Talk:Exponentiation/Archive 2009

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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 0⁰ 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 x². Dmcq (talk) 18:26, 14 October 2008 (UTC)

Every book and paper that defines 0⁰ makes the definition that 0⁰=1, so there is no controversy as to how 0⁰ is defined. Many books and papers do not define 0⁰. For example the bible and the qur'an do not define 0⁰. This does not imply that wikipedia must state the fact that some authors do not define 0⁰. 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^-1/x2)^x2 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 " 0⁰ is indeterminate " reflects the fact that the function x^y cannot extended from {(x,y):x>0,y>0} to {(x,y):x≥0,y≥0} by continuity, but it does not imply that 0⁰ is, or must remain, undefined. Nor does it define it of course. To Dmcq: the function (2^−1/x2)^x2 is undefined for x=0 because 1/0 is undefined. The limit lim_x→0((2^−1/x2)^x2) =2⁻¹ is defined for real x, but not for complex x, while (lim_x→0(2^−1/x2))^{lim_x→0(x2)} =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 0⁰, 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^−1/x2 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^y, and does not define 0⁰, and is unaware that 0⁰ has been defined elsewhere, then he will in good faith publish that 0⁰ is undefined. But in this article, where 0⁰ 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^−1/x2 with the value 0 at x = 0 raised to the power of x². When x = 0 then this is of the form 0⁰, 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 0⁰ = 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 0⁰ 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 f₁ defined by f₁(x)=2^−1/x2 has f₁(0) = 2^−1/02 = 2^−1/0 which is undefined because you cannot divide by zero. The definition may be supplemented by defining the function f₂ by f₂(x)=f₁(x) for x≠0 and f₂(0)=0. This function f₂ is continuous for x=0 because lim_x→0(f₂(x)) = 0 and f₂(lim_x→0x) = 0 so lim_x→0(f₂(x)) = f₂(lim_x→0x). Consider the function f₃ defined by f₃(x)=(f₂(x))^x2. This function f₃ is discontinuous for x=0 because lim_x→0(f₃(x)) = lim_x→0(2⁻¹) = 2⁻¹ while f₃(lim_x→0x) = 0⁰ = 1, so lim_x→0(f₃(x)) ≠ f₃(lim_x→0x). So consider the function f₄ defined by f₄(x) = f₃(x) for x≠0 and f₄(0) = 2⁻¹. This function f₄ is continuous for x=0. Care must be taken when handling discontinuous functions, and it is not solved by either redefining or undefining 0⁰. 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 0⁰ 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 0⁰. 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 0⁰ 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 0⁰ ill-defined or un-defined? I don't think you sincerely suggest redefining 0⁰ to be 1/2 based on you calculation above, because another calculation, (a^1/x2)^x2=a for 0<a<1, will provide 0⁰=a when x=0. It is an error to assume continuity of x^y. It is not an error to define a discontinuous function. If it "is perfectly okay for 0⁰ to be always 1 when talking about integers", then 0⁰=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 0⁰=1 if anyone would. In his introduction to Blissard calclus, what's more commonly called Umbral calculus he says "A sequence a₀, a₂, a₂, ... may be replaced by a⁰, a¹, a², ... with the exponents treated as powers during all formal operations, and only restored as indexes when operations are completed; note that a⁰ 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 0⁰ 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 0⁰=1 if anyone would" and then you quote him for writing that "a⁰ 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 0⁰ 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 0⁰ 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 0⁰ 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 (aⁿ)^1/n=a for all integers n>0, and so the limit for n→∞ is lim((aⁿ)^1/n)=a, while (lim(aⁿ))^lim(1/n)=0⁰ . The wrong assumption that lim((aⁿ)^1/n)=(lim(aⁿ))^lim(1/n) leads to the wrong result that a=0⁰. The convention that 0⁰ is undefined leads to the wrong result that a is undefined. The convention that 0⁰=1 leads to the wrong result that a=1. The error is not in the definition of 0⁰, but in assuming that x^y is continuous. Conclusion: undefining 0⁰ 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 0⁰ 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 0⁰=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 0⁰ undefined rather than indeterminate. The words are: "justifications for treating 0⁰ 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 0⁰ is defined and indeterminate form at the same time, than that 0⁰ 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^x as x approaches 0 is 1. So it sure makes sense to define 0⁰ as 1. Also, by repeated application of the abovementioned rule it can be shown that x^xx has a limit of 0 as x approaches 0 and that x^xxx 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^x is 1, for other functions f(x) and g(x) both going to zero f(x)^g(x) 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)

0⁰ (yet again)

I don't mean to start a whole 'nuther discussion on 0⁰, but I do want to make the observation that if 0⁰ = 1 while 0^x = 0 for all x > 0, doesn't this define a discontinuous function? | Loadmaster (talk) 19:58, 30 December 2008 (UTC)

Yes. The function x^y is necessarily discontinuous at x=y=0, no matter how one defines 0⁰. You get different limits for 0^t and t⁰ 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^Talk 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^Talk 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^Talk 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^ε 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^Talk 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^Talk 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^Talk 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^Talk 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^Talk 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^Talk 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 0⁰, the section Exponentiation#Zero to the zero power which deals with the explicit case of 0⁰, 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 0⁰ = 0⁄0, you could use similar logic to equate 0⁰ = 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)

Well, how about this: 0^4 = 0^5/0^1 = 0/0. (I'm just repeating my argument above. If your reasoning were correct, then 0^4 would be indeterminate too. But it isn't: 0^4 is just 0.) --FactSpewer (talk) 22:16, 16 August 2009 (UTC)

Good point. Robo37 (talk) 23:10, 16 August 2009 (UTC)

x^x

I have been looking for an article on the function x^x, 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^x·log(x). 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 0⁰ 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 0⁰ 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 0⁰ 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 0⁰, because middle expression is type indeterminacy 1^0/0, 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 0⁰ 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^b/0.

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! 0⁰ 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 0⁰=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)^fg(x). 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 0⁰ 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)

Formula help - x+y²

If I have the following formula;
x+y^e
What formula would be used to return x, and what would the formula be to return y?
For example,
z=x+y^e

x=the formula to return x, only z and e allowed as an variable
y=z-x
or possibly
y=the formula to return x, only z and e allowed as an variable
x=z-y

Please help me.--84.216.39.153 (talk) 10:31, 23 July 2009 (UTC)

Haha, I solved it myself, it was easier than I thought.

Solution:
y=(√z rounded down)²
x=z-y

D --84.216.39.153 (talk) 11:24, 23 July 2009 (UTC)

The Maths reference desk is where questions like this should be asked. Dmcq (talk) 15:30, 23 July 2009 (UTC)

Suggestion to Math Nuts And Devout Math Studiers

To save your progress on this ever-changing internet, why doesn't anyone "lock" the page (protect from unauthorized edits)? It would save many headaches and random editing troubles. —Preceding unsigned comment added by 76.247.183.31 (talk) 06:07, 4 December 2009 (UTC)

Please put questions like this on a more appropriate forum,see WP:HELP. As to this one old versions of pages are saved and can be seen using the history tab. Progress would be stopped if edits were not allowed. Dmcq (talk) 06:23, 4 December 2009 (UTC)