0

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?

Koenraad van Duin
  • 3,156

2 Answers2

2

[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 $x=\alpha_k$, $1\le k\le n$.

Gerry Myerson
  • 179,216
  • I understood everything you said aside from the last sentence – Koenraad van Duin Aug 29 '14 at 13:24
  • In the equality that you are trying to prove, if you let $x=\alpha_k$, then the left side is $\alpha_k^r$, and, by the calculation I gave, so is the right side. – Gerry Myerson Aug 29 '14 at 23:00
  • mmm, does that mean that we need to show that $\sum_{j\neq k}\alpha_j \ = \ 0$, or am I mistaking? – Koenraad van Duin Aug 30 '14 at 07:50
  • I don't know why you think we need to show that sum is zero. Incidentally, you never mention $r<n$ in your question, but as Ewan pooints out the equation you want to prove is false without that assumption. That means my answer really is complete, since we get two polynomials of degree $r$ agreeing at more than $r$ places. – Gerry Myerson Aug 30 '14 at 08:54
1

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 $c_j=\frac{\alpha_j^r}{h_j(\alpha_j)}= \frac{\alpha_j^r}{f'(\alpha_j)}$.

Ewan Delanoy
  • 61,600