Derivative of a function defined by the divided difference of another function.

Question

Given a function $f$ of class $C$ $^{n+2}$ in an interval $[a,b]$ and $x_{0}=a<x_1<x_2 ... <x_n = b$ a subdivision of $[a,b]$ into $n+1$ points. Given another function $g$ defined in the same interval $[a,b]$ by the divided difference such that $g(x) = f[x_0, x_1, ... , x_n, x]$. Prove that $g'(x)=f[x_0, x_1, ... , x_n, x, x]$.

Answer 1

The divided difference is the same whatever the arrangement of the inputs i.e.

$f[x_0, x_1, ... , x_n, x, x]=f[x,x_0, x_1, ... , x_n, x]$

and $f[x_0, x_1, ... , x_n, x]=f[x,x_0, x_1, ... , x_n]=g(x)$

$g[u,v]=\cfrac{g(u)-g(v)}{u-v}$

Also $\displaystyle \lim_{(u,v)->(x,x)}$$g[u,v]=g'(x)$

we have $g(u)=f[u,x_0, x_1, ... , x_n]$ and $g(v)=f[x_0, x_1, ... , x_n, v]$, then it follows that $g'(x)=\displaystyle\lim_{(u,v)->(x,x)}g[u,v]=\lim_{(u,v)->(x,x)}\cfrac{f[u,x_0, x_1, ... , x_n]-f[x_0, x_1, ... , x_n, v]}{u-v}$

since $\cfrac{f[u,x_0, x_1, ... , x_n]-f[x_0, x_1, ... , x_n, v]}{u-v}$$=f[u,x_0, x_1, ... , x_n, v ]=f[x_0, x_1, ... , x_n, u,v ]$

$\displaystyle g'(x)=\lim_{(u,v)->(x,x)}f[x_0, x_1, ... , x_n, u,v ]=f[x_0, x_1, ... , x_n, x, x]$