applying elementary symmetric polynomials

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

,

…

,

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

λ

1

⁢

x

2

λ

2

⁢

⋯

⁢

x

n

λ

n

where

λ

1

≥

λ

2

≥

…

≥

λ

n

≥

0

and

λ

1

+

λ

2

+

…

+

λ

n

=

d

.

Then

P

=

Q

⁢

(

p

1

,

p

2

,

…

,

p

n

)

=

∑

i

m

i

⁢

p

1

λ

1

-

λ

2

⁢

p

2

λ

2

-

λ

3

⁢

⋯

⁢

p

n

-

1

λ

n

-

1

-

λ

n

⁢

p

n

λ

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

,

0

,

0

;

3

,

1

,

0

;

2

,

2

,

0

;

2

,

1

,

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

=

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

+

9

⁢

a

+

3

⁢

b

=

9

+

3

⁢

b

,

yielding

b

=

-

1

.  Hence we have the result

P

=

p

1

2

⁢

p

2

-

2

⁢

p

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

+

…

+

x

n

4

.  If we suppose that

n

≧

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

=

…

=

x

n

=

0

and in as follows.

1

∘

.

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

∘

.

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

=

4

⁢

b

+

d

=

8

+

d

,  giving

d

=

-

4

.

3

∘

.

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

+

4

⁢

a

+

b

=

18

+

4

⁢

a

,  i.e.

a

=

-

4

.

4

∘

.

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

-

12

⁢

c

,  or

c

=

4

.  Consequently, we obtain from (3) the result

P

=

p

1

4

-

4

⁢

p

1

2

⁢

p

2

+

2

⁢

p

2

2

+

4

⁢

p

1

⁢

p

3

-

4

⁢

p

4

.

(4)

Although it has been derived by supposing

n

≧

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

-

4

⁢

p

1

2

⁢

p

2

+

2

⁢

p

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

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