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Let $P$ be a $n$-degree real polynomial having $n$ simple roots $x_1, x_2, \cdots,x_n$.

The problem asks to prove that

$$ \displaystyle\sum_{k=1}^n \frac{P''(x_k)}{P'(x_k)} =0$$

I tried writing $P$ as the product of $x−x_i$ and differentiating it twice but the calculations are too hard to carry out. Thanks in advance.

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marked as duplicate by Martin R, Song, Carl Schildkraut, Cesareo, Parcly Taxel Mar 1 at 1:14

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  • $\begingroup$ What have you tried so far ? $\endgroup$ – Martin Hansen Feb 28 at 18:29
  • $\begingroup$ I tried writing $P$ as the product of $x-x_i$ and differentiating it twice but the calculations are too hard to carry out $\endgroup$ – ahmed Feb 28 at 18:32
  • $\begingroup$ @ahmed Can you differentiate it once? Notice how simple $P’$ looks when you evaluate it at an $x_i$. Can you see a way to simplify $P’’$ also? (it will probably still have about $n$ terms but it’s manageable) $\endgroup$ – Erick Wong Feb 28 at 18:42
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An example :

Let $$P(x)=(x-1)(x+1)(x-2)$$ Then $$P'(x)=1.(x+1).(x-2)+(x-1).1.(x-2)+(x-1).(x+1).1$$ and $$P''(x)=1.1.(x-2)+1.(x+1).1+1.1.(x-2)+(x-1).1.1+1.(x+1).1+(x-1).1.1$$ $$P''(x)=2 [(x-1)+(x+1)+(x-2)]$$ $$P''(1)=2[2-1] =2 : P''(-1)=2[-2-3]=-10 : P''(2)=2[1+3]=8$$ And $$2-10+8=0$$ So, now it's about generalising this process...

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