Are the algebraic functions dense in the space of analytic functions? I had a quick google, and couldn't ascertain a answer to the question
'Are the algebraic functions dense in the space of analytic functions over the interval [0,1]?' 
This is functions over one variable and coefficients in the real numbers.
I am sorry if this question is obvious, my functional analysis isn't that strong.
 A: By the Weierstrass approximation theorem, polynomials are dense in $C[0,1]$, the space of continuous functions with the norm $\|f\|_{C^0}=\sup_{[0,1]}|f|$. In particular, every real-analytic function on $[0,1]$ can be uniformly approximated by polynomials. 
Furthermore, for every positive integer $k$ polynomials are dense in $C^k[0,1]$, the space of $k$ times continuously differentiable functions with the norm $\|f\|_{C^k}=\sum_{j=0}^k\sup_{[0,1]}|f^{(j)}|$. To prove this, use Weierstrass's theorem to approximate $f^{(k)}$ by a polynomial $p$ and then integrate $k$ times to get a polynomial $q$ such that $q^{(k)}=p$ and $q^{(j)}(1/2)=f^{(j)}(1/2)$ for $j=0,\dots,k-1$. It's an exercise with the Mean Value theorem to show that $\|f-q\|_{C^k}$ is small.
To go further, we need some complex analysis. A real-analytic function $f$ on $[0,1]$ can be extended (using its power series) into an open set $\Omega\subset \mathbb C$ containing the segment $[0,1]$. Pick $d<\operatorname{dist}([0,1],\partial \Omega)$.  Let $K$ be the closed $d$-neighborhood of $[0,1]$. By Mergelyan's theorem there is a sequence of polynomials $p_n$ converging to $f$ uniformly on $K$. By the Cauchy estimate for derivatives,
$$\sup_{[0,1]}|p_n^{(k)}-f^{(k)}|\le \frac{k!}{d^k}\sup_K |p_n-f|,\quad k=1,2,\dots$$
Hence, the sequence $\{p_n\}$ approximates $f$ in the sense of uniform convergence of all derivatives. In other words, polynomials are dense in the space of analytic functions equipped with the topology induced by  seminorms $\sup_{[0,1]} |f^{k}|$, $k=0,1,2,\dots$.
