Computation of partial fraction decomposition

Question

Suppose the polynomial $Q(x)$ be $$Q(x)=(x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_{n+1})$$, where $\alpha_1, \alpha_2, \cdots , \alpha_{n+1}$ are distinct real numbers and $n \in \mathbb{N}$. Show that if $P(x)$ is a polynomial with degree less than $n+1$ then $$\frac{P(x)}{Q(x)}=\sum^{n+1}_{k=1}\frac{P(\alpha_k)}{Q'(\alpha_k)(x-\alpha_k)},$$ where $Q'(x)$ is the derivative of the polynomial $Q(x)$.

The generalisation of partial fractions decomposition is strongly related to Lagrange polynomial, should I start proving Lagrange polynomial before the statement above? Or I should have another approach to prove this?

Answer 1

Two polynomials of degree $n$ are equal if they have the same values at $n+1$ distinct points. Multiply RHS and LHS of your equation by $Q(x)$ and compute the obtained LHS and RHS at any $\alpha_j$. If LHS($\alpha_j$)=RHS($\alpha_j$) are the same for all $j=1,...,n+1$ then the equality is true. Do not forget that $$ (\prod_{j=1}^{n+1}(x-\alpha_j))'=\sum_{j=1}^{n+1}\prod_{i\neq j}(x-\alpha_i). $$

Answer 2

hint

The decomposition gives

$$\frac{P(x)}{Q(x)}=\sum_{k=1}^{n+1}\frac{A_k}{(x-\alpha_k)}$$

with

$$A_k=\lim_{x\to\alpha_k}\frac{P(x)(x-\alpha_k)}{Q(x)}$$ and $$Q'(\alpha_k)=\lim_{x\to\alpha_k}\frac{Q(x)-0}{x-\alpha_k}.$$

You can take it.