Talk:Absolute value

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Format issue (old discussion prior to Feb 25, 2002)

I call that ugly and further, the symbols are no longer symmetric and it is harder to read for someone not accustomed to the greater than and equal sign. I can't imagine why you changed it. RoseParks

why dont you just slob on a nob


It's knob*! — Preceding unsigned comment added by 72.54.81.241 (talk) 19:00, 1 August 2011 (UTC)

I made the major changes to this page, mainly to wikify what appeared to be HTML-based text and to make it more readable from my own esthetic perspective, so if you think there is something that doesn't work, I'd be happy to fix it, but you'll have to be more specific. I don't know what you mean by "the symbols are no longer symmetric", and I don't know what you mean by someone being "unaccustomed" to the ≥ sign. We have to express that idea somehow, and the only two reasonable ways are "≥" and ">=". But I think the former is more readable for students of mathematics, while the latter is more readable for computer geeks. I can't imagine the article being of any use at all to someone who doesn't know the basic symbols of elementary mathematics; if you think it might be useful for readers that elementary, perhaps the article could contain links to other basic articles explaining the comparison operators? Are you perhaps talking about the alignment of the blockquote? How do you think it should appear? --LDC---- I had used and to match "<" and ">." Actually, someone else took the font size out before you. And, I just put it back...:-(..RoseParks

OK. That's a very font-specific thing (on my machine the ≥s look a bit too small), and it makes the text a real pain to edit, but if you think it makes a real readability difference on your machine, and the text doesn't need to be edited much, go for it. I won't remove any further ones I see. I would hesitate to make that a standard practice for math pages here in general unless we do end up using something like TtH to do the conversions automatically so we won't have to edit all those font commands. I do think agree that esthetic details can make a big difference in the readability of math formulas (I really wish there were an easy way to vertically center the internal ||s within the larger enclosing ||s in rule 3 above), but the limitations of HTML are pretty severe, so you can't have everything you might want, and what works on one machine might not work on others. --LDC

bar notation

what's the difference between $|x|\,$ and $\|x\|\,$? It should be specified in the article. - Omegatron 18:04, Sep 26, 2004 (UTC)

The latter notation is usually reserved to represent some kind of norm, such as in a normed linear space. Revolver 03:50, 27 Sep 2004 (UTC)in the bars it all maters from absolute value. The absolute value of x, denoted "| x |" (and which is read as "the absolute value of x"), is the distance of x from zero. This is why absolute value is never negative; absolute value only asks "how far?", not "in which direction?". This means not only that | 3 | = 3, because 3 is three units to the right of zero, but also that | –3 | = 3, because –3 is three units to the left of zero.

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I've changed the following:

In mathematics, the absolute value (or modulus) of a number is the difference between that number and 0. Simply speaking, it is the number without a negative sign. So, for example, 3 is the absolute value of both 3 and −3.

to:

In mathematics, the absolute value (or modulus) of a number is its numerical value without regard to its sign. So, for example, 3 is the absolute value of both 3 and −3.

My reasons are that I find the former both confusing and misleading. For example, for the number three, "the difference between that number and 0", can be construed as either "0 - 3" or "3 - 0". And saying that "it is the number without a negative sign" leads to the misimpression that |-a| = a.

Paul August 13:46, Apr 17, 2005 (UTC)

I think the change is good. We should mention the misimpression |-a| = a. I suspect it may not be obvious to those with little math knowledge. -- Taku 16:42, Apr 17, 2005 (UTC)

Is the seventh property ok?

Is is ok that |x|=sqrt(x^2) as it is written in the article? As far as I understand, the square root has two solutions: one that is positive and the other one that is negative (except for sqrt(0)). That's why I would like to fix the article but I'm not completely sure about it because the one who wrote it must have had some idea in mind. (Unsigned comment by User:Erast)

Yes, the seventh property is correct. While it is true that for every non-negative real number x, there are two (provided x ≠ 0) numbers whose square equals x, the symbol "$\sqrt x$" denotes the principal square root of x, that is the non-negative real number whose square is x see: Square root. Paul August 15:46, Apr 19, 2005 (UTC)

Rewrite of article

I've just completed a major revision of this article in which I have:

1. Expanded slightly the lead, better placing (I hope) the absolute value in context.
2. Combined the "Definitions" and "Properties" sections into one section called "Real numbers".
3. Named, organized, reformatted, and added to the properties in the "Real numbers" section.
4. Moved all the content about the complex absolute value to one section called "Complex numbers", and expanded.
5. Moved content pertaining to absolute value functions to its own section called "absolute value functions", and expanded.
6. Created new section about absolute value in ordered rings.
7. Created new section called "Distance", describing the relationship between the absolute value and distance.
8. Created new section about absolute value in fields.
9. Moved content about norms and vector spaces to new section called "Vector spaces", and expanded.
10. Added a "References" section, and some notes.

I'd appreciate any comments/criticisms anyone might have. Thanks, Paul August 19:49, July 16, 2005 (UTC)

Wikipedia logo appears on top of article text in the following browsers: Mozilla 1.7.8 (Debian Sarge), Konqueror (Debian Sarge), Netscape 7.1 (Windows XP). --Trovatore 21:20, 16 July 2005 (UTC)

Travatore: Can you isolate what the problem is? Paul August 21:24, July 16, 2005 (UTC)
For the record, it looks ok to me in Safari, Firefox and IE (MacOS X). Paul August 21:31, July 16, 2005 (UTC)
Experiments in my sandbox suggest that it's the Ent and Rf templates. Removing just the Rf's doesn't solve it, but removing both Ent and Rf does. I haven't tried removing just Ent. --Trovatore 21:42, 16 July 2005 (UTC)
I've replaced each template with it's substitution, did that fix it? Do other articles use thes templates like Demetrius of Pharos have this problem? Paul August 21:54, July 16, 2005 (UTC)
Didn't fix it. It must be a very strange bug--removing any 2 of the 3 Ent templates fixes the problem, but removing 1 of 3 (no matter which one) does not. --Trovatore 21:58, 16 July 2005 (UTC)
Demetrius of Pharos looks fine. --Trovatore 21:59, 16 July 2005 (UTC)
I've made another change, added ending div tags, does this help? I'm afraid I'm flying blind, grasping at straws and mixing my metaphors (when i'd much rather be mixing my drinks). Paul August 22:17, July 16, 2005 (UTC)
Yes, it worked --Trovatore 22:21, 16 July 2005 (UTC)

The last change (inserting the <div>s) worked --Trovatore 22:15, 16 July 2005 (UTC)

The older version Trovatore mentioned at Wikipedia talk:WikiProject Mathematics still works fine for me in Mozilla 1.6 and in Konqueror 3.2.2-4 on Fedora Linux. Oleg Alexandrov 22:51, 16 July 2005 (UTC)
Hm. I don't know why that would be. It certainly doesn't seem likely that it's a peculiarity of my machine, given that I observed the problem under both Linux and Windows, and in both a Gecko and a non-Gecko browser (I don't think Konqueror uses Gecko, does it?). --Trovatore 22:54, 16 July 2005 (UTC)

Thanks Oleg, and Trovatore, for helping out with this problem. I would very much like to get to the bottom of it, particularly since I'm the author of the "rf" and "ent" templates, and I use them a lot, so any further help would be greatly appreciated.

Just tried it on Netscape 4.79 (WinXP), but none of the Wikipedia really works very well there, so it's hard to tell.--Trovatore 23:36, 16 July 2005 (UTC)
One thing that strikes me, though: If Wikipedia is not putting on the closing </div> tags, then it's not producing compliant HTML, and that's something that should be fixed, irrespective of problems in any particular extant browsers. Sounds like a developer issue. --Trovatore 23:41, 16 July 2005 (UTC)

Regarding the content of the article, do either of you have in comments?

Paul August 23:13, July 16, 2005 (UTC)

Articles should ideally not be in proposition-definition form (proofs are fine, it's just the layout of content that's the issue), but that's just my POV. Dysprosia 01:37, 17 July 2005 (UTC)

Yes, I don't usually use this "proposition" style, but I chose to in this case, because I wanted to group and name the properties of the real absolute value, so I could refer to them further down in the article. Proposition 1, is the identity, which can be seen to be related to the definition of the complex absolute value, Proposition 2, is the group of properties used to motivate the definitions a metric, and of the absolute value in the case of fields and vector spaces, and Proposition 3, is a group of properties which "come for free" as a consequence of adopting the properties in Proposition 2 as axioms. Also I think listing them in this layout helps to show the various structural relationships. Having said all that, feel free, of course to make or suggest any changes you think appropriate. I'm very open to better ways of presenting this material. Paul August 02:46, July 17, 2005 (UTC)
Perhaps giving the results names instead of merely calling them propositions would help. It would also be advantageous to possibly tie the idea of norms and inner products in slightly, in some of the cases (esp. the real case). Dysprosia 06:38, 17 July 2005 (UTC)
Dysprosia, could you be more specific? Paul August 16:45, July 17, 2005 (UTC)
In the real case, we can define an inner product over R (which is a vector space) to be regular multiplication of reals. The absolute value of a number is then the same as the vector norm with respect to the inner product just defined. Likewise for the complex case. It would be nice to mention this in relation to the absolute value. In a sense the article does so, but it doesn't seem too unified. This may also take care of the first problem of mentioning things as "propositions" instead of giving them more illustrative names. Dysprosia 04:49, 18 July 2005 (UTC)
Well we could make the relationship between inner products and absolute values more clear, but I would be inclined to do that in the "Vector space" section. I would prefer to keep the "Real numbers" section as basic as possible, so that it could, for example, be read by a high school student. As you say the article does mention this relationship in the "vector space" section, but I agree, it is a bit disjointed. I will/may try to rewrite that section. Any suggestions on how best to do that (or doing it yourself) would be appreciated. However, this doesn't address your "propositions" concern. Could you explain your suggestion about giving "them more illustrative names"? I'm not sure I understand. Could you give examples? Paul August 15:04, July 18, 2005 (UTC)
I'm not saying rewrite the real numbers section completely in terms of vector space terminology but merely mentioning it somewhere to give a view of the big picture and to give the page more of a unified feel. Instead of saying "Proposition 1" one could say "Alternate definition", for "Proposition 2" one could say "Axiomatic properties" (thus linking with the properties of the norm), for "Proposition 3" one could say "Consequent properties", perhaps. Dysprosia 12:44, 19 July 2005 (UTC)

Shouldn't the fact that absolute value generalizes into norm be mentioned (more prominently) in the heading? I have also heard that absolute value on reals is related to valuation (I don't know any details, I may be completely wrong). Samohyl Jan 09:29, 17 July 2005 (UTC)

I've added a mention of the norm in the lead. Paul August 16:45, July 17, 2005 (UTC)
The article suggests that absolute values and valuations are synonyms, but this is wrong. The usual absolute value on the rationals does not correspond to any valuation.--Gwaihir 10:45, 17 July 2005 (UTC)
Well as defined in the article, a real-valued function on a field which is non-negative, positive-definite, multiplicative, and subbadditive is called both an "absolute value" and a "valuation" according to Eric Schechter in Handbook of Analysis and Its Foundation pp 259-260. see also PlanetMath: "Valuation", and the absolute value for the rationals is a "valuation, under this definition. I'm unfamiliar with the definition given in valuation (mathematics). Paul August 16:45, July 17, 2005 (UTC)
MathWorld has both, and IIRC Bourbaki (Algèbre commutative) uses the logarithmic definition.--Gwaihir 09:17, 20 July 2005 (UTC)

One thing that doesn't seem to make sense: In the section on distance, I believe there was a mistake. |a+b| doesn't equal $sqrt{a^{2}+b^{2}}$ It's $sqrt{(a+b)^2}=sqrt{a^2+ab+b^2}$ Exponentation doesn't distribute additively.He Who Is 22:09, 27 April 2006 (UTC)

terminology should be precise

1. an "absolute value in vector spaces" seems completely out of scope to me. A norm is not an absolute value, it does not verify the multiplicative identity (except in some algebras). Although there should be a reference to norm (insisting on the necessity of having an abs value on the scalars), there should not be so much details about norms here, since it is out the subject.
2. the computer functions calculating the absolute value has many other names than abs(), e.g. fabs(), dabs(), mmabs(), ieeeabs()...
3. there IS a general definition for an abs value, which should be found here
4. a valuation is NOT an absolute value: it has another definition (see e.g. Lang's "Algebra").
5. there should be a note about important results that cam be deduced from the existence of an absolute value on a ring or on a field (e.g. that the latter contains the rationals).

MFH: Talk 13:35, 28 September 2005 (UTC)

MFH: I've numered your points above. I will rely by the numbers ;-) — Paul August 15:56, 28 September 2005 (UTC)

1. I believe several authors refer to a norm as an "absolute value", but I don't have any references ready to hand. Perhaps we should de-emphasize that somewhat. However the norm is certainly a generalization of the idea of absolute value. You are correct, a norm in general does not satisfy the multiplicativeness property of the real and complex absolute values. It's "corresponding" property is positive homogeneity (also called positive scalability): $\|a \mathbf{v}\| = |a| \|\mathbf{v}\|$. I think there is a good reason to have a thorough discussion and explication of the generalizations of the idea of absolute value here. — Paul August 15:56, 28 September 2005 (UTC)
2. Feel free to add those. — Paul August 15:56, 28 September 2005 (UTC)
3. What is that general definition? — Paul August 15:56, 28 September 2005 (UTC)
4. Eric Schechter's Handbook of Analysis and Its Foundation pp 259-260 and PlanetMath: "Valuation", define valuation in the same way that it is defined here, as a real-valued function on a field which is non-negative, positive-definite, multiplicative, and subbadditive. I don't have a copy of Lang's book, what is his definition? — Paul August 15:56, 28 September 2005 (UTC)
5. Feel free to add those too ;-) — Paul August 15:59, 28 September 2005 (UTC)

tvexiall?

I've removed from the article twice the following:

The V-shaped graph of an absolute value function is called a tvexiall.

This was first added by 68.162.147.182, and later by Emorgasm. I've never heard of this term and neither has Google. Anyone know anything about this? Paul August 17:01, 18 November 2005 (UTC)

Never. Sounds like some kind of bizarre neologism to me. It seems they intended it to be a term along the same lines as "parabola", describing the shape of the curve. It might be a misspelling of something, but I have no clue what. Deco 01:05, 12 December 2005 (UTC)

Needs disambiguation

It seems to me that this page needs disambiguation as absolute values is a term that is also used in philosophy. —Preceding unsigned comment added by 85.230.64.73 (talkcontribs) 11 December 2005

Multiplicative Absolute Values

Since for all x and y the actual 'difference of x and y can be represented as |x-y|, and that equals x-y if x is the greater, it only stands to reason thata similaroperator should be defined multiplicitively, such that:

$MAV\left( \frac{x}{y} \right) = \left\{ \begin{matrix} \frac{x}{y} & \mbox{if } x > y \\ \frac{y}{x} & \mbox{if } y > x \end{matrix} \right.$

This would give no results that lie on the interval [-1,1], 1 being the multiplicitive identity. This is analogous to the absolute value giving no results less than 0, whitch is the additive identity. Graphed, this would hyperbolic over [-x,x] and linear outside that. Anyone care to comment on the subject?

He Who Is 00:07, 27 April 2006 (UTC)

Here are my thoughts on your idea. A formula which suggests itself:
$MAV(x,y)=\exp \left | \log \left( \frac{x}{y}\right) \right |.$
Note that I have denoted it as being a function of two variables. If you want to define it as a function of one variable, as your notation suggests, then its definition is slightly problematic for irrational numbers and negative numbers. However, I don't think this formula is really what you want. You want a function like this:
$f(x) = \begin{cases} x & |x| >1\\ \frac{1}{x} & |x|<1 \end{cases}$
and then you can define MAV(x,y) = f(x/y). This seems to extend naturally to all complexes without difficulty:
$f(z) = \begin{cases} z & |z| >1\\ \frac{1}{\overline{z}} & |z|<1 \end{cases}$
and I take the complex conjugate to make the number's argument stay the same, though you don't have to do that if you don't want to. This is actually a bit reminiscent of the concept of radial ordering, which is used in quantizing fields on the worldsheet in string theory. -lethe talk + 00:40, 27 April 2006 (UTC)

The only problem I have with making MAV a function of two variables is that it loses its analogy to ABS in that it is no longer graphable on a requalar two-dimensional carteisian plane. Because of this, while it can be said that -5, for instance has an ABS of 5, -5 doesn't have a Multiplicitive Absolute Value unless it is paired with another number. And as for problems concerning negatives and irrationals, as far as I can tell MAV(x) = MAV(-x), and if you could elaborate on the problems with irrationals, that would be helpful, as they're not apparent to me. He Who Is 01:57, 27 April 2006 (UTC)

If you want a function of a single variable, see my second suggestion above. The problem with your definition is, –1/2 = (–1)/2 = 1/(–2). According to the first representation, –1 < 2, so MAV(–1/2) = –2. According to the second representation, 1 > –2, so MAV(–1/2) = –1/2. For positive rationals, each positive rational can be reduced to a fraction of the form p/q, where p and q are relatively prime. Irrational numbers generally can be written as fractions many different ways, and there's no way to choose one. The best thing to do is forget about fractions, and just check whether the real number is bigger than 1 or less than 1, which gives you the definition I gave. -lethe talk + 02:21, 27 April 2006 (UTC)

I think you mean check it it's absolutely greater than one. Also, in your example, you were comparing two numbers within a fraction, but the need not be represented as a fraction at all. Also, you said that "Irrational numbers generally can be written as fractions many different ways, and there's no way to choose one." This is inherently untrue, as by definition an rational number is:$\mathbb{Q} = \left\{\frac{a}{b}|\left\{ a,b\right\} \sub \mathbb{Z}\right\}$ and an irrational is it's complement on $\mathbb{R}$. Therefore, Irrationals cannot be represented as any fractions. —The preceding unsigned comment was added by He Who Is (talkcontribs) .

I think you've finally understood my point, which is this: forget about fractions. I'm slightly bothered however by the fact that now you're representing this point as your own, and trying to teach it to me. It is not me who was comparing two numbers in a fraction, that was you. I was the one telling you that you should not be doing that, because some numbers can be represented more than one way as a fraction. In my definition, I compare a single number to the number 1. No fractions in sight. As for your suggestion that irrational numbers cannot be represented as fractions, let me show you a counterexample: π=π/1=3π/3 = –π/(–1). -lethe talk + 19:50, 27 April 2006 (UTC)

I apologize, as I must have misspoke. I meant that irrationals cannot be represented as fractions of integers. Also in what case was I comparing components of a fraction? If I was I hadn't meant to, unless you mean representing a reciprocal as 1/x, which is the only way I know of in which they can be represented. I fear this entire conversation may have just come from a few misunderstandings.He Who Is 20:06, 27 April 2006 (UTC)

You defined MAV(x) as x/y if x>y, in your first post in this thread (scroll up to see for yourself). Here, you take a real number, and compare the numerator with the denominator. For various reasons I mention, this is problematic. One way to get around it is to view the MAV function as a function of two variables; then you're simply comparing the first variable to the second. Another is to forget about fractions, and simply compare the real number to the number 1. I suggested to you that the second method was the better way: forget about fractions. You indicated that you didn't understand the difficulties with fractions, so I explained further. A few posts later, you tried making the same point to me, which struck me as odd, since it was the same point I made to you in my first post to this thread. Finally, your point about irrationals admitting no representations as fractions of integers is true, but is no longer a correction to anything I wrote. -lethe talk + 20:14, 27 April 2006 (UTC)

Oh... You're right. And I now see what you mean. Inherently, this is the same as your notation posted thereafter, but as irrationals cannot be represented as fractions of rationals, it presents a problem. I suppose I had forgotten how I represented it in my first post, and for that reason was confused by your subsequent replies.
This aside, I was recently considering this and something strange struck me. Where x plus its additive inverse equals zero, the additive identity, and x times its multiplicitive inverse equals 1, the multiplicitive identity, a theoretical exponential inverse seems slightly more complicated. It follows that 1 would also be an "exponential identity", but for all complex numbers z, z^0 = 1. This implies an exponential inverse of 0 for all numbers. But this does not follow in that x minus -x = 2x, x divided by 1/x = x^2, but the 0th root of x does not equal x^^2, as would implied.which seems contridictory... He Who Is 20:32, 27 April 2006 (UTC)

Yes, I see what you mean. Well, let's look more closely: z^1=z for all z, so 1 is a right-identity, meaning that when we put 1 on the right-hand side of the exponentiation, we get back what we started with. Unfortunately, it is not also a left-identity, since 1^z≠z, for all z. I guess there is no left-identity. When there are both a left-identity and a right-identity, then they have to be equal (at least when the operation is associative), which allows us a flexible definition of inverse. We don't have that, and exponentiation is not associative (since 1^(2^3) = 1 ≠ 8 = (1^2)^3), so our inverse is also in trouble. Nevertheless, we could try to imagine 0 as some sort of "exponential right-inverse". It's not an "exponential left-inverse", since 0^z≠1 for all z. But check this out, there are two inverse operations of exponentiation: extracting roots, and taking logarithms with a given base. Under addition, we get 0–x = -x, and under multiplication we get 1÷x = x–1, and under exponentiation, we get logx 1 = 0 (which is the exponential inverse of x). This doesn't work with root extraction: 11/x≠0. Thus, what we really want, instead of taking roots, is taking logarithms. Trying your equation again, I get logx 0, which is probably some infinity. Alright, you're right. It just doesn't work.
It's a phenomenon that I've thought about in the past: addition is a nice operation. Iterate it, and you get multiplication, another nice operation. Iterate again, and you get exponentiation, which is not so nice. What happened in the second iteration that didn't happen in the first iteration? I never came up with a satisfactory answer to this question. -lethe talk + 21:19, 27 April 2006 (UTC)

Yes. It's a strange phenomenon. Why does exponentiation lack the commutativity of addition and mutiplication? This subject is closely related to the hyper operator, which I've been doing some work with lately. Namely, to generalize for things like $2^{(1.5)}3$,which essentially would ask to perform the operation halfway between addition and multiplication. Perhaps understanding this could give a better reason for the differences from exponentation on up. And another related question: where does commutativity disappear? Does it exist only for addition and multiplication, or for everything in between them also? If so, this would mean that the functions hyper(a,x,b) and hyper(b,x,a) coincide from 1 to 2, but seperate outside that which doesn't seem to make sense.

And for the record, I believe logx0 is in fact negative infinity, which makes sense looking at an exponential graph, as it diverges from 0.He Who Is 21:59, 27 April 2006 (UTC)

To try to understand this question, I turned to abstract algebra. Starting with an abelian group operation (addition), by iterating you get something that, while similar to multiplication, is not precisely a multiplication. That is to say, you get a left Z-module, mutliplication by "numbers" on the left by integers. For general modules, the left R-module is not isomorphic to the right R-module. Over abelian rings, they are isomorphic. So viewed in this way, the commutativity of multiplication is actually a restatement of the commutativity of addition. This framework provides a little bit of insight as to why iterated addition is still commutative, but it doesn't really support any definition of exponentiation. Perhaps this is the reaon, though I don't find this explanation completely satisfying.
As for logx 0, over the reals, that quantity is undefined. Over the extended reals, negative infinity is indeed the natural definition for values of x greater than 1, while positive infinity is a natural definition for values of x between 1 and 0, while I'm not sure what to do with 0, 1, or negative numbers. I thought perhaps some insight might be provided by the projective real line, where there is only an unsigned infinity. In the end, I just gave up and left the vague phrase "some infinity". -lethe talk + 22:49, 27 April 2006 (UTC)

For negative numbers, the answers are simply complex. For 1, it in undefined even over complex numbers, because 1 does not change with exponentation, and as for zero, zx is zero for all complex numbers, so that could be anything.He Who Is 20:30, 28 April 2006 (UTC)

The logarithm is single-valued on its Riemann surface, infinite-valued on the complex plane (because log –1 = iπ + 2iπn for any integer n), and undefined on the reals. There is, as far as I know, no way to add signed infinities to the complexes, so if you want log 0 to be negative infinity, you can't also want log –1 to be complex. (Maybe you could add an infinity along every line through the origin in the complex plane, getting the extended complex numbers, homemomorphic to a closed disk?) May I make a suggestion? You should be more careful with the domains of your functions. I think that's the third time in this thread that your disregard for the domains of your functions has led you to make statements which might get you into trouble. -lethe talk + 21:26, 28 April 2006 (UTC)

I apologize for that. I should be more careful. But either way I think a more formal definition for MAV(x) could be found (anologous to the sqrt(x2) for |x|) if another function for which the value at x and 1/x are equal could be found. Then that nested within its own inverse would equal to MAV(x). Do you know of any such function? He Who Is 00:17, 13 May 2006 (UTC)

Absolute value a degenerate hyperbola?

I have a question... Correct me if I'm wrong, but isn't an absolute value graph just a hyperbola of a circle radius 0? Because if

$y^2 = \sqrt x^2$

Then this would lead to the conclusion that

$y^2 - x^2 = 0$

Thereby being a hyperbola, albeit a degenerate one.

Fisrt of all, you really should sign your messages. But either way, I see three problems in your above statement. First of all, $y^2 = \sqrt x^2$ becomes $y=\sqrt x$, not $y^2 - x^2 = 0$. But even simpler that you seem to have misread the formula. Absolute value is graphes as $y=\sqrt x^2$ not $y^2 = \sqrt x^2$. Finally, y^2-x^2=0 is not a hyperbola, and a hyperbola does not have a radius. In fact if you were to simplfy that, it bould be equivalent to $\pm y= \pm x$, which would be a large x, made up of x=y, and -x=y.He Who Is 02:21, 18 May 2006 (UTC)

I give my apologies for both not signing my post and also for making a syntactical error. I meant to say $y=\sqrt x^2$, the squaring of y was an error. This explains both the first and second fallacies in my logic; as for the third, isn't a geometric point considered a circle radius zero? If this is true, then absolute value would, I believe, be the hyperbola of said circle. --ThatOneGuy 07:19, 19 May 2006

Ah. I'm sorry. I read hyperbola and thought parabola. As for your error, You ar in fact correct, come to think of it. The reason this doesn't show in the graph at first glance may lie in the difference between sqrt(x^2-r^2) and sqrt(r^2-x^2). One is your hyperbola, and the other is the circle of the same radius. If the absolute value is taken before extracting the squared root, you get both at once. In the case of the absolute value, the radius is zero, and the hyperbola's slope starts at $\pm 1$, it does not converge, and simple creates a right angle at the nonexistently small circle that normally seperates both parts. In the case of absolute value, the squared rootis taken to mean the principal root, so only the top half is shown, and because with a redius of zero, neither the circle nor the hyperbola are made up of complex points, they both show up. (even though since it is degenerate, as you said, the circle cannot be seen.) I've never thought about absolute value that way.He Who Is 00:55, 20 May 2006 (UTC)

Defferentiability

I think a section, or at least a mention, of differential and antiderivitive of |x| should be added. It's differentibility is mentioned, but not it's actual derivitive. (That is, $\frac{d}{dx}|x| = sgn{x} = \frac{x}{|x|}$. Also, $\int |x| dx = x|x|$. Which is x2 for all x greater than or equal to 0, and -x^2 otherwise.)He Who Is 22:08, 23 May 2006 (UTC)[/itex]

I think those would be good additions to the article. The derivative of the absolute value might actually be a useful formula (mention the Heaviside step function), but I don't think I've ever seen the integral used. Of course you have to make sure to mention that it's not differentiable at the origin. -lethe talk + 00:20, 24 May 2006 (UTC)

Added it. It could probobaly use some work, but I think its good for now.He Who Is 01:27, 24 May 2006 (UTC)

I copy-edited the new addition. It's mostly OK, but I also decided to get rid of the stuff about multiple integrals, which seems a bit weird to me (not to mention badly formed). -lethe talk + 01:53, 24 May 2006 (UTC)

Division

I'm tempted to remove a certain statement, but decided to see what other wikipedians have to say about it first. It is the statement that claims preversation of division is equivalent to multiplicativeness. This is because of the following: (Remember that this assuming the person performing the calculations doesn't know about preversation of division in absolute value.)

$\frac{|a|}{|b|} = |a| \cdot \frac{1}{|b|}$

You see, preversation of division is a result of multiplicativeness as well as the fact that 1/|b| = |1/b|, which is not implied by multiplicativness. Otherwise you don't have the product of two absolute value, you have an absolute value times the reciprocal of another. -- He Who Is[ Talk ] 22:30, 20 July 2006 (UTC)

But $|b||1/b| = |b \cdot 1/b| = |1| = 1$ by multiplicativity too. Dmharvey 22:54, 20 July 2006 (UTC)
Just so. Paul August 04:22, 21 July 2006 (UTC)

Another Definition of Absolute Value

I am not at all advanced in Math, or absolute values, so please correct me if I am wrong. I would like to know if it makes more sense to define Absolute Value as "An integer's distance from zero". Yeah, it's basic, but absolute value isn't just disregarding the integer's sign, is it? --Chrishy Man 02:02, 29 September 2006 (UTC)

First note that absolute value is defined for all real (and complex) numbers not just integers. Second the two definitions amount to the same thing, but I think the article's previous definition is better, since it doesn't involve a geometric interpretation of the real numbers. I've reverted to the original definition. Paul August 21:15, 16 October 2006 (UTC)

Hey guys, the definition "modulus of a real number is its numerical value without regard to its sign" is 1) childish; and 2) misleading. It is good for the first introductory mathematical lesson when you noramlly are too young to be aware of the vectors and complex numbers, but it does not address the whole basic and great idea of the modulus: "Modulus is a length of a (radius) vector". This is brief and universal. The widely known complex and real numbers are included as vectors on the complex plain and number line correspondingly. I agree that the professional math things I'm not aware of must be left out of the definition, but covering the complex numbers is important to give the essence of absolute value, which relies not in the sign, but is determined by the distance from the origin. It took me ages to understand why the absolute value of a plain number and of complex number and vector -- all have the name modulus and math notation |a|. Therefore, I strive to reduce the confusion caused by the children definition. Not to speak that the module was introduced first for the complex numbers. --Javalenok (talk) 21:13, 10 November 2008 (UTC)

Absolute value for rings

User:AlainD made this edit, which added that the definition given for the absolute value of a field in the "Fields" section, is also used for rings. Since I have some questions about this edit I have reverted it. Is there a source for this? Do the given equivalences for non-Archimedean still hold? Paul August 19:44, 8 February 2007 (UTC)

Well, it took me a long time to choose between "ring" and "integral domain", and I am ready to review my decision. It is possible to define absolute values for general rings, but as |x|=0 => x=0 and |xy|=|x|.|y|, they are simply no absolute values on rings with zero divisors. So the decision turn on to be a choice between no harm and no interest, sorry between generality and vacuity.
I have sources. But may I suggest a look at the intersting remark at the end of Absolute_value_(algebra)#Valuations before sorting contaradictory definitions.
What do you call non-Archimedean: to be not Archimedean or to be ultrametric?
AlainD 11:51, 9 February 2007 (UTC)

Hi Alain, thanks for the reply. Yes this is all a bit more complicated than I thought, and will take some sorting. More later. Paul August 22:58, 10 February 2007 (UTC).

Redundancy in complex definition

Mentioning these two definition seperatley seems redundant to me:

where x and y are real numbers, the absolute value or modulus of $z$ is denoted $|z|,$ and is defined as

$|z| = \sqrt{x^2 + y^2}.$

...

[if]

$\bar{z} = x - iy$

is the complex conjugate of $z$, then it is easily seen that

$|z| = r\,$
$|z|=|\bar{z}|$
$|z| = \sqrt{z\bar{z}}.$

$|z| = \sqrt{x^2 + y^2}$ is derived from :$|z| = \sqrt{z\bar{z}}.$ by simply mulitiplying the conjugates out. The latter is simply the factored version. Why have both mentioned seperately as if they are two different definitions or something? Brentt 05:49, 10 February 2007 (UTC)

$|z| = \sqrt{x^2 + y^2}$ is the definition, $|z| = \sqrt{z\bar{z}}$ is simply an identity which follows from that definition and the definition of conjugation. Paul August 23:10, 10 February 2007 (UTC)

Wrong example code

I removed the last three code samples for computing absolute value. They were wrong because while the highest bit is the sign bit, changing doesn't necessarily produce negated version of the number. For example, -2 is 0xfffffffe in two's complement and ANDing it with 0x7fffffff produces 0x7ffffffe which is 2147483646, the correct result being 0x00000002. C version can work, but only assuming the machine uses sign-magnitude representation of signed integers. x86 assembly version is fully irrelevant, as x86 always uses two's complement. Coriakin 09:58, 30 September 2007 (UTC)

You are right in every point, thanks for reverting this change, I don't know whwat I was thinking about while adding this. 158.75.90.90 (talk) 11:31, 21 February 2008 (UTC)

It should include the person who created this symbol

It should include the person who created this symbolDieknight1 (talk) 13:28, 9 January 2008 (UTC)

Which symbol? What is "it"134.29.231.11 (talk) 20:42, 31 January 2011 (UTC)

What is the impact and significance of absolute value?

I took physics for 5 years in college and other than handy way to magnitude and distance equations sensible. I never really understood it's importance in mathematics in general? What are the mathematical ramifications and implications of absolute value... is it just an operator to make numbers positive, or make solutions symmetric? I don't know if this question deserves a section, but I would like to know just for personal curiosity.--Sparkygravity (talk) 04:27, 11 January 2008 (UTC)

It's used in the formal definition of a limit, for one thing. —Preceding unsigned comment added by 207.245.46.103 (talk) 18:18, 24 April 2008 (UTC)

For one thing, yes; for another, it also makes sense if you want to stay in real number scope and have to take a square root of a - b and you don't know what value will be the higher one. So in programming, it would be totally awkward to say "if a > b then sqrt(a-b) else sqrt (b-a)" but instead you can say "sqrt (abs(b-a))" and do not have to care about what number is greater: you will have less code, and the result will be the same. -andy 217.50.60.205 (talk) 16:30, 11 April 2012 (UTC)

Hannibal

Some of these concepts are very simple: |x|=root(x^2). The problem is that the differentiation to n^n where n is a positive integer exceed the length to a 64 bit signed integer, of how to express it is a waste of time. There are Java plug ins that allow the writer to express math symbols as they would on a chalk (see 1*) or white board analysis is a waste of time.

Some problem occur; To solve this follow the following 1) Remember your audience, keep the discussion within their limits 2) Being unintelligible is a most heinous crime against science, when it's deliberate 3) Oscar Wilde, 'Brevity is the sign of intelligence' 4) Asking someone to marry you in binary is bound to fail :) 5) Remember kindergarten and primary school, diagrams are good, a picture tells a thousand word or as I prefer 'a picture summarizes a multidimensional massive array of data to the point that a human can understand'.

Have you discovered Eugen, or Flume?

(1*) I'm old school to drinking with the guys who drank with Boyle and Hook :) Sorry about riding a heard of Elephants through your site. Geoff B.Surv PGradDip Applied Math Computing (Completed last century) —Preceding unsigned comment added by 220.253.12.221 (talk) 01:38, 14 April 2008 (UTC)

ή↓ЭВ† —Preceding unsigned comment added by 207.179.153.36 (talk) 19:31, 12 June 2008 (UTC)

Category:Numeration

What is the justification for Category:Numeration for this article? Numeration is about notations and names for numbers; how is absolute value relevant to that? --Macrakis (talk) 02:05, 9 July 2008 (UTC)

It isn't. I've removed it. Paul August 02:13, 9 July 2008 (UTC)

Unclear Explanation of Antiderivative of |x|

$\int\frac{d|x|}{dx}xdx=x|x|-\int|x|dx$
This part of the proof that the indefinite integral of the absolute value of |x| is equal to x|x|/2 is unclear. How does one get from the first expression to the second expression? The article should show an intermediate step. Metroman (talk) 01:27, 14 February 2010 (UTC)

Python program example error

The simple function for getting the absolute value seems to generate a syntax error. Is this general, or is this example viable under Python 3.0?

If else I suggest the following:

>>> def absolute_value(x):
...     if x < 0:
...         return -x
...     else:
...         return x
>>> absolute_value(2)
2
>>> absolute_value(-75)
75
>>> absolute_value(-5.63)
5.6299999999999999


It's less elegant, and the float isn't exact. But a novice should understand it. Thaum1el (talk) 04:49, 2 November 2010 (UTC)

*ness and #tivitiy

Multiplicativeness is not a word, nor does it sound good(note, there may be weaselwords in the previous statement) Should we try to change descriptions of properties to the styles “"blank" property” rather than adding the postfixes ness and tivity?134.29.231.11 (talk) 20:40, 31 January 2011 (UTC)

Algorithm section should be deleted

because:

• wikipedia is not a manual WP:NOTMANUAL of computer languages
• computer language descriptions are prone to error (see some of the above sections)
• logically, if we give examples from 3 programming languages, we should give examples from about a dozen others, which is ridiculous.

If this was a more complex algorithm (e.g. the FFT), I think we could justify a single description in pseudo code. But for the absolute value function, I think it's unnecessary. Adpete (talk) 06:05, 4 March 2011 (UTC)

Done. Adpete (talk) 05:02, 8 March 2011 (UTC)

Derivative of |x| = x sgn(x)

It seems to follow from the definition $|x| = x \sgn (x)$ that $\tfrac{d |x|}{d x} = \sgn(x) \tfrac{d x}{d x} + x \tfrac{d \sgn(x)}{d x}$, which, from the sign function page, is simply $\tfrac{d |x|}{d x} = \sgn(x) + 2 x \delta(x)$ (where $\delta(x)$ is the Dirac delta function). Interestingly, this definition of the first derivative of the absolute function is not given in the article. Why?KlappCK (talk) 16:48, 17 October 2011 (UTC)

I added it to the derivatives section.KlappCK (talk) 17:50, 26 October 2011 (UTC)
The recent changes to the derivative section try to blend the real and complex cases, and the classic and generalized function cases. But this doesn't make sense. There really are three distinct cases:
• As a classical function of a real variable, abs has a derivative everywhere except at 0, where the derivative is undefined. I don't see an advantage in referring to the sgn derivation for 0.
• If we extend to generalized functions, the derivative at zero is indeed the delta function, but that is not part of classical analysis.
• Finally, as a function of a complex variable, the derivative is nowhere defined; abs violates the Cauchy-Riemann equations. The derivation directly from the definition of the derivative is shorter and more elementary than the sgn derivation. I will work on supplying this.
The new introductory sentence "The derivative of the real absolute value function is hard to define for all values of its argument, particularly because it does not exist because at every point in the complex plane[1], so a concise rigorous definition of the complex derivative does not exist." is completely garbled, both syntactically and mathematically. What does it mean to say that "the derivative of the real absolute value function is hard to define"? The real-function derivative is easy to define. And what does "a concise rigorous definition of the complex derivative does not exist" mean? In fact, it is not the definition that does not exist (and rigor has nothing to do with it); the complex derivative of abs is simply everywhere undefined. Finally, why does this sentence try to connect the real and the complex cases? That just complicates a very simple thing. --Macrakis (talk) 18:39, 26 October 2011 (UTC)
Agree. Paul August 19:51, 26 October 2011 (UTC)

I'm surprised no one has pointed out that the distribution $x\delta(x)$ is identically zero. I have reverted this addition. It's rather ironic that the same person who insisted on a citation for the standard fact that the absolute value function is not differentiable at zero is the same person adding (without any citations) this sort of nonsense. I have reverted to the old revision. Sławomir Biały (talk) 22:45, 26 October 2011 (UTC)

If any of you think you can do better, fix it, that's what I tried to do; simply saying that the $\tfrac{d|x|}{dx}=\sgn(x)$ is undefined at $x=0\,$ when $\sgn(x)\,$ is defined at $x=0\,$, without sourcing or proving that assertion, particularly when the sign page contradicts this assertion, is unsatisfactory. Furthermore, I did source the work, and no, the $x\delta(x)\,$ is not identically zero, if you think I'm wrong, prove it or source it.(KlappCK (talk) 12:55, 31 October 2011 (UTC)) Don't be a dick and assume good faith.KlappCK (talk) 12:42, 27 October 2011 (UTC)
x delta(x) is indeed equal to the zero distribution. So you're wrong. If you can find a source that says otherwise (doubtful), then it's also wrong. Pairing it with any compactly supported test function gives zero. Sławomir Biały (talk) 14:08, 27 October 2011 (UTC)
And, for what it's worth, since |x| is Lipschitz, its almost-everywhere derivative (which is sgn(x)) is equal to its distributional derivative. Sławomir Biały (talk) 14:42, 27 October 2011 (UTC)
See Graham's Hierarchy of Disagreement (I'm not going to argue like a child with you about this). Since you were incapable of doing the research or politely explaining yourself, I did you a favor and found your argument for you: you could be right, insofar as, according to Wolfram, $x \tfrac{d\delta(x)}{dx} = - \delta(x)$, and, therefore, $\tfrac{d}{dx}x \delta(x) = \delta(x) + x \tfrac{d\delta(x)}{dx} = \delta(x) - \delta(x) = 0$. So, the derivative is identically zero for all $x \,$ and $\left.x \delta(x)\right|_{x \ne 0} = 0$. However, insofar as L'Hôpital's rule is applicable to this "function", in the weak limit sense, $\lim_{x \rightarrow 0} x \delta(x) = \lim_{x \rightarrow 0} \tfrac{x}{\tfrac{1}{\delta(x)}} = \lim_{x \rightarrow 0} \tfrac{1}{-\tfrac{\delta'(x)}{\delta(x)^2}}$ how do you intend to show that that limit is zero? Could it be undefined? I don't know. If we use one of the many asymptotic limit definitions of the delta function, $\delta(x) = \tfrac{1}{\pi} \lim_{\epsilon \rightarrow 0} \tfrac{x}{x^2+\epsilon^2}$, one can show that the aformentioned $x \delta(x)\,$ limit is equivalent to $\lim_{\{x,\epsilon\}\rightarrow \{0,0\}} \tfrac{-x^2 (x^2+\epsilon^2)^2}{3x^2+\epsilon^2} = 0$ (by my math). Is that proof that you are right and I am wrong? I don't know. I hope that this elucidation inspires you to do better at explaining your own views on the subject matter. Most of all, I hope that, if you don't like my feable work, you'll fix it yourself.KlappCK (talk) 15:38, 27 October 2011 (UTC)
Um... I just told you how to show that x delta(x)=0. Pair it with a test function. You get zero. Anyway, this is a standard exercise in distribution theory. See any textbook if you don't believe me. For instance, first book off my shelf: Hoermander, Analysis of Linear PDOs Volume 1, exercise 3.1.7. It's also in Vladimirov Equations of mathematical physics, chapter 2, section 9. Sławomir Biały (talk) 18:21, 27 October 2011 (UTC)
I believe you are, in fact, right, as I am now convinced by my own reasoning, even thought the result is surprising to me. I have stricken my original assertion from this thread of discussion. Indeed, $x\delta(x) = 0 \, \forall \, x \in \Re$. My apologies. Perhaps you will take the time to demonstrate your point using a test function, as I am not familiar with this method as you (briefly) describe it.KlappCK (talk) 19:44, 27 October 2011 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── See distribution (mathematics). Sławomir Biały (talk) 20:19, 27 October 2011 (UTC)

I feel like I'm beating a dead horse here, but pointing to such a large article without even a link to a specific section is less than elucidating. If you can find the time to keep making it seem so easy to demonstrate, what exactly is stopping you from writing it out mathematically? Do you not know how to use LaTeX?KlappCK (talk) 14:17, 1 November 2011 (UTC)
Once you know what a distribution is, then hopefully you'll understand my reply. If you get that material under your belt and still have trouble, then I'd be happy to explain. But I'm not here to give free lessons in standard material (especially not to someone who has been so consistently rude). Learn it yourself, please. Sławomir Biały (talk) 01:40, 2 November 2011 (UTC)
I'm sorry we to have given you a bad impression of my character. It is perhaps too easy to be terse when exchanging messages on Wikipedia. Though, for the sake of this article, I encourage you to demonstrate your proof (if not here, then on the absolute value page itself). After all, we are all working toward a common goal here, and in my mind, saying $\int_{-\infty}^{\infty} e^x x \delta(x) \text{d} x = 0$ and saying "that's why the function is montonically zero", seems like hand waving to me.KlappCK (talk) 04:17, 9 February 2012 (UTC)
$\langle x\delta(x),\phi(x)\rangle=0$ for all compactly supported smooth functions $\phi$. Therefore the distribution (mathematics) $x\delta(x)$ is zero, by the definition of distributions! Sławomir Biały (talk) 13:58, 9 May 2012 (UTC)
$\langle x\delta(x),\phi(x)\rangle=0$[citation needed] <--- This is the hand waving to which I was referring earlier, see mathematical proof.KlappCK (talk) 18:49, 23 May 2012 (UTC)

Math gurus please speak up! Can you really say "magnitude" in C?

I've currently picked this up off a JavaScript book giving me a hand in how to implement a utility function set for complex numbers. As far as I know, in C scope you'd normally call the absolute value (i. e. Re² + Im²) the modulus and NOT the "magnitude". So I'd like to ask you guys: is the term "magnitude" (apart from "modulus") really common for complex numbers or isn't it? The article merely "allows" the terms modulus and absolute value. -andy 217.50.60.205 (talk) 16:39, 11 April 2012 (UTC)

It is not common for the field of the complex numbers, and not said, in the article, to be equivalent to modulus. "Magnitude" is said to be equivalent to "absolute value" only for vector spaces and fields. For vector spaces "magnitude" is widely used in physics, while mathematicians use mainly "norm". For absolute values in fields, I have never seen any usage of "magnitude". I have therefore added a "citation needed" template. D.Lazard (talk) 17:30, 11 April 2012 (UTC)
See Shechter, p. 260. I've add this citation, and removed the "citation needed" templates. Paul August 21:12, 23 May 2012 (UTC)

Series Approximations Of Absolute Value Function

I recently removed two sections of original research that seems to have been fairly blatant. The first section was pure "not even wrong" original research. The second was sourced entirely to a calculation done by a computer algebra system. Even if this were a reliable source (in the sense of WP:RS), which is debatable, it is not a reliable secondary source (WP:PSTS). There has, at any rate, long been consensus that referencing computer algebra calculations is not considered acceptable sourcing, and material that is so referenced is often routinely removed from articles. Sławomir Biały (talk) 13:30, 19 May 2012 (UTC)

Yes agree, and besides which the results of a user request to an engine like that have zero zilch nada weight even if they were correct so we shouldn't fill Wikipedia with them.,m We need to show somebody was remotely interested in it besides some editor. Dmcq (talk) 15:18, 19 May 2012 (UTC)
I see all the approximations have been removed. I'm sure I've seen a couple of books showing it can be approximated so I'll try and find something along those lines. Dmcq (talk) 15:33, 19 May 2012 (UTC)
It might also be worth mentioning somewhere its derivative when it is considered as a hyperfunction. Dmcq (talk) 15:39, 19 May 2012 (UTC)
How is its derivative as a hyperfunction remarkable? It's just the hyperfunction induced by the classical derivative. Sławomir Biały (talk) 21:48, 19 May 2012 (UTC)
I see the basic idea has just been put in for the second derivative saying as a generalized function. I was just saying hyperfunction as a reasonably well defined version of generalized function. The first derivative, the step function, is also a generalized function but people may not resort to calling it that till they get to distributions, I feel hyperfunctions make that more obvious Dmcq (talk) 00:30, 20 May 2012 (UTC) Dmcq (talk) 00:30, 20 May 2012 (UTC)
I was just copying the generalized function link from our Dirac delta function article but I don't have strong feelings about generalized functions vs hyperfunctions. —David Eppstein (talk) 00:36, 20 May 2012 (UTC)
It seems fine to me as it is, I wasn't really pushing them specifically. Dmcq (talk) 00:52, 20 May 2012 (UTC)
Two questions: how is the most widely used computer algebra system in the world's reliability "debatable", any more than say, the content of Abramowitz & Stegun? To what documented consensus are your referring regarding computer algebra systems and material that is so referenced, aside from this example? I'd be happy to relent on this matter provided you can direct me to some notable 'neutral' demonstration of your argument. Short of that, I see no difference between the fallibility of Mathematica and that of the 'Table of Integrals, Series, and Products', short of the fact that Mathematica doesn't provide a source to some obscure material so that you can sleep at night.KlappCK (talk) 16:01, 22 May 2012 (UTC)
It's not about fallibility, it's about WP:NOR. Sławomir Biały (talk) 16:58, 22 May 2012 (UTC)
IMO, in this kind of question, a computation may be a reliable source only in this kind of writing:
As every function of some type, the function absolute value may be expanded in series of Chebishev, Legendre, ... polynomials, converging in the interval ]-1,1[.[citation needed] This expansion may be computed with someone's algorithm.[citation needed]. The implementation of this algorithm in Mathematica gives the following result.
Only this way of sourcing make sense for an encyclopedia. Referring to a computation without any indication of the underlying algorithm is not a reliable sourcing. D.Lazard (talk) 17:22, 22 May 2012 (UTC)
And exactly why would we be quoting something Mathematica outputs unless someone though it was worth writing down is what I'd like to know. Which is another way of saying it would be WP:original research. It would just have no WP:WEIGHT. If a paper quoted the Mathematica result then I'd consider it was possibly worth sticking in. Dmcq (talk) 17:30, 22 May 2012 (UTC)
So, Swalomir, if it is indeed about no original research, why the need to appeal to past consensus? D. Lazard, I understand your argument, but would we not, as extension of the necessity of an underlying algorithm need, to toss out the entirety of the list of integrals of trigonometric functions since no source or algorithm is given? This ties in well with Dmcq's point, which I assume is about secondary sources, because the linking to WP:WEIGHT is itself superfluous given the depth of the article, and Mathematica is as good a primary source as any other Mathematical text. To argue that noone has yet observed this identity and, indeed, written it down, seems foolish to me. I am sure you would just as quickly toss out such an equation as $\frac{\sin(x)}{x} = \prod_{k = 1}^\infty \cos\left(\frac{x}{2^k}\right)$, if I did not point out that it is simple transformation of identity 1.439.1 in the Table of Integrals, Series and Products. Just because Wolfram|Alpha can provide the result does not mean that it is not sourced. I am sure that close relatives of those identities exist in either the Handbook of Mathematical Functions or the Table of Integrals, Series and Products, if one cared to look. I'm sure someone will argue that is on me to provide such sources via WP:SOFIXIT, but by that same reasoning, I should follow Swalomir's lead and just delete all the content of the aforementioned articles (after creating a talk section intitled "removal of blatant original research") since I am too lazy to look up source for the integrals myself and (I can only assume) too skeptical of their validity to leave the content alone because it adds to meaningful content of the article. I am sure that someone will try to point me to WP:WAX in response to this argument, but that only deflects the thrust of the argument; all I ask is that we make a reasonable attempt to be even-handed with this content in light of it's verifiability, particularly given the glaring lack of precedence provided by the community regarding such source material.KlappCK (talk) 18:44, 23 May 2012 (UTC)
Who said it's not about original research? It's about original research, and weight, and RS. As I've said, calculations from computer algebra systems have never been considered acceptable sourcing. I don't see any reason to make an exception here. Sławomir Biały (talk) 19:15, 23 May 2012 (UTC)
Clearly, you misread what I had written. I said, "if it is indeed about no original research, why the need to appeal to past consensus?". Since you mentioned this consensus regarding computer algebra systems again, perhaps you can move a step up the on Graham's Hierarchy of Disagreement and provide some evidence, because right now you're just contradicting me, not refuting my arguments. I implore you to provide evidence for your arguments. I am not arguing in favor of this material for my own amusement. I want this resolved, even if it is in your favor.KlappCK (talk) 19:54, 23 May 2012 (UTC)
This is just pointless Wikipedia:Wikilawyering without substance. It is original research to add identities supported only by computer algebra system calculations, and it is past consensus that it is inappropriate to do so. So we have two reasons not to do it; they are not in contradiction. You do not have a right to have your arguments responded to in the proper hierarchical format you desire. What you need to do is to persuade the other editors that these calculations are appropriate for this article. So far you have been failing to do so. —David Eppstein (talk) 20:42, 23 May 2012 (UTC)
Yup. Paul August 21:25, 23 May 2012 (UTC)
KlappCK, I'm not going to Wikilawyer with you about this. This is the second time you've appealed to Graham's hierarchy, but the basic fact is that (in both cases) the burden is on you (the person adding contentious material) to convince others, not the other way around. So if it seems that I've been short in my replies, this is part of the reason. I have said that it has long been accepted that WP:NOR applies to calculations done with the aid of a computer algebra system. I indicated the relevant part of that policy (WP:PSTS), and stated that Wolfram Alpha is a primary source, and so not allowed to support this kind of content. That seems fairly straightforward, and I honestly can't see how anyone can in good faith read our WP:NOR policy and come away with any other impression, but obviously you would like this to be clarified. In fact, the same policy contains WP:CALC which allows routine calculations. If the output of a sophisticated computer algorithm were considered acceptable, you would think that this would be a good place to point it out, but it doesn't. Instead, the purpose of this section is to provide an exception to an overly zealous reading of the WP:NOR policy for this particular kind of computatation. If you honestly misunderstand the policy to allow the kind of thing you're insistent upon adding, I encourage you to post to WP:NOR/N to get clarification on the issue. There seems to be no lack of local consensus here, however, based not only in the WP:NOR policy but also WP:NPOV, so the removal should be allowed to stand. You are welcome to start a formal request for comment if you are unhappy with this decision. Sławomir Biały (talk) 00:25, 24 May 2012 (UTC)
Conceding that this is most likely a primary source, I can only find one passage on the page you mention regarding primary sources that seems relavent to me: Unless restricted by another policy, primary sources that have been reliably published may be used in Wikipedia, but only with care, because it is easy to misuse them. Any interpretation of primary source material requires a reliable secondary source for that interpretation. A primary source may only be used on Wikipedia to make straightforward, descriptive statements of facts that any educated person, with access to the source but without specialist knowledge, will be able to verify are supported by the source. I note a distinct lack of references to calculations done with the aid of a computer algebra system and I take issue with being accused of original research as well as being accused of wikilawyering without proper context. Nevertheless, the consensus among editors of this page appears to be that the primary source is insufficient because it is ostensibly the work of a computer algebra system. Assuming I have that right (and please correct me if I do not), then I concede that the material should be removed until a relaible secondary source can be found. As a matter of good faith, what I would like to see and still have not, is evidence of this often mentioned prior consensus regarding the exclusion of computer algebra systems. If such a policy does indeed exist, given all the links the community has been able to provide to other wikipedia policies, it should not be much trouble to provide a link to that policy as well.KlappCK (talk) 15:05, 24 May 2012 (UTC)
It's remarkable to me that the WP:NOR policy can, in any way, be construed as allowing things to be sourced to the output of a computer algebra system. This is the policy I would have you read, and, if it is still unclear to you, then you should ask at WP:NOR/N for clarification rather than here. Editors have routinely applied this in the past to remove precisely the sort of thing that is under discussion. There is no clear statement against including the output of computer algebra systems because editors generally know that this is not acceptable (WP:BEANS). There have been discussions in the past, pertaining for instance to WP:CALC, including lengthy discussions at ArbCom, making very clear what the community view of WP:NOR is regarding calculations of this kind. Sławomir Biały (talk) 15:21, 24 May 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── It is not really a primary source so much as not a source at all. If you work out something yourself with pencil and paper rather than finding it in a publication, that is original research. If you ask a friend to do it and she works it out herself with pencil and paper rather than finding it in a publication, that is still original research. And if you ask a symbolc algebra system to do it and it works it out itself rather than finding it in a publication, that is still original research. —David Eppstein (talk) 15:33, 24 May 2012 (UTC)

So going to | Wolfram|Alpha and typing in "Abs[x]" is original research? If so, then is not all the content contained therein original research? Where do we draw the line here?KlappCK (talk) 15:41, 24 May 2012 (UTC)
Probably you shouldn't be trying to use Wolfram Alpha as a source. Since this is supposed to be an encyclopedia, and this article is supposed to be about a very common, well-understood function, everything in it should appear in many sources. We don't have to reference those sources, but out article should reflect the relative weight that various aspects of the subject as they appear in multiple independent reliable sources. If you find it impossible to add something without resorting to obscure sources or computer algebra systems, chances are that it doesn't belong in the article. Sławomir Biały (talk) 15:55, 24 May 2012 (UTC)
I'm starting to sloth out on this one...Although the source is generally considered trustworthy within the mathematical community, it does seem obscure and since I can't find a published source, so I'll just leave the identities here should anyone else have an interest in the material...

──────────────────────────────────────────────────────────────────────────────────────────────────── Infinite series: The absolute value function can be written as various infinite series convergent for −1 < x < 1: (according to Wolfram|Alpha)

• $|x| = \frac{2}{\pi} \left(1 - 2 \sum_{k=1}^\infty \frac{(-1)^k}{4k^2-1}T_{2k}(x)\right)$, where $T_{n}(x)\,$ are the Chebyshev polynomials of the first kind.
• $|x| = \frac{-1}{2\sqrt{\pi}} \sum_{k=0}^\infty \frac{(-1)^k \left(2k + \tfrac{1}{2}\right)\Gamma\left(k-\tfrac{1}{2}\right)}{\left(k+1\right)!}P_{2k}(x)$, where $P_{n}(x) \,$ are the Legendre polynomials.
• $|x| = \frac{-1}{2\pi} \sum_{k=0}^\infty \frac{(-1)^k \Gamma\left(k-\tfrac{1}{2}\right)}{(2k)!} H_{2k}(x)$, where $H_{n}(x)\,$ are the Hermite polynomials.

Note that $x! = \Gamma(x+1)\,$ are the factorial and gamma functions, respectively.

Sound fair?KlappCK (talk) 16:09, 24 May 2012 (UTC)

+ ──────────────────────────────────────────────────────────────────────────────────────────────────── It is not really a primary source so much as not a source at all. If you work out something yourself with pencil and paper rather than finding it in a publication, that is original research. If you ask a friend to do it and she works it out herself with pencil and paper rather than finding it in a publication, that is still original research. And if you ask a symbolc algebra system to do it and it works it out itself rather than finding it in a publication, that is still original research. —David Eppstein (talk) 15:33, 24 May 2012 (UTC) −

(This post has been written before reading the last two posts of KlappCK.) First, my previous post was about reliability of computer algebra results, not about WP:OR. I am sorry if it has been misunderstood. About reliability of WolframAlpha in this case, one may add that the question is about series expansions at the origin and the first answer of WolframAlpha is a (trivial) series expansion around 1 given with a silly big O approximation of the error. This is not serious.
About WP:OR. IMO, this is not the computation by itself which is original research, but the choice of the question (here asking the series expansion of the absolute value in term of orthogonal polynomials), the fact of giving worth to the resulting identities and, over all, the decision of publishing them in Wikipedia when they have not been published elsewhere. D.Lazard (talk) 16:30, 24 May 2012 (UTC)

Simpler antiderivative proof

Why such a convoluted proof? What's wrong with the plain standard one:

• For $x > 0$, $|x|=x$ therefore $\int |x| dx = \frac{x^2}{2} + C_1$.
• For $x < 0$, $|x|=-x$ therefore $\int |x| dx = -\frac{x^2}{2}+C_2$.
• Joining the two expressions you get $\sgn(x)\frac{x^2}{2} + C=\frac{x|x|}{2}+C$ (for $x\neq 0$), as claimed. It should also be noted that the "constant" $C$ here is only locally constant. FilipeS (talk) 11:22, 13 June 2012 (UTC)
Your proof it not at all standard. It is not correct because is does not defines the antiderivative at zero, although it is perfectly defined (This is a theorem that every continuous function has an antiderivative, and the absolute value is continuous, even at zero). Moreover, the absolute value and its antiderivatives being all continuous, your assertion that the constant is only locally contstant is a mathematical nonsense. D.Lazard (talk) 11:43, 13 June 2012 (UTC)

You are right about the absolute value having an antiderivative in intervals containing zero. Well spotted. Nevertheless, it's trivial to correct the proof as follows:

• For $x > 0$, $|x|=x$ therefore $\int |x| dx = \frac{x^2}{2} + C$.
• For $x < 0$, $|x|=-x$ therefore $\int |x| dx = -\frac{x^2}{2}+C$.
• Combining the two expressions yields the function $f(x)=\sgn(x)\frac{x^2}{2} + C=\frac{x|x|}{2}+C$, which is differentiable at $x=0$ as well, with $f'(0^+)=f'(0^-)=0=|0|$, so $\int |x| dx = \sgn(x)\frac{x^2}{2} + C=\frac{x|x|}{2}+C$.

Standard or not, this is simpler and clearer than what the article currently has. FilipeS (talk) 12:07, 13 June 2012 (UTC)

We didn't need a proof in the first place and we certainly don't need a longer one. One line for things that are unnecessary is enough for me. Dmcq (talk) 12:23, 13 June 2012 (UTC)
This is now unnecessary, but after a little more thought it occurs to me that actually this proof is easy to make rigorous: |x| is continuous, so by the Fundamental Theorem of Calculus it has an antiderivative given by $F(x) = \int_0^x |t|\, dt$. Now compute this integral depending on the sign of x to get the desired expression. Finally, observe that all other antiderivatives differ from this one by a constant (again, because |x| is a continuous function). The definite integral does the stitching-together naturally. (This all said, I think the article is better without a proof.) --Joel B. Lewis (talk) 13:44, 13 June 2012 (UTC)

In fact the present proof is also dubious at x = 0, since it involves division by |x| (or, equivalently, the derivative of |x| at 0). Frankly I prefer FilipeS's version for this reason. (Actually, though, maybe the best thing is to remove the proof altogether.) --Joel B. Lewis (talk) 13:04, 13 June 2012 (UTC)

There shouldn't be a proof at all. This is not a textbook. Sławomir Biały (talk) 13:11, 13 June 2012 (UTC)
I have removed the proof as uncited and disputed - problem gone. Dmcq (talk) 13:16, 13 June 2012 (UTC)

I agree with omitting the proof. The reader can easily differentiate the antiderivative himself, to check that it's right. FilipeS (talk) 11:18, 14 June 2012 (UTC)

Other expression for the antiderivative

When I added the expression for the antiderivative in terms of the signum function to the article, ‎D.Lazard removed it with the comment "Sgn function being discontinuous, this expression is misleading, suggesting wrongly that antiderivative is not continuous". His caution is however misguided, as the product between a discontinuous function and another function need not be discontinuous, and indeed is not discontinuous in this case. FilipeS (talk) 10:47, 16 June 2012 (UTC)

Sure, the expression using the sign function that you have introduced is true. But trueness is not sufficient for deserving to appear in Wikipedia. You have not provided any reason for introducing this alternative expression. On the other hand there are, at least, three reasons to not introducing it:
• Sign function does not appear before in the article. Its relationship with absolute value appears only in next section.
• Passing from one expression for the antiderivative to the other is absolutely obvious, using the next formula in the article.
• But the main reason for my revert is the following: It is not obvious for most reader that the antiderivative is continuous; even for you, if one consider some of your edits which have been reverted. It is not obvious either to recognize when the product of a discontinuous function by a continuous one is continuous. Thus the occurrence of a discontinuous function in the expression for the derivative may (and certainly will) give to many readers the wrong impression that the antiderivative is discontinuous. This has to be avoided in an Encyclopedia.
D.Lazard (talk) 12:50, 16 June 2012 (UTC)
Unrelated to this, I moved the "Relationship to other functions" before "derivative" and "antiderivative" sections on the theory that it's more fundamental. I'm indifferent to the inclusion of the form FilipeS suggests. --Joel B. Lewis (talk) 14:44, 16 June 2012 (UTC)

Proposed merge

I think Absolute value (algebra) should be merged into this article. There are already various generalizations here. If the two articles are kept separate for some reason, at least there should be a clear connection between the two, like a section summarizing Absolute value (algebra) here with a link to the main article. Isheden (talk) 11:58, 26 August 2012 (UTC)

I do not see any problem here (but some users searching for problems). "Absolute value (algebra)" is a well-defined thing, a self-sufficient one (outside any quasi-physical "magnitude" context) and does not require such thing as a "main article". Why necessarily "absolute value"? It may be function (mathematics) or valuation (algebra), indeed. For a reader familiar only with arithmetic, certainly, absolute value is the better known concept. Other readers, probably, do not need analogies with normed spaces or so. Incnis Mrsi (talk) 13:05, 26 August 2012 (UTC)
I guess it's clear that absolute value is the logical target for a merge, since the term is the same in algebra and in arithmetic, and since this article already has a subsection Fields under Generalizations. If the typical reader of this article does not need analogies in abstract algebra, then it might be better to move at least the definition of "Archimedian" to Absolute value (algebra). Isheden (talk) 13:40, 26 August 2012 (UTC)
I hardly believe that a sane reader may complain that "[non-]Archimedean" is insufficiently defined in the "Absolute value (algebra)". Isheden, why do you search for duplicates where distinct articles title and title (disambiguation) are present?! If someone made "title (disambiguation)" then s/he probably knew what s/he is doing, isn't it? There are probably thousands duplicates or even non-articles with ambiguous, incomplete, obscure, unscientific and even outright mistaken titles which are not easily detectable. Please, divert your energy to searching and merging/{{dab}}ifying a real crap, not established articles edited and watched by numerous experts. Incnis Mrsi (talk) 18:08, 26 August 2012 (UTC)
As you may have noticed, I've already withdrawn the merge proposal for lack of support. I never claimed that the articles are duplicates. I also didn't claim that "[non-]Archimedean" is insufficiently defined in the "Absolute value (algebra)". And I don't see what would motivate such an aggressive tone. Isheden (talk) 19:21, 26 August 2012 (UTC)
I understood the insinuation about "Archimedian" in this way; does anybody see something else in it? BTW Archimedean property is a notable topic itself and deserves a separate discussion. What do you perceive as "an aggressive tone" is my disgust to mergers-because-of-the-same-word, either explicit or implicit. I really think that such users, trying to save one or two reader's clicks on disambiguations, make Wikipedia less structured and more confusing. Incnis Mrsi (talk) 07:54, 27 August 2012 (UTC)
It's pointless to let other people judge what could have been meant by a certain remark in a text-only medium that conveys nuances poorly. Feel free to express your disgust to mergers, but please try to avoid personal remarks. Isheden (talk) 09:45, 27 August 2012 (UTC)
I agree with Isheden that the aggressive and attacking nature of your latter comments is unnecessary and that you should tone it down, not least because the substantive question was already resolved when you started. --JBL (talk) 13:14, 27 August 2012 (UTC)
The current situation, in which the algebraic object is mentioned but not dealt with in full detail and has its own article, seems exactly right to me. I don't think there should be any merge. --JBL (talk) 14:02, 26 August 2012 (UTC)

Derivative of Absolute Value

Section 3.2 gives the derivative of the absolute value function. In the cases, one of them says "undefined if x = 0". This is true but is there any need to write this? Can't you simply just not define anything at x = 0 by not giving a case for x = 0? Pratyush Sarkar (talk) 02:17, 5 December 2012 (UTC)

"undefined" does not means "nobody has defined it", but "the derivative does not exist". Maybe, it would be worthy to replace "undefined" by "does not exist". In fact, "undefined" applies correctly to the fraction in the middle, while "does not exists" applies only to the derivative. Thus a fully correct formulation would need a more complicated edit. D.Lazard (talk) 06:53, 5 December 2012 (UTC)
What do you mean? The derivative of course does not exist at x = 0, so the derivative function is not defined at x = 0. Not defined means a value has not been assigned to some point for the function (In this case, no value is assigned to x = 0 for f'). So, not defined and does not exist mean the same thing.Pratyush Sarkar (talk) 15:23, 7 December 2012 (UTC)
“Undefined” suggests that the nonexistence of the value of the function is only a matter of definitional conventions, that someone could come and say, “I’ll define $f'(0)=42$”. This is wrong here, the fact that the absolute value function is not differentiable at 0 is a theorem, not a definition. To answer the original question, if one takes a function $f\colon X\to\mathbb R$ and asks for its derivative, it is expected the derivative will be given for all points of the domain of the original function, which is $X=\mathbb R$ here. Therefore it needs to be explicitly said that the derivative does not exist at 0. If it were simply omitted, it would look as if this case was overlooked and the information is missing, not that the derivative does not exist.—Emil J. 16:00, 7 December 2012 (UTC)
The following answer was in edit conflict with Emil's
When you define a function, like $x/|x|$, you define it in some domain, and the function is not defined outside this domain. In the case of $x/|x|$, the value at 0 should be 0/0, which is not defined as a value. But, like for Heavyside function, you are allowed to chose any value that you want for the function $x/|x|$ at 0 (In fact $x/|x|=(H(x)+1)/2$, and the arbitrary choice made for H is convenient for $x/|x|$, see Heavyside function). On the other hand, when you compute the derivative at a point, you have nothing to define, but you have to apply an existing definition, which, here, implies to compute a limit. It is a theorem that this limit does not exist for the derivative of the absolute value at zero. Thus you are not allowed to define a value for it. In other words, "undefined" means "no definition has been given" and does not suggest any theorem, while "the derivative does not exist" is a theorem that may be proven and does not suggest any definition. D.Lazard (talk) 16:16, 7 December 2012 (UTC)

Replacement of a with x

The main reason is that a conflicts with English indefinite article. It makes it appearance amid the text potentially confusing. I would not object if somebody replaced x with something third. Incnis Mrsi (talk) 12:30, 2 February 2013 (UTC)

Properties

1. |a - b| ≤ ε iff b - ε ≤ a ≤ b + ε
1. if |a| is less than ε for all ε positive, then a = 0.
1. |x| = max{ x, -x} --190.118.16.55 (talk) 04:38, 14 March 2014 (UTC)--190.118.16.55 (talk) 14:50, 14 March 2014 (UTC)