 3 June 2009 (UTC)

That quote is about Knuth's ideas and not about the history, that is all dealt with in the previous section. The section here is from Two Notes on Notation page 6 dealing with the history, which is what this section is about. Donald Knuth then launches into his bit about 0<sup>0</sup> being 1 which is something William Kahan also wants, but that is the previous section and both point of views are treated there as they should be in an encyclopaedia because they are both notable. Personally I think having that in the context of limits is silly and error prone but I'm happy to quote the opposition as it were. Dmcq (talk) 15:20, 3 June 2009 (UTC)

Quote from Knuth's "Two Notes on Notation"

"On the other hand, Cauchy had good reason to consider~$0^{0}$ as an undefined {limiting form}, in the sense that the limiting value of $f(x)^{g(x)}$ is not known {\a priori\} when f(x) and g(x) approach~0 independently. In this much stronger sense, the value of~$0^{0}$ is less defined than, say, the value of $0+0$. Both Cauchy and Libri were right, but Libri and his defenders did not understand why truth was on their side." Now we have answer, because limit is functional, and works with functional. In Cauchy's time, what is limits and how operate with theirs, did not understand yet. Therefore, "S" was mistaken, but in Libri's time cannot prove it, and debates were stopped.

Dear Dmcq, And in Knuth's text is not existing "unconvincing" and "erroneously" words. These is not culture words. U restricted Knuth's and added that he did not write. Therefore, sense of said exchanged on opposite and result is wrong. Must be full or true text, or text without own grades. —Preceding unsigned comment added by 62.181.43.110 (talk) 15:58, 3 June 2009 (UTC)

What is a functional? How is it different than a function? — Carl (CBM · talk) 16:02, 3 June 2009 (UTC)

Light words, function has on inputs values/numbers(!) and on outputs one value/number(!). Functions reduced into composition way $f\circ g$ with possible different results depended from decomposition ways. Functional has on inputs functions/functionals (not value/number!) and on outputs one value/number. Functional is not composition of functions, therefore, independed from calculation way. U can imagine functional like function with infinity arguments by each input.

For example, you cannot calculate integral over decomposition of functions, $\int _{a}^{b}f(g(x))dx\neq \int _{a}^{b}f(y)dx\circ g(x)$ and $\int _{a}^{b}f(x)g(x)dx\neq (\int f(x)dx\int g(x)dx)|_{a}^{b}$! Integral has only one value independently from ways of calculation. In integral, You must x*1/x = 1 always. You cannot calculate in beginning 1/x, further x and multiplicate its. I was explaining it badly, but better i am not able.--62.181.43.110 (talk) 16:46, 3 June 2009 (UTC)

I still do not understand what you are saying. The function f(x) = x is a function. The function g(x) = 1/x is a function. And the indeterminate form for $\lim _{x\to 0}f(x)g(x)$ is $0\cdot \infty$. What does any of this have to do with "functionals"? All that matters, for the purposes of computing the indeterminate form, are the limits of f(x) and g(x). — Carl (CBM · talk) 16:52, 3 June 2009 (UTC)

Limit is not function, and functional! Function - U can calculate by parts, functional - U cannot. To function is rightly composition rule $f(g(x))=f(x)\circ g(x)$ (U calculate g(1) and result set into f, U calculate g and f independently), to functional is not rightly composition rule $f(g(x))\neq f(x)\circ g(x)$ by definition!

By definition you find the indeterminate form of a limit $\lim _{x\to a}h(x)j(x)$ by multiplying $\lim _{x\to a}h(x)$ by $\lim _{x\to a}j(x)$. See indeterminate form. The algebraic limit theorem gives a way to compute many limits in terms of the limits of subexpressions. If the subexpressions have limits but these do not determine the overall limit, the result is an indeterminate form. — Carl (CBM · talk) 17:20, 3 June 2009 (UTC)

Far from convincing and unconvincing are equivalent. Having one of ones works quietly dropped from collected works is pretty much equivalent to them being considered mistaken. The bit you quote is from his recent ideas, it is not history. Other people disagree with Knuth. Knuth and Kahan are more concerned with getting results in computer science calculations rather than mathematics itself, but there have been a number of complaints about the power function from Kahan which is implemented in the standard libraries and many people would prefer separate real and integer versions where the real version gave the result undefined and the integer version gave the result 1, never mind that it spends too long determining if the arguments are integers. Dmcq (talk) 17:36, 3 June 2009 (UTC)