# Location of the roots of $f'$ (Laguerre's theorem)

Let $f \in \mathbb{R}[X]$ be a polynomial of degree $n$ having $n$ distinct roots $a_1,...,a_n$. Let $b_1<...<b_{n-1}$ be the roots of its derivative $f'$ (note that $b_i \in ]a_{i}, a_{i+1}[$ by Rolle's theorem). Show that for every $j$, $b_j$ can't be in the interval $A_j$ included in $[a_j,a_{j+1}[$, containg $a_j$ and having length $\frac{a_{j+1} - a_j}{n}$.

We could maybe use the following relation : let $x$ be a root of $f'$, then $$\frac{1}{x-a_1} + ... + \frac{1}{x - a_n} = 0$$

My plan was to suppose that there is a $b_j \in A_j$ and to derive a contradiction, but I couldn't find it. Help ?

Edit. This can be found in the book Problems in real analysis (Rudinescu, Andreescu), pb 5.6.42. Apparently it's called "Laguerre's theorem".

• Do you want to prove that $b_j \notin A_j$ for all $j$ or for at least one $j$? – Martin R May 2 '15 at 8:54
• I don't understand: if $A_j$ is included in $(a_j,a_{j+i})$, how can it contain $a_j$? – Emilio Novati May 2 '15 at 8:56
• @Martin : For all $j$... I guess. The question is stated in the book exactly as in my post. There is (little) ambiguity, but I really think this is for all $j$. – Tlön Uqbar Orbis Tertius May 2 '15 at 8:56
• @Emilio $A_j = [a_j, a_j + ...[$ (ok there was a mistake in the side of the bracket) – Tlön Uqbar Orbis Tertius May 2 '15 at 8:57

Well, I've worked it out. Note $d_i = |a_i - a_{i+1}|$ for simplicity. Suppose that there is $j$ such that $b_j \in A_j$ and note $x=b_j$. Then for every $i>j$ we have $a_i - x > d_i\frac{n-1}{n}$, so we'll have $$\frac{1}{x - a_i}>- \frac{n}{(n-1)d_i}$$ On the other side, if $i<j$ then $\frac{1}{x - a_i}>0$. So when we put everything together we have the strict inequality $$\sum_{i=1}^{n-1}\frac{1}{x-a_i} > \frac{n}{d_i} - \frac{n}{(n-1)d_i}- ... \frac{n}{(n-1)d_i}$$ where there are at most $n-1$ terms in the last sum. The right side is equal to
$$\frac{n}{d_i}\left( 1 - \frac{1}{n-1} - ... - \frac{1}{n-1}\right)\geqslant \frac{n}{d_i}\left(1 - (n-1)\frac{1}{n-1}\right) =0$$ and finally, $$\sum_{i=1}^{n-1}\frac{1}{x-a_i} > 0$$ which is in contradiction with the fact that $$\sum_{i=1}^{n-1}\frac{1}{x-a_i}=0$$ therefore, $b_j \notin A_j$ for every $j$.