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 polynomialMathworldPlanetmathP(x1,x2,,xn)=P  of degree d be homogeneousPlanetmathPlanetmathPlanetmath (  Starting from the highest term of P we form all productsMathworldPlanetmathPlanetmath



λ1λ2λn 0andλ1+λ2++λn=d.


P=Q(p1,p2,,pn)=imip1λ1-λ2p2λ2-λ3pn-1λn-1-λnpnλn, (1)

in which the coefficientsMathworldPlanetmath mi are determined by giving some suitable values to the indeterminates xj.

Example 1.  Express the polynomialMathworldPlanetmathPlanetmathP=x13x2+x13x3+x23x1+x23x3+x33x1+x33x2  in the elementary symmetric polynomials

p1=x1+x2+x3,p2=x2x3+x3x1+x1x2,p3=x1x2x3. (2)

We have four

4, 0, 0;3, 1, 0;2, 2, 0;2, 1, 1,

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


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


Using  x1=x2=1  and  x3=0  makes  p1=2,  p2=1  and  p3=0, when

P= 2= 4+a+0,

implying  a=-2.  Using similarly  x1=x2=x3=1  we get  p1=p2=3,  p3=1, which give

P= 6= 27+9a+3b= 9+3b,

yielding  b=-1.  Hence we have the result




Example 2.  Let  P=x14+x24++xn4.  If we suppose that  n4,  the possible highest terms are


whence we may write

P=p14+ap12p2+bp22+cp1p3+dp4. (3)

For determining the coefficients, evidently we can put  x5=x6==xn=0  and in as follows.
1.  x1=1,  x2=-1,  x3=x4=0.  Then we have  P=2,  p1=0,  p2=-1,  p3=p4=0.  Thus (3) gives  b=2.
2.  x1=x2=1,  x3=x4=-1.  Now  P=4,  p1=0,  p2=-2,  p3=0,  p4=1,  whence (3) reads  4=4b+d=8+d,  giving  d=-4.
3.  x1=x2=1,  x3=x4=0.  We get  P=2,  p1=2,  p2=1,  p3=p4=0 .  These yield  2=16+4a+b=18+4a,  i.e.  a=-4.
4.  x1=x2=2,  x3=-1,  x4=0.  In this case,  P=33,  p1=3,  p2=0,  p3=-4,  p4=0,  whence  33=81-12c,  or  c=4.  Consequently, we obtain from (3) the result

P=p14-4p12p2+2p22+4p1p3-4p4. (4)

Although it has been derived by supposing  n4 (= 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  p3=p4=0,  which changes the to the form  P=p14-4p12p2+2p22.

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