Four Questions about this weird sequence

Let's start with Q2. As discussed above, we can formulate the recurrence purely in terms of $S_n$ as $S_{n+1}=S_n+\left(1-S_n \right) c^2+c \sqrt{\left(1-S_n \right)^2 c^2 + S_n \left(2-S_n \right)}$

Consider the recurrence as an iterated function scheme. Does it have any fixed points? Say $x$ is a fixed point; then

$\begin{aligned} x&=x+(1-x) c^2+c \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x) c^2 &= -c \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x) c &= - \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x)^2 c^2 &= (1-x)^2 c^2 + x (2-x) \\ x(2-x)&=0 \end{aligned}$

In other words, the system has two fixed points, $x=0$ and $x=2$. However, plugging in $x=0$ to the original expression, we find it is not a fixed point (the reason for this apparent contradiction is we lost some information when we squared both sides of the expression.)

It's easy to check that $x=2$ is a fixed point.

Let $f(x)=x+(1-x) c^2+c \sqrt{(1-x)^2 c^2 + x (2-x)}$

Differentiating, we find $f'(2)=0$

so that $x=2$ is in fact a stable fixed point, and the iterated function system does indeed converge to $S_n=2$.

Now Q1 is simple; since $S_n \to 2$, $S_{n}-S_{n-1}=a_n \to 0$

For q3, we can use small angle approximations. The thing to keep track of is $T_n=2-S_n$. As above, we get $T_{n+1}=T_n+\left(1-T_n \right) c^2-c \sqrt{\left(1-T_n \right)^2 c^2 + T_n \left(2-T_n \right)}$

With $c=\cos\theta \approx 1-\frac12 \theta^2$, neglecting terms in $\theta^3$ and above, this becomes $\begin{aligned} T_{n+1} &\approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{\left(1-T_n \right)^2 (1-\theta^2) + T_n \left(2-T_n \right)} \\ &=T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{1-\left(1-T_n \right)^2 \theta^2} \end{aligned}$

Since both $\theta$ and $1-T_n$ are small, we can also approximate the square root: $T_{n+1} \approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \left(1-\frac12 \left(1-T_n \right)^2 \theta^2 \right)$

After tidying up, this (amazingly) becomes $T_{n+1} \approx \frac12 \theta^2 T_n^2$

I'd appreciate some double-checking here, but this seems pretty accurate. If we take $\theta=0.2$ (which is not even that small; it's about $11.5^\circ$), and compare the exact and approximated $T$ values, the errors are very small.

So, this approximate $T$ has a much more simple recurrence, and it's solvable; since $T_0=2$, we get $T_n=2\theta^{2^{n+1}-2}$

This still looks a bit crazy, but hopefully you'll recognise the exponents:

Perhaps you can write the problem on a sheet and upload a snapshot. That way, it will be readable.

What's the derivation of this question? Did you find this recurrence from another calculation? If so that might be useful.

No solution yet but some observations:

• The form of the recurrence looks a lot like the solution of a quadratic; in fact, $a_{n+1}$ is one of the roots of $\frac{x^2}{c^2} - 2(1-S_n)x - S_n (2-S_n) = 0$

• we can reformulate this purely in terms of $S_n$; we have $S_1=2c^2$ and $S_{n+1}=S_n+\left(1-S_n \right) c^2+c \sqrt{\left(1-S_n \right)^2 c^2 + S_n \left(2-S_n \right)}$

• if we let $T_n=2-S_n$ and substitute in (with the idea of proving that $T_n \to 0$ in order to answer Q1 and Q2), we get $T_{n+1}=T_n+\left(1-T_n \right) c^2-c \sqrt{\left(1-T_n \right)^2 c^2 + T_n \left(2-T_n \right)}$ which is really similar to the relation for $S_n$

• alternatively, letting $R_n=1-S_n$ and $s=\sqrt{1-c^2}$, we find $R_{n+1}=s^2 R_n - c\sqrt{1-s^2 R_n^2}$

which looks a little like some sort of elliptic function. (That choice of $s$ is to make it equal to $\sin \theta$).