Parameterization of simple closed curve A curve $Z$ in a two dimensional space is parametrized by $0\leq t < 1$ , and satisfies $Z(t) = Z(t+1)$. If it is sufficiently well behaved, it can be represented using a Fourier series with a basic frequency of $1$.
So far so good, but this parametrization still allows the curve to intersect itself. I'm looking for a parametrization which allows only for simple closed curves. Any ideas as to how I might be able to enforce this analytically? (this is to be used for an optimization problem in which a few points are known to lay on a simple closed curve)
 A: A parametrized curve Z is called closed if there exists a parametrization $p:[a,b] \rightarrow \mathbb{R}^2$ of Z such that $p(a)=p(b)$. According to what you wrote Z is the curve and also its parametrization. This is confusing, so I prefers call the Z parametrization $p$. A closed curve can of course intersect itself. 
If the curve is closed and regular we have $p(a)=p(b)$, $p'(a)=p'(b)$, $p''(a)=p''(b)$, etc.
In our case you are looking for a closed simple curve, i.e. a closed curve with no self-intersections except at the endpoints.
A more topological approach involves using a circle $\mathbb{S}^1$ as the domain for the map of $p$.The circle $\mathbb{S}^1$ is defined as the set [0,1] with the points 0 and 1 identified. Then an open neighbothood V of a point m, contains for some number $\epsilon$ the subsets
\begin{cases}
(m-\epsilon,m+\epsilon)  \enspace with  \enspace \epsilon<min\{m, 1-m\}  \enspace if  \enspace m\neq 0\\
[0,\epsilon) \mp (1-\epsilon,1]  \enspace if  \enspace m=0 \enspace or  \enspace m=1
\end{cases}
We say that Z is closed if there is a continuous surjective (onto) function $\psi:\mathbb{S}^1  \rightarrow Z$. We sy that a closed curze Z is simple if there exists a bijection $\psi:\mathbb{S}^1  \rightarrow Z$ that is continuous and such that its inverse is continuous.
Alternatively If you want just find a curwe which is simpàle you can restrict yourself to closed regular convex curves. A closed regular convex plane curve Z is convex if and only if its curvature $k_g$ does not change sign.
For instance if Z is parametrized by $\{x(t),y(t) \}$ the curvature is:
\begin{equation}
k_g(t) = \frac{x'(t)y''(t)-x''(t) y'(t)}{(x'(t)²+y'(t)²)\frac{3}{2}}
\end{equation}
