# Find a sum that consists of roots of polynomial and it's derivative polynomial

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 $$\mu_j$$.

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.

• 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? Nov 14, 2023 at 21:02
• @ALiCeP.: If $f(x) = c(x-\alpha_1)\cdots (x-\alpha_n)$ with distinct roots $\alpha_i$ then $f'(\alpha_i) \ne 0$. Nov 14, 2023 at 21:15