# sum of reciprocals of derivative of polynomial at its roots

If $P(x)$ is a polynomial of degree $n > 1$ with only simple roots $a_1,\ldots,a_n$, is it true that $\frac 1{P'(a_1)} + \cdots + \frac 1{P'(a_n)} = 0$, and, if so, what is the proof? I can see this directly for $n = 2,3,4$ with some brute force for $4$.

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$$\frac{1}{P(x)}= \sum_{i=1}^n \frac{c_i}{x-a_i}$$
Multiplying by $x-a_j$ yields
$$\frac{x-a_j}{P(x)}= \sum_{i=1}^n \frac{c_i(x-a_j)}{x-a_i} (*)$$
from where $c_j =\frac{1}{P'(a_j)}$. [ED: $P(x)=(x-a_j)Q(x)$, sub it back up and observe that $P'(a_j)=Q(a_j)] Hence $$1=\sum_{i=1}^n \frac{1}{P'(a_i)} \frac{(x-a_1)\cdots(x-a_n)}{x-a_i}$$ The coefficient of$x^{n-1}$is $$0= \sum_{i=1}^n \frac{1}{P'(a_i)}$$ - It's the residue theorem, applied to the function$g(z):={1\over f(z)}$and a large circle$\gamma_R:=\partial D_R$: As$p:={\rm deg}(f)\ge2$one has $$\left|\int_{\gamma_R}{1\over f(z)}\ dz\right| \leq\ C\ {1\over R^p}\ 2\pi R \to 0\qquad (R\to\infty)\ .$$ This implies that the sum of the residues of${1\over f}$at the poles$a_k$of${1\over f}$is zero. Since all zeros of$f$are supposed to be simple we can write$f$near a zero$a_k$as$f(z)=(z-a_k)g(z)$with$g$analytic near$a_k$and$g(a_k)=f'(a_k)\ne0$. It follows that the residue of${1\over f}$at$a_k$is $${1\over g(a_k)}={1\over f'(a_k)}\ .$$ Altogether we see that $$\sum_{k=1}^n {1\over f'(a_k)}\ =\ 0\ .$$ - Consider the meromorphic differential form$\omega (z)= \frac {1}{P(z)}dz$. Its residue at infinity is$Res(\omega;\infty)=0$and its residue at$a_j$is$Res(\omega;a_j)= \frac {1}{P'(a_j)}$(a standard basic result following from the definition) Now if you remember that the sum of the residues of a meromorphic forms on the extended plane$\hat { \mathbb C}$[= Riemann sphere$\mathbb P^1(\mathbb C)$] is zero, you get as required $$\Sigma_{P\in {\hat {\mathbb C}}} Res(\omega; P)=0+ \Sigma Res(\omega; a_j)=0+ \Sigma \frac {1}{P'(a_j)}=0$$ Edit: an optional exercise Just as a little exercise in residues of differential forms, let's analyze what happens in degree 1 if$P(x)=c\cdot (x-a)$and$\omega (z)=\frac {1}{c\cdot (z-a)} dz$(this is the case, correctly excluded by Joe, where the formula$\Sigma \frac {1}{P'(a_j)}=0$is obviously false). We still have$Res(\omega;a)= \frac {1}{P'(a)}=\frac {1}{c}$. At infinity we write$z=\frac {1}{t}$and$\omega(z)dz= \frac {1}{c\cdot (z-a)}\cdot dz=\frac {t}{ c(1-at)} \cdot\frac{-1}{t^2} dt$Hence$Res(\omega:\infty)=\frac{-1}{c} $by the "standard basic result " evoked above. The sum of the residues of$\omega$is again zero, as it must:$\frac {1}{c}+(-\frac {1}{c})=0$- In fact, we have $$\sum_{j=1}^n\frac{f(a_j)}{p'(a_j)}=f[x_1,\ldots,x_n],$$ where$f=1$and$f[x_1,\ldots,x_n]$is the$n$-th divided difference of$f\$. Here is a proof of the used identity.