Let $f:\Bbb R\to\Bbb R$ be a $\cal C^\infty$ function. Consider the power series $$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$ and call $R$ its radius of convergence. Then, is it true that in $(-R,R)$ $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\,?$$
If we call $g$ the function defined on $(-R,R)$ such that $g(x)=\sum\limits_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!}x^n$, then I know that $g$ is $\cal C^\infty$ on $(-R,R)$ and $\dfrac{g^{(n)}(0)}{n!}=\dfrac{f^{(n)}(0)}{n!}$ for all $n$. So I know that every derivative of $g$ and $f$ are equal in $0$. But does it imply that $g=f$?