[seqfan] Re: Sums of squares of polynomials
Max Alekseyev
maxale at gmail.com
Fri Dec 5 02:48:29 CET 2008
On Thu, Dec 4, 2008 at 11:49 AM, <franktaw at netscape.net> wrote:
> Not every polynomial with integer coefficients can be represented
> as the sum of squares. Any such polynomial must obviously have
> even degree; almost as obvious, the coefficients of odd powers
> of x must be even. An additional constraint is that the polynomial
> must be non-negative on the entire real number line (which
> actually implies that the degree is even). I don't know if these
> conditions are sufficient, but this is not the main focus of my
> question.
Just aside note: the above conditions are sufficient, although
polynomials in squares may not necessary have integer coefficients.
Proof is simple but a bit tricky.
Let f(x) be a polynomial of degree 2n that is positive and have no real roots.
Since the field of complex numbers is algebraically closed, f(x) has
2n complex roots.
Moreover, each root of f(x) comes with its complex conjugate.
Let complex numbers r_1, r_2, ..., r_n and their conjugates r'_1,
r'_2, ..., r'_n be the roots of f(x).
Consider a polynomial of degree n:
(x-r_1)(x-r_2)...(x-r_n) = g(x) + I*h(x)
where polynomials g(x) and h(x) with real coefficients represent the
real and imaginary parts of this polynomial.
It is clear (by taking conjugates) that
(x-r'_1)(x-r'_2)...(x-r'_n) = g(x) - I*h(x)
Therefore,
f(x) = ( g(x) + I*h(x) ) * ( g(x) - I*h(x) ) = g(x)^2 + h(x)^2
as required.
Regards,
Max
More information about the SeqFan
mailing list