The interactive example above was produced using the
Theory of Functional Connections(TFC) , a mathematical framework discovered by
Daniele Mortari in 2017 with the seminal paper on the subject,
The Theory of Connections: Connecting Points. Over the next five years this theory was expanded, generalized, and refined into a textbook published by
Leake, C., Johnston, H, and Mortari, D, “The Theory of Functional Connections: A Functional Interpolation Framework with Applications,” Lulu (2022), ISBN: 9781716816642.
At its heart, Theory of Functional Connection is a generalization of the
Lagrange interpolating polynomial, but where the former produces a single interpolating function, TFC generates a family of interpolating functions represented through a functional-functional Interpolation. Additionally, TFC is not just limited to points and the theory has explored a wide range of constraints.
So how does that relate to our current example? Below you will see a "green" dot and a "red" dot (you can drag them around). Let's call the "green" dot \(\boldsymbol{p}_i = [x_i, y_i]\) which defines the initial \(x\) and \(y\) values of the dot within the frame. The same goes for the "red" dot, \(\boldsymbol{p}_f = [x_f, y_f]\). Additional, we will define the varible \(t \in [0, 1]\), and \(t_i = 0\) and \(t_f = 1\)
For this, the TFC
constrained express would take the form:
$$\boldsymbol{f}(t, \boldsymbol{g}(t)) = \boldsymbol{g}(t) + \frac{t_f - t}{t_f - t_i} (\boldsymbol{p}_i - \boldsymbol{g}(0)) + \frac{t - t_i}{t_f - t_i} (\boldsymbol{p}_f - \boldsymbol{g}(1))$$
You may notices that \(t_f - t_i = 1\) and could be simplfied in the equation, but we will kept it there for now (for a reason). Now, looking at this equations you should be able to see that \(\boldsymbol{g}(t)\) can be any function. If \(t=0\)
$$\boldsymbol{f}(\textcolor{red}{0}, \boldsymbol{g}(\textcolor{red}{0})) = \boldsymbol{g}(\textcolor{red}{0}) + \frac{t_f - \textcolor{red}{0}}{t_f - t_i} (\boldsymbol{p}_i - \boldsymbol{g}(0)) + \frac{\textcolor{red}{0} - t_i}{t_f - t_i} (\boldsymbol{p}_f - \boldsymbol{g}(1))$$
and then remembering that \(t_i = 0\) and \(t_f = 1\) and replacing those
$$\boldsymbol{f}(\textcolor{red}{0}, \boldsymbol{g}(\textcolor{red}{0})) = \boldsymbol{g}(\textcolor{red}{0}) + \frac{\textcolor{red}{1} - \textcolor{red}{0}}{\textcolor{red}{1} - \textcolor{red}{0}} (\boldsymbol{p}_i - \boldsymbol{g}(0)) + \frac{\textcolor{red}{0} - \textcolor{red}{0}}{\textcolor{red}{1} - \textcolor{red}{0}} (\boldsymbol{p}_f - \boldsymbol{g}(1))$$
where the first fraction goes to \(1\) and the second fraction goes to \(0\).
$$\boldsymbol{f}(\textcolor{red}{0}, \boldsymbol{g}(\textcolor{red}{0})) = \boldsymbol{g}(\textcolor{red}{0}) + (\boldsymbol{p}_i - \boldsymbol{g}(0))$$
but the \(\boldsymbol{g}(0)\) also cancel each other out leaving exactly the coordinates for our "green" dot starting point.
$$\boldsymbol{f}(\textcolor{red}{0}, \boldsymbol{g}(\textcolor{red}{0})) = \boldsymbol{p}_i = [x_i, y_i]$$
You can do the same thing for \(t=1\) and should arrive at the position of the "red" dot. But what is \(g(t)\)? It can be anything! Any function that you can imagine you can use as \(g(t)\). So what does that mean? It means that our functions \(f(t, g(t))\) represents all paths starting at "green" dot and ending at "red" dot.
Okay, so what if \(g(t)\) is just the function \(0\)? Well, now we come to an interesting realization:
$$\boldsymbol{f}(t, \textcolor{red}{0}) = \textcolor{red}{0} + \frac{t_f - t}{t_f - t_i} (\boldsymbol{p}_i - \textcolor{red}{0}) + \frac{t - t_i}{t_f - t_i} (\boldsymbol{p}_f - \textcolor{red}{0})$$
which simplfies to
$$\boldsymbol{f}(t, \textcolor{red}{0}) =\frac{t_f - t}{t_f - t_i} \boldsymbol{p}_i + \frac{t - t_i}{t_f - t_i} \boldsymbol{p}_f$$
which is simply the
Lagrange interpolating polynomial through two points.