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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)

JMCF125
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user1834069
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  • In other words, a complex function $t\in [0,1)\mapsto \mathbb{C}$ that you are representing as a series, has to be an injective function. I don't think there's an easier way of saying that. – orion Mar 29 '14 at 13:20
  • Please plot real, imaginary parts of a typical Z(t). – Narasimham Dec 25 '14 at 20:33
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    Don't know if this fits the bill somehow, but isolines / contours of some "neat" function $f(x,y)$ can never self intersect; i.e. solve $(x,y)$ from $f(x,y) = c$ where $c$ is a constant with $\min(f(x,y)) < c < \max(f(x,y))$ (everything real-valued). Then parametrize the result. I routinely do such things numerically, though. – Han de Bruijn Dec 29 '16 at 15:14
  • Perhaps also useful to know : Can a polygon with minimal perimeter self-intersect? . Where a polygon may be considered as the discretization of a closed curve. Or the closed curve is "continuization" of a polygon. – Han de Bruijn Dec 30 '16 at 10:44

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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}

Upax
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