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Let $m_\lambda(X_1(t),X_2(t),...X_N(t))$ be a monomial symmetric function with partition $\lambda$.

For example:

$$ m_{(3,1,1)}(X_1(t),X_2(t),...X_N(t)) =X_1^3X_2X_3 + X_1X_2^3X_3 + X_1X_2X_3^3 $$

  1. If $X_j=e^{i \omega_j t}$, is it possible to prove that there is at least one root $m_\lambda(t)=0$?

  2. If not, is it possible with some restrictions on the set $\{\omega_j\}$?

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This is $$\sum_{J}\exp\left(it\sum_{j\in J}\lambda_j w_j\right),$$ where $J$ ranges over all sequences of indices from $\{1,\cdots,N\}$ of length equal to that of the partition $\lambda$. –  anon Jul 12 '12 at 22:57
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What you get is an exponential polynomial, that is, a finite sum of terms of the form $a_ie^{b_iz}$. There is a lot of literature on the zeros of such things. Typing $$\rm zeros\ exponential\ polynomials$$ into the internet should get you started. –  Gerry Myerson Jul 12 '12 at 23:44
    
@anon, if your comment is a question, use the (?)-tag. But the answer is yes. –  draks ... Jul 13 '12 at 5:04
    
@GerryMyerson, thanks for the hint, I'll come back, when I've done my homework... –  draks ... Jul 13 '12 at 5:06

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