trigonometric series with nonnegative coefficients and prescribed values Given real numbers $x_1$, $\dots$, $x_n$ and $r_1$, $\dots$, $r_n$ (with reasonable restrictions), is there a trigonometric series $T(x)=\sum_k a_k \cos(kx)$ with $a_k\ge 0$ such that
$$
|T(x_i)-r_i|<\epsilon
$$
for all $i$?
 A: [EDITED] $\def\conv{\mathop{\rm conv}\nolimits}$The answer is yes if $x_1,\dots,x_n\in(0,\pi]$ are pairwise distinct (are  these `reasonable restrictions'?). 
Consider a set $C=\{x(t)=(\cos tx_1,\dots,\cos tx_n)\colon t\in\mathbb Z\}\subset \mathbb R^n$. Let $D$ be the closure of the convex hull of $C$.
I claim that $0$ is the interior point of $D$. This yields that for every sufficiently small $r_1,\dots,r_n$ there are required coefficients $a_j\geq 0$ (at most $n+1$ of them nonzero) summing up to $1$. An appropriate scaling then completes the proof 
Firstly, notice that 
$$
  \frac1N\sum_{t=1}^N x(t)\to 0, \quad N\to+\infty,
$$
so $0\in D$. The only case to exclude is that $0$ is a boundary point of $D$, or --- that there esists a supporting hyperplane of $D$ passing through $0$, or --- that there exist some nontrivial coefficients $\alpha_i$ such that
$$
  f(t)=\sum_{j=1}^n\alpha_j\cos tx_j\geq 0
$$
for all $t\in\mathbb Z$. 
So we need to show that it is impossible. Assume first that $f$ is not identically zero. Then there exists some $n$ such that $\delta=\sum_{t=1}^n  f(t)>0$. This yields that $\sum_{t=1}^N f(t)\geq \delta$ for all $N\geq n$. This is false, since 
$$
  \sum_{t=1}^N f(t)
  =\sum_j\alpha_j\sum_{t=1}^N\cos tx_j
  =\sum_j\alpha_j\frac{\sin(N+1/2)x_j-\sin x_j/2}{\sin x_j/2},
$$
and all fractions can simultaneously be made very small by Kronecker's theorem.
We are left to show that $f$ is not identically zero. Indeed, it is a nontrivial linear combination of the sequences $(\exp(\pm itx_j))$ which are independent by Vandermonde.
