1
$\begingroup$

The uniqueness theorem for analytic functions states that suppose two series $\sum_{n=0}^\infty s_nx^n$ and$\sum_{n=0}^\infty t_nx^n$ converges in the interval $(-R,R)$. If the set of $x$ that satisfies $$\sum_{n=0}^\infty s_nx^n=\sum_{n=0}^\infty t_nx^n$$ has a limit point in the interval, then $s_n=t_n$ for all $n \in N$. I know there are lots of functions which are infinitely differentiable but not analytic, such as the one in the wikipedia http://en.wikipedia.org/wiki/Non-analytic_smooth_function, but I do not know how to use these functions to give a counterexample of the uniqueness theorem. Can anyone give me a hint?

$\endgroup$
3
  • 2
    $\begingroup$ Use the $f(x)$ given on the Wikipedia page and consider $f(x)$ and $f(x/2)$. They're equal on all of $(-\infty,0]$, but none of $(0,\infty)$. $\endgroup$ May 3, 2015 at 3:34
  • $\begingroup$ @Meelo Oh I see, thx! Then I can find a sequence with limit point that f(x)=f(x/2) but their Taylor series are definitely not the same. Am I right? $\endgroup$
    – user194201
    May 3, 2015 at 3:38
  • 1
    $\begingroup$ Hmm I'm not sure what question you are asking. Any power series that converges on $(-R,R)$ is analytic with convergence radius $R$, which is why John gave you his answer. If you are asking whether there are two $C^\infty$ functions that agree on a convergent sequence but are not the same (as a function, and not comparing any coefficients), then Meelo's comment would do it. $\endgroup$
    – user21820
    May 3, 2015 at 3:43

1 Answer 1

0
$\begingroup$

You cannot use that to give a counterexample. The theorem assume already that the two series converge in an interval. Thus they are both analytic functions (Not only smooth) on that interval.

$\endgroup$

You must log in to answer this question.