$f_k(x)=x^k$ for $k=1,\dots ,5$ is linearly independent.
Proof:
Assume that $ax+bx^2+cx^3+dx^4+ex^5 =0$ ıf some $a,b,c,d,e$ are not $0$ then the polynomials could have at most $5$ zeros.However,our expression for infinitely whenever $x\in R$.
Hence, $a=b=c=d=e=0$ so it is linearly independent.
Can you explain this proof ? I don't understand.
On
Suppose $p(x) = \sum_{k=0}^n p_k x^k $ is a polynomial with $p_n \neq 0$. Then for $x \neq 0$, $p(x)= x^n ( p_n + \sum_{k=0}^{n-1} p_k {1 \over x^{n-k}})$ and so, if $|x| \ge 1$, we have $|p(x)| \ge |x|^n ( |p_n| - \sum_{k=0}^{n-1} |p_k| {1 \over |x|^{n-k}}) ) \ge |x|^n ( |p_n| - {1 \over |x|}\sum_{k=0}^{n-1} |p_k| )$. In particular, if $|x| \ge 2 {1 \over |p_n|} \sum_{k=0}^{n-1} |p_k| $ we see that $p(x) \neq 0$. (I should also have $|x| \ge 1$, but it looks ugly inside a $\max$.)
This is a long winded way of saying that if $p$ is not the zero polynomial, then it is non zero somewhere.
In particular, if $p(x) = ax+bx^2+cx^3+dx^4+ex^5 =0$ for all $x$ then $p$ must be the zero polynomial and $a= \cdots = e = 0$.
I am note sure about the correctness of the proof. But here is an easy way to prove it.
Proof: Suppose that the list $x, x^2, \dots, x^5$ is linearly dependent. Then one of the vectors $x^j$ is in the span of the previous vectors. Thus, there exists scalars $a_1, \dots , a_{j-1}$ such that $$a_1x+ \dots + a_{j-1}x^{j-1} = x^j$$
Differentiating both sides $j$ times we get a contradiction as the left hand side is $0$ and the right hand side is non-zero. Thus, the list $x, x^2, \dots, x^5$ is linearly independent.