# Computation of partial fraction decomposition

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?

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 $$(\prod_{j=1}^{n+1}(x-\alpha_j))'=\sum_{j=1}^{n+1}\prod_{i\neq j}(x-\alpha_i).$$

• So I have to make $r(x)=P(x)-\sum^{n+1}_{k=1}\frac{\alpha_k}{Q'(\alpha_k)(x-\alpha_k)} \cdot Q(x)=0$ holds true for all $\alpha_k$, where $k=1,2, \cdots, n+1$. Prove that $r(x)$ with degree $n$ has $n+1$ roots, concludes that $r(x)$ must be a zero polynomial, i.e. $r(x)=0$, so the conclusion follows? – weilam06 Dec 13 '18 at 14:55
• Yes! If a polynomial of degree $n$ has at least $n+1$ roots then it is $0$. Just a remark: put $P(\alpha_k)$ instead of $\alpha_k$ in the numerator in your comment. – AAK Dec 13 '18 at 18:10
• Oh okay, I have mistyped it. Thanks. – weilam06 Dec 14 '18 at 8:36

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.

• I don't understand how to relate with MVT. – weilam06 Dec 14 '18 at 2:46