Simplifying a polynomial expression

Question

Let $P$ be a polynomial of degree $n$ with real coefficients. Is there a more elegant (concise) form of the expression $$\sum_{i=1}^{n+1}P(x+\lambda_i)\prod_{j=1, \\ \ \\ j\neq i}^{n+1} (\lambda_i-\lambda_j)^{-1}$$ where $\lambda_1<\lambda_2<...<\lambda_{n+1}$ are arbitrary reals?

Answer 1

I prove below that whenever $P$ is a polynomial of degree $\leq n$, your expression is simply $a_n$, the coefficient of degree $n$ in $P$.

The polynomial $$ Q=P(x+T)-a_n(T-\lambda_1) \ldots (T-\lambda_n) \tag{1} $$

has degree $ \leq n-1$ in $T$. By partial fraction expansion, we can write

$$ \frac{Q(T)}{(T-\lambda_1)\ldots (T-\lambda_{n})}= \sum_{i=1}^{n} \frac{b_i}{T-\lambda_i} \tag{2} $$

where the $b_i$ are constants. In other words,

$$ \frac{P(x+T)}{(T-\lambda_1)\ldots (T-\lambda_{n})}= a_n+\sum_{i=1}^{n} \frac{b_i}{T-\lambda_i} \tag{3} $$

A well-known trick to compute the $b_i$ is to multiply both sides of (3) by $(T-\lambda_i)$ and then evaluate at $T=\lambda_i$, which yields

$$ b_i=\frac{P(x+\lambda_i)}{\prod_{j\neq i}(\lambda_i-\lambda_j)} \tag{4} $$

Injecting (4) into (3), and evaluating at $T={\lambda}_{n+1}$ finishes the proof.