Mean value theorem (divided differences)

In mathematical analysis, the mean value theorem for divided differences generalizes the mean value theorem to higher derivatives.

Statement of the theorem

For any n + 1 pairwise distinct points x₀, ..., x_n in the domain of an n-times differentiable function f there exists an interior point

[math]\displaystyle{ \xi \in (\min\{x_0,\dots,x_n\},\max\{x_0,\dots,x_n\}) \, }[/math]

where the nth derivative of f equals n ! times the nth divided difference at these points:

[math]\displaystyle{ f[x_0,\dots,x_n] = \frac{f^{(n)}(\xi)}{n!}. }[/math]

For n = 1, that is two function points, one obtains the simple mean value theorem.

Proof

Let [math]\displaystyle{ P }[/math] be the Lagrange interpolation polynomial for f at x₀, ..., x_n. Then it follows from the Newton form of [math]\displaystyle{ P }[/math] that the highest term of [math]\displaystyle{ P }[/math] is [math]\displaystyle{ f[x_0,\dots,x_n](x-x_{n-1})\dots(x-x_1)(x-x_0) }[/math].

Let [math]\displaystyle{ g }[/math] be the remainder of the interpolation, defined by [math]\displaystyle{ g = f - P }[/math]. Then [math]\displaystyle{ g }[/math] has [math]\displaystyle{ n+1 }[/math] zeros: x₀, ..., x_n. By applying Rolle's theorem first to [math]\displaystyle{ g }[/math], then to [math]\displaystyle{ g' }[/math], and so on until [math]\displaystyle{ g^{(n-1)} }[/math], we find that [math]\displaystyle{ g^{(n)} }[/math] has a zero [math]\displaystyle{ \xi }[/math]. This means that

[math]\displaystyle{ 0 = g^{(n)}(\xi) = f^{(n)}(\xi) - f[x_0,\dots,x_n] n! }[/math],

[math]\displaystyle{ f[x_0,\dots,x_n] = \frac{f^{(n)}(\xi)}{n!}. }[/math]

Applications

The theorem can be used to generalise the Stolarsky mean to more than two variables.