Limits of recursions like $f(n+2)=\frac{1}{f(n)} - \frac{1}{f(n+1)}$ !?

Question

Consider the sequences

$$f(0)=1,f(1)=2$$

$$f(n+2)=\frac{1}{f(n)} - \frac{1}{f(n+1)}$$

$$g(0)=1,g(1)=2,g(2)=3,g(3)=4$$

$$g(n+4)=\frac{3}{g(n)} - \frac{3}{g(n+1)} + \frac{3}{g(n+2)} - \frac{3}{g(n+3)}$$

Then as $n$ goes to infinity :

$$\lim f(n)^2 = 2$$ $$\lim g(n)^2 = 12$$

Notice that one can not simply plug in the limits $L_1$ or $L_2$, because

$$L_1=\frac{1}{L_1} - \frac{1}{L_1} = 0 ??$$

does not make sense and

$$L_2=\frac{3}{L_2} - \frac{3}{L_2} + \frac{3}{L_2} - \frac{3}{L_2} = 0 ??$$

does not make sense either.

So we must keep in mind that it behaves different; double limits or multiple limits or limits converging at different rates.

But also noteworthy is this :

The initial values $f(0),f(1),g(0),g(1),g(2),g(3)$ do not matter much as long as most of them are distinct like $f(0) \neq f(1)$ for instance. We get the same limits anyways.

Also keep in mind that replacing $-$ with $+$ in the formulas will also work BUT changing the positions of $+$ or $-$ will NOT.

However although most initial conditions will work for both $f$ and $g$ , in particular if no division by zero occurs , some exceptional ones might not work.

And what if we start with say gaussian integers or eisenstein integers or complex numbers ?

Despite it almost always converges to these limits , what are the exceptional set of exceptions ?

Are the exceptions related to continued fractions, julia sets or irrationality measures ?

I considered iterating the equations a few times to get expressions for $f(n+5),g(n+5)$ and such and although it has its benefits and seems to make more sense in a way ( dealing with the + and - ) it also becomes more complicated fast. ( somewhat remind me of somos or modular equations )

Also want to note these equations are part of a large family, I could easily describe a similar $h(n)$ depending on 6 or 10 previous terms converging to algebraic numbers.

I also considered reducing the dependancy on previous values, by special case starting positions or by trying to relate it to pure iteration ( of one previous value ) of other functions ( like fibonacci is related to exp ).

But it seems non-trivial.

It is not clearly related to typical numerical methods such as newton's method or continued fractions or the babylonian method.

How do we even prove it attracts ?

Notice limits like $\frac{f(2n)^2 - 2}{f(2n+2)^2 - 2}$ and the alike also do not seem to converge in an easy pattern, so this sets it apart from most numerical methods. This is probably due to the divisions.

I considered comparing to median methods and other averaging methods but it just seems different fundamentally.

Any insight in this would be appreciated.

edit

I found this question as a "+ analogue" for the case $f(n)$.

For that case there are starting values that get different outcomes.

So that suggests maybe here also ?

This was asked in the comments and mentioned in the OP, so I add it now :

Why does the process defined with $a_{n+2} = \frac{1}{a_n} + \frac{1}{a_{n+1}}$ converge to $\pm\sqrt{2}$ for most choices of the starting values?

and the case for $g(n)$ is probably more complicated.

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Answer 1

Firstly, let $a_n = f(n)$. Now, notice: $$1\le a_n \le2 \ \ \& \ -2 \le a_{n+1} \le-1 \implies 1\le \frac{1}{a_n} - \frac{1}{a_{n+1}} \le2$$ Then we have: $$-2\le a_{n+1} \le-1 \ \ \& \ 1 \le a_{n+2} \le2 \implies -2\le \frac{1}{a_{n+1}} - \frac{1}{a_{n+2}} \le-1$$ A little computation shows that: $$1\le a_6 \le2 \ \ \& \ -2 \le a_7 \le-1$$ So we can see inductively that past this point, the sign of $a_n$ is alternating. This then tells us:

$$\lvert a_{n+2} \lvert \ = \ \lvert \frac{1}{a_n} - \frac{1}{a_{n+1}} \lvert \ = \, \frac{1}{\lvert a_n \lvert} \ + \ \frac{1}{\lvert a_{n+1 }\lvert} $$ (Check this if you are unsure).
Then letting $b_n = \lvert a_n \lvert$, we have the same conditions as in this question: limit of : $a_{n+2} =\frac{1}{a_n} + \frac{1}{a_{n+1}}$
So we know: $\lvert a_n \lvert \to \sqrt{2} \ $ hence $ \ a_n^2 \to 2$
You were wondering why just plugging in the limits was giving $L_1 = 0$. This is because you were not considering the alternating sign. We have $a_{2n} \to \sqrt{2} \ $ and $ \ a_{2n+1} \to -\sqrt{2} $, but in your attempt you assume $\lim a_{2n} = \lim a_{2n+1}$ which is false.

As for the second part, we can again notice and prove an alternating sign similiarly, then if we let $L = \lim{\lvert g(n) \lvert} $ and plug into the recursion, we get: $$L = \frac{3}{L} + \frac{3}{L} + \frac{3}{L} + \frac{3}{L} \implies L^2 = 12 $$ Unfortunately I wasn't able to prove that $\lvert g(n) \lvert$ converges. Hopefully you can make a good attempt at this with what's given above, or someone else can shine some insight on this problem. Good luck!