Extending Taylor's theorem to differentiable, but not continuously differentiable functions

Question

Taylor's theorem says that if $f:I\to\mathbb R$, $I\subseteq\mathbb R$ an open interval, is $n$ times continuously differentiable, then for $x_0\in I$ there exists a continuous function $R_{n,x_0}:I\to\mathbb R$ with $\frac{R_{n,x_0}(x)}{(x-x_0)^n}\to0$ as $x\to x_0$ such that

$f(x)=\sum\limits_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k~+~R_{n,x_0}(x)$.

For the case $n=1$, the requirements of continuity, or even differentiability on all of $I$, can be dropped: a function $f$ like above is differentiable in $x_0\in I$ iff there exist a number $f'(x_0)$ and a function $R:I\to\mathbb R$ with $\frac{R(x)}{x-x_0}\to0$ as $x\to x_0$ such that

$f(x)=f(x_0)+f'(x_0)(x-x_0)+R(x)$.

This is essentially Taylor's theorem without requiring differentiability or even continuity anywhere other than $x_0$. As a consequence, the remainder $R$ is also no longer continuous.

My question is: can I extend this to higher order Taylor polynomials? I would like a statement like the following:

If $f$ is $n$-times differentiable in $x_0$ (the $n$-th derivative need not be continuous in $x_0$, or even existent outside of $x_0$). Then there exists a function $R_{n,x_0}:I\to\mathbb R$ with $\frac{R_{n,x_0}(x)}{(x-x_0)^n}\to0$ such that

$f(x)=\sum\limits_{k=0}^n\frac{f^{(k)}}{k!}(x-x_0)^k~+~R_{n,x_0}(x)$.

Context: I want to find a definition of higher order derivatives using polynomial approximation. The above case with $n=1$ is a useful definition of differentiability, which essentially says that a function is differentiable iff there is a linear polynomial and a small remainder such that $f=\textrm{polynomial}+\textrm{remainder}$. I want to define higher order derivatives by saying that $f$ is $n$-times differentiable if there is a polynomial of degree $n$ and a small remainder such that $f=\textrm{polynomial}+\textrm{remainder}$. If my question from above can be answered with yes, then this definition would be a generalization of higher order differentiability, otherwise it would just be something different and thus useless.

Answer 1

The pointwise form of Taylor's theorem you propose is true; in fact the same statement holds almost word-for-word even in a higher dimensional case for functions $f: \Bbb{R}^n \to \Bbb{R}^m$ (or even more generally, for functions $f:V\to W$ where $V$ and $W$ are real Banach spaces, not necessarily finite-dimensional). See for example this answer for a statment and outline of proof in the general case of several variables (but by making necessary simplifications you can easily restrict yourself to the $1$-dimensional case, and the proof should be just as easy to carry out).

However, in the "context" part of your question you say

I want to define higher order derivatives by saying that $f$ is $n$-times differentiable if there is a polynomial of degree $n$ and a small remainder such that $f=\text{polynomial+remainder}$. If my question from above can be answered with yes, then this definition would be a generalization of higher order differentiability, otherwise it would just be something different and thus useless.

Unfortunately the situation isn't quite as nice as you might hope. Taylor's theorem says that "if $f$ is $n$-times differentiable at a point $x_0$, then $f$ is a polynomial plus a small remainder with $\frac{R_{n,x_0}(x)}{(x-x_0)^n} \to 0$". What you seem to be asking for is the converse. But a direct converse is completely incorrect. Just because you have $f$= polynomial+ "small remainder", it DOES NOT (for $n\geq 2$) imply $n$-times differentiability at $x_0$. For example, with $n \geq 2$, the function $f: \Bbb{R} \to \Bbb{R}$ defined as \begin{align} f(x) &= \begin{cases} x^{n+1} & \text{if $x$ irrational}\\ 0 & \text{if $x$ rational} \end{cases} \end{align} satisfies $f(0+h) = 0 + o(|h|^n)$ (i.e it equals the zero polynomial to order $n$), yet fails to even be twice differeniable. Take a look at this answer where I present this counter example (which I learnt from Spivak's Calculus) and give links for partial converses to Taylor's theorem (all these answers assume slightly stronger hypotheses on the form of the remainder).