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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?

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

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  • $\begingroup$ 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? $\endgroup$ – weilam06 Dec 13 '18 at 14:55
  • $\begingroup$ 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. $\endgroup$ – AAK Dec 13 '18 at 18:10
  • $\begingroup$ Oh okay, I have mistyped it. Thanks. $\endgroup$ – weilam06 Dec 14 '18 at 8:36
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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.

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

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