mathjam0990 said:
Express r12+r22+...+rn2 as a polynomial in the elementary symmetric polynomials s1, s2, . . . ,sn.
I'm sure the equation we are dealing with is (r1+r2+...+rn)2 which is very large to factor out but should yield r12+r22+...+rn2+(other terms)
I believe s1=r1+r2+...+rn
s2=Σri1ri2 for 1≤i1≤i2≤n
s3=r1r2⋅⋅⋅⋅⋅⋅rn
So the answer should be r12+r22+...+rn2 = s12 - (something with s2,...sn) Sorry I am not sure what to employ here to break this all the way down.
If there is anyone who could provide an explanation, that would be amazing. Thank you!
It's easier to illustrate what the $s_i$ are with a particular $n$. Let's use $n = 4$. Suppose our variables are $r_1,r_2,r_3,r_4$. Then our polynomials are in the polynomial ring:
$F[r_1,r_2,r_3,r_4]$ (where $F$ is our underlying field).
Given $f \in F[r_1,r_2,r_3,r_4]$, we can have $\sigma \in S_4$ operate on $F[r_1,r_2,r_3,r_4]$ by:
$\sigma(f(r_1,r_2,r_3,r_4)) = f(r_{\sigma(1)},r_{\sigma(2)},r_{\sigma(3)},r_{\sigma(4)})$.
We say a polynomial $f \in F[r_1,r_2,r_3,r_4]$ is *symmetric* if $\sigma(f) = f$.
The ELEMENTARY symmetric polynomials are the "basic" symmetric polynomials of any given degree > 0. Let's see how they are constructed, for $n = 4$, by seeing what they ought to be (note we don't care about constant terms, since $\sigma$ never affects them, so for simplicity's sake, we'll always assume all constant terms are 0).
For degree $1$, our polynomials are just linear combinations of the $r_i$. So the "simplest" symmetric combination would be:
$s_1 = r_1 + r_2 + r_3 + r_4$.
This can't be improved on, eliminating any term would break the symmetry.
for degree $2$, our polynomials would be linear combinations of $r_ir_j$ ($i = j$ might happen) + terms of lower degree. since we can capture symmetric terms of lower degree by the polynomial above, we will only look at symmetric combinations of $r_ir_j$.
At first, it might seem the best candidate would be:
$f = r_1^2 + r_1r_2 + r_1r_3 + r_1r_4 + r_2^2 + r_2r_3 + r_2r_4 + r_3^2 + r_3r_4 + r_4^2$ (all possible degree 2 terms summed).
However, note that:
$s_1^2 = (r_1 + r_2 + r_3 + r_4)^2 =$
$ r_1^2 + r_1r_2 + r_1r_3 + r_1r_4 + r_2r_1 + r_2^2 + r_2r_3 + r_2r_4 + r_3r_1 + r_3r_2r_3^2 + r_4r_1 + r_4r_2 + r_4r_3 + r_4^2$
$= r_1^2 + r_2^2 + r_3^2 + r_4^2 + 2(r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4)$
If we call the "cross-terms" $2g$, we have:
$f = s_1^2 - 2g + g$ where the $s_1^2 - 2g$ gives us just the degree 2 terms in $f$ that are squares.
Thus all we need to get $f$ is $s_1$ and $g$ (both of these are still symmetric) and $g$ is "more minimal" than $f$ (fewer terms). So we should pick $g$ as $s_2$:
$s_2 = r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4$
The same sort of logic applies to the higher terms (but the algebra is horrendous), so:
$s_3 = r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4$
(Do you see the pattern? We pick $s_k = \sum\limits_{i_1 < i_2 < \cdots < i_k} r_{i_1}r_{i_2}\cdots r_{i_k}$)
and finally $s_4 = r_1r_2r_3r_4$.
