sum of derivatives in roots of a polynomial of odd degree - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:51:38Z http://mathoverflow.net/feeds/question/41666 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41666/sum-of-derivatives-in-roots-of-a-polynomial-of-odd-degree sum of derivatives in roots of a polynomial of odd degree Fedor Petrov 2010-10-10T09:25:13Z 2010-10-10T17:46:42Z <p>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.</p> http://mathoverflow.net/questions/41666/sum-of-derivatives-in-roots-of-a-polynomial-of-odd-degree/41674#41674 Answer by Gjergji Zaimi for sum of derivatives in roots of a polynomial of odd degree Gjergji Zaimi 2010-10-10T12:01:27Z 2010-10-10T12:01:27Z <p>If I'm not mistaken this is basically the same question as <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366673&amp;sid=0f77743cc628257f609bb649b7ecfc84#p366673" rel="nofollow">this</a> 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 <a href="http://en.wikipedia.org/wiki/Schur%27s_inequality" rel="nofollow">Schur's inequality</a> in many variables.</p> http://mathoverflow.net/questions/41666/sum-of-derivatives-in-roots-of-a-polynomial-of-odd-degree/41684#41684 Answer by Denis Serre for sum of derivatives in roots of a polynomial of odd degree Denis Serre 2010-10-10T15:23:24Z 2010-10-10T17:46:42Z <p>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.</p> <hr> <p>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 $x_1&lt; x_2&lt; x_3$, the expression $y_1+y_2+y_3$ must keep a constant sign, which we may calculate with $f(x)=x(x^2-1)$.</p>