2
$\begingroup$

Today my students asked me the following problem:

Define polynomials $P_j$ with real coefficients $$P_{j}=\sum_{i_{1},\cdots,i_{s}}a^{(j)}_{i_{1}\cdots i_{s}}X^{i_{1}}_{1}\cdots X^{i_{s}}_{s}, \qquad (1\le j\le s)$$ where $s$ is a given positive integer.

Assmue that $a^{(1)}_{1,0,0,\ldots,0}, a^{(2)}_{0,1,0,\ldots,0}, \ldots, a^{(s)}_{0,0,\ldots,1}$ are negative, and such that $$2\min\{|a^{(1)}_{1,0,\ldots,0}|,|a^{(2)}_{0,1,\ldots,0}|,\ldots,|a^{(s)}_{0,0,\ldots,1}|\}>\sum_{1\le j\le s,i_{1},i_{2},\ldots,i_{s}}|a^{(j)}_{i_{1},\ldots,i_{s}}|$$

Show that this system of polynomial equations $$P_{j}(X_{1},X_{2},\ldots,X_{s})=0, \qquad(1\le j\le s)$$ has a solution $(X_{1},X_{2},\ldots,X_{s})\in \mathbb{R}^{s}$

I know this result: Criteria to determine whether a real-coefficient polynomial has real root?

But for a system of real polynomials to have a real solution, I can't find any similar results in the literature. Maybe this result is old? Thank you for you help.

$\endgroup$

1 Answer 1

1
$\begingroup$

The system can be put in the form $Y=LX-G(X)$. Here $X$ is the column vector with the unknowns $X_1,\ldots,X_s$. $Y$ is the column vector with components $Y_i=-a_{0,\ldots,0}^{(i)}$. $L$ is the matrix describing the linear part of your system, namely, $$ L_{ij}=a_{0,\ldots,0,1,0,\ldots,0}^{(i)} $$ with the $1$ at the $j$-th position. Finally $$ G_i(X)=-\sum_{j_1,\ldots,j_s} a_{j_1,\ldots,j_s}^{(i)} X_1^{j_1}\cdots X_s^{j_s} $$ where you only keep terms of degree at least 2.

Then your criterion and the Gershgorin circle theorem guarantee that the matrix $L$ is invertible so you can try to solve iteratively $$ X=L^{-1}Y+L^{-1}G(X) $$ $$ X=L^{-1}Y+L^{-1}G(L^{-1}Y+L^{-1}G(X)) $$ etc. In the fact one can write the end result as an explicit series in terms of trees. See for instance Theorem 1 in my article "The Jacobian Conjecture as a Problem of Perturbative Quantum Field Theory". It is possible to write explicit bounds which imply the convergence of this series and therefore the existence of a real solution. Such bounds essentially mean that $Y$ and $G$ are sufficiently small. The question is whether your criterion implies such bounds. In the above paper the particular case treated is that where $G$ is homogeneous. To see how the formulas look like in the nonhomogenous case you can see my other article "Feynman Diagrams in Algebraic Combinatorics".

$\endgroup$
1
  • $\begingroup$ Oh,It's Nice! $Y=LX-G(X)$!!+1 $\endgroup$
    – math110
    Commented Jan 16, 2015 at 16:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .