Formal power series with all derivatives zero I have the following question.
Suppose I have a formal power series $f(x)=\sum\limits_{i=0}^\infty c_ix^i$ with real coefficients. Suppose that all the derivatives $f'(1),f''(1),\dots,f^{(n)}(1),\dots$ of $f(x)$ at the point $x=1$ are zero. 
What can I say about the coefficients $c_i$? I want to say that they all must be zero, but when I try to write down the system of equations, I get infinite systems, and each equation involves infinitely many variables. Is there a nice way to solve this problem?
Thank you!
 A: A "formal power series" can't be evaluated at $x=1$, just $x=0$.  If you mean that $\sum_{i=0}^\infty c_i x^i$ is the Maclaurin series of a function $f(x)$ which is analytic in a connected open set $D$ that contains both $0$ and $1$, then
indeed all the coefficients are $0$, because $f$ is constant on $D$.  On the other hand, you might consider $f(x) = \exp(-1/(x-1)^2)$ with $f(1) = 0$, which is analytic on ${\mathbb C} \backslash \{1\}$ and $C^\infty$ on $\mathbb R$, 
with all derivatives $0$ at $x=1$.  Its Maclaurin series is 
$$  e^{-1} - 2 e^{-1} x - e^{-1} x^2 + \dfrac{2}{3} e^{-1} x^3 + \ldots$$
A: You cannot plug in $1$ in a formal power series. You have to regard it as an analytic function to do so (therefore, check for convergence and so on). However, any calculation from the analytical viewpoint will give you a correct identity for formal power series if both sides admit an interpretation as formal power series, and any calculation from the formal viewpoint will give you a correct identity for analytic series if both sides are indeed convergent.
