From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: first problem: more variables
Date: 19 Aug 2000 05:42:53 GMT
Newsgroups: sci.math
Summary: [missing]

In article <e2n7sakg8mv8@forum.mathforum.com>,
Zakir F. Seidov <seidov@bgumail.bgu.ac.il> wrote:
>Let (n=4):
>
>a=x+y+z+w,
>b=x^2+y^2+z^2+w^2,
>c=x^3+y^3+z^3+w^3,
>d=x^4+y^4+z^4+w^4.
>
>Then:
>
>e=x^5+y^5+z^5+w^5=
>  (-a^5 + 10*a^3*b - 15*a*b^2 - 20*a^2*c + 20*b*c + 30*a*d)/24.
>
>May anyone derive general formula for n variables?


You want the Newton formulae. 

Given a finite set of numbers  x_i  we may form the infinite sequence of
the sums of their powers:  p_n = \sum_i  (x_i)^n  .  We may also compute
their elementary symmetric functions, only finitely many of which are
non-zero: expand  \product_i (1 + x_i T) = \sum_n  s_n  T^n  . (So s_0 = 1.)
These two collections of numbers are related by the Newton formulae:
	0= p_1 s_0 -   1 s_1
	0= p_2 s_0 - p_1 s_1 +   2 s_2
	0= p_3 s_0 - p_2 s_1 + p_1 s_2 -   3 s_3
	0= p_4 s_0 - p_3 s_1 + p_2 s_2 - p_1 s_3 +   4 s_4
and so on.

These formulae can be used in sequence to express the  p_n  as integer 
polynomials in the  s_n.  They may similarly be used to express the  s_n 
as rational polynomials in the  p_n  (assuming we are working in 
characteristic zero).

Since  s_n = 0  for  n > N  (the count of the numbers  x_i)  we may
combine these two sorts of computations to express each  p_n  as a 
rational polynomial in  {p_1, ..., p_N}.  

These calculations lead to the formula shown for  N=4; for N=5  I get  120*p6=
    40*p3^2 + 90*p2*p4 - 15*p2^3 + 144*p5*p1 - 120*p2*p3*p1 + 
    45*p2^2*p1^2 - 90*p4*p1^2 + 40*p3*p1^3 - 15*p2*p1^4 + p1^6
Of course, there's no reason except exhaustion to stop here.

By the way, once we know that  N! * p_{N+1}  is an integer linear combination
of the appropriate monomials in the other  p_n, we need only compute the
coefficients; it suffices to pick sufficiently many sets of the  {x_i}
to generate enough linear equations relating these coefficients, then
solve (using integer arithmetic!) 

dave