In Klenke's Probability Theory one can read
[…] $f$ can be expanded about any point $t\in \mathbb R$ in a power series with radius $\geq R$. In particular it is analytic and is hence determined by the coefficients of its power series about $t=0$.
Why is $f$ uniquely determined by its derivatives at $0$ ? I get that $f$ has a power expansion at each $x_0$ over the domain $(x_0-R,x_0+R)$ as $f(x)=\sum_{i=0}^\infty a_i (x-x_0)^i$, but I don't know why the $a_i$ (which depend on $x_0$) are related to the derivatives at $0$.
Best Answer
For the expansion around $0$ we have $a_i=\frac {f^{(i)}(0)} {i!}$ so $f$ must vanish in some interval around $0$. Now verify that $\{x: f=0 \, \text {in some neighborhood of }\, $x$ \}$ is open and closed in $\mathbb R$. By connectedness of $\mathbb R$ we can conclude that $f \equiv 0$.