Confused about this lemma for power series

Question

I find this difficult to believe, a consequence would be that any analytic function has no non trivial roots? What about $\sin(x), \cos(x)$ or any polynomial not in the form $x^n$?

I don't particularly understand this either, why can we conclude the RHS is $0$ if $x \neq y$. Any insight appreciated.

Answer 1

The statement is just false: You have already mentioned a counterexample. Namely $$\cos(x) = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!}.$$ E.g. $$0 = \cos(\pi/2) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{\pi^k}{2^k}.$$ Of course, if $a_n \geq 0$, then the statement is true. The statement is also true, if $\sum_{n=0}^\infty a_n x^n=0$ for all $|x| < \delta$ for some $\delta >0$.

Answer 2

The statement above is clearly false ! I think it should read as follows: if $y \ne 0$ and if $\sum_{n=0}^{\infty}a_n x^n = 0$ for all $x$ with $|x|<|y|$, then $a_n=0$ for all $n$.

I am right ?