If $\displaystyle \sum_{n=0}^{\infty}c_{n}x^{n}=0$, show that $c_n= 0$ for any $n$ Suppose that $f(x)=\displaystyle\sum_{n=0}^{\infty}c_{n}x^{n}$ for all $x$ with the radius of convergence $R>0$. If $\displaystyle\sum_{n=0}^{\infty}c_{n}x^{n}=0$, show that $c_n=0$ for any $n$. 
 A: Hint $c_n=f^{(n)}(0)/n!$ ${}{}{}{}{}{}{}{}{}{}{}{}$
A: Hint: It can be proved by induction on $n$. First, $c_0=f(0)=0$. Then $g(x):=\frac{f(x)}{x}= \sum\limits_{n=1}^{+ \infty} c_nx^{n-1}$, hence $g(0)=c_1=0$, etc.
A: Neither differentiability considerations nor a proof by induction are necessary. Instead, assume that some $c_n$ are not zero, then consider $N=\inf\{n\mid c_n\ne0\}$ and note that $f(x)\sim c_Nx^N$ when $x\to0$ (in the sense that $f(x)/x^N\to c_N$ when $x\to0$).
In particular there exists some positive $\varepsilon$ such that $f(x)\ne0$ for every $x\ne0$ such that $|x|\lt\varepsilon$. This contradicts the hypothesis hence $c_n=0$ for every $n$.
A: This is essentially the same answer as the others, this just makes a connection with the derivative.
$f(0) = c_0 = 0$.
Inside the radius of convergence, the function is smooth and $f'(x) = \sum_{n=1}^\infty n c_n x^{n-1}$. Since $f'(x) = 0$, we have $f'(0) = c_1 = 0$.
Continuing in this way, we have $f^{(k)}(x) = 0$ which gives $f^{(k)}(0) = k! c_k = 0$ from which we get $c_k = 0$.
