TITLE: Is there a mean value theorem for higher order differences?
QUESTION [9 upvotes]: The standard mean value theorem tells us $\frac{f(x+h)-f(x)}{h} = f'(c)$ for some $c$ between $x$ and $x+h$. Rewriting this, we may see it as $\frac 1h\Delta_h f(x) = f'(c)$. This makes me wonder if there is a similar formula for higher order differences.
Here $$\Delta_h f(x) =f(x+h)-f(x)$$
Can I say something like $$\frac{1}{h^n}\Delta_h^nf(x) = f^{(n)}(c)$$ for some $c$ between $x$ and $x+nh$?

REPLY [4 votes]: I refer to Wikipedia for notation.
I will give you an exact representation formula for finite differences, in the same spirit of the Taylor theorem with remainder in integral form.
The key point is that finite differences and derivatives commute: $D\Delta_h=\Delta_hD$.
For $f\in C^1(\mathbb R)$ you can compute
$$
\frac1h\Delta_h[f](x) = \frac{f(x+h)-f(x)}h
= \frac1h \int_0^h D[f](x+x_1) \,dx_1
$$
For $f\in C^n(\mathbb R)$, iterating the above formula, you get
$$
\begin{split}
\frac1{h^n}\Delta_h^n[f](x)
&= \frac{\frac1{h^{n-1}}\Delta_h^{n-1}[f](x+h)
 - \frac1{h^{n-1}}\Delta_h^{n-1}[f](x)} {h} \\
&= \frac1h \int_0^h \frac1{h^{n-1}}D\Delta_h^{n-1}[f](x+x_1) \,dx_1 \\
&= \frac1{h^2} \int_0^h \int_0^h
 \frac1{h^{n-1}} D^2\Delta_h^{n-2}[f](x+x_1+x_2) \,dx_1dx_2 \\
&= \,\cdots \\
&= \frac1{h^n} \int_0^h\dotsi\int_0^h D^n[f](x+x_1+\dotsb+x_n) \,dx_1\dotsm dx_n .
\end{split}
$$
The last integral is a weighted average of $D^n[f]$ over the segment $[x,x+nh]$. More precisely, let $X_1,\dotsc,X_n\sim\mathrm{Uniform}(0,h)$ be i.i.d. and $S_n=X_1+\dotsb+X_n$. Then the previous expression can be viewed as
$$
\frac1{h^n}\Delta_h^n[f](x)
= \mathbb{E}\left[D^n[f](x+S_n)\right].
$$
This can be also written as
$$
\frac1{h^n}\Delta_h^n[f](x) = \int_0^{nh} D^n[f](x+t) \sigma_n(t) \,dt
$$
where $\sigma_n$ is the probability density function of $S_n$, which is a rescaling of the Irwin–Hall distribution (refer to here for a beautiful derivation of the PDF).
The mean value theorem for integrals applies and tells us that there exists $x^*\in(x,x+nh)$ such that
$$
\frac1{h^n}\Delta_h^n[f](x) = D^n[f](x^*).
$$