An equality in some extension field

Question

Let $L=K(\alpha)$ be a seperable field extension, end write $f \in K[X]$ for the minimal polynomial of the $\alpha$, let $\alpha_1, \cdots \alpha_n$ denote the roots. Prove the following equality:

$$ x^r \quad = \quad \sum_{1 \leq k \leq n} \frac{f(x) \cdot \alpha_k^r}{(x-\alpha_k)f'(\alpha_k)} $$

---

What I tried myself

First of all I rewrote $f$ and $f'$:

$$ f \ = \ \prod_{1\leq j \leq n} (X - \alpha_j) \quad \quad \text{and} \quad \quad f' \ = \ \sum_{1\leq j \leq n} \prod_{i \neq j}(X-\alpha_i) $$ Now this is what the big sum looks like if I am not mistaking: $$ \sum_{1\leq k \leq n } \frac{\alpha_k^r \cdot \prod_{1\leq j \leq n} (X - \alpha_j)}{(x-\alpha_k)\sum_{1\leq j \leq n} \prod_{i \neq j}(\alpha_k-\alpha_i) } $$ Some of those products in the denominator might vanish, but that is all I can see. Can you give me a small hint to go on?

Answer 1

[Not quite a complete answer, but some progress]$${f'(x)\over f(x)}=\sum{1\over x-\alpha_j}$$ $${(x-\alpha_k)f'(x)\over f(x)}=1+\sum_{j\ne k}{x-\alpha_k\over x-\alpha_j}$$ Evaluating at $x=\alpha_k$, we get $${(x-\alpha_k)f'(x)\over f(x)}\Biggl|_{x=\alpha_k}=1$$ So the left and right sides of the equality agree at $x=\alpha_k$, $1\le k\le n$.

Answer 2

This is very simple and explained in the "Procedure" section on Wikipedia.

Consider the rational fraction $g(x)=\frac{x^r}{f(x)}$. Its degree is $r-n<0$, so it admits a partial fraction decomposition of the form

$$ g(x)=\sum_{k=1}^{n}\frac{c_k}{x-\alpha_k} $$

where the $c_k$ are constants. Putting $h_j(x)=\prod_{k\neq j}(x-\alpha_k)$, we have

$$ \frac{x^r}{h_j(x)}=\frac{x-\alpha_j}{g(x)}=c_j+ \sum_{k\neq j}\frac{c_k(x-\alpha_j)}{x-\alpha_k} $$

Setting $x=\alpha_j$ above, we obtain $c_j=\frac{\alpha_j^r}{h_j(\alpha_j)}= \frac{\alpha_j^r}{f'(\alpha_j)}$.