How to show that the set of functions $1,x,x^2,x^3...$ is linearly independent? Show that the set of functions $1,x,x^2,x^3...x^n...$ is linearly independent on any interval $[a,b]$.
If $$c_1+xc_2+x^2c_3+x^3c_4...=0$$ we should show $$c_i=0,\quad i=1,2, \ldots$$
how could I start?
My second question: is it linearly independent on $C[0,1]$?
 A: Consider the polynomial $p(x)=c_1+c_2x+\dots+c_nx^{n-1}$. How many zeros can a polynomial of order $n-1$ have? How many zeros does your polynomial $p$ have?
A: Note that if coefficients $a_{0},...,a_{n}$ exist s.t
$$
p(x)=a_{0}+a_{1}x+...+a_{n}x^{n}=0;\,\forall x\in[a,b]
$$
Then since a polynomial is a continues function it means that 
$$
p(x)=0;\,\forall x\in\mathbb{R}
$$
We now prove by induction that $a_{m}=0$ for all $m$.
Our base case will be $m=n$ and we proceed similarly for $m=n-1$
etc'.
If $a_{n}\neq0$ then 
$$
\lim_{x\to\infty}p(x)=\begin{cases}
\infty & a_{n}>0\\
-\infty & a_{n}<0
\end{cases}
$$
and thus $a_{n}=0$.
When we reach the step $$a_{n}=a_{n-1}=...=a_{1}=0$$ we will be left
with $p(x)=a_{0}$ and since $$a_{0}=p(0)=0$$ we reach the conclusion
$p\equiv0$.
This shows that if we have a linear combination that is zero then
all coefficients are also zero, which shows the independence by definition.
A: Pick any finite set of them $\{x^{i_1},\ldots, x^{i_n}\}$ and consider the linear combination
$$p(x)=c_{i_1}x^{i_1}+\ldots+c_{i_n}x^{i_n}=0$$
This means that $p(\alpha)=0$ for every $\alpha\in \mathbb R$. But you know that a non constant polynomial has at most a finite number of roots (here you don't need the fundamental theorem of algebra, but you must only know how the polynomial multiplication works with the degrees).
It follows that $p(x)$ is a constant polynomial and in particular it is the zero polynomial.
note: Remember that you only need to verify the independence for every finite set of elements. An infinite sum is not defined!  From the question it seems that you're trying to operate with infinite linear combinations.
