This paper presents an elementary proof of Lagrange's form of the remainder for the Taylor series of a real function.

The following is proved.

Proposition. Suppose $f$ has an $(n+1)^\text{th}$ derivative in $[a,b]$ and that $m\leq f^{(n+1)}(x)\leq M$ for all $a<x<b$. Then for all $a\leq x\leq b$ we have $$\frac{m}{(n+1)!}(x-a)^{n+1}\leq R_nf(x)\leq \frac{m}{(n+1)!}(x-a)^{n+1}.$$

The authors then proceed to formulate and prove the Lagrange form as follows.

Theorem. Suppose $f$ has an $(n+1)^\text{th}$ derivative in $[a,b]$. Then there is some $a\leq c\leq b$ such that $R_nf(b)=\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}$.

Proof. Choose $m=\inf_{[a,b]}f^{(n+1)},M=\sup_{[a,b]}f^{(n+1)}$ (if $f^{(n+1)}$ is unbounded we allow $m,M=\pm \infty$). Thus by the proposition, $R_nf(b)=\frac{k}{(n+1)!}(b-a)^{n+1}$ for some $m\leq k\leq M$. If one of the equalities holds, then the result is immediate from the proposition. Otherwise it follows directly from Darboux's intermediate value theorem.

If neither equality holds, i.e $m<k<M$, then I see Darboux's theorem takes care of things. However, I don't understand the preceding sentence.

Question. If one of the equalities holds, i.e $k\in \left\{ m,M \right\}$, then why is the assertion immediate from the proposition?

At first I thought perhaps a bounded derivative attains its extrema, but this answer gives a counterexample.

  • 1
    $\begingroup$ I don't see how to deduce it immediately from the proposition, but if you are looking for a simple proof not necessarily directly from the proposition, it's easy to see that $k \in \{m,M\}$ implies that $f^{(n+1)}$ is constant. $\endgroup$ Dec 13, 2019 at 16:06
  • $\begingroup$ @DanielFischer why does $k \in \{m,M\}$ imply constancy? $\endgroup$
    – Arrow
    Dec 13, 2019 at 16:22
  • $\begingroup$ Consider $$g(x) = f(x) - \sum_{r = 0}^n \frac{f^{(r)}(a)}{r!}(x-a)^r - \frac{k}{(n+1)!}(x-a)^{n+1}\,.$$ Then $g^{(r)}(a) = 0$ for $0 \leqslant r \leqslant n$, and $g(b) = 0$. Also, since $k \in \{m,M\}$ we either have $g^{(n+1)}(x) \geqslant 0$ for all $x \in (a,b)$ or $g^{(n+1)}(x) \leqslant 0$ for all $x \in (a,b)$. But then all derivatives of lower order would satisfy the same, and that works only if they all identically vanish. $\endgroup$ Dec 13, 2019 at 16:27

1 Answer 1


In the paper there is a 4th lemma that shows that $k$ is $M$ (or $m$) iff the $(n+1)$-th derivative is constant.


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