Show that $\mathcal{A}$ is dense in $C([0,2016])$. Exercise: Let $\mathcal{A}$ be the family of functions $f:[0,2016]\to\mathbb{R}$ that can be written as $$f(x) = a_0 + a_1x^4 + a_2x^8 + a_3x^{12} + \ldots + a_nx^{4n}$$ with $a_0,a_1,\ldots,a_n\in\mathbb{R}$ and $n\in\mathbb{N}$.
Formulate the algebra version of the Stone-Weierstrass approximation theorem and show that $\mathcal{A}$ is dense in $C([0,2016])$.
What I've tried: The algebra version of the Stone-Weierstrass approximation theorem says the following: Let $X$ be a compact metric space and let $\mathcal{A}$ be a subalgebra of $C(X)$. If $\mathcal{A}$ separates points in $X$ and vanishes at no point in $X$ then $A$ is dense in $C(X)$. So I think I need complete the following steps:
1) Show that $[0,2016]$ is compact.
2) Show that $\mathcal{A}$ is a subalgebra of $C([0,2016])$
3) Show that $\mathcal{A}$ separates points in $X$.
4) Show that $\mathcal{A}$ vanishes at no point in $X$.
Every closed interval in $\mathbb{R}$ is compact so $[0,2016]$ is compact, this satisfies 1). I don't know how to satisfy 2) unfortunately. To show that $\mathcal{A}$ separates points in $X$ consider the function $f(x) = a_0 + a_1x^4$; for every $x,y\in[0,2016], x\neq y$ we have that $f(x) \neq f(y)$. To show that $\mathcal{A}$ vanishes at no point in $X$ consider the function $f(x) = a_0$ with $a_0 \neq 0$. Then $f(x) \neq 0$ for every $x\in[0,2016]$.
Question: How do show that $\mathcal{A}$ is dense in $C([0,2016])$ by showing that $\mathcal{A}$ is a subalgebra of $C([0,2016])$?
Thanks!
 A: Hint. In order to show that  such special set of polynomials (which are continuous over $[0,2016]$),
$$\mathcal A=\{a_0 + a_1x^4 + a_2x^8 + \ldots + a_nx^{4n}: a_0,a_1,\ldots,a_n\in\mathbb{R}, n\in\mathbb{N}\}=\{P(x^4): P\in\mathbb{R}[x]\}$$
 is a subalgebra of $C([0,2016])$ prove that it is closed under linear combinations and multiplication of functions. 
P.S. To separate points in $[0,2016]$ take $x^4\in \mathcal A$ (which is injective for $x\geq 0$). To show that $\mathcal A$ vanishes at no point in $[0,2016]$ take $1\in \mathcal A$. 
A: *

*Correct, $[0,2016]$ is compact.

*You only need to show that $\mathcal A$ is closed under sum and multiplication.

*Your example $a_0+a_1x^4$ is not specific enough. We need $a_1\ne 0$ to have the separation. A more specific example that separates is $f(x)=x^4$.

*Your argument seems tautological. We need to prove here that for any $x\in[0,2016]$ there is at least one $f\in\mathcal A$ such that $f(x)\ne 0$. 
Well, for this, if $x$ is given, pick any $f\in\mathcal A$. If $f(x)\ne 0$, we're ready, and if $f(x)$ happens to be $0$, then take $f+1$.

