Is this even possible to compute?

Question

One of my homework questions was the following (please see image). However, I ask myself if it is even possible to compute it without any extra knowledge about the sequence itself.

There were some subquestions about this exercise, which for example assumed beta to be 1; however, I do not think that it is allowed to assume such things for this certain question.

But if there is no way to solve this question without any other assumptions than the one in the image, then I guess I will be having to use the assumptions of the other subquestions regarding this sequence.

Thank you very much in advance.

Answer 1

There's a clue in the question, which is entitled "Monotone convergent sequences". Here's the most mechanical way of doing it:

• Find conditions under which the sequence is monotone. (You will have various cases depending on $\beta$.)
• Prove that the sequence [almost always] eventually reaches a point where it becomes monotone for the rest of time.
• Prove that the sequence is [almost always] bounded below.
• Prove that the sequence is [almost always] bounded above.
• Deduce that the sequence [almost always] converges.
• Find the two possible limits, and show that the sequence in fact [almost always] only converges to one of them.

Note that there are cases where the sequence fails to be even well-defined, so you need to rule those out first.

Answer 2

The map $T:\>x\mapsto 4-{2\over x}$ can be viewed as restriction of a Moebius transformation $T:\>\bar{\mathbb C}\to\bar{\mathbb C}$ to $\bar{\mathbb R}$. The fixed points of $T$ are obtained by solving the equation $x=4-{2\over x}$, and are $2\pm\sqrt{2}$. This suggests to introduce in $\bar {\mathbb C}$ the new complex projective coordinate $$u={x-2-\sqrt{2}\over x-2+\sqrt{2}}=:S(x)\ .$$ With respect to $u$ the given $T$ computes to $$u\mapsto \hat T(u)=S\circ T\circ S^{-1}(u)=(3-2\sqrt{2})u\ .$$ Since $3-2\sqrt{2}=0.1716$ it follows that in the $u$-system the point $u_1=0$ is an attracting fixed point of $\hat T$, and $u_2=\infty$ is a repelling fixed point of $\hat T$. In terms of the variable $x$ this means that $x_0=2+\sqrt{2}$ is attracting, and $x_1=2-\sqrt{2}$ is repelling. Concerning the sequence $(a_n)_{n\geq0}$ in the question we therefore can say that it converges to $2+\sqrt{2}$, whatever the initial point $a_0$, as long as $a_0\ne2-\sqrt{2}$.