Express the strange symmetric polynomial in terms of elementary symmetric polynomials

Question

How can I express the symmetric polynomal? $$ \sum_{1\leq i_1 <i_2<i_3\leq n} x_{i_1}^2x_{i_2}^2x_{i_3}^2 $$

for n=4 it looks much more difficult. I tried to group them, but I was defeated

it is clear that once the degree of each element is equal to 2. This means that I only need to use elementary polynomials of the first or second degree

the sum of the degrees is 6, and the largest degree is 2 It means, that degrees could be only 2-2-2; 2-2-1-1; 2-1-1-1-1 or 1-1-1-1-1-1 right?

Answer 1

Use $$\sum_{i=1}^Na_i^2=\left(\sum_{i=1}^Na_i\right)^2-2\sum_{1\leq i<j\leq N}a_ia_j.$$

Let $\sum\limits_{i=1}^nx_i=nu$, $\sum\limits_{1\leq i_1<i_2\leq n}x_{i_1}x_{i_2}=\frac{n(n-1)}{2}v^2$, $\sum\limits_{1\leq i_<i_2<i_3\leq n}x_{i_1}x_{i_2}x_{i_3}=\binom{n}{3}w^3$,

$\sum\limits_{1\leq i_<i_2<i_3<i_4\leq n}x_{i_1}x_{i_2}x_{i_3}x_{i_4}=\binom{n}{4}t^4$, $\sum\limits_{1\leq i_<i_2<i_3<i_4<i_5\leq n}x_{i_1}x_{i_2}x_{i_3}x_{i_4}x_{i_5}=\binom{n}{5}s^5$ and $\sum\limits_{1\leq i_<i_2<i_3<i_4<i_5<i_6\leq n}x_{i_1}x_{i_2}x_{i_3}x_{i_4}x_{i_5}x_{i_6}=\binom{n}{6}p^6.$

For $n=4$ and variables $a$, $b$, $c$ and $d$ we obtain: $$\frac{1}{6}\sum_{sym}a^2b^2c^2=\sum_{cyc}a^2b^2c^2=\left(\sum_{cyc}abc\right)^2-$$ $$-2(abc\cdot abd+abc\cdot acd+abc\cdot bcd+abd\cdot acd+abc\cdot bcd+acd\cdot bcd)=$$ $$=\left(\frac{1}{6}\sum_{sym}abc\right)^2-2\cdot\frac{1}{4}\sum_{sym}a^2b^2cd=16w^6-12v^2t^4.$$ For $n=5$ and variables $a$, $b$, $c$, $d$ and $e$ we obtain: $$\frac{1}{12}\sum_{sym}a^2b^2c^2=\left(\frac{1}{12}\sum_{sym}abc\right)^2-$$ $$-2\left(\frac{1}{12}\sum_{sym}(abc\cdot abd+abc\cdot abe+abd\cdot abe)+\frac{1}{24}\sum_{sym}(abc\cdot ade+abd\cdot ace+abe\cdot acd)\right)$$ $$=\left(\frac{1}{12}\sum_{sym}abc\right)^2-\frac{1}{2}\sum_{sym}a^2b^2cd-\frac{1}{4}\sum_{sym}a^2bcde=$$ $$=\left(\tfrac{1}{12}\sum_{sym}abc\right)^2-2 \left(\tfrac{1}{24}\sum_{sym}abcd\cdot\tfrac{1}{12}\sum_{sym}ab-4\cdot\tfrac{1}{24}\sum_{sym}a^2bcde\right)-\tfrac{1}{4}\sum_{sym}a^2bcde=$$ $$=100w^6-2(5t^4\cdot10v^2-4\cdot5us^5)-30us^5=100w^6-100v^2s^4+10us^5.$$ For $n\geq6$ we can use the similar reasoning.

Can you end it now?

Answer 2

Here is a systematic way to deal with the matter.

Let $$ p(X)=\prod_{1}^{n}(X-x_i)=\sigma_0 X^n-\sigma_1 X^{n-1}+\sigma_2 X^{n-2}-\dots+(-1)^{n}\sigma_n, $$ and $$ P(Y)=\prod_{1}^{n}(Y-x_i^2)=\Sigma_0 Y^n-\Sigma_1 Y^{n-1}+\Sigma_2 Y^{n-2}-\dots+(-1)^{n}\Sigma_n. $$

(Of course $\sigma_0=\Sigma_0=1$.)

Then $$ P(X^2)=(-1)^n p(X)p(-X). $$

From this we can read off the coefficients of $P$ in terms of those of $p$:

$\Sigma_0=\sigma_0^2$,

$\Sigma_1=\sigma_1^2-2\sigma_0 \sigma_2$,

$\Sigma_2=\sigma_2^2-2\sigma_1 \sigma_3+ 2\sigma_0 \sigma_4$,

and then $\Sigma_3=\dots$: the pattern is clear.