Finding a function f with information of its power series

Question

Suppose f has a power series representation at $0$ that converges on (−1, 1). Assume we have a sequence of distinct elements ($x_n$) such that $x_n → 0$ and $$f(x_n) = 0$$ for all $x_n$. Find $f$.

Answer 1

Let $f(x)= \sum_{n=0}^{\infty}a_nx^n$ for $x \in (-1,1).$

We prove by induction that $a_n=0$ for all $n \ge 0$. Consequence: $f=0$ on $(-1,1).$

Base case: $f$ is continuous in $(-1,1)$, hence $a_0=f(0) = \lim f(x_n)=0.$

Now assume that $n \in \mathbb N_0$ and $a_0=a_1=...=a_n=0$.

Then we have $f(x)=a_{n+1}x^{n+1}+a_{n+2}x^{n+2}+....$ for $-1 <x<1$.

Are you now in a position to show that $a_{n+1}=0 ?$

Remark: the result that $f=0$ on $(-1,1)$ is called the "identity theorem for power series".