What is the Condition for Convergence of Sequence $p_{n}$?

[Francolino]

Let $ p_{n} = (x_{n},y_{n}) $ be a sequence that verifies:

$$ x_{n+1} = \sqrt{x_{n}^2 + y_{n}^2}\left ( \frac{1}{2}x_{n} - \frac{\sqrt{3}}{2}y_{n} \right ), \quad y_{n+1} = \sqrt{x_{n}^2 + y_{n}^2}\left ( \frac{\sqrt{3}}{2}x_{n} + \frac{1}{2}y_{n} \right ), \quad n \geq 0 $$

Which is the necessary and suficient condition that $ x_{0} $ and $ y_{0} $ have to verify to be sure that $ p_{n} $ is convergent?

I literally don't know how to solve this one. I hope you can help me. :)

 

[Opalg]

As a hint, try expressing $p_n$ in polar form $(r_n, \theta_n)$ instead of $(x_n,y_n)$.

 

[Francolino]

First of all, thank you for answering!

I did what you suggested:

$$ \left\{\begin{matrix}
\displaystyle r_{n+1}\cos(\theta_{n+1}) = r_{n}\left ( \frac{1}{2}r_{n}\cos(\theta_{n}) - \frac{\sqrt{3}}{2}r_{n}\sin(\theta_{n}) \right )\ \\
\displaystyle r_{n+1}\sin(\theta_{n+1}) = r_{n}\left ( \frac{\sqrt{3}}{2}r_{n}\cos(\theta_{n}) + \frac{1}{2}r_{n}\sin(\theta_{n}) \right )\ \\
\end{matrix}\right. $$

Even though, I still don't see what's the matter with this. :/

 

[Opalg]

First of all, thank you for answering!

I did what you suggested:

$$ \left\{\begin{matrix}
\displaystyle r_{n+1}\cos(\theta_{n+1}) = r_{n}\left ( \frac{1}{2}r_{n}\cos(\theta_{n}) - \frac{\sqrt{3}}{2}r_{n}\sin(\theta_{n}) \right )\ \\
\displaystyle r_{n+1}\sin(\theta_{n+1}) = r_{n}\left ( \frac{\sqrt{3}}{2}r_{n}\cos(\theta_{n}) + \frac{1}{2}r_{n}\sin(\theta_{n}) \right )\ \\
\end{matrix}\right. $$

Even though, I still don't see what's the matter with this. :/

The next step is to think geometrically. The numbers $\dfrac12$ and $\dfrac{\sqrt3}2$ ought to remind you of things from trigonometry. Can you use that to give a geometric description of $p_{n+1}$ in terms of $p_n$?

 

[Francolino]

I must be blind, but I cannot see it.

Anyway, the fact that $ \cos(\pi/3) = 1/2 $ and $ \sin(\pi/3) = \sqrt{3}/2 $ seems very curious to me. Is $ \pi/3 $ important in this context?

 

[Opalg]

Anyway, the fact that $ \cos(\pi/3) = 1/2 $ and $ \sin(\pi/3) = \sqrt{3}/2 $ seems very curious to me.

Right, so $\frac{1}{2}r_{n}\cos(\theta_{n}) - \frac{\sqrt{3}}{2}r_{n}\sin(\theta_{n}) = r_n\bigl(\cos(\pi/3\cos(\theta_{n}) - \sin(\pi/3)\sin(\theta_{n}) \bigr) = \ldots$ (use a trig. addition formula). Can you continue from there?

 

[Francolino]

I came up to this:
$$ \left\{\begin{matrix}
r_{n+1}\cos(\theta_{n+1}) = r_{n}\cos(\theta_{n} + \pi/3) \\
r_{n+1}\sin(\theta_{n+1}) = r_{n}\sin(\theta_{n} + \pi/3)
\end{matrix}\right. $$
(I hope I didn't make any silly mistake).

 

[Opalg]

I came up to this:
$$ \left\{\begin{matrix}
r_{n+1}\cos(\theta_{n+1}) = r_{n}\cos(\theta_{n} + \pi/3) \\
r_{n+1}\sin(\theta_{n+1}) = r_{n}\sin(\theta_{n} + \pi/3)
\end{matrix}\right. $$
(I hope I didn't make any silly mistake).

Correct except that the $r_n$s should be squared: $$ \left\{\begin{matrix}
r_{n+1}\cos(\theta_{n+1}) = r_{n}^2\cos(\theta_{n} + \pi/3) \\
r_{n+1}\sin(\theta_{n+1}) = r_{n}^2\sin(\theta_{n} + \pi/3)
\end{matrix}\right. $$
What that says is that $r_{n+1} = r_n^2$ and $\theta_{n+1} = \theta_n + \pi/3$. In other words, to get from $p_n$ to $p_{n+1}$, you square its modulus and you rotate it through an angle $\pi/3$. What condition will you then need to ensure that the sequence $(p_n)$ converges to a limit?

 

[Francolino]

Well, clearly the rotation won't decide the convergence. Then, if $ r_{n} \geq 1 $ it'll diverge. So $ r_{n} < 1 $ it'll converge (and we don't care about $ \theta_{n} $?).

So if $ r_{n} < 1 $ then $ r_{n}^2 < 1 $. Now, squaring and summing both equations (the ones I used to convert into polar coordinates*) I've got:
$$ x_{n}^2 + y_{n}^2 = r_{n}^2\cos^2(\theta_{n}) + r_{n}^2\sin^2(\theta_{n}) = r_{n}^n < 1 $$

Then, the codition is that $ x_{0}^2 + y_{0}^2 < 1 $?

---

* $ x_{n} = r_{n}\cos(\theta), y_{n} = r_{n}\sin(\theta) $

 

[Opalg]

Then, the condition is that $ x_{0}^2 + y_{0}^2 < 1 $?

Yes, exactly! If $ x_{0}^2 + y_{0}^2 < 1 $ then $r_n\to0$ as $n\to\infty$ and so the sequence converges to $0$.

If $ x_{0}^2 + y_{0}^2 > 1 $ then $r_n\to\infty$ and so the sequence will diverge to infinity.

Finally, if $ x_{0}^2 + y_{0}^2 = 1 $ then the sequence will rotate around the unit circle and will not converge.