Function with Taylor series of order $k,$ which is not twice differentiable Find a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that it has order $k$ Taylor expansion at $0$ but $f"(0)$ does not exist.
It is a problem I am dealing with and for which I can't find any example. I would appreciate if someone could give me any example. Thanks in advance.
 A: There is no such thing. If $f$ has derivatives up to order $n\geq0$ at $0$ then the $n^{\rm th}$ order Taylor polynomial of $f$ at $0$ is given by
$$j^n_0f\,(x):=\sum_{k=0}^n{f^{(k)}(0)\over k!}\>x^k\ ,$$
period. One then can deliberate how well this polynomial approximates $f$ in a neighborhood of $x=0$.
If $f''(0)$ does not exist then there is no second order Taylor polynomial of $f$ at $0$. 
A: I assumed OP meant there exists $a_0,a_1,\dots,a_k$ such that around $0$, $f(x)=a_0+a_1x+\dots+a_kx^k+o(x^k)$ but $f''(0)$ does not exist.
Hint :
For $k=2$ consider : $$\begin{align}f:&\mathbb{R}\rightarrow\mathbb{R}\\&x\mapsto 0\text{ if } x=0\\&x\mapsto x^3\sin(x^{-1}) \text{ if } x\neq0 \end{align}$$
A: Differentiability is a local property. The key question will be where is it not twice differentiable. Taylor series and more generally Laurent series will be expandable with convergence radius up to it's closest pole. For example the function
$$f(x) = \frac{1}{1-x^2} = \frac{1}{(1-x)(1+x)}$$
Will be Taylor expandable around $x=0$ with convergence radius $\min\{|0-(-1)|,|0-1|\} = 1$. We won't be able to calculate second derivative at $\pm1$, so we would fail if we tried centering the expansion around any of those two points.
Even if the pole of a function is right on the point we want to expand around the we can still find a Laurent expansion, that is a power series allowing negative integer exponents:
$$f(x) = \sum_{-\infty}^{\infty}c_kx^k$$
But with only positive exponents as the Taylor expansion has, it would not be possible.
