Talk:Absolute value

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)

is your defintion of the max function correct( I have no idea I am asking ) WorldDiagram837 (talk) 05:54, 26 January 2025 (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)

Relationship to the max and min functions

Perhaps some connecting links in that section of the article would be helpful ? WorldDiagram837 (talk) 05:55, 26 January 2025 (UTC)

mhm should have replied after the changes were done(thank you). but perhaps somebody could explain to me how the formula was derived(later in that section)? WorldDiagram837 (talk) 14:01, 26 January 2025 (UTC)

It's pretty straight forward, one has just to consider two cases: a) $s>t$ and b) $t>s$ each for the maximum and minimum. For the minimum and $s>t$ the right hand side is $-2t+s+t=s-t$ and $|t-s|=(t-s)\operatorname {sgn}(t-s)=s-t$. The rest follows the same way--Tensorproduct (talk) 14:17, 26 January 2025 (UTC)

yes I did try this, but wouldn't there be something "starter idea" where this came from(or the derivation started from) WorldDiagram837 (talk) 15:37, 26 January 2025 (UTC)

by the way, can we use the first Talk topic(Absolute value as a maximum) in any way for the article? WorldDiagram837 (talk) 15:39, 26 January 2025 (UTC)