# 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$P(x_{1},\,x_{2},\,\ldots,\,x_{n})=P$  of degree $d$ be homogeneous (http://planetmath.org/HomogeneousPolynomial).  Starting from the highest term of $P$ we form all products

 $x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{n}^{\lambda_{n}}$

where

 $\lambda_{1}\;\geq\;\lambda_{2}\;\geq\;\ldots\geq\;\lambda_{n}\;\geq\;0\quad% \mbox{and}\quad\lambda_{1}\!+\!\lambda_{2}\!+\ldots+\!\lambda_{n}\;=\;d.$

Then

 $\displaystyle P\;=\;Q(p_{1},\,p_{2},\,\ldots,\,p_{n})\;=\;\sum_{i}m_{i}p_{1}^{% \lambda_{1}-\lambda_{2}}p_{2}^{\lambda_{2}-\lambda_{3}}\cdots p_{n-1}^{\lambda% _{n-1}-\lambda_{n}}p_{n}^{\lambda_{n}},$ (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

 $\displaystyle p_{1}\;=\;x_{1}\!+\!x_{2}\!+\!x_{3},\quad p_{2}\;=\;x_{2}x_{3}\!% +\!x_{3}x_{1}\!+\!x_{1}x_{2},\quad p_{3}\;=\;x_{1}x_{2}x_{3}.$ (2)

We have four

 $4,\,0,\,0;\quad 3,\,1,\,0;\quad 2,\,2,\,0;\quad 2,\,1,\,1,$

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

 $p_{1}^{4},\quad p_{1}^{2}p_{2},\quad p_{2}^{2},\quad p_{1}p_{3},$

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

 $P\;=\;p_{1}^{2}p_{2}\!+\!ap_{2}^{2}\!+\!bp_{1}p_{3}.$

Using  $x_{1}=x_{2}=1$  and  $x_{3}=0$  makes  $p_{1}=2$,  $p_{2}=1$  and  $p_{3}=0$, when

 $P\;=\;2\;=\;4\!+\!a\!+\!0,$

implying  $a=-2$.  Using similarly  $x_{1}=x_{2}=x_{3}=1$  we get  $p_{1}=p_{2}=3$,  $p_{3}=1$, which give

 $P\;=\;6\;=\;27\!+\!9a\!+\!3b\;=\;9\!+\!3b,$

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  $P=x_{1}^{4}\!+\!x_{2}^{4}\!+\ldots+\!x_{n}^{4}$.  If we suppose that  $n\geqq 4$,  the possible highest terms are

 $x_{1}^{4},\quad x_{1}^{3}x_{2},\quad x_{1}^{2}x_{2}^{2},\quad x_{1}^{2}x_{2}x_% {3},\quad x_{1}x_{2}x_{3}x_{4}$

whence we may write

 $\displaystyle P\;=\;p_{1}^{4}\!+\!ap_{1}^{2}p_{2}\!+\!bp_{2}^{2}\!+\!cp_{1}p_{% 3}\!+\!dp_{4}.$ (3)

For determining the coefficients, evidently we can put  $x_{5}=x_{6}=\ldots=x_{n}=0$  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

 $\displaystyle P\;=\;p_{1}^{4}\!-\!4p_{1}^{2}p_{2}\!+\!2p_{2}^{2}\!+\!4p_{1}p_{% 3}\!-\!4p_{4}.$ (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 ApplyingElementarySymmetricPolynomials 2013-03-22 19:10:07 2013-03-22 19:10:07 pahio (2872) pahio (2872) 16 pahio (2872) Application msc 13B25 msc 12E10