A few hours ago, I asked a question about using Taylor expansion of two analytic functions on $\mathbb{R}$ to determine whether these two functions are linearly independent. Basically I was trying to express every analytic function as an element in $\mathbb{R}[[x]]$.
A contributor commented that though $\{1,x,x^2,x^3,x^4,\dots\}$ is a basis of $\mathbb{R}[x]$, it is not a basis of $\mathbb{R}[[x]]$.
My question is, since $\forall a(x)\in\mathbb{R}[[x]]$, $\exists! \{a_n\}_{n=0}^\infty$ s.t. $a(x)=\sum_{n=0}^\infty a_nx^n$, isn't $\{x^n\}_{n=1}^\infty$ a Schauder basis of $\mathbb{R}[[x]]$? Is the user referring to Hamel bases when he says that it is not a basis of $\mathbb{R}[[x]]$?
Thanks in advance. Any help will be appreciated.
A Schauder basis of a vector space $V$ over a field $k$ is a sequence $b_0,b_1,\ldots$ of vectors in $V$ such that every element of $V$ can be expressed uniquely as a convergent series of the form $\sum_{i=0}^\infty a_i b_i$ where each $a_i$ is in $k$. For this to make any sense, there must be a topology on $V$ that gives meaning to the notion of convergence. See https://en.wikipedia.org/wiki/Schauder_basis .
For any field $k$, the ring $k[[x]]$ of formal power series is a vector space over $k$, and has a natural topology as the product of countably many copies of $k$, with the discrete topology on each. See https://en.wikipedia.org/wiki/Formal_power_series .
A sequence of formal power series converges if, for every $n$, the coefficient of $x^n$ stabilizes. (That is, eventually, as we go further in the sequence, that coefficient never changes again.)
So: Yes $\{1,x,x^2,\ldots\}$ is a perfectly good (Schauder) basis for $k[[x]]$.