Proof of the divided-difference mean value lemma

Expert intro: This lemma is a standard bridge between interpolation theory and repeated applications of Rolle’s theorem.

Detailed walkthrough

What the lemma says

The statement connects a higher-order divided difference to a higher derivative at some interior point. In practical terms, it says that the divided difference of $f$ at $k+1$ distinct points behaves like a scaled $k$th derivative somewhere inside the interval.

Formally, if $f\in C^k[a,b]$ and $x_0,\ldots,x_k$ are distinct points in $[a,b]$, then there exists $\xi\in(a,b)$ such that

$f[x_0,x_1,\ldots,x_k]=\frac{f^{(k)}(\xi)}{k!}.$

Interpolation polynomial and the error function

Let $P_k(x)$ be the interpolating polynomial of degree $<k$ satisfying

$P_k(x_i)=f(x_i),\qquad i=0,1,\ldots,k.$

Define the interpolation error

$e_k(x)=f(x)-P_k(x).$

Because $P_k$ matches $f$ at each node $x_0,\ldots,x_k$, the error has at least $k+1$ distinct zeros:

$e_k(x_0)=e_k(x_1)=\cdots=e_k(x_k)=0.$

This is the key setup, since repeated zeros allow repeated use of Rolle’s theorem.

Repeated application of Rolle’s theorem

Since $e_k$ is differentiable on $(a,b)$, Rolle’s theorem implies that between each pair of adjacent zeros of $e_k$ there is at least one zero of $e_k'$. With $k+1$ zeros, this gives at least $k$ zeros of $e_k'$. Applying the same argument to $e_k'$ gives at least $k-1$ zeros of $e_k''$, and continuing in this way yields at least one zero of $e_k^{(k)}$.

So there exists $\xi\in(a,b)$ such that

$e_k^{(k)}(\xi)=0.$

Since $e_k=f-P_k$,

$0=e_k^{(k)}(\xi)=f^{(k)}(\xi)-P_k^{(k)}(\xi).$

Therefore

$f^{(k)}(\xi)=P_k^{(k)}(\xi).$

Why the derivative of $P_k$ equals the divided difference term

A standard Newton form of the interpolating polynomial is

$P_k(x)=\sum_{i=0}^k f[x_0,\ldots,x_i]\prod_{j=0}^{i-1}(x-x_j).$

The highest-degree term comes from $i=k$:

$f[x_0,\ldots,x_k]x^k+\text{terms of degree }<k.$

When you take the $k$th derivative, every term of degree less than $k$ vanishes, and the leading term contributes

$P_k^{(k)}(x)=k!\,f[x_0,\ldots,x_k].$

Substituting this into the earlier identity gives

$f^{(k)}(\xi)=k!\,f[x_0,\ldots,x_k],$

which rearranges to the desired result.

Final conclusion

The proof works because interpolation creates an error function with many zeros, and Rolle’s theorem converts those zeros into a zero of the $k$th derivative. The Newton form then identifies the $k$th derivative of the interpolating polynomial with the divided difference coefficient.

That is the exact mechanism behind the lemma, and it is also why divided differences are so useful in error estimates for polynomial interpolation.

💡 Pitfall guide

A frequent mistake is to stop after applying Rolle’s theorem once and assume the result is already enough. The argument needs to be repeated $k$ times to reach $e_k^{(k)}$. Another common issue is confusing the degree of the interpolating polynomial in the proof. Because the polynomial interpolates at $k+1$ points, its degree is at most $k$, so the $k$th derivative is the first derivative order that isolates the leading divided-difference term. Skipping that observation makes the final step look mysterious.

🔄 Real-world variant

A useful variation is the case $k=1$. Then the lemma reduces to the familiar mean value theorem in divided-difference form: for distinct $x_0,x_1$, there exists $\xi$ such that

$f[x_0,x_1]=f'(\xi).$

If you increase to $k=2$, the statement becomes a second-order analogue involving the quadratic interpolant and $f''(\xi)/2$. These low-order cases help show that the lemma is a direct extension of standard mean value ideas.

🔍 Related terms

divided differences, Rolle's theorem, Newton interpolation