Talk:Exponentiation/Archive 2013

Archive 2010

Archive 2011

Archive 2012

Archive 2013

(a^b)^c = a^(b*c)

The following identities hold, provided that the base is non-zero whenever the integer exponent is not positive:

$(b^{m})^{n}=b^{m*n}$

Using this property..

$-1/2=((-1/2)^{2})^{3/6}=(-1/2)^{2*3/6}=((-1/2)^{2})^{3/6}=(1/4)^{1/2}=1/2$

What's going on here..67.1.3.194 (talk) 22:28, 23 April 2013 (UTC)

It would be more appropriate to ask this question at the mathematics reference desk, WP:RD/Math. This is an article talk page and is intended for discussions aimed at improving the article, not general discussion of the subject matter. --Trovatore (talk) 22:55, 23 April 2013 (UTC)

Yes I understand the purpose of this page, but I thought maybe something was misstated in the article because of what's in my original post. However I now see that I neglected the "integer exponent" part. This can be deleted. Unless someone else found this unclear? I think this is a common quandary people without formal maths experience have. -206.207.225.104 (talk) 23:05, 23 April 2013 (UTC)

Note the bit about "integer exponent" your exponents are not all integers. The problem you have is discussed at the end of the section Exponentiation#Rational exponents.--Salix (talk): 06:26, 24 April 2013 (UTC)

The OP got it, when they wrote: "However I now see that I neglected the 'integer exponent' part".

Anyways, I can't understand why they had to write down that a long proof for the pseudo-contradiction. They wrote $-1/2=\dots \dots \dots \dots \dots =(1/4)^{1/2}=1/2$, so they could simply write $-1=1^{1/2}=1$, thus getting to the same pseudo-contradiction - without using the rule $(b^{m})^{n}=b^{m*n}$. To sum up, this rule has nothing to do with the pseudo-contradiction indicated by the OP.

As for the pseudo-contradiction itself: One can avoid it if and only if one avoids attaching the sign "=" to powers involving non-integer exponents, unless the domain of discourse is a sub-set of the set of non-negative real numbers. This simple instruction is valid whenever one uses powers, even without using the rule indicated by the OP. HOOTmag (talk) 07:20, 24 April 2013 (UTC)

Removed statement makes a worthwhile point

The article currently contains the statement

So 0⁰ is an indeterminate form. This behavior shows that the two-variable function x^y, though continuous on the set {(x,y): x > 0}, cannot be extended to a continuous function on any set containing (0,0), no matter how 0⁰ is defined. However, under certain conditions, such as when f and g are both analytic functions and f is positive on the open interval (0,b) for some positive b, the limit approaching from the right is always 1.

This revert removed the content

If on the other hand, you approach $0^{0}$ on $x^{y}$ from any angle other than $0^{y}$, such as $(ax^{x})$ the value readily converges on 1. Also, the shape of the curve of $a^{x}$ shows that as x approaches zero, the curve sharply climbs to 1. The argument that $0^{0}=0$, simply amounts to $0/0$, and therefore is to be considered less reliable than approaches from any other angle (which does not rely on 0/0.

which makes a significant point (albeit not well-worded) not obviously contained in the statement quoted above. In particular, a two-variable function x^y extended to 0⁰ being defined as 1 is continuous on the set {(x,y): x > εy, x < y/ε} for any ε > 0. This could be phrased as a noteworthy and "certain condition". In particular, it may be worth mentioning that very marginally constraining the domain (as opposed to defining a path for the limit) ensures continuity, which may also be seen in the adjacent diagram. — Quondum 11:27, 14 June 2013 (UTC)

It could be stated in a correct manner; as written, it's incomprehensible. The correct statement that $\lim _{x\rightarrow 0^{+}}x^{ax}=1$ follows in a more coherent matter from the statement involving analytic functions below. (If we don't force the base to be positive, then we need say something like "any continuous branch".) The part left behind is about what can be said which is both correct, not an opinion ("sharply"), and relevant. Of course, if the statement had appeared in an (otherwise) reliable source, it could be added. — Arthur Rubin (talk) 14:52, 14 June 2013 (UTC)

Conceded. I did not read the statement that is in the article closely enough; it covers a pretty broad set of cases including the one that was removed. — Quondum 15:25, 14 June 2013 (UTC)

Section on Negative exponents

My removal of the section Exponentiation#Negative exponents with edit summary (Undoing 2 edits: not particularly coherent, and detracting from the quality.) has been reverted by the author of the section, with the edit summary (there needs to be a "negative exponents" subsection, so let's increase the quality not remove the contribution. thanks.) Since the content of this section is more than adequately covered under the section Exponentiation#Arbitrary integer exponents, it should be removed. All it contains is effectively a repetition of one statement from this latter section, and then simply adds some trivial examples. I do not want to get involved in an edit war and will leave it to other editors to preserve the GA status. — Quondum 03:43, 25 July 2013 (UTC)

I agree. The Negative exponents section duplicated a case already adequately covered in the article, so I have removed it. Gandalf61 (talk) 08:54, 25 July 2013 (UTC)

Article structure

I think this article would benefit from some restructuring and from more consequent usage of terminology. I would suggest developing the case of positive real base b first with increasingly general (probably up to real) exponents. After that, the case of base 0 could be treated, and finally negative bases leading to the most general case where both base and exponent may be complex. Also, the usage of the term power is sometimes not clear. For example, does "Complex power of a complex number" refer to w^z being complex or to z being complex, or both? Isheden (talk) 09:39, 25 July 2013 (UTC)

Perhaps you could be more explicit, what would be the changes to the index be for instance? I don't see your problem about'complex power o a complex number'. Why should one be referring to an already decided answer when posing a question? The question is (complex power) of a (complex number), it is not that one gets a (power of a (complex number)) and then has a section only about the answers which are complex. If you are finding that difficult I am rather worried about what else you are thinking of doing. Dmcq (talk) 14:55, 25 July 2013 (UTC)

I agree with Isheden that "Complex power of a complex number" is an unfortunate phrase, and I would argue to replace it with something that does not tax the reader's powers (excuse the pun) of disambiguation, perhaps something like "Complex base and complex exponent". Because the word "power" is ambiguous in this context, care should be exercised in its use, and it should be avoided where another phrasing allows it to be avoided. — Quondum 15:12, 25 July 2013 (UTC)

Complex base and complex exponent would work okay. Some people always try to make things difficult for themselves, I was worried about some long winded phrasing with even more opportunities for people that way inclined. Dmcq (talk) 15:30, 25 July 2013 (UTC)

I've changed it to "Complex exponents with complex bases" – this matches several other similar section headings. I have also changed another section heading along the same lines. — Quondum 20:22, 25 July 2013 (UTC)

a × a × … = 1 × a × a × …

Why not to put the multiplicative identity in front of series? This would answer what is the answer in case you have no items to multiply together. You still have 1 in result of such product because here is intuitive definition of the product. Likewise, the repeated sum a×n is must be defined as 0+a+a+a+, w.r.t. to additive identity 0. This is the true meaning of the exponentiation. Why not to claim that explicitly? This would resolve a lot of confusion right away. I suppose that in continuous analysis we have something different from 0⁰. Why do we say that we have zeroes there if they are infinitesimals and not true zeroes actually? --Javalenok (talk) 10:43, 11 September 2013 (UTC)

Despite agreeing wholeheartedly with you (and being tempted to present a more general argument in favour), this debate has raged back and forth on this page, and what will win is that Wikipedia should present the view of notable secondary sources. Which it probably does pretty well, as well as doing a (IMO very) good job of explaining the debate. This talk page should not tackle the subject matter, but rather the article's content. So, this debate does not belong here. With adequately notable sources, via the empty product could be presented as an alternative definition for integer exponents. — Quondum 12:17, 11 September 2013 (UTC)

We're supposed to document things as they are, not make up things. What you are talking about would cause problems with the usual way limits are worked out. How it puts it in the article is best I think. Personally I'd just do it the way the pown and powr IEEE functions do as explained in the article, i.e. just say that exponentiation by an integer is a different function and gives different results from exponentiation by a real. Dmcq (talk) 13:20, 11 September 2013 (UTC)

Strictly speaking I shouldn't do this, but I'll indulge myself in an extremely brief response on the merits. Javalenok, no one (well, hardly anyone) disagrees that the product of no copies of 0 is 1. The disagreement is on whether that's the right interpretation of 0⁰ in a real-number context — certainly, 4^1/2 is not interpreted as the product of half a copy of 4. --Trovatore (talk) 18:28, 11 September 2013 (UTC)

I agree with Javalenok. When the exponent, n, is a nonnegative integer, the recursive definition

aⁿ=1 when n=0

aⁿ=a×aⁿ⁻¹ when n=1, 2, 3, &c.

is indeed appropriate. Trovatore's argument about 4^1/2 is invalid because 1/2 is not an integer. Trovatore's talking about the real-number context is nonsense because zero is an integer as well as a real number. The present quite unsatisfactory state of the article on exponentiation does not reflect consensus but merely the fact that I don't make edit wars and Trovatore did. Dmcq's argument that exponentiation by an integer is different from exponentiation by a real refers to the computer science meaning of a 'real'= floating point number, which is different from the mathematical concept of a real number. In computer science zero may be represented in several ways, such as fixed point binary and floating point binary. Mathematically zero a single object. Bo Jacoby (talk) 20:51, 6 December 2013 (UTC).

Travatore's argument about 0⁰ is justified;

• $\lim _{x,y\rightarrow 0}x^{y}$ doesn't exist.
• $\lim _{y\rightarrow 0^{+}}0^{y}=0.$
• $\lim _{x\rightarrow 0}x^{0}=1.$

Which one is appropriate depends on the context. — Arthur Rubin (talk) 02:54, 7 December 2013 (UTC)

The discontinuity $\lim _{y\rightarrow 0^{+}}0^{y}\neq \lim _{x\rightarrow 0}x^{0}$ cannot be helped. The constant term of the polynomial $\sum _{i=0}^{n}a_{i}x^{i}$ is $a_{0}$ even if x=0. Undefining 0⁰ has a high cost but no benefit. Bo Jacoby (talk) 04:02, 7 December 2013 (UTC).

Anyone who wants to debate this can do so endlessly on USEnet. What we have currently in the article is a discussion that reasonably reflects the state of the literature and the usage of working mathematicians. --Trovatore (talk) 04:05, 7 December 2013 (UTC)

The overwhelming consensus is 0^0 is 1, and wikipedia should say so. Try to find a textbook that gives the derivative of x^n, while listing n=1 as a separate case! If we can't find one, it means that everyone implicitly uses 0^0=1, and it is OK for wikipedia to say so. As for the debate and the controversy, there is just as much controversy about Monty Hall's problem, and nevertheless, wikipedia simply gives the correct answer there without giving merit to persistent incorrect views. The limit argument only shows that 0^(approximate 0) is not defined. Applying a limit argument to 0^(exact 0) is not logically sound, because the continuous version of x^y is only defined after x^y is defined for integer values of y. We should not let an unsound argument interfere with presenting what is clearly (at least implicitly, with polynomial and power series representation, and many other formulas) the overwhelming consensus. Mark van Hoeij — Preceding unsigned comment added by 128.186.104.253 (talk) 15:49, 17 December 2013 (UTC)

Your first sentence is just not true; that is not in fact the consensus at all.

The "derivative" example is a good argument for the real-to-integer exponential function, but not for the real-to-real one. --Trovatore (talk) 18:57, 17 December 2013 (UTC)

I still maintain that there consensus for 0^0=1 throughout math books, papers, journals, as well as throughout wikipedia, with this page being the only exception on wikipedia, all by itself surrounded by an ocean of 0^0=1. For example, click on [power series], and you see in the very first formula that (x-c)^0 = 1, with no restrictions on x. Likewise for the pages on the binomial theorem, the exponential function, they all contain the formulas in standard form that hold only when 0^0 is interpreted as 1. Mark van Hoeij — Preceding unsigned comment added by 71.229.28.197 (talk) 21:39, 17 December 2013 (UTC)

That's real-to-integer exponentiation. "Implicitly", yes, those give you 1. But "implicitly", it's a different function from real-to-real exponentiation. --Trovatore (talk) 21:59, 17 December 2013 (UTC)

There is nothing wrong with pointing out that the textbooks that leave 0^0 undefined still contain formulas that do assume the reader to interpret it as 1! While it is true that not everyone agrees with defining 0^0 to be 1, once and for all, readers should know that calculus textbook contain numerous formulas that implicitly assume 0^0 to be 1. Mark van Hoeij — Preceding unsigned comment added by 128.186.104.253 (talk) 19:06, 17 December 2013 (UTC)

OK, we have to be very careful about explaining what the "implicit" assumptions are. Wikipedia has very strict rules about this — see Wikipedia:No original research.

However, on the talk page, we can say what we think the implicit assumptions are, as a way of guiding interpretation of the sources. My view would be that the sources "implicitly" take real-to-real exponentiation to be a distinct function from integer-to-integer or real-to-integer exponentiation.

Here's one place they do: You'll probably find texts that define exponentiation as $x^{y}=\exp(y\ln x)$, and also define the exp function in terms of its power series. I don't know which specific sources, haven't looked, but you agree it's a plausible combination and probably occurs somewhere?

But if there's only one exponential function, then this is circular, because the power series uses exponents!

The resolution is, the power series uses the real-to-integer function, which is not defined as $x^{y}=\exp(y\ln x)$, but rather in terms of the inductive definition (or algebraic extension to negative integer exponents, not actually needed here).

It's a different function. It has a different meaning. It happens to be notated the same, and the obvious diagram commutes for values where everything is defined, but it's a different function. --Trovatore (talk) 19:18, 17 December 2013 (UTC)

That sounds reasonable, but would you agree then that the corresponding section is called something like "Real and complex exponents", or "approximate exponents", and not something like "analysis". Because even in analysis, many formulas still assume 0^0 to be 1. The issue only arises when the exponent is not assumed to be an exact integer. Mark van Hoeij. — Preceding unsigned comment added by 128.186.104.253 (talk) 19:36, 17 December 2013 (UTC)

That's a fair point, I think. --Trovatore (talk) 19:46, 17 December 2013 (UTC)

The obvious fact that some books don't define 0⁰ is unimportant. No mathematician, save Trovatore, defines a function f(x) such that the definition of f(0) depends on whether 0 is 'in a real-number context'. Our article needs improvement. Bo Jacoby (talk) 17:48, 7 December 2013 (UTC).

So you see a person here disagreeing with you and come to that conclusion. Sounds like you are an astronomer rather than a mathematician according to Mathematical joke#Stereotypes of mathematicians. Add me in with Trovatore. Dmcq (talk) 19:01, 17 December 2013 (UTC)

Calculus textbooks list the derivative of x^n, the binomial theorem, etc., in their usual formulations, which are valid if and only if 0^0 is interpreted as 1. This is relevant information that deserves to be the main page, calculus students need it to correctly interpret the formulas. Also, since the first subsection was called "For discrete exponents" it makes sense that the second subsection has a related title (I propose "approximate exponents" or "real exponents"). Please reverse some my changes that you deleted. I do understand that the 0^0 debate is everlasting, and I do want to leave room for opposing viewpoints, but I disagree with removing relevant information. Mark van Hoeij — Preceding unsigned comment added by 128.186.104.253 (talk) 19:25, 17 December 2013 (UTC)

The bit about integer exponents is in the article. We should not be debating 0^0. We are supposed to base what is in the article on what is in the main sources. Wikipedia does not pretend to be a textbook nor is it written from a particular point of view. It is an encyclopaedia. Dmcq (talk) 23:36, 17 December 2013 (UTC)

Most main sources do use the convention that 0^0=1. Most wikipedia pages in which the issue arises also use the 0^0=1 convention (e.g. read the first sentence of the article on power series). Placing 0^0=undefined on an equal footing with 0^0=1 is not neutral but is giving it undue weight. I have a number of calculus books in my office (publishers send them to me on a regular basis), they all contain numerous formulas that are false without 0^0=1. Mark van Hoeij — Preceding unsigned comment added by 71.229.28.197 (talk) 00:35, 18 December 2013 (UTC)

Again, your first sentence is just not true. You're taking them to be "implicitly" using that convention. That doesn't count. They have to say it explicitly, or explicitly use a definition from which it follows. The most common definition for real-to-real exponentiation is $x^{y}=\exp(y\ln x)$, and it doesn't follow from that. --Trovatore (talk) 00:44, 18 December 2013 (UTC)

If it makes a difference, I'll count tomorrow to see how many spell it out explicitly and how many do not.

0^0 for power series is the integer case. Saying there is a lot of those says nothing about the not integer case. Next we'll be getting people saying that e^x is the same whether e is a real or a complex number. Dmcq (talk) 01:21, 18 December 2013 (UTC)

I checked 5 calculus textbooks. None of them stated explicitly what the value of 0^0 is. But they all contain formulas (like the binomial theorem, stated without exceptions) that are correct if and only if the reader interprets 0^0 as 1. Without 0^0=1 all our textbooks contain errors. Textbooks that leave 0^0 undefined either (a) have errors in their formulas, or (b) expect their readers to interpret 0^0 as 1, neither of which supports the "undefined" point of view. Of course, wikipedia can't choose convenience or correctness, it has to go with the consensus. But the context in which the controversy arises is narrow (limits in calculus) whereas x^0=1 for all x, without restrictions, is the norm in a much larger context (discrete math, polynomials, power series, rings with identity, etc.). — Preceding unsigned comment added by 128.186.104.253 (talk) 13:56, 18 December 2013 (UTC)

I am deliberately not getting involved in the whole discussion at this stage, but will make only the following suggestion: In Exponentiation#Arbitrary integer exponents, it says The case of 0⁰ is controversial and is discussed below. Since this is in the context of integer exponents, and in this context usage is apparenty universally such that 0⁰=1, it might be better to rephrase this particular statement something more along the lines of The case of 0⁰ in the context of integer exponents is generally taken as 1, but in some contexts is controversial as discussed below'. I'm tempted to argue that WP does allow unsourced statements when usage is so prevalent that it is rarely discussed. —Quondum 16:23, 18 December 2013 (UTC)

I don't want to drag on the debate and the edits, but I made an edit that I think is a good compromise. Rather than simply defining 0^0 to be 1, we can state (without controversy or originality) that there are formulas that require 0^0 to be interpreted as 1. The previous text is not OK because it simply says that 0^0=1 reduces the number of cases, but fails to mention that 0^0=1 is needed for the correct interpretation of these formulas. Mark van Hoeij — Preceding unsigned comment added by 71.229.28.197 (talk) 18:39, 18 December 2013 (UTC)

Something like this is possibly acceptable, but I think the sentence needs to clarify that this is (as the section heading currently says, but not the sentence itself) specifically in the natural-number-exponent context. (I really do prefer to say "natural number" rather than "integer"; it occurred to me yesterday that the justification gets iffier in the integer-exponent context, and all the examples being presented are for natural-number exponents, where of course I count 0 as a natural number.) --Trovatore (talk) 18:45, 18 December 2013 (UTC)

This is a good compromise. All the relevant information is there (there are formulas that require 1, not all sources define it as 1, and one has to be careful with limits) (none of which is controversial). Mark — Preceding unsigned comment added by 71.229.28.197 (talk) 18:52, 18 December 2013 (UTC)

So just to clarify, you're good with my tweak? I made it before you made the above comment, so I assume you saw it. --Trovatore (talk) 19:30, 18 December 2013 (UTC)

merging three definitions

Trovatore wrote: "The most common definition for real-to-real exponentiation is $x^{y}=\exp(y\ln x)$, and it doesn't follow from that". So, according to Trovatore et al., we must also undefine e.g. (−1)² as ln(−1) is not defined as a real number. Bo Jacoby (talk) 23:34, 18 December 2013 (UTC).

In real-to-real exponentiation, that is correct. The point (-1,2) is not in the domain of that function. --Trovatore (talk) 01:55, 19 December 2013 (UTC)

Ah, now. If we can distinguish functions with identical notation like this (i.e. treat them as separate functions), the whole debate about the definition at specific points as in 0⁰ evaporates. The domain is determined by which function it is. The entire problem seems to arise from the premise that it is the same function, only with its domain restricted. —Quondum 04:19, 19 December 2013 (UTC)

Exactly. --Trovatore (talk) 04:20, 19 December 2013 (UTC)

We have three real functions, f₁ and f₂ and f₃ defined by

f₁(x,y)=1 for y=0

f₁(x,y)=x⋅f₁(x,y−1) for y=1,2,3,...

and

f₂(x,y)=1 for y=0

f₂(x,y)=f₁(x,y+1)/x for x≠0 and y=−1,−2,−3,...

and

f₃(x,y)=exp(y⋅log(x)) for x>0.

The domain of f₁ is D₁ = ℝ×{0,1,2,3,...}.

The domain of f₂ is D₂ = (ℝ×{0})∪((ℝ\{0})×{−1,−2,−3,...})

The domain of f₃ is D₃ = ℝ₊×ℝ.

If (x,y) ∈ D_i ∩ D_j then f_i(x,y)=f_j(x,y) for i,j=1,2,3.

Then define x^y:

x^y=f_i(x,y) if (x,y) ∈ D_i for i=1,2,3.

This definition covers the real cases of x^y. I hope very much that we can eventually get rid of the ridiculous nonsense that 0⁰ is defined or undefined depending on whether 0 is an integer or a real. Bo Jacoby (talk) 21:51, 20 December 2013 (UTC).

You think saying 2+i0 > 1+i0 is okay don't you? Dmcq (talk) 19:45, 20 December 2013 (UTC)

Yes, 2+i0 = 2 is true. 2 > 1 is true. 1 = 1+i0 is true. So 2+i0 > 1+i0 is true. The fact that the ordering of reals is not extended to non-real complex numbers does not undefine the ordering of reals. Do you think that 2+i0 > 1+i0 is false? And why? Bo Jacoby (talk) 21:51, 20 December 2013 (UTC).

It's not false, just meaningless. But guys, we really shouldn't be discussing this here. --Trovatore (talk) 21:54, 20 December 2013 (UTC)

The point is if you think 2 and 2+i0 are exactly the same thing then even things like 2^x starts giving problems as shown by Clausen's paradox in the article. I'm just showing that saying that there is a real difference between integers and reals and complex numbers. We can't sweep the differences under a rug and get a unified view of exponentiation. Dmcq (talk) 23:50, 20 December 2013 (UTC)

You don't solve Clausen's paradox by making 2≠2+i0. The point is that a number is not either real or complex; a real number is both. Bo Jacoby (talk) 20:02, 21 December 2013 (UTC).

This is not really the case at all. A real number (in the usual formalization) is a Dedekind cut of rational numbers. A complex number (in the usual formalization) is an ordered pair of real numbers. Because a Dedekind cut is not an ordered pair of Dedekind cuts, a real number (in the usual formalization) is not a complex number. What is true is that there is an injective ring homomorphism from the reals to the complex numbers. So, strictly speaking, what we have is a sequence of embeddings

$\mathbb {N} \to \mathbb {Z} \to \mathbb {Q} \to \mathbb {R} \to \mathbb {C}$

None of these embeddings is an identity map (that is, none of them is an inclusion). — Carl (CBM · talk) 20:26, 21 December 2013 (UTC)

However, even if we treat the reals as a subset of the complex numbers, that does not affect the point that exponentiation of real numbers is not the same operation as exponentiation of complex numbers. There are two different operations (for example, exponentation of reals is often defined as a continuous extension of another function, while exponentiation for complex numbers is defined using a branch of the complex logarithm). If they happen to have the same value for certain inputs, that is a theorem, not a definition. — Carl (CBM · talk) 20:30, 21 December 2013 (UTC)

Not every formalization is the usual one. Other formalizations are equally valid models. Real numbers can be defined as complex numbers of the form x+i0, and so on. Then we have the sequence of inclusions ℕ⊂ℤ⊂ℚ⊂ℝ⊂ℂ. Exponentiation of positive real (complex) numbers is not the same operation as exponentiation of negative or nonreal (complex) numbers. It is still ridiculous nonsense to claim that 2+i0≠2. Bo Jacoby (talk) 21:16, 21 December 2013 (UTC).

Are you saying that in the particular case of the complex part of a complex number being zero then real exponentiation should be used? In that case I would be intrigued to see how you explain Clausen's paradox. Dmcq (talk) 17:17, 30 December 2013 (UTC)

Clausen's paradox: The identity (e^x)^y = e^xy holds for real numbers x and y, but assuming its truth for complex numbers leads to . . . paradox, should be rephrased: . . . assuming its truth for non-real (complex) numbers leads to . . . paradox. Bo Jacoby (talk) 00:13, 31 December 2013 (UTC).