How to expand undifferential function as power series? If a function has infinite order derivative at $0$ and $\lim_{n\to \infty}(f(x)-\sum_{i=1}^{n} a_{n}x^n)=0$ for every $x \in (-r,r)$,then it can be expand as power series$\sum a_{n}x^n$,
My question is if this function is not differential at $0$,how to expand it as $\sum a_{n}x^n$ satisfied with $\lim_{n\to \infty}(f(x)-\sum_{i=1}^{n} a_{n}x^n)=0$ for every $x \in (-r,r)$?Is it unique ?

If $\sum_{j=0}^\infty a_j x^j$ converges for every $x$ in an interval $(-r,r)$, then the radius of convergence of the series is at least $r$, and the sum is analytic in the disk $\{z \in {\mathbb C}: |z| &lt r\}$.  So if $f(x)$ is not analytic in $(-r,r)$, in particular if it is not differentiable at $0$, there is no way to represent it as $\sum_n a_n x^n$ with
$\sum_{j=0}^n a_j x^j \to f(x)$.  
However, you can try a series $\sum_n a_n x^n$ such that some subsequence of partial sums $P_N(x) = \sum_{j=0}^N a_j x^j$ converges to $f(x)$.  Suppose $f$ is continuous on $[-r,r]$ except possibly at $0$.  I'll let $a_0 = f(0)$ and $N_0 = 0$.  Given $a_j$ for $0 \le j \le N_k$, let $g_k(x) = (f(x) - P_{N_k}(x))/x^{N_k}$ for $x \ne 0$, $g_k(0) = 0$. Since $g_k$ is continuous on $E_k = [-r, -r/(k+1)] \cap \{0\} \cap [r/(k+1), r]$, Stone-Weierstrass says there is a polynomial $h_k(x)$ with $|g_k(x) - h_k(x)| &lt r^{-N_k}/(k+1)$ on $E_k$.  Moreover we can assume $h_k(0) = g_k(0) = 0$.  Let $N_{k+1} = N_k + \deg(h_k)$, and let
$a_j$ be the coefficient of $x^j$ in $x^{N_k} h_k(x)$ for $N_k &lt j \le N_{k+1}$.  Thus
$P_{N_{k+1}}(x) = P_{N_k}(x) + x^{N_k} h_k(x)$ so that 
$|P_{N_{k+1}}(x) - f(x)| = |x|^{N_k} |g_k(x) - h_k(x)| &lt 1/(k+1)$ for $x \in E_k \backslash \{0\}$ (we already know $P_{N_{k+1}}(0) = f(0)$).
Since the union of the $E_k$ is all of $[-r,r]$, the partial sums $P_{N_k}(x)$ converge to $f(x)$ pointwise on $[-r,r]$.         
