Recreational Math Problem about unknown amount of rotating vectors

Question

I created this recreational math problem that where it looks like solver doesn't have enough information to solve it. This kind of algebra problem I haven't really seen in the wild before, as you don't know the number of variables that you are given. Along the way of solving, there are several mathematical gems that you come across, (at least through the method that I solved it). I deleted a post of this exact problem with a physics narrative, but I figured that math by itself is more appropriate here.

Please have fun solving!

Problem

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A countable amount of vectors in $\mathbb{R}^2$ rotate about the origin at their own constant speeds. Each vector $\vec{v}$ has a magnitude that varies in time $t$ described by $|\vec{v}(t)| = a \cdot b ^ t$, for their own values of $a,b\in \mathbb R_{>0}$. The sum of the vectors at time $t$ is notated by $\vec{u}(t)$.

There's a time $t_c$ where each vector equals $\vec{v_c}$ and:

1. $|\vec{u}(t_c-1)| = 1+\frac{1}{|\vec{v_c}|}$.
2. $\vec{u}(t_c+1) = -\vec{v_c}$.
3. $\vec{u}(t_c+2) = 3 \vec{v_c}$.
4. $\vec{u}(t_c+n) = \vec{u}(t_c + n - 2) - \vec{u}(t_c + n - 1), \quad \forall n\in\mathbb{N}_{>2}$

What is the numerical value of $|\vec{v_c}|$?

Answer 1

The complex plane is basically ismorphic to $\mathbb{R}^2$, right? Let's set this in the complex plane. Then each of the vectors can be defined as $a \cdot b^t \cdot \exp(i\tau k(t-t_c) + i\tau m)$, where $k$ and $m$ depend on the period and phase. Also, I'm setting $t_c = 0$ to save on a bunch of annoying $(t-t_c)$ factors like we just saw.

Equations 2, 3, and 4 are $\vec v_c$ times the definition of the alternating Lucas sequence, i.e., $(1, -3, 4, -7, 11, -18, \cdots)$. That implies that $\vec u(0) = -2\vec v_c$, and $\vec u(-1) = -\vec v_c$. Why this implication? Because the three equations generate a recurrence, we can find an explicit closed form for the values of the magnitude over time. The characteristic polynomial is $x^2 +x-1$, which has the two roots $r_1, r_2 = -\phi, \phi -1 = 1/\phi$. (Here $\phi$ is the golden ratio, i.e. $\phi = \frac12(1+\sqrt 5)$.) Then the closed form is $$u(t) = ar_1^{t} + br_2^{t}$$ Then, we can solve this system in order to determine $a$ and $b$: $$\begin{cases} u(1) = v_c = a(-\phi)^1 + b\phi^{-1} \\ u(2) = -3 v_c = a(-\phi)^2 + b\phi^{-2} \end{cases} \\ \\ \implies a = b = -v_c \implies \color{green}{u(t) = -v_c(-\phi)^{t} - v_c\phi^{-t}} $$ We still need to fit with equation 1, but that's actually really easy. What number is equal to its own reciprocal plus one? Why $\phi$ of course! Since we have absolute values in 1, it could also be $-\phi$, but we'll assume not for the moment. This suggests we have two vectors: $$\vec v_1 = \phi \cdot \phi^t \exp \left(\frac{i\tau t + i\tau}{2} \right) = \phi^{(t+1)} \exp \left(\frac{i\tau t+i\tau }{2}\right)$$ $$\vec v_2 = \phi \cdot\phi^{-t} \exp \left(\frac{i\tau}{2} \right) = \phi^{(-t+1)} \exp\left(\frac{i\tau}{2}\right)$$ Note that $\vec v_1$ rotates once per $2$ time units, $\vec v_2$ is stationary, and at $t=0$, both are pointed along the negative real axis. These two vectors account for all of the constraints... except one. We're told that at $t=0$, all of the vectors are equal to $\vec v_c$, and at $t=1$, their sum is $\vec v_c$. Yet it's clear that at $t=0$ they both have value $-\phi$, but at $t=1$ their sum has value $\phi$. But here's the thing: any linear combination of vectors which produces the series after $t=1$ must be equivalent to the combination given above (or the same thing, with $\vec v_1$ turning clockwise rather than counterclockwise). Hence, the combination must produce a value of $u(0) = -2\phi$ rather than $u(0) = 2\phi$, as implied by the question.

I look forward to seeing what the correcting factor is that I've missed.

Oh! The actual answer to the question is, of course, $\phi$, and I don't think it depends on the recurrence at all; only on equation 1.