A proof of Taylor's Peano Remainder using little o notation

Solution 1:

Let's state more or less the statement you want to prove:

Suppose $f$ is $n-1$--times differentiable in a neighborhood of a point $a$, and that $f^{(n-1)}$ is differentiable at $a$, then $$r(x):=f(x)-\sum^n_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k = o\big((x-a)^n\big)$$

-Notice that no continuity on the $n$--th derivative is assumed and that only two things are required: (1) all derivatives up to order $n-1$ in an interval around $a$ exists, (2) the $n$--th derivative at $a$ exists.

Proof: Set $g(x)=(x-a)^n$. Notice that

• $r$ and $g$ are $n$--times differentiable at $a$,
• $r^{(k)}(a)=0$ for all $0\leq k\leq n$,
• and $g^{(k)}(x)\neq0$ for all $x\neq a$ and $0\leq k\leq n$.

Recall the Cauchy mean value theorem for two differentiable functions that states that for any differentiable functions $F$ and $G$ in an interval $[\alpha,\beta]$, there is a point $\alpha<\xi<\beta$ such that $$ G'(\xi)(F(\beta)-F(\alpha))=F'(\xi)(G(\beta)-G(\alpha))$$

Using this theorem repeatedly, we obtain that there are points $\xi_1,\ldots,\xi_n$ such that $\xi_k$ is between $a$ and $\xi_{k-1}$ ( define $\xi_0=x$), such that \begin{aligned} \frac{r(x)}{g(x)}&=\frac{r(x)-r(a)}{g(x)-g(a)}=\frac{r'(\xi_1)}{g'(\xi_1)}=\frac{r'(\xi_1)-r'(a)}{g'(\xi_1)-g'(a)} =\frac{r''(\xi_2)}{g''(\xi_2)}\\ &=\ldots=\frac{r^{(n-1)}(\xi_{n-1})}{g^{(n-1)}(\xi_{n-1})}=\frac{r^{(n-1)}(\xi_{n-1}) - r^{(n-1)}(a)}{g^{(n-1)}(\xi_{n-1})-g^{(n-1)}(a)}=\frac{1}{n!}\frac{r^{(n-1)}(\xi_{n-1})-r^{(n-1)}(a)}{\xi_{n-1}-a} \end{aligned} The $\xi$'s depend on $x$ and \begin{aligned} a<\xi_{n-1}<\ldots<\xi_1<x,&\qquad\text{if}\quad a<x\\ x<\xi_1<\ldots<\xi_{n-1}<a, &\qquad\text{if}\quad x<a \end{aligned} Thus, as $x\rightarrow a$, so do all $\xi_k\rightarrow a$, and $$\lim_{x\rightarrow a}\frac{r(x)}{(x-a)^n}=\lim_{\xi_{n-1}\rightarrow a}\frac{1}{n!}\frac{r^{(n-1)}(\xi_{n-1})-r^{(n-1)}(a)}{\xi_{n-1}-a}=\frac{1}{n!}r^{(n)}(a)=0$$ where the lat limit holds by the fact that $f^{(n-1)}$, and hence $r^{(n-1)}$, is differentiable at $a$.

Solution 2:

I have already given some details in the comments but it appears that it is better to combine them to form an answer.

Taylor's theorem with Peano's form of remainder uses minimal assumptions of the function and then gives a simple estimate for the remainder. A proof is given here in another answer. I have myself discussed this topic in this thread.

The proof provided in the question uses integrals and thus assumes more than what is needed. In particular the assumption here is that n-th derivative $f^{(n)} $ is continuous in a neighborhood of $x_0$. Also the proof contains a subtle error. The author writes $$f^{(n)} (x_{0}+t(x-x_0))-f^{(n)}(x_0)=o(|x-x_0|)$$ This should be replaced by $$f^{(n)} (x_{0}+t(x-x_0))-f^{(n)}(x_0)=o(1)$$ (this is an immediate consequence of continuity of $f^{(n)} $) and then the later part of the proof works fine.

I don't know if the mistake by author is a typo.