Let :
- $n \geq 1$
- $E_n= \mathbb{R}^n[X]$
- $\forall ~ 0 \leq j \leq n, \quad L_{n,k}(j)= \delta_{k,j}$
- Scalar product $<P_1,P_2> =\sum_{k=0}^n P_1(k)P_2(k)$
Let $P \in E_{n-1}^{\perp}$ and $\alpha_0 \dots \alpha_n$ its coordinates in on the basis $(L_{n,0} \dots L_{n,n})$ we want to show that : $$\alpha_k =\frac{(-1)^{n-k}}{k! (n-k)!} \sum_{j=0}^{n}P(j) j^n$$
My attempt : Let $0 \leq k \leq n$, we can find $\lambda_k$ such that $X^n + \lambda_k L_{n,k} \in E_{n-1}$ $L_{n,k}$ has for dominant term $ (-1)^{n-k} \binom{n}{k} \frac{1}{n!} X^n$. We choose $\lambda_k= -(-1)^{n-k} \binom{n}{k} \frac{1}{n!}$ so $ X^n + \lambda_k L_{n,k}$
$ \begin{align*} nX^n &= \sum_{k=0}^{n} X^n \\ &= \sum_{k=0}^{n} ( X^n + \lambda_k L_{n,k} - \lambda_k L_{n,k} ) \\ & = \sum_{k=0}^{n} - \lambda_k L_{n,k} + \sum_{k=0}^{n} ( X^n + \lambda_k L_{n,k}) \\ \end{align*} $
We have a decomposition of $nX^n$ on $ E_{n-1} \bigoplus E_{n-1} ^ { \perp }$