Fourier series of a function with boundary conditions

Question

Let's consider any parametrizable curve $g : R \rightarrow C$

For that I thought of creating Fourier series $g(t)=\sum_{n=-\infty}^{\infty} c_ne^{int}$ which can possibly cover any curve on $C$ plane.

For the sake of optimization problem I have a differentiable loss function

$F: C^m \rightarrow R$

$F(c_{-n},...,c_{-1}, c_{0}, c_{1}, ..., c_{n})$

It takes the parameters of $g$ and returns some real value.

I would like to adjust $c_{n}$ with gradient descent to find the optimal path $g$ (reach the minimum of $F$).

The problem starts when the boundary conditions come in.

A curve must start at $g(0) = z_0$ and end on $g(t_0) = z_1$. Can the Fourier series represent any curve starting meeting these conditions? If that's the case, how?

What are the other approaches to this problem not necessarily using Fourier series?

Answer 1

One solution is create a polynomial instead of Fourier series as @TravorLZH stated.

$W(t) = F(t) + P(t)$

$F$ would regulate the conditions $W(0) = z_0, W(t_0) = z_1$

and $P$ would be the one to optimize on, but it has to keep the requirement

$t \epsilon \{0, t_0\} \rightarrow P(t) = 0 $

To solve this, first I'll start with $f$. It can be a simple linear function such that:

$F(t) = ax+b F(0) = z_0$ thus $b = z_0$ and $F(t_0) = z_1$ thus $a = \frac{z_1 - z_0}{z1}$

Then when it comes to $P(t)$, I divided it into

$P(t) = (t)(t-t_0)P_1(t)$

I'd create a random polynomial $P_t(t) = c_0*t^0 + c_1*t^1+ ... + c_n*t^n$, on which I can optimize the problem by manipulating $c_0, c_1, ..., c_n$ Note that by adjusting these coefficients we can preserve the boundary conditions.

The final polynomial is the $W(t)=F(t)+P(t) = (\frac{z_1 - z_0}{z1}t^2 +z_0t) + (t)(t-t_0)(c_0*t^0 + c_1*t^1+ ... + c_n*t^n)$.

Answer 2

Theorem: Let $f : [0,2\pi]\rightarrow\mathbb{C}$ be a function of bounded variation. Then the Fourier series for $f$ converges pointwise everywhere to the mean of the left- and right-hand limits of $f$ for all $0 < t < 2\pi$, and converges to $\frac{1}{2}(f(0^+)+f(2\pi^-))$ at $0,2\pi$. (These limits exist for such an $f$.) Furthermore, the convergence is uniform on any interval $[a,b]$ where $f$ is continuous at every point of the interval, including $a,b$.