# Taylor $k$-differentiability of a real function at a point

I am interested in the standard name for the following weak form of $$k$$-differentiability.

Definition. A function $$f:\mathbb R\to\mathbb R$$ is called Taylor $$k$$-differentiable at a point $$x_0$$ if there are real numbers $$a_0,\dots,a_k$$ such that $$f(x)=\sum_{n=0}^k\tfrac1{n!}a_n(x-x_0)^n+o(|x-x_0|^k).$$The numbers $$a_0,\dots,a_k$$ are unique (if exist) and can be called the derivatives $$f^{(0)}(x_0),\dots,f^{(k)}(x_0)$$ of $$f$$ at $$x_0$$.

The Taylor formula says that each $$k$$-differentiable function at a point $$x_0$$ is Taylor $$k$$-differentiable. The converse is not true: the Dirichlet-like function $$f(x)=\begin{cases}x^{k+1}&\mbox{if x is rational};\\ 0&\mbox{if x is irraional} \end{cases}$$ is Taylor $$k$$-differentiable at zero (with all derivatives $$f^{(0)}(0)=\dots=f^{(k)}(0)=0$$) but is discontinuous at all non-zero points, so is not $$k$$-differentiable in the standard sense.

So, my question:

Has the Taylor $$k$$-differentiability some standard name, accepted in the literature? If yes, what is a suitable reference? Thanks.

I would not recommend "Taylor $$k$$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $$x_0$$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $$f$$ has a polynomial expansion of order $$k$$ at $$x_0$$, where "order $$k$$" refers to the form of the remainder, $$o(|x-x_0|^k)$$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since I think Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this subject. In particular, the implication you mention and the form of the remainder $$o(x-x_0)^k$$ is due to Peano (it's a one-page note, Une nouvelle forme du reste dans la formule de Taylor, 1889). This polynomial expansion has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $$C^k$$ function.
• Thank you for your answer. Peano $k$-differentiable also sounds good for me. So I will use this terminology. – Taras Banakh Oct 27 '19 at 17:49
It's called the “de la Vallée-Poussin derivative” or “Peano derivative” (both terms seem to have been used, sometimes with a difference between them; the latter is perhaps slightly more standard): see this answer and the references it contains for a proof of the fact that if the $$k$$-th Peano derivative exists and is continuous on an interval then it is, in fact, a $$k$$-th derivative.