System of equations and recurrence relation

Question

I am trying to find the general solution for $N$ of the following system of equations

$$ \begin{cases} (x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\ {x_n}^2 + {y_n}^2 = 1 \end{cases} $$

with the initial values $x_0 = 1$ and $y_0 = 0$ and the following

• $\theta$ is a constant and $0 \leqslant \theta \leqslant 2$
• $N$ is a constant and we want to find the terms $(x_N, y_N)$

Using substitution with respect to $N$, we have

$$ \begin{align} x_0 = 1 \quad & ; \quad y_0 = 0 \\ x_1 = -\frac{\theta^2 - 2N^2}{2N^2} \quad & ; \quad y_1 = -\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2} \\ x_2 = \frac{\theta^4 - 4N^2\theta^2 + 2N^4}{2N^4} \quad & ; \quad y_2 = \frac{(\theta^3 - 2N^2\theta) \sqrt{4N^2 - \theta^2}}{2N^4} \\ x_3 = -\frac{\theta^6 - 6N^2\theta^4 + 9N^4\theta^2 - 2N^6}{2N^6} \quad & ; \quad y_3 = -\frac{(\theta^5 - 4N^2\theta^3 + 3N^4\theta) \sqrt{4N^2 - \theta^2}}{2N^6} \end{align} $$

By using substitution, it becomes very difficult with $N \geqslant 2$.

Answer 1

Using complex numbers as well, it is obvious from the second equation that the $n^{th}$ point can be written

$$z_n=e^{i\alpha_n}.$$

Then

$$|e^{i\alpha_n}-e^{i\alpha_{n-1}}|=|e^{i(\alpha_n-\alpha_{n-1})}-1|=|e^{i\alpha}-1|=\frac\theta N.$$

It is easy to show that this equation has at most two solutions in $e^{i\alpha}$, which are conjugate. If we always keep the same sign of $\alpha$, then by induction $$z_n=e^{i\alpha_n}=e^{i\alpha_{n-1}}e^{i\alpha}=e^{in\alpha}.$$

Answer 2

The second equation expresses that the points $(x_n,y_n)$ remain on the unit circle, and the first, that the successive points form chords of constant length, subtending an angle $\alpha=2\arcsin\frac\theta{2N}$.

Hence $$(x_n,y_n)=(\cos n\alpha,\sin n\alpha).$$

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By the way, the system has in fact $2^n$ distinct solutions, as from every intermediate point, the chord can be drawn in two directions.