Show that $\sum_{j=1}^{n}\dfrac{g(x_j)}{f'(x_j)} =1$ for a monic polynomial with degree $n$ with distinct zeroes.

Let $f(x)$ be a monic polynomial with degree $n$ with distinct zeroes $x_1,x_2,...,x_n$. Let $g(x)$ be any monic polynomial of degree $n-1$. Show that $$\sum_{j=1}^{n}\dfrac{g(x_j)}{f'(x_j)} =1$$

I am totally clueless about how to proceed please give hints or solutions. Thank you.

• Do you know about Lagrange interpolation?
– Pig
Sep 4, 2016 at 23:47
• Yeah I know it, thanks for the hint.
– user366265
Sep 5, 2016 at 0:00

We have $f(x) = \displaystyle\prod_{i=1}^n (x-x_i)$, differentiating both sides w.r.t $x$ we have $f'(x) = \displaystyle\sum_{i=1}^n \displaystyle\prod_{j\neq i} (x-x_j)$.
Therefore $f'(x_i) = \displaystyle\prod_{j\neq i} (x_i-x_j)$. Apply the Lagrange Interpolation Formula to the polynomial $g$ and points $x_i$. We deduce $\displaystyle\sum_{i=1}^n g(x_i) \dfrac{\prod_{j\neq i} (x-x_j)}{\prod_{j\neq i} (x_i-x_j)} = g(x)$. Comparing the coefficients of $x^{n-1}$ on both sides of the equation gives the desired result.