Proving $∆^nf(x_0;h_1,\cdots,h_n)=f^{(n)}(ξ)h_1\cdots h_n$

Let's define finite differences of order $$n$$ in $$x_0$$ for a function $$f$$ as: $$\Delta ^1 f(x_0;h_1)= f(x_0+h_1)-f(x_0)$$ \begin{align*}\Delta ^2 f(x_0;h_1,h_2)&= \Delta f^1 (x_0+h_2;h_1)-\Delta f(x_0;h_1)\\ &=f(x_0+h_2+h_1)-f(x_0+h_2)-\left [ f(x_0+h_1)-f(x_0) \right ]\end{align*} $$\vdots$$ $$\Delta ^n f(x_0;h_1,\cdots,h_n)= \Delta f^{n-1} (x_0+h_n;h_1,\cdots,h_{n-1})-\Delta^{n-1} f(x_0;h_1,\cdots,h_{n-1})$$

Let's suppose now that $$f\in\mathcal{C}^{n-1}([a,b])$$ and $$\exists f^{(n)}$$ in $$(a,b)$$. I would like to show that if $$x_0$$, $$x_0+h_1$$, $$x_0+h_2$$, $$x_0+h_1+h_2$$, $$\cdots$$, $$x_0+h_1+\cdots+h_n$$ are always inside $$[a,b]$$), then there exists a point $$\xi$$ inside the smallest interval containing all of them such that $$\Delta^nf(x_0;h_1,\cdots,h_n)=f^{(n)}(\xi)h_1\cdots h_n$$ I tried by induction but I can't.

Are there other ways that doesn't use the concept of function of more than one variable? Thanks.

EDIT: By induction I succeeded in proving only that $$\Delta^nf(x_0;h_1,\cdots,h_n)\leq f^{(n)}(\xi)h_1\cdots h_n$$

• Did you mean $C^{n}$ instead of $C^{n-1}$? – parsiad Mar 1 at 16:21
• No, $n-1$. I just edited the answer to correct a mistake. – Nameless Mar 1 at 16:23
• @parsiad how would you show the statement if the hypothesis were what you say? – Nameless Mar 1 at 20:14

$$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$$Lemma: For any differentiable function $$g$$ and any $$0 \leqslant m \leqslant n$$,$$\frac{\d}{\d x}(∆^m g(x; h_1, \cdots, h_m)) = ∆^m g'(x; h_1, \cdots, h_m).$$

Proof: It will be proved by induction. For $$m = 0$$, there is $$\dfrac{\d}{\d x}(∆^0 g(x)) = \dfrac{\d}{\d x}(g(x)) = g'(x) = ∆^0 g'(x)$$. Now assume that it holds for $$m$$, then by the induction hypothesis,\begin{align*} &\peq \frac{\d}{\d x}(∆^{m + 1}g(x; h_1, \cdots, h_{m + 1})) = \frac{\d}{\d x}(∆^m g(x + h_{m + 1}; h_1, \cdots, h_m) - ∆^m g(x; h_1, \cdots, h_m))\\ &= ∆^m g'(x + h_{m + 1}; h_1, \cdots, h_m) - ∆^m g'(x; h_1, \cdots, h_m) = ∆^{m + 1} g'(x; h_1, \cdots, h_{m + 1}). \end{align*} End of induction.

Now back to the problem. For $$1 \leqslant k \leqslant n$$, define$$F_k(x) = ∆^k f^{(n - k)}(x; h_1, \cdots, h_k),\quad G_k(x) = ∆^{k - 1} f^{(n - k)}(x; h_1, \cdots, h_{k - 1}),$$ and also define $$F_0(x) = f^{(n)}(x)$$, then the lemma implies that $$G_k' = F_{k - 1}$$.

Next it will be proved by induction on $$n \geqslant m \geqslant 0$$ that $$F_n(x_0) = F_m(ξ_m) \prod\limits_{k = m + 1}^n h_k$$ for some $$ξ_m$$. For $$m = n$$, it suffices to take $$ξ_n = x_0$$. Assume that it holds for $$m$$ ($$m \geqslant 1$$). The mean value theorem implies that there exists $$ξ_{m - 1}$$ satisfying$$F_m(ξ_m) = G_m(ξ_m + h_m) - G_m(ξ_m) = G_m'(ξ_{m - 1}) h_m = F_{m - 1}(ξ_{m - 1}) h_m,$$ thus $$F_n(x_0) = F_m(ξ_m) \prod\limits_{k = m + 1}^n h_k = F_{m - 1}(ξ_{m - 1}) \prod\limits_{k = m}^n h_k$$. End of induction.

Finally, taking $$m = 0$$ yields$$∆^n f(x_0; h_1, \cdots, h_n) = F_n(x_0) = F_0(ξ_0) \prod_{k = 1}^n h_k = f^{(n)}(ξ_0) \prod_{k = 1}^n h_k.$$

• $$F_m(ξ_m) = G_m(ξ_m + h_m) - G_m(ξ_m) = G_m'(ξ_{m - 1}) h_m = F_{m - 1}(ξ_{m - 1}) h_m,$$ Here, $G_m'$ is w.r.t. $x$ (the first point), but $ξ_m$ depends on $x$, so I have a composite derivative, no? – Nameless Mar 19 at 22:36
• @Nameless No, $G_m'$ is with respect to its independent variable, be it $G_m'(x)$, $G_m'(y)$ or $G_m'(ξ)$. That's why it's not written as $\dfrac{\mathrm d}{\mathrm dx}(G_m(ξ_m+h_m))$ here. – Saad Mar 20 at 0:29
• Okay, I see. Thanks. – Nameless Mar 20 at 21:14

I had an idea that was longer than comment. But not a complete solution. I felt it may be helpful to you so I’ve provided it below:

This is essentially applying the mean value theorem with induction.

Let’s define the “divided difference” of a function as

$$D_h(f) = \frac{f(x+h)-f(x)}{h}$$

If we consider the interval $$[x_0, x_0+h]$$ (and the order here might be wrong since h could be negative, so we should really say “minimum interval containing both”) and let $$f$$ be differentiable over this interval then the mean value theorem applies and we have that there exists a $$\eta \in [x_0, x_0+h]$$ such that

$$f’(\eta) = D_h(f)(x_0) = \frac{f(x_0 + h)-f(x_0)}{h}$$ Or in your notation (and you’ll see why we are doing this)

$$h f’(\eta) = f(x_0 + h)-f(x_0) = \Delta f(x_0 : h)$$

You can go further and say that there is some function $$\eta(x)$$ such that:

$$h f’(\eta(x)) = \Delta f(x:h)$$

And we have that $$\eta(x)$$ is always contained in the smallest interval containing $$x$$,$$x+h$$

So now the to continue building our intuition let’s Say we have some g(x) = $$\Delta f(x: h)$$. It is the case that over the minimum interval containing $$x_0, x_0+h, x_0+h_2$$ that there is some $$c$$ in the interval such that

$$\Delta(g:h_2)(x_0) = h_2 g’(c)$$

Filling in the definition of g we have that:

$$\Delta(f: h_1 : h_2)(x_0) = h_2 ( \Delta f(x: h))’(c)$$

$$\Delta(f: h_1 : h_2)(x_0) = h_2 h_1 (f’(\eta(x)))’(c)$$

Which means:

$$\Delta(f: h_1 : h_2)(x_0) = h_2 h_1 f’’(\eta(c)) \eta’(c)$$

Now the problem is that $$\eta’(c)$$ at the very end. There should be some way to absorb it by showing that $$f’’(\eta(x)) \eta’(x) = f’’(m)$$ for some suitable choice of m, at which point you can then use induction on the strategy we have demonstrated to cover the arbitrary case.

• Thank you too @frogeyedpeas (+1). – Nameless Mar 20 at 21:16