Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I currently study special relativity and some authors write stuff like: $$ r^\mu = \left(ct, \vec x\right) $$

This is awful since $\mathsf r$ is a vector, and $r^\mu$ ist just a single component of that vector.

Now I am wondering whether something I write occasionally is similarly wrong: As far as I know, I can define a function $f$ like $f\colon x \mapsto x^2$ or $f(x) = x^2$. Sometimes I just write $f = x^2$. Would that be incorrect in a strict sense?

share|cite|improve this question
Yes, unless it refers to the associated polynomial $x^2 \in K[x]$. – Siméon Jan 15 '13 at 9:22
@StefanHansen: I think you meant “ambigious“ when you wrote “unambiguous”. – Martin Ueding Jan 15 '13 at 9:51
I updated the question: What happens to functions of other functions that have the same variable? – Martin Ueding Jan 15 '13 at 9:57
it does not make sense to edit your post and add an additional question after a lot of people hafe written ansers adn comments. Now it is completely unclear to what questions the statements refer to. I think you should rollback your edit and pose another question. – miracle173 Jan 15 '13 at 10:05
The notion $f=x^2$ is ambiguous. It could refer to both the polynomial $x\mapsto x^2$ and the constant function $y\mapsto x^2$. I would recommend that you use one of the two notions you mention yourself. – Stefan Hansen Jan 15 '13 at 10:12
up vote 1 down vote accepted

Yes, that would be incorrect. If $f$ is - say - a continous function from $\mathbb{R}$ to $\mathbb{R}$. Then $f$ would be an element of $C(\mathbb{R},\mathbb{R})$ and $x$ as well as $x^2$ would be an element of $\mathbb{R}$. It makes no sense to identify them. $f$ is a function, $f(x)$ is the value of this function at a point $x$, namely $x^2$.

share|cite|improve this answer
Not quite - $x$ is not some value, but instead a variable. In that case, $x^2$ can be seen as an element of $\mathbb{R}[x]$, the ring of polynomials, which is isomorphic to a subset of the ring of continuous functions. – akkkk Jan 15 '13 at 10:30
@akkkk This is just another way to view it. Yet, if $f: \mathbb{R} \to \mathbb{R}$ one can (for $x \in \mathbb{R}$) still view $f(x)$ as an element of $\mathbb{R}$. – mjb Jan 15 '13 at 14:05
So why is your view correct? Under my view, the notation is correct (albeit confusing, since I am leaving away all the isomorphisms), but under your view, it is not. – akkkk Jan 15 '13 at 14:30
@akkkk I'm not sure I understand what you mean. As Ju'x pointed out in his comment to the original question: for a function $f : \mathbb{R} \to \mathbb{R}$ it is correct to write $f=x^2 \in \mathbb{R}[x]$, but not $f(x)=x^2 \in \mathbb{R}[x]$, since clearly $f(x) \in \mathbb{R}$ for $x \in \mathbb{R}$, no? – mjb Jan 15 '13 at 16:47
if, by $x$, the /variable/ is meant, then I do not see how that would be incorrect. You are right however, that (in words) "plugging in a value" produces a new "value". – akkkk Jan 15 '13 at 17:00

Yes, strictly speaking it would mean that $f$ is the square of whatever $x$ is. So if $x$ was a real number then $f$ would be the square of that number.

share|cite|improve this answer

The notation $r^\mu = \left(ct, \vec x\right)$ is not awful. This notation is defined in relativity. Upper indices denote contravariant components.

share|cite|improve this answer
The problem I see with it that $r$ is the four-vector and $r^\mu$ is one of its components. It does not make sense to me to refer to a single component $\mu$ while specifying a tuple. It is kind of like writing $\vec a = a_i$, I think. – Martin Ueding Jan 16 '13 at 15:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.