Taylor's Theorem with Peano's Form of Remainder

Question

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.

Answer 1

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.

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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.

Answer 2

A slightly more efficient version of Hardy's argument, which at the same time proves Taylor's theorem with the Lagrange form of the remainder goes as follows:

Assume that $f$ is $n$ times differentiable at $a \in \mathbb R$ (so that it is $n - 1$ times differentiable on some neighborhood of $a$). Then consider, for $C$ any constant, $$ G(t) = f(a+t)-\sum_{j=0}^{n-1}\frac{1}{j!}f^{(j)}(a)t^k - \frac{1}{n!}Ct^n $$ Note that for each $k$, $0\leq k \leq n-1$ we have $$ G^{(k)}(t) = f^{(k)}(a+t) - \sum_{j=k}^{n-1}\frac{1}{(j-k)!}f^{(j)}(a)t^{j-k}- \frac{1}{(n-k)!}Ct^{n-k} $$ and so in particular, $G^{(k)}(0)=0$ for all $k\in \{0,1,\ldots,n-1\}$.

Now for any $h\neq 0$, we may choose the constant $C$ so that $G(h)=0$ -- that is, we set $$C(h):=\frac{n!}{h^n}\left(f(a+h)-\sum_{j=0}^{n-1}\frac{1}{j!}f^{(j)}(a)h^j\right)$$ Then for this choice of $C$ we have $G(0)=G(h)$, so that there is some $h_1$ in the open interval between $0$ and $h$ with $G'(h_1)=0$. Since $G'(0)=0$, this implies there is some $h_2$ between $0$ and $h_1$ with $G''(h_2)=0$. Continuing in this way for each $k=1,2,\ldots,n-1$ we find that there is some $h_{n-1}$ lying between $0$ and $h$ with $G^{(n-1)}(h_{n-1})=0$.

However, $$G^{(n-1)}(t) = f^{(n-1)}(a+t) - f^{(n-1)}(a)-C(h)t,$$ and hence, $$0 = G^{(n-1)}(h_n) = f^{(n-1)}(a+h_n) - f^{(n-1)}(a)-C(h)h_n,$$ meaning that $$C(h)=h_{n-1}^{-1}\left(f^{(n-1)}(a+h_{n-1})-f^{(n-1)}(a)\right).$$ Now if $f$ is $n$-times differentiable at $a$, then since $h_{n-1}$ lies between $0$ and $h$, $\lim_{h\to 0} C(h) = f^{(n)}(a)$. But now recall that $C(h)$ was chosen so that $G(h)=0$, hence $$ f(a+h) = \sum_{j=0}^{n-1} \frac{1}{j!}f^{(j)}(a)h^k +\frac{1}{n!}C(h)h^n $$ and we have just shown that $C(h)\to f^{(n)}(a)$ as $h \to 0$, hence $$f(a+h)- \sum_{j=0}^n \frac{1}{j!}f^{(j)}(a)h^j = \frac{h^n}{n!}\left(C(h) - f^{(n)}(a)\right) = o(h^n),$$ as required.

Note that if one knows that $f$ is $n$-times differentiable in a neighbourhood of $a$, then the fact that one has $G^{(n-1)}(0) = G^{(n-1)}(h_{n-1})$ implies that there is some $h_n$ between $0$ and $h_{n-1}$ with $G^{(n)}(h_n)=0$. Since $G^{(n)}(t) = f^{(n)}(a+t)-C(h)$ it follows that $C(h)= f^{(n)}(a + h_n)$, and one thus obtains Taylor's theorem with the Lagrange form of the remainder.

Answer 3

The following argument is for $n=2$, but it can be extended to higher derivatives without much pain. Suppose $f:(\mathbb R\supseteq )D\to\mathbb R$ is twice differentiable at $a\in D$ (open), then $f'$ exists in a neighborhood (without loss of generality call it $D$), it is thus gauge (or Kurzweil-Henstock) integrable (see Lamoreaux & Armstrong (1998) for an undergraduate level discussion) and satisfies \begin{equation} f(a+h)=f(a)+\int_0^1 f'(a+th) \operatorname dt\,h. \end{equation} Differentiability of $f'$ at $a$ is equivalent to say for all $k:a+k\in D$ we have \begin{equation} f'(a+k)=f'(a)+f''(a)k+\hat g(k) \text{ where } \frac{\hat g(k)}k=:\bar g(k)\to0 \text{ as }k\to0. \end{equation} Note that since $f'(a+k)$, $f'(a)$ and $f''(a)k$ are gauge integrable in $k$ (the first by the cited article and the latter two are Riemann integrable) also $\hat g(k)$ is gauge integrable in $k$. Taking $k=ht$ in the first equation we see \begin{equation} \begin{split} f(a+h)-f(a) &= \int_0^1f'(a)+f''(a)th+\hat g(th)\operatorname dt\,h \\ &= f'(a)h+(\smallint_0^1t\operatorname dt)f''(a)h^2+\int_0^1\hat g(th)\operatorname dt\,h \\ &= f'(a)h+\frac12f''(a)h^2+g(h) \end{split} \end{equation} where \begin{equation} g(h):= h\int_0^1\hat g(th)\operatorname dt =\int_0^h\hat g(k)\operatorname dk =\int_0^h\bar g(k)k\operatorname dk \end{equation} and $\bar g(k)\to0$ as $k\to0$. Taylor's theorem (as stated in the question) will follow from \begin{equation} \frac{g(h)}{h^2}\to0 \text{ as }h\to0. \end{equation} To see this, suppose $\epsilon>0$ and let $\delta>0$ such that $|\bar g(k)|<2\epsilon$ if $|k|<\delta$, and use monotonicity for the gauge integral (summarised in Heikkilä (2011)) to get \begin{equation} g(h)\leq\frac{h^2}\epsilon \text{ if } |h|<\delta. \end{equation}

In the general case (including this one) all we need is the gauge integrability of $f^{(n-1)}$ in a neighborhood of $a$, but this is guaranteed by the fact that $f^{(n-1)}$ is the derivative of $f^{(n-2)}$ in a whole such neighborhood.

The concept of gauge integral may seem sophisticated, but in fact it is quite elementary and many people teach it to second and even first year undergrads. A letter underwritten by many analysis researchers and teachers, calling for a review of the "standard" calculus curriculum, has been around for some years.