Taylor's Theorem with Peano's Form of Remainder

The following form of Taylor’s Theorem with minimal hypotheses is not widely popular and goes by the name of Taylor’s Theorem with Peano’s Form of Remainder:

Taylor’s Theorem with Peano’s Form of Remainder: If $f$ is a function such that its $n^{\text{th}}$ derivative at $a$ (i.e. $f^{(n)}(a)$) exists then $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f”(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})$$ where $o(h^{n})$ represents a function $g(h)$ with $g(h)/h^{n} \to 0$ as $h \to 0$.

One of the proofs (search “Proof of Taylor’s Theorem” in this blog post) of this theorem uses repeated application of L’Hospital’s Rule. And it appears that proofs of the above theorem apart from the one via L’Hospital’s Rule are not well known. I have asked this question to get other proofs of this theorem which do not rely on L’Hospital’s Rule and instead use simpler ideas.

BTW I am also posting one proof of my own as a community wiki.

Solutions Collecting From Web of "Taylor's Theorem with Peano's Form of Remainder"

We will prove the result for $h \to 0^{+}$ and the argument for $h \to 0^{-}$ is similar. The proof is taken from my favorite book A Course of Pure Mathematics by G. H. Hardy.


Since $f^{(n)}(a)$ exists it follows that $f^{(n – 1)}(x)$ exists in some neighborhood of $a$ and $f^{(n – 2)}(x)$ is continuous in that neighborhood of $a$. Let $h \geq 0$ and we define another function $$F_{n}(h) = f(a + h) – \left\{f(a) + hf'(a) + \frac{h^{2}}{2!}f”(a) + \cdots + \frac{h^{n – 1}}{(n – 1)!}f^{(n – 1)}(a)\right\}\tag{1}$$ Then $F_{n}(h)$ and its first $(n – 1)$ derivatives vanish at $h = 0$ and $F_{n}^{(n)}(0) = f^{(n)}(a)$. Hence if we write $$G(h) = F_{n}(h) – \frac{h^{n}}{n!}\{f^{(n)}(a) – \epsilon\}\tag{2}$$ where $\epsilon > 0$, then we have $$G(0) = 0, G'(0) = 0, \ldots, G^{(n – 1)}(0) = 0, G^{(n)}(0) = \epsilon > 0\tag{3}$$ Since $G^{(n)}(0) > 0$ it follows that there is a number $\delta_{1} > 0$ such that $G^{(n – 1)}(h) > 0$ for all values of $h$ with $0 < h < \delta_{1}$. Using mean value theorem and noting that $G^{(n – 1)}(0) = 0$ we can see that $G^{(n – 2)}(h) > 0$ for all $h$ with $0 < h < \delta_{1}$. Applying the same argument repeatedly we can see that $G(h) > 0$ for all $h$ wih $0 < h < \delta_{1}$. Thus $$F_{n}(h) > \frac{h^{n}}{n!}\{f^{(n)}(a) – \epsilon\}\tag{4}$$ for $0 < h < \delta_{1}$. Similarly we can prove that $$F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{5}$$ for all $h$ with $0 < h < \delta_{2}$.

Thus for every $\epsilon > 0$ there is a $\delta = \min(\delta_{1}, \delta_{2}) > 0$ such that $$\frac{h^{n}}{n!}\{f^{(n)}(a) – \epsilon\} < F_{n}(h) < \frac{h^{n}}{n!}\{f^{(n)}(a) + \epsilon\}\tag{6}$$ for all values of $h$ with $0 < h < \delta$. This proves the theorem for $h \to 0^{+}$.

Slight care has to be taken when dealing with negative values of $h$ for the case $h \to 0^{-}$ because here the nature of inequalities will depend on whether $n$ is odd or even and thus we need to handle both the cases of even $n$ and odd $n$ separately.