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I'm trying to solve this task:

$f(x)$ is a polynomial of degree $n$, it has different roots $\alpha_1, \alpha_2, ..., \alpha_n$. Let $\mu_1, \mu_2, ..., \mu_{n-1}$ be the roots of the derivative polynomial $f'(x)$. Find $\sum_{i=1}^{n} \sum_{j=1}^{n-1} \frac{1}{\alpha_i - \mu_j} (*) $.

I was trying to use this equation:

$\frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{1}{x - \alpha_i}$ (where $\alpha_i$ are roots of $f(x)$)

It looks like I can represented the sum $(*)$ in this way:

$\sum_{i=1}^{n} \frac{f''(\alpha_i)}{f'(\alpha_i)}$

However, I am not sure if it correct to use second derivative here or if some $\alpha_i$ wouldn't be equal to some $\mu_j$.

Could somebody please give a hint? Thanks in advance.

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It is correct that $$ \frac{f'(x)}{f(x)} = \sum_{i=1}^{n} \frac{1}{x - \alpha_i} $$ for all $x \notin \{ \alpha_1, \ldots, \alpha_n \} $. Since $f$ has $n$ distinct roots, $f$ and $f'$ have no common roots, and we can set $x = \mu_j$ in that formula: $$ 0 = \frac{f'(\mu_j)}{f(\mu_j)} = \sum_{i=1}^{n} \frac{1}{\mu_j - \alpha_i} $$ for $j=1, \ldots, n-1$. It follows that $\sum_{i=1}^{n} \sum_{j=1}^{n-1} \frac{1}{\alpha_i - \mu_j}$ is zero.

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  • $\begingroup$ Could you please explain in a bit more detail why $f$ and $f'$ will not have common roots? Is it because for the root to be both the root of $f$ and $f'$ it must be root of degree at least 2? $\endgroup$
    – ALiCe P.
    Nov 14, 2023 at 21:02
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    $\begingroup$ @ALiCeP.: If $f(x) = c(x-\alpha_1)\cdots (x-\alpha_n)$ with distinct roots $\alpha_i$ then $f'(\alpha_i) \ne 0$. $\endgroup$
    – Martin R
    Nov 14, 2023 at 21:15

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