Talk:Absolute value

Absolute number listed at Redirects for discussion

An editor has asked for a discussion to address the redirect Absolute number. Please participate in the redirect discussion if you wish to do so. — Arthur Rubin (talk) 10:43, 19 May 2019 (UTC)

Short description

@D.Lazard: The purpose of a short description is to aid disambiguation, not to be a comprehensive mini-definition of the subject. Sometimes mini-definitions can work, but in this case it'd require going over the 40-character target (the previous description was 60 characters). {{u|Sdkb}} ^talk 19:49, 14 May 2021 (UTC)

"Concept" is even shorter and is almost as useful as "Mathematics concept". Moreover, "Mathematics concept" is gramatically dubious and, in any case, ambiguous: it may mean "concept of mathematics" or "concept in mathematics", but certainly not "concept used in mathematics" which seems to be the intended meaning. Also, the limit of 40 characters is a recommendation, not a policy.

This being said the current formulation is not very good, not because its number of characters, but because it is too cumbersome. I'll try for a better formulation. D.Lazard (talk) 20:35, 14 May 2021 (UTC)

Another possibility might be "distance of a number from zero" (30 characters). —David Eppstein (talk) 20:56, 14 May 2021 (UTC)

I like that a lot. {{u|Sdkb}} ^talk 21:09, 14 May 2021 (UTC)

Me too. Paul August ☎ 00:58, 15 May 2021 (UTC)

Before reading David's suggestion, I have changed the sd into "Magnitude of a possibly negative number" (39 character). IMO, the two versions are acceptable. I have thought to a version using "distance", but I have rejected it because this use of "distance" may be confusing for a non-mathematician. Another reason of my choice is that it emphasizes that the absolute value makes sense only if non-positive numbers are considered. I'll not be bothered if there is a consensus for David's version. D.Lazard (talk) 08:50, 15 May 2021 (UTC)

Absolute value as a maximum

I have just seen the equation $|x|=\max(x,-x)$ and I am absolutely baffled that I had not come across this until now! Surely this should be included somewhere in the article, as it is much more compact than the piecewise definition – indeed it "piggybacks" off of the piecewise definition of the maximum, $\max(a,b)={\begin{cases}a,&a\geq b\\b,&a<b\end{cases}}$, so that one only has to define one of $|\ |$ and $\max(\ ,)$ by a piecewise formula, not both. The fact that this formula is not in the article suggests to me that many other people also have not seen this property (of course it's obvious once you have seen it, but without seeing it most people wouldn't think of it). Joel Brennan (talk) 22:26, 18 April 2022 (UTC)

|x| <= a doesn't imply -a<= x <= a

From -a <= x <= a, we would have x = a and x = -a, a contradiction. Maybe the intention was to write:

|x| < a iff -a < x < a.

I never saw a book that explains this equivalences. I mean, though intuitive, what is the logical justification why we get the logical "and" in a situation and the logical "or" in another? For example:

|x| = a iff x = a or x = -a

|x| < a iff x < a and x > -a

|x|> a iff x > a or x > -a

The explanation I gave to myself is that the definition of abs is equivalent to:

(x>=0 and |x| := x) or (x<0 and |x| := - x).

For example |x| < a iff

(x >= 0 and x < a) or ( x < 0 and -x < a).

And then it's just interval calculations and distributions of logical operations. Sr cricri (talk) 18:32, 2 April 2023 (UTC)

Perhaps rather than trying to convince us by drawn-out explanations you could provide an explicit example of two numbers $x$ and $a$ for which $|x|\leq a$ but not $-a\leq x\leq a$. If you could do that, it would make for a more convincing argument. —David Eppstein (talk) 18:36, 2 April 2023 (UTC)

Oh, I see now, you are right David. I didn't see that |x| <= a is x < a or x = a. Maybe it is better to delete this topic discussion. Sr cricri (talk) 18:56, 2 April 2023 (UTC)

Absolute Value = Modulus? Is this entirely accurate?

"In mathematics, the absolute value or modulus" - are these terms equal or does it depend on context?

This discussion came up in a stack overflow thread which referenced this article as evidence that absolute value = modulus. However, some additional reading suggests that these may not be entirely interchangeable terms, perhaps making this article a source of misinformation. It's been about 20 years since my last math class, so I'm hoping someone more familiar with this subject matter might be able to clarify this as it is causing discussion and confusion elsewhere:

https://stackoverflow.com/questions/664852/which-is-the-fastest-way-to-get-the-absolute-value-of-a-number

https://math.stackexchange.com/questions/472856/what-is-the-difference-between-modulus-absolute-value-and-modulo Oudent (talk) 05:36, 18 July 2023 (UTC)

This is going to depend on some specific source's precise definitions. As with most things of mathematics, conventions are not entirely standardized and vary a bit from country to country, year to year, source to source. Some sources surely treat these as interchangeable synonyms. Other sources might use "absolute value" for real numbers and "modulus" for complex numbers, vectors, matrices, or other kinds of quantities. This article seems fine though. –jacobolus (t) 06:14, 18 July 2023 (UTC)

n-fold derivative

My edit correcting the formula of the derivative was undone. The formula given currently is: ${d^{n} \over dx^{n}}f(\left\vert x\right\vert )={x \over |x|}({d^{n} \over dx^{n}}f(|x|))$

which has the typographical mistake that the derivative of the function f(|x|) is conflated with the derivative of the function f that is since postcomposed with |x|. My remedy was to write it instead as:

${d^{n} \over dx^{n}}f(\left\vert x\right\vert )={x \over |x|}(f^{(n)}(|x|))$

But if there is a dislike of Newton notation, one could write it as:

${d^{n} \over dx^{n}}f(\left\vert x\right\vert )={x \over |x|}({d^{n}f \over dx^{n}}(|x|))$

as well. Regardless the current formula is incorrect. Qsdd (talk) 14:08, 12 October 2023 (UTC)

The problem is that the two versions are wrong, since the right formula is $${d^{n} \over dx^{n}}f(\left\vert x\right\vert )=\left({x \over |x|}\right)^{n}{d^{n}f \over dx^{n}}(|x|).$$ The formula that follows is also wrong, and the sentence that introduces these two formula is nonsensical. Moreover, AFAIK, higher derivatives of the absolute value are rarely considered, and I am thus unable to provide reliable sources for these formulas.

For these reasons, I have removed the two formulas and their introduction. Feel free to provide the right formulas if you are able to provide a reliable source. D.Lazard (talk) 18:26, 12 October 2023 (UTC)