Formula for $1/f(x)$ where $f$ is a polynomial

Question

Let $f$ be a polynomial having $n$ distinct real roots: $$f(x)=(x-x_1)(x-x_2)\dots(x-x_n)$$
Prove that $$\frac{1}{f(x)}=\sum_{k=1}^n \frac{1}{f'(x_k)(x-x_k)}, \: \forall x \in \mathbb{R} - \{x_1,x_2,\dots,x_n \} $$

I don't know much about partial fractions, but this looks very much like them. I tried to use induction for this, but I couldn't really make the jump from $n-1$ to $n$ and I honestly wouldn't have been satisfied even if I had solved it that way, because I would really like to see where this formula comes from. I'm sorry if this is actually very easy, I simply don't know how to approach it

Answer 1

Have you tried decomposing $\frac{1}{f(x)}$ into partial fractions. By that i mean writing $\frac{1}{f(x)} = \frac{A_1}{x-x_1}+...+\frac{A_n}{x-x_n}$ and solving for each $A_i$? Then i think the first fraction would look something like $\frac{1}{(x-x_1)(x_1-x_2)(x_1-x_3)...(x_1-x_n)}$. Then calculating $f'(x_1)$ should get you the result you want.

Answer 2

Let $s(x)$ be the summation and notice that

$$s(x)f(x)$$ is the Lagrangian interpolation polynomial* on the points $\left(x_k,y_k\right)$ where $$y_k:=\frac{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}{f'(x_k)}.$$

But we precisely have

$$f'(x_k)=\prod\limits_{\substack{i\ne k}}(x_k-x_i)$$

so that $$s(x)f(x)$$ is identically $1$.

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*Identify with

$$\sum_{k} y_k\frac{\prod\limits_{\substack{i\ne k}}(x-x_i)}{\prod\limits_{\substack{i\ne k}}(x_k-x_i)}.$$

For conciseness the ranges $1, n$ have been left implicit.

Answer 3

$1/(x-x1)(x-x2)..(x-xk)..=A/(x-x1)+B/(x-x2) ...K/(x-xk)..$

So When you multiply both sides by denominator you get, $1=A(x-x2)..(x-xk..(x-xn)+B(x-x1)(x-x3)..(x-xk)..(x-xn) + K(x-x1)..(x-(k-1))(x-(k+1)..(x-xn)$

To find Constant K substitute $x=xk$

After cancelling all terms you end up with, $K=1/(xk-x1)..(xk-(k-1))(xk-(k+1)..(xk-xn)$

We can express $f(x)$ as follows,

$f(x)=(x-xk)*[(x-x1)(x-x2)..(x-(k-1))(x-(k+1)..(x-xn)]$

say

$g(x)=[(x-x1)(x-x2)..(x-(k-1))(x-(k+1)..(x-xn)]$

(also note that $K=1/g(xk)$)

Now $f(x)=(x-xk)*g(x)$

$f'(x)=(x-xk)g'(x)+g(x)*1$

when $x=xk$ is substituted, $f'(xk)=g(xk)$

So K th term is $1/(f'(xk)*(x-xk)$