Is there a mean value theorem for higher order differences?

Question

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 $x+nh$?

Answer 1

Given a function $f(x)$ defined on $[a, a+nh]$ continuously differentiable up to $n$ times.
Let $P(x)$ be a polynomial of degree $n$ coincides with $f(x)$ on the $n+1$ points $a, a+h, \ldots, a+nh$. It is clear

$$\left. \Delta_h^n ( f(x) - P(x) )\right|_{x=a} =\sum_{k=0}^n (-1)^{n-k} \binom{n}{k}( f(a+kh)-P(a+kh) ) = 0 $$ This implies $$\left.\Delta_h^n f(x)\right|_{x=a} = \left.\Delta_h^n P(x)\right|_{x=a} = h^n P^{(n)}(a)\tag{*1}$$

On the other hand, $f(x) - P(x)$ vanishes on $n+1$ points $a, a+h, a+2h, \ldots, a+nh$.
By Rolle's theorem, we can find $n$ points $x_1, x_2, \ldots, x_n$:

$$a < x_1 < a+h < x_2 < a+2h < \cdots a + (n-1)h < x_n < a+nh$$

such that $f'(x) - P'(x)$ vanishes on these $x_i$. Repeat applying Rolle's $(n-1)$ more times, we can conclude there is a $b \in (a, a+nh)$ such that

$$f^{(n)}(b) - P^{(n)}(b) = 0 \quad\iff\quad f^{(n)}(b) = P^{(n)}(b)\tag{*2}$$

Since $P(x)$ is a polynomial of degree $n$, $P^{(n)}(x)$ is a constant. This means $P^{(n)}(a) = P^{(n)}(b)$ and by combining $(*1)$ and $(*2)$, we get:

$$\frac{1}{h^n}\left.\Delta_h^n f(x)\right|_{x=a} = f^{(n)}(b)\quad\text{ for some }b \in (a, a + nh)$$

Answer 2

A quick idea - we can say the following using Taylor's theorem

$$\frac{1}{h^{2}}\Delta_{h}^{2}f(x) = 2f''(d) - f''(c)$$

where $c$ lies between $x$ and $x+h$ and $d$ lies between $x$ and $x + 2h$. I'm not sure if we can improve this and I'd be interested to find out.

Details of the above.

First expand about $x$ with a step size of $h$ and $2h$ using Taylor.

$$f(x+h) = f(x) + hf'(x) + \frac{f''(c)}{2!}h^{2}$$ $$f(x+2h) = f(x) + 2hf'(x) + \frac{f''(d)}{2!}(2h)^{2}$$

Subtract the second equation from twice the first. Rearrange terms and rewrite using the $\Delta_{h}$ notation.