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Are all single valued functions, of one variable, symmetric. Where symmetric is taken to mean invariant under permutation of the distinct $n$ variables.

Presumably as there is no variable to permute, any function of one variable is symmetric in this sense? ie, $F(x)=F(x)\,\text{where}\,,F(x)\in \mathbb{R}$.

Secondly, what about the 1st degree, elementary symmetric functions of one variable. Is this just taken the the $k$th (first power sum of the one and only variable) so that $F(x)=x^k=x$ as $k=1$ ; ie does the permutation just mean, (in the one variable case), just swapping the 1st element of the domain with the first element of the domain (ie,itself), so that given the permutation, $F(x)=x$ ?, which is the original function value, and thus, $F$ is symmetric?.

I know that,that this is trivial if this is what is meant, so am just checking that I am, correct that permutation symmetry is not re-defined to mean something else, in the case of a single variable function, simply because there does not exist a distinct permutation of the domain elements, to begin with?

Ie, does a permutation, of an element of the domain, $x$ (or in general), have to be distinct,to count as a permutation of that $x$ ie is $\langle x \rangle$, considered, non-distinct, permutation of $\langle x\rangle$, formally?. That is, formally, is it acceptable to use the term, 'a permutation of $x$', in the case that where the permutation, simply returns '$x$', that is leaves the entity permuted, invariant. Which, incidentally, will always be the case for a one variable function, and thus will be symmetric as per my question . And am I correct, that this applies in general. For example,in cases where each element of the domain, is an ordered set of 2 or more elements of reals, that, it is still acceptable to call the identity transformation, of some element of the domain, a 'permutation', of that element of the domain?

ie $\langle x,y\rangle$, a 'permutation' of $\langle x,y\rangle$, despite the fact that it returns the very same ordered set, must a 'permutation' return a distinct ordering of the elements of that ordered set, for the term, 'permutation of that ordered set', to be correctly used?

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    $\begingroup$ Yes, but the "symmetry" of symmetric functions of one variable has what use? $\endgroup$
    – quasi
    Commented May 4, 2017 at 6:39
  • $\begingroup$ presm-ably none? I think that was the point of my question. I presume that by the elementary symmetric functions that of one variable; it is meant the identity linear function F(x). I dont know if this is necessarily a good definition of symmetry in general. Should be call permutation symmetry I sppose $\endgroup$
    – William Balthes
    Commented May 4, 2017 at 7:17

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Are all functions of one variable, symmetric. Where symmetric is taken to mean invariant under permutation of the distint n variables.

This is true for $n=1$ but not for $n>1$. For example, $f(x,y)=x$ is not a symmetric function of two variables $x$ and $y$.

Secondly What about the 1st degree, elementary symmetric functions of one variable. Is this just taken the the kth (first power sum of the one and only variable) so that F(x)=x^k=x as k=1 ; ie does this denote F(x)=x

If you're trying to ask if $x$ is the only elementary symmetric polynomial of degree one in one variable (namely $x$), then yes.

One reason it could be useful to include functions of one variable in the class of symmetric polynomials - besides the fact that they obviously are, albeit a degenerate case - is as a base case in induction hypotheses or algorithms. One example is a constructive algorithm used in a proof of the fundamental theorem of symmetric polynomials.

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  • $\begingroup$ What about when the function is not single valued, by which I do not mean a function that for every real in the domain, it returns a single set of reals, which is arguably sinle valued if its range is defined as a set ,of sets of 2 reals: $\text{ranF(x)}=\mathbb{R}\times\mathbb{R}$. (ie, the function value at a point in the domain, is a 'single' set of 2 elements, not two inconsistent sets, nor both of the elements themselves, which would not be single valued (but then its range would be defined as the reals) $\endgroup$
    – William Balthes
    Commented Nov 28, 2018 at 8:47
  • $\begingroup$ Dear Songmorning, so long as the domain has more than one element of reals, its not just a single number ie: $\text{dom(F)}=x\,, |\text{dom(F)}|=1$,so that, $F(x,y)=x \land F(y,x)=y$, where, $y=x$. Apart from that case, ie, when there is more than one element in the domain, the function would be symmetric, though as there would be $(y,x)$ where $y\neq x$ $\endgroup$
    – William Balthes
    Commented Nov 28, 2018 at 8:48

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