TITLE: Is there a specialized formula for Lagrangian interpolation on equispaced points?
QUESTION [9 upvotes]: If we know $f(0),f(1),f(2),\cdots f(n)$, is there a specialized version of the Lagrangian interpolation formula and a shortcut to compute the coefficients ?
(Stability is not a concern.)

REPLY [5 votes]: For the $n+1$ points
$\{\left(x_0,f(x_0)\right),...,\left(x_n,f(x_n)\right)\}$
the Lagrange interpolating polynomial is defined as
$$P(x)=\sum_{j=0}^nP_j(x)=\sum_{j=0}^n f(x_j)\prod_{k=0\\{k\ne j}}^{n}\frac{x-x_k}{x_j-x_k}$$
Now since the $x_j=j,\; (j=0,1,...,n)$ the product can be simplified.
Let $p_n(x):=x(x-1)\cdots(x-n)$ then
$$\begin{align}\prod_{k=0\\{k\ne j}}^{n}\frac{x-x_k}{x_j-x_k}&=\prod_{k=0\\{k\ne j}}^{n}\frac{x-k}{j-k}
\\&=\frac{p_n(x)}{x-j}\prod_{k=0\\{k\ne j}}^{n}\frac 1{j-k}\\
&=\frac{p_n(x)}{x-j}\cdot\frac 1{j!(-1)^{n-j}(n-j)!}\\
&=\frac{p_n(x)}{n!}{{n}\choose{j}}\frac{(-1)^{n-j}}{x-j}
\end{align}$$
Therefore
$$P(x)=\frac{p_n(x)}{n!}\sum_{j=0}^n(-1)^{n-j}{{n}\choose{j}}\frac{f(j)}{x-j}$$