applying elementary symmetric polynomials

The method used in the proof of fundamental theorem of symmetric polynomials may be applied to concrete instances as follows.

We assume the given a symmetric polynomial  $P(x_1,x_2,\mathrm{\dots },x_n)=P$  of degree $d$ be homogeneous (http://planetmath.org/HomogeneousPolynomial).  Starting from the highest term of $P$ we form all products

$$x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\mathrm{\cdots }{x}_n^{{\lambda }_n}$$

where

$${\lambda }_1\ge {\lambda }_2\ge \mathrm{\dots }\ge {\lambda }_n\ge \mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\lambda }_1+{\lambda }_2+\mathrm{\dots }+{\lambda }_n=d.$$

Then

$P=Q(p_1,p_2,\mathrm{\dots },p_n)={\displaystyle \sum _i}{m}_i{p}_1^{{\lambda }_1-{\lambda }_2}{p}_2^{{\lambda }_2-{\lambda }_3}\mathrm{\cdots }{p}_{n-1}^{{\lambda }_{n-1}-{\lambda }_n}{p}_n^{{\lambda }_n},$ (1)

in which the coefficients $m_i$ are determined by giving some suitable values to the indeterminates $x_j$.

Example 1.  Express the polynomial  $P=x_1^3{x}_2+x_1^3{x}_3+x_2^3{x}_1+x_2^3{x}_3+x_3^3{x}_1+x_3^3{x}_2$  in the elementary symmetric polynomials

$p_1=x_1+x_2+x_3,p_2=x_2{x}_3+x_3{x}_1+x_1{x}_2,p_3=x_1{x}_2{x}_3.$ (2)

We have four

$$4,\mathrm{\hspace{0.17em}0},\mathrm{\hspace{0.17em}0};3,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}0};2,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}0};2,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}1},$$

for which the corresponding $p$-products of the sum (1) are

$$p_1^4,p_1^2{p}_2,p_2^2,p_1{p}_3,$$

respectively.  Apparently, the first one is out of the question.  Therefore, clearly

$$P=p_1^2{p}_2+a{p}_2^2+b{p}_1{p}_3.$$

Using  $x_1=x_2=1$  and  $x_3=0$  makes  $p_1=2$,  $p_2=1$  and  $p_3=0$, when

$$P=\mathrm{\hspace{0.33em}2}=\mathrm{\hspace{0.33em}4}+a+0,$$

implying  $a=-2$.  Using similarly  $x_1=x_2=x_3=1$  we get  $p_1=p_2=3$,  $p_3=1$, which give

$$P=\mathrm{\hspace{0.33em}6}=\mathrm{\hspace{0.33em}27}+9a+3b=\mathrm{\hspace{0.33em}9}+3b,$$

yielding  $b=-1$.  Hence we have the result

$$P=p_1^2{p}_2-2p_2^2-p_1{p}_3,$$

i.e.

$$x_1^3{x}_2+x_1^3{x}_3+x_2^3{x}_1+x_2^3{x}_3+x_3^3{x}_1+x_3^3{x}_2={(x_1+x_2+x_3)}^2(x_2{x}_3+x_3{x}_1+x_1{x}_2)-2{(x_2{x}_3+x_3{x}_1+x_1{x}_2)}^2-(x_1+x_2+x_3)x_1{x}_2{x}_3.$$

Example 2.  Let  $P=x_1^4+x_2^4+\mathrm{\dots }+x_n^4$.  If we suppose that  $n\geqq 4$,  the possible highest terms are

$$x_1^4,x_1^3{x}_2,x_1^2{x}_2^2,x_1^2{x}_2{x}_3,x_1{x}_2{x}_3{x}_4$$

whence we may write

$P=p_1^4+a{p}_1^2{p}_2+b{p}_2^2+c{p}_1{p}_3+d{p}_4.$ (3)

For determining the coefficients, evidently we can put  $x_5=x_6=\mathrm{\dots }=x_n=0$  and in as follows.
$1^{\circ }$.  $x_1=1$,  $x_2=-1$,  $x_3=x_4=0$.  Then we have  $P=2$,  $p_1=0$,  $p_2=-1$,  $p_3=p_4=0$.  Thus (3) gives  $b=2$.
$2^{\circ }$.  $x_1=x_2=1$,  $x_3=x_4=-1$.  Now  $P=4$,  $p_1=0$,  $p_2=-2$,  $p_3=0$,  $p_4=1$,  whence (3) reads  $4=4b+d=8+d$,  giving  $d=-4$.
$3^{\circ }$.  $x_1=x_2=1$,  $x_3=x_4=0$.  We get  $P=2$,  $p_1=2$,  $p_2=1$,  $p_3=p_4=0$ .  These yield  $2=16+4a+b=18+4a$,  i.e.  $a=-4$.
$4^{\circ }$.  $x_1=x_2=2$,  $x_3=-1$,  $x_4=0$.  In this case,  $P=33$,  $p_1=3$,  $p_2=0$,  $p_3=-4$,  $p_4=0$,  whence  $33=81-12c$,  or  $c=4$.  Consequently, we obtain from (3) the result

$P=p_1^4-4p_1^2{p}_2+2p_2^2+4p_1{p}_3-4p_4.$ (4)

Although it has been derived by supposing  $n\geqq 4$ (= the degree of $P$), it holds without this supposition.  One has only to see that e.g. in the case  $n=2$,  one must substitute to (4) the values  $p_3=p_4=0$,  which changes the to the form  $P=p_1^4-4p_1^2{p}_2+2p_2^2.$

Title	applying elementary symmetric polynomials
Canonical name	ApplyingElementarySymmetricPolynomials
Date of creation	2013-03-22 19:10:07
Last modified on	2013-03-22 19:10:07
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	16
Author	pahio (2872)
Entry type	Application
Classification	msc 13B25
Classification	msc 12E10