# Finding a power series for a piece-wise analytical function.

Let $$f(x)=\begin{cases}x^2,& x\ge0,\\-x^2,& x\le 0.\end{cases}$$

Does there exist a power series $\sum_{n\ge 2} a_n x^n$ and a real number $R>0$ such that $\forall x\in(-R,R)$,

$$f(x)=\sum_{n\ge 2} a_n x^n.$$

What I did was this, but I don't know if this is correct. First checked if $f(x)$ is differentiable at $x=0$. By the use of

$$\lim_{h\to 0+}\frac{ f(x+h)-f(x)}{h}=\lim_{h\to 0-}\frac{ f(x+h)-f(x)}{h},$$

it was differentiable, so $f '(0)$ exists. Again checked if $f ' (x)$ was differentiable at $x=0$. It wasn't. Hence $f ''(0)$ doesn't exist. Now supposed that there exist a power series such that $$f(x)=\sum_{n\ge 2} a_n x^n.$$

Then find $f '(x)$ and $f ''(x)$.

$$f ''(x) = \sum_{n\ge 2} n(n−1)a_n x^{n−2}.$$

Then $f '' (0)= 2a_2$, but this is a contradiction as $f ''(0)$ does not exist. Thus there is no such power series.

• Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. – Zev Chonoles Jul 2 '13 at 8:44

Formal manipulations are correct, but a very important piece is missing: you did not justify passing from $f(x)=\sum_{n\ge 2} a_n x^n$ to $f''(x)=\sum_{n\ge 2} n(n-1)a_n x^{n-2}$. The justification can be a reference to a standard textbook theorem:
Actually, once you have this, there is no need for computations with sums. It follows from the theorem that a function represented by a power series has derivatives of all orders in the interval $(-R,R)$. Since your function $f(x)=x|x|$ does not (its first derivative, $f'(x)=|x|$, is not differentiable at $0$), it cannot be represented by a power series.