Every polynomial of degree $\le m$ can be written in a form using its $m+1$ values I'm not sure if the title is comprehensible. What I mean is this:
I've found here the following corollary:
Let $x_1, ..., x_{m+1}$ be $m+1$ different points in $\mathbb{R}$, and for $i = 1,2,...,m+1$ let $g_i$ be the unique polynomial of degree at most $m$ with $g_i(x_j)= \delta_{ij}$, where $\delta_{ij}=0$ for $i \neq j$ and $\delta_{ii}=1$.
Then every polynomial $f$ of degree at most $m$ is of the form $f(x)= \sum _{i=1} ^{m+1} f(x_i)g_i(x)$ for each $x \in \mathbb{R}$.
Could you show me how to prove that?
 A: This is Lagrange interpolation, which is a special case of the Chinese Remainder Theorem.
A: The Lagrange polynomials $(g_1,\ldots,g_{m+1})$ where
$$g_i(x)=\prod_{j=1,j\not=i}^{m+1}\frac{x-x_j}{x_j-x_i}$$
is a basis of $\mathbb{R}_m[x]$
and  we verify easly that $g_i(x_j)=\delta_{ij}$ and 
then every polynomial $f$ of degree at most $m$ is of the form $f(x)= \sum _{i=1} ^{m+1} f(x_i)g_i(x)$ for each $x \in \mathbb{R}$.
A: If you assume the lemma given in your link, then you know that, given $m+1$ distinct values $x_1,...,x_{m+1}$ and values $y_1,...,y_{m+1}$, there is exactly one polynomial $h$ of degree $\le m$ such that $h(x_i) = y_i$ for all $i$.
Now set $y_i$ to be $f(x_i)$ and note that $f(x)$ and $\sum_{i=1}^{m+1} f(x_i) g_i(x)$ are two polynomials of degree at most $m$ which map the $x_i$ to $y_i$. The uniqueness in the lemma implies $$f(x) = \sum_{i=1}^{m+1} f(x_i) g_i(x).$$
A: Every polynom $f$ can be expressed by its convolution. If this polynom is a linear combination of another polynoms, its convolutionis a same combination of their convolutions and vice versa. Here you have polynoms with independent convolution vectors: $(1, 0, ..., 0)$, ..., $(0, 0, ..., 1)$. So the convolution of f, which is $(f(x_1), f(x_2), f(x_3), ...)$, can be expressed as linear combination of them with coefficients $f(x_1), f(x_2), ...~$. Then $f$ is the combination of $g_i$ with same coefficients.
A: Two polynomials of degree $m$ that agree on $m + 1$ points are identical (just set up the $m + 1$ equations for the $m + 1$ differences of coefficients, you'll see they only have the solution all zeros, see Vandermonde's determinant). Then the Lagrange interpolation polynomial:
$$
p(x) = \sum_{0 \le r \le m}
         y_r \prod_{\substack{0 \le s \le m \\ s \ne r}}
               \frac{x - x_s}{x_r - x_s}
$$
has each $y_r$ multiplied by a factor that is 1 for $x = x_r$ and 0 if $x = x_s$ for $s \ne r$. It is of degree $m$, and it agrees at $m + 1$ points.
