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

I have two questions relating to a proof I found of the Peano remainder form: $$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(x_{0})}{k!}(x-x_{0})^{k}+\mathrm{o}\left(|x-x_{0}|^{n}\right)$$. Keep in mind that the little o notation here is that if $$f \in \mathrm{o}(t)$$ then: $$\lim_{\substack{t\rightarrow0\\ t\neq0} }\frac{f(t)}{t}=0$$

I am not interested in general proofs of Peano's remainder form, but particularly in how the little o notation is manipulated. So this is more of a little o notation question than a Taylor's theorem proof question.

This linked proof begins using the integral form of the remainder and the assumption that $$f \in C^n$$, so the proof has an extra assumption compared to the normal Peano remainder proofs, which usually only assume $$n$$ times differentiability (so $$f^{(n)}$$ is not necessarily continuous).

My first issue is with the proof itself, which says that \begin{aligned}\frac{(x-x_{0})^{n}}{(n-1)!}\int_{0}^{1}\left[f^{(n)}(x_{0}+t(x-x_{0}))-f^{(n)}(x_{0})\right](1-t)^{n-1}dt.\end{aligned} can be simplified to $$\mathrm{o}\left(|x-x_{0}|^{n}\right)$$.

The last line says that since $$f^{(n)}$$ is assumed continuous, we have that $$\left[f^{(n)}(x_{0}+t(x-x_{0}))-f^{(n)}(x_{0})\right] \in \mathrm{o}\left(|x-x_{0}|\right)$$. I am trying to show that this is true, but can't seem to justify it.

I think that continuity is necessary so that the limit as $$x \rightarrow x_0$$ of $$f^{(n)}(x_{0}+t(x-x_{0}))$$ is $$f^{(n)}(x_{0})$$ and so the $$\left[f^{(n)}(x_{0}+t(x-x_{0}))-f^{(n)}(x_{0})\right]$$ term goes to $$0$$. But I am not sure how this implies that $$\left[f^{(n)}(x_{0}+t(x-x_{0}))-f^{(n)}(x_{0})\right]$$ goes to $$0$$ faster than $$(|x-x_{0}|)$$, since we have no details about $$f^{(n)}$$. I am thinking it could be since $$t \in [0,1]$$ and so for a given increment $$x - x_0$$, $$f^{(n)}(x_{0}+t(x-x_{0}))$$ is closer to $$f^{(n)}(x_{0})$$ than $$f^{(n)}(x_{0}+(x-x_{0}))$$ otherwise would have been, but I'm unsure if this is valid reasoning. It also seems more like a 'big O' case than 'little o'.

In addition, I am wondering how \begin{aligned}\frac{(x-x_{0})^{n}}{(n-1)!}\int_{0}^{1}\left[\mathrm{o}\left(|x-x_{0}|\right)\right](1-t)^{n-1}dt.\end{aligned} is simplified to $$\mathrm{o}\left(|x-x_{0}|^{n}\right)$$, since there is an $$(x-x_{0})^{n}$$ multiplied by $$\mathrm{o}\left(|x-x_{0}|\right)$$, which I think should be $$\mathrm{o}\left(|x-x_{0}|^{n+1}\right)$$ and not $$\mathrm{o}\left(|x-x_{0}|^{n}\right)$$ as the remainder should appear as.

So those are my main questions, and my confusion mostly pertains to the little o notation and it's manipulations. I know there are other proofs for Taylor's theorem but wanted to understand this notation a bit better. Thanks in advance!

• This theorem is already discussed in this thread. – Paramanand Singh Oct 24 '19 at 6:28
• My question is less about the theorem and more about how the little o notation is used and how it works – Slade Oct 24 '19 at 11:47
• Assume that we are dealing with some limiting behavior of functions of $x$ as $x\to a$ ($a$ can be $\pm\infty$ also). The little-o notation assumes that this information is provided implicitly or via context. Then $f(x) = o(g(x))$ is just another way of saying that $f(x) /g(x) \to 0$. If someone writes $f(x) =g(x) +o(h(x))$ then it means $f(x) - g(x) =o(h(x))$ and this further means that $(f(x) - g(x)) /h(x) \to 0$. Thus every statement which contains this little-o is in reality a statement about limits. – Paramanand Singh Oct 24 '19 at 14:01
• The proof of Taylor's theorem with Peano's form of remainder is not done via integrals as that approach requires more hypotheses than actually needed. Moreover integrating things like $o(f(x))$ are tricky. Thus for example if $f(x) =o(x^3)$ as $x\to 0$ then $\int_{0}^{x}f(t)\,dt=o(x^4)$ but this is mainly a consequence of L'Hospital's Rule. – Paramanand Singh Oct 24 '19 at 14:06
• Thanks a lot for looking into this. Do you have any idea why if $f^{(n)}$ is continuous then the author implies $\left[f^{(n)}(x_{0}+t(x-x_{0}))-f^{(n)}(x_{0})\right] \in \mathrm{o}\left(|x-x_{0}|\right)$, (for $t \in [0,1]$). I can't seem to show this. My other issue was with \begin{aligned}\frac{(x-x_{0})^{n}}{(n-1)!}\int_{0}^{1}\left[\mathrm{o}\left(|x-x_{0}|\right)\right](1-t)^{n-1}dt.\end{aligned} since it is said to be $\mathrm{o}\left(|x-x_{0}|^{n}\right)$, but looks $\mathrm{o}\left(|x-x_{0}|^{n+1}\right)$ to me – Slade Oct 24 '19 at 14:12

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.

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 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 $$a$$.

• Nice proof directly using Cauchy Mean Value Theorem instead of its infamous consequence L'Hospital's Rule. +1 – Paramanand Singh Oct 24 '19 at 14:10
• It is in a sense the way one typically proves L'Hospital rule. – Oliver Diaz Oct 24 '19 at 19:36