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In mathematical analysis, the mean value theorem for divided differences generalizes the mean value theorem to higher derivatives.^[1]

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

ξ∈(min{x0,…,xn},max{x0,…,xn}){displaystyle xi in (min{x_{0},dots ,x_{n}},max{x_{0},dots ,x_{n}}),}

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

f[x0,…,xn]=f(n)(ξ)n!.{displaystyle f[x_{0},dots ,x_{n}]={frac {f^{(n)}(xi )}{n!}}.}

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

Proof

Let P{displaystyle P} be the Lagrange interpolation polynomial for f at x₀, ..., x_n.
Then it follows from the Newton form of P{displaystyle P} that the highest term of P{displaystyle P} is f[x0,…,xn](x−xn−1)…(x−x1)(x−x0){displaystyle f[x_{0},dots ,x_{n}](x-x_{n-1})dots (x-x_{1})(x-x_{0})}.

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

0=g(n)(ξ)=f(n)(ξ)−f[x0,…,xn]n!{displaystyle 0=g^{(n)}(xi )=f^{(n)}(xi )-f[x_{0},dots ,x_{n}]n!},

f[x0,…,xn]=f(n)(ξ)n!.{displaystyle f[x_{0},dots ,x_{n}]={frac {f^{(n)}(xi )}{n!}}.}

Applications

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

References