(I am reposting this from physics stack exchange since they seem to not accept this type of question, and in my case it is also probably more math-focused.)
I am trying to solve a 3D ballistics problem: given initial kinematic conditions, find the minimum required initial velocity of a projectile (subject to acceleration) such that it hits the target. Details can be found here (post was closed for "not being focused", thought I think the problem description is quite specific).
I have thought of a solution like this (all the variables denote initial conditions, at time $0$):
Using kinematic equations derived from differential equations, we get that for each x, y, z component of the 3-dimensional vector, $$x_p+ \dot{x_p} \cdot t+ \frac{1}{2} \ddot{x} \cdot t^2= \sum_{k=0}^{n} \frac{x^{(k)}_t t^k}{k!}$$ where $x^{(k)}$ is the $k$-th derivative of position. $p$ subscript denotes projectile, $t$ subscript denotes target. $n$ is the highest non-zero position derivative.
Moving everything to one side of the equation (say, let $x$ be $x_p - x_t$): $$\sum_{k=0}^{n} \frac{x^{(k)}t^k}{k!}=0.$$ At this stage, every variable is known except for time $t$ and projectile velocity $\dot{x}_p$.
I extended the reasoning used on Wikipedia to solve for time. Since there is only one minimum velocity possible, then we constrain the equation to have only one solution. In the Wikipedia page, they set the discriminant of the polynomial to 0 to find the point at which all the possible paths converge at the minimum velocity's one. This solution has worked for constant acceleration when I tried it in Unity. I figured a similar approach could work if I applied this to other equations, setting the multiplicity of the root to the degree of the polynomial. However, it was not feasible for higher-degree polynomials since the discriminant gets more complicated. I approached it like this:
By setting the discriminant to $0$, we are basically trying to find the point where all the points converge to a single root of multiplicity $n$, the degree of the polynomial. Therefore, we impose the equation with the form $$\frac{x^{(n)}}{n!}(t-t_0)^n=\sum_{k=0}^{n} \frac{x^{(k)}t^k}{k!}=0$$ My idea was to expand using the binomial theorem and compare coefficients. However, what I got was $$t_0=\sqrt[n]{\frac{n\cdot x}{x^{(n)}}}$$ and $$\dot{x}_p=\frac{x^{(n)}}{(n-1)!}t_0^{n-1}+\dot{x}_t$$ This equality does not make any sense. It is saying that the initial velocity does not depend on any other initial conditions than the initial positions, as well as the velocity and highest derivative of the target. So what exactly went wrong? This approach works and gives the correct solution for the case of constant acceleration as covered on Wikipedia, providing the same answer. However, I do not understand why this does not extend to higher derivatives. Why does matching the coefficients give this result? After all, polynomials are uniquely identified by their coefficients, so we can compare them, and I followed the procedure listed to find the minimum velocity in the case of constant acceleration.
Also, if anyone is able to provide a correct solution, that would be appreciated. I would like to know how to properly solve this problem.