Given odd positive integer $n$ and a monic polynomial $f(x)=(x-x_1)\dots (x-x_n)$ with $n$ distinct real roots. Is it always true that $\sum f'(x_i) > 0$? I may prove it for $n=3$ and $n=5$ and it looks plausible.
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asked Oct 10, 2010 at 9:25
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1$\begingroup$ A possible reformulation: given an unreduced tridiagonal matrix $\mathbf{T}$ of odd order with characteristic polynomial $f(x)$, is it true that $\mathrm{trace}(f^{\prime}(\mathbf{T}))>0$? $\endgroup$– J. M. isn't a mathematicianCommented Oct 10, 2010 at 11:06
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2 Answers
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If I'm not mistaken this is basically the same question as this question from the international mathematical olympiad in 1971. The statement is only true for 3 and 5 variables showing that there is no obvious generalization to Schur's inequality in many variables.
answered Oct 10, 2010 at 12:01
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Instead, the sum of reciprocals $$\sum\frac{1}{f'(x_j)}$$ vanishes. This is because of the formula $$\frac{1}{f(x)}=\sum_j\frac{a_j}{x-x_j},\qquad a_j:=\frac{1}{f'(x_j)},$$ together with the asymptotics as $x\rightarrow\infty$. This is valid for every degree, odd or even.
When $n=3$, this gives an amazing proof of the property that you quote. Denote $y_j=f'(x_j)$. Then $y_1y_2+y_3y_1+y_2y_3=0$, which means that $y=(y_1,y_2,y_3)$ belongs to a quadric whose intersection with the plane $y_1+y_2+y_3=0$ reduces to $(0,0,0)$, not equal to $y$. By continuity and connexity of the parameter space $x_1< x_2< x_3$, the expression $y_1+y_2+y_3$ must keep a constant sign, which we may calculate with $f(x)=x(x^2-1)$.
answered Oct 10, 2010 at 15:23