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 $x+nh$?

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  • @Stahl If $f(x)=x^2$ then $\Delta^3_h f(x) = 0$ for all $x$, I think you might be misunderstanding the definition. $\Delta^n_h f = \Delta_h \Delta^{n-1}_h f(x)$ – nullUser Sep 17 '13 at 04:40
  • You're absolutely right; looks like I'm too tired to be doing math right now! – Stahl Sep 17 '13 at 04:42

3 Answers3


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:

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

achille hui
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  • Why this is true: $\left.\Delta_h^n P(x)\right|_{x=a} = h^n P^{(n)}(a)\tag{*1}$? – Ma Joad Dec 05 '19 at 09:52
  • @Jethro When $P(x)$ is a polynomial of degree $n$, $\Delta_h P(x)$ is a polynomial of degree $n-1$, $\Delta_h^2 P(x)$ is a polynomial of degree $n-2$ and so on. In particular $\Delta_h^n P(x)$ is a polynomial of degree $0$, i.e. a constant independent of $x$. Since $P^{(n)}(x)$ is also a constant. $\Delta_h^n P(x)$ equals to $P^{(n)}(x)$ up to some constant independent of $x$ (and explicit form of $P(x)$). It just turns out that constant is $h^n$. – achille hui Dec 05 '19 at 11:14
  • O I see. Thanks. – Ma Joad Dec 05 '19 at 12:17

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 $$ \frac1{h^n}\Delta_h^n[f](x) = D^n[f](x^*). $$

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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.