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.

  • $\begingroup$ Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. $\endgroup$ – 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:

Theorem. A power series can be differentiated term by term in the interior points of its interval of convergence. The differentiated series has the same radius of convergence.

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.

  • $\begingroup$ Are you still here on MSE? $\endgroup$ – bryanj Jan 21 '14 at 5:25

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