The polynomial of degree $n$ taking the values
\begin{equation}
\label{eq:27.13.22}
y_0(x=0),y_1(x=1),\cdots,y_n(x=n)
\end{equation}
is given by the formula
\begin{equation}
\label{eq:27.13.24}
f(x)=y_0+\frac{x}{1}\Delta y_0+\frac{x(x-1)}{1\cdot
2}\Delta^2y_0+\cdots+\frac{x(x-1)\cdots(x-n+1)}{1\cdot
2\cdots\cdot n}\Delta^ny_0
\end{equation}
Proof:First,I want to prove that $\forall 0\leq i\leq n,i\in\mathbf{N}$,
\begin{equation}
\label{eq:27.15.32}
y_i=C_i^0 y_0+ C_i^1\Delta y_0+C_i^2\Delta^2y_0+\cdots+ C_i^{i}\Delta ^iy_0
\end{equation}
According to this scheme,
this is obvious(why?hint:细心观察会发现这是一个倒置的杨辉三角1;1,2,1;1,3,3,1;……).So $\forall 0\leq i\leq n$,when $x=i$,\ref{eq:27.13.24}hold.This means that
\begin{equation}
\label{eq:27.20.24}
f(0)=y_0, f(1)=y_1,f(2)=y_2,\cdots,f(n)=y_n
\end{equation}
According to Gauss's elimation method(How?),every coefficient of this polynomial has been determined(If exists).
注:该结果只是牛顿多项式插值的一个特例.