$\sum_{n=0}^{\infty}a_n \cdot x^n=0$ why do we need $a_0=a_1=a_2=......=0$? prove it To have
$$\sum_{n=0}^{\infty}a_n \cdot x^n=0$$
for all $x$, we need $a_0=a_1=a_2=......=0$
I don't quite see the logic here, why only $(0,0,0,......)$ can satisfy this?
Proof by contradiction:
Suppose $a_i\neq 0$
But then we can say nothing about other coefficients...
Or if we assume $a_{i\neq i}=0$ then for $x=1$, then for $x=1$ this is not $0$. But this is only a special case...
or we can solve the linear system
$$1,1,1,1,.......$$
$$1,2,4,8,.......$$
$$1,3,9,27,......$$
$$...$$
$$...$$
 A: Assume otherwise and let $n$ be the smallest index with $a_n \neq 0$. Then you can write the series as $x^n g(x)$, where $g(x) = \sum_{k=0}^{\infty} a_{n+k} x^k$ is non-zero near $x = 0$ (since $g(0) = a_n \neq 0$ and $g$ is continuous near $0$). This shows that $x = 0$ is the only zero of $x^n g(x)$ near $x = 0$, which contradicts the assumption.
A: $f(x) = 0$ is an analytic function, hence the only valid power series representation is the Taylor series at $x=0$. In particular, we have $a_n = \frac{ f^{(n)} (0) }{n!} = 0$ for all $n$.

We can also prove it using the following theorem:
Let $E$ be the set of all $S$ at which 
$$\sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} b_n x^n.$$
If $E$ has a limit point in $S$, then $a_n = b_n $ for $n = 0, 1, 2, \dots$.
Here we have the $b_n = 0$ for all $n$, and they are equal in $S = \mathbb{R}$, so $a_n = 0$ for all $n$ a swell.
A: If the series converges for all real $x$, the sum is an entire function.  The generalized Cauchy integral formula then shows that the coefficients are all $0$.
