I am doing some research into the movement of robots executing a given algorithm, and have come up with the following recursive equations that describe the movement of the robots at each iteration:

\begin{equation} \theta_i = \frac{3}{4}\theta_{i-1} \end{equation}

\begin{equation} L_i=\frac{1}{2}\sqrt{L_{i-1}^2+A_{i-1}^2} \end{equation}

\begin{equation} A_i=L_i\tan\theta_i \end{equation}

Where the initial conditions are $\theta_1=45^{\circ}$, $\;L_1=\frac{1}{2}d$, and $\frac{1}{2}d$.

It is clear that the first recurrence relation can simply be replaced with the following closed-form solution:

\begin{equation} \theta_i = \big(\frac{3}{4}\big)^{i-1}\;\theta_1 \end{equation}

But my problem is trying to obtain closed-form solutions to the other two equations that are defined in terms of each other. Is there a way to do this?

  • $\begingroup$ $L_i=\frac{1}{2}\sqrt{L_{i-1}^2+A_{i-1}^2}=\frac12 L_{i-1}\sqrt{1+\tan\theta_{i-1}}$, ... $\endgroup$ – Jens Schwaiger Jul 15 '18 at 3:09
  • $\begingroup$ You can eliminate $\,A_i\,$ and get $\displaystyle L_i^2 = \frac{L_{i-1}^2}{4 \cos^2 \theta_{i-1}}$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form. $\endgroup$ – dxiv Jul 15 '18 at 3:13

I believe if you replace the $A_{i-1}$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (\frac12)^{i-1}\prod_{k=1}^{i-1}sec(θ_k)L_1$$ and that consequently, $$A_i = (\frac12)^{i-1}\prod_{k=1}^{i-1}sec(θ_k)L_1tan(θ_i)$$ where you have already given a closed form formula for $θ_i$.


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