On Wikipedia it is said that the error when interpolating a function $f(t)$ at $n+1$ distinct points $x_0, x_1, …, x_n$ using a polynomial $P_n(t)$ of degree $n$, the error term is given by:
$$f(t) – P_n(t) = f[x_0, …, x_n, t] \prod_{i=0}^{n} (t-x_i)$$
How would I go about proving this statement ? So far, the only thing I've figured out is that at $x_i$, the error is $0$, which satisfies what it means for a polynomial to interpolate a function at $x_i$.
Best Answer
Use Newton interpolation of degree $n+1$ of $f$ at the points $x_0,...,x_n,t$, \begin{align} P_{n+1}(x)&=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+... \\&\qquad...+f[x_0,...,x_n](x-x_0)...(x-x_{n-1})+f[x_0,...,x_n,t](x-x_0)...(x-x_{n-1})(x-x_n) \\ &=P_n(x)+f[x_0,...,x_n,t](x-x_0)...(x-x_n). \end{align} By construction, this is exact at $x=t$, resulting in $$ f(t)=P_{n+1}(t)=P_n(t)+f[x_0,...,x_n,t](t-x_0)...(t-x_n), $$ which is the claimed formula. Non of this uses derivatives, only the function values of $f$.
Of course, in a possible next step connecting the divided differences to derivatives, one would require differentiability of $f$ of a sufficiently high order.