Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
A possible reformulation: given an unreduced tridiagonal matrix $\mathbf{T}$ of odd order with characteristic polynomial $f(x)$, is it true that $\mathrm{trace}(f^{\prime}(\mathbf{T}))>0$? –  J. M. Oct 10 '10 at 11:06
add comment

2 Answers 2

up vote 16 down vote accepted

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.

share|improve this answer
add comment

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 $f(x)=x(x^2-1)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.