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Let $f(x) = (x-a_1)(x-a_2)...(x-a_n)$.

Find a decomposition into simple fractions of $\frac{f'}{f}$.

Where $f'$ is a derivative of our polynomial.

As I understand, we have to find a pretty-format of $f'$ and then task will be easy. But I have no idea how to find this.

Note that some $a_i$ can be equal to each other.

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  • $\begingroup$ Suggestion: Compute it for several small $n$ and look for a pattern. $\endgroup$
    – coffeemath
    Commented Nov 22, 2020 at 20:23
  • $\begingroup$ It might help to recognize that (fgh)'=f'gh+fg'h+fgh', and I'm sure you can see the generalization to n factors. $\endgroup$ Commented Nov 22, 2020 at 20:26
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    $\begingroup$ Hint: $f'/f = (\log f)'$ $\endgroup$
    – Martin R
    Commented Nov 22, 2020 at 20:27

1 Answer 1

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Following Martin's idea,

Let for $ x\ne a_1,a_2,...,a_n $, $$g(x)=\ln(|f(x)|)=$$ $$\sum_{k=1}^n\ln(|x-a_k|)$$

thus

$$g'(x)=\frac{f'(x)}{f(x)}$$ $$=\sum_{k=1}^n\frac{1}{x-a_k}$$

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