Re: Polynomial Question


Clashton@xxxxxxxxx wrote:
|Let m>=2 be an integer and consider the polynomial
|
|f(x) = product_{i=1}^{m} (x - a_1) - product_{i=1}^{m} (1 - a_i) .

I think in the first term you want a_i instead of a_1.

|For an arbitrary positive integer m, I would like to find expressions
|for a_1, ..., a_m in terms of
|indeterminates such that f(x) factors completely.

So if we let g(x) = product_{i=1}^m (x-a_i), and h(x)
= product_{i=1}^m (x-b_i), you want an identity of the
form g(x)-g(1) = h(x). The choice of "1" here is not crucial,
so you're essentially looking for cases where g(x) and h(x)
differ by a constant. If need be we can translate the
solution so that one of the b_i becomes 1, and get a
solution to the problem as you've defined it.

To put it a different way, you want equations among the
elementary symmetric polynomials (which give the
coefficients):

a_1+...+a_m = b_1+...+b_m,
a_1*a_2+a_1*a_3+...+a_{m-1}*a_m =
b_1*b_2+...+b_{m-1}*b_m,

etc. This in turn is equivalent to finding a_i and b_i such
that for k=1,...,m-1, the sum of the k-th powers of a_i and
the sum of the k-th powers of the b_i are the same.

I think in that form, it's a well-known problem, although
I'm not familiar with what's known about it. I think the
two sets of numbers are sometimes thought of as measures
that are sums of point masses, and then the sum of the
k-th powers corresponds to a moment of the measure.

|For example for m=5 and tinkering around with Mathematica I found the
|following example:
|For indeterminates a and k, if
|
|a_1= 1 + a*k,
|
|a_2 =1 + k + 3*a*k,
|
|a_3 = 1 + 4*k + 14*a*k + 12*a^2*k,
|
|a_4 =1 + 2*k + 10*a*k + 8*a^2*k,
|
|a_5 = 1 + 3*k + 10*a*k + 8*a^2*k

The problem is invariant under translation of the a_i and
b_i, so we might as well leave off the 1 terms. It's also scale
invariant, so we might as well divide through by k. That
leaves us with:

a_1 = a,
a_2 = 1 + 3*a,
a_3 = 4 + 14*a + 12*a^2,
a_4 = 2 + 10*a + 8*a^2,
a_5 = 3 + 10*a + 8*a^2.

Your solution corresponds to

b_1 = 0,
b_2 = 2 + 4*a,
b_3 = 1 + 7*a + 8*a^2,
b_4 = 4 + 13*a + 8*a^2,
b_5 = 3 + 14*a + 12*a^2.

The first four moments are indeed the same.

Keith Ramsay

--
Maarten Bergvelt
Mathematics Department, University of Illinois
Urbana-Champaign, IL 61801
Ph: 217-333-6326 email: bergv@xxxxxxxxxxxxx
.

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