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For $a\in R$, let $x_1 = a$ and $4x_{n+1} = x_n^2 + 3$ for all $n \in N$. Show that $x_n$ converges iff $|a|\le3$. Moreover, find the limit of the sequence when it converges.

To figure out the monotonicity of the sequence, I subtracted $4x_n$ on both sides. The RHS can now be easily factorised into $(x_n - 1)(x_n - 3)$ This didn't help, as the sign of RHS depends on what interval $x_n$ lies in.

Next, I tried to play with the ratio of $\frac{x_{n+1}}{x_n}$ ,which is clearly just $\frac{x_n}{4} + \frac{3}{4x_n}$. Using the AM GM inequality on the RHS, we can say that the ratio under consideration is always greater than or equal to $\frac{\sqrt3}{2}$.This doesn't help either, as it is only a lower bound (on the other hand, an upper bound might have helped)

So, could someone please help me with the solution, or possibly point me in the right direction? Extensions of the two approaches I've taken until now would be great too, if possible! Thanks!

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We first note that if the sequence converges to a limit $L\in\mathbb{R}$ then $L$ satisfies the equation $4L=L^2+3$ and we may conclude that $L\in\{1,3\}$.

Now let $f(x):=\frac{x^2+3}{4}$, so that $x_{n+1}=f(x_n)$, and in order to discuss the monotonicity of the sequence we consider a few cases.

1) If $x_1=1$ or $x_1=3$ then the sequence is constant (and convergent). If $x_1=-1$ then $x_2=1$ and the sequence becomes constant. If $x_1=-3$ then $x_2=3$ and the sequence becomes constant.

2) If $x\in (1,3)$ then $$1<f(x)=\frac{x^2+3}{4}< x.$$ which implies that if $x_1\in(1,3)$ then $(x_n)_n$ is strictly decreasing and bounded in $(1,3)$ and therefore it has finite limit $L=1$.

3) If $x_1\in (-3,-1)$, then $x_2\in(1,3)$ and again we have the convergence to $L=1$.

4) If $x\in(-1,1)$ then $$1> f(x)=\frac{x^2+3}{4}> x$$ which implies that if $x_1\in(-1,1)$ then $(x_n)_n$ is strictly increasing and bounded in $(-1,1)$ and therefore it has finite limit $L=1$.

Finally, if $|x|>3$ then $$f(x)=\frac{x^2+3}{4}> x$$ which implies that if $|x_1|>3$ then $(x_n)_n$ is strictly increasing and unbounded (why?).

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  • $\begingroup$ @arya_stark Any further doubt? $\endgroup$ – Robert Z Aug 15 at 7:38
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From $$4x_{n+1}=\frac{x_n^2}{4}+\frac{3}{4}$$ we get $$x_{n+1}-x_n=\frac{1}{4}(x_n^2-4x_n+3)=\frac{1}{4}(x_n-2)^2-\frac{1}{4}$$

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  • $\begingroup$ What then? How does it help me bound $x_1$, I wonder. $\endgroup$ – arya_stark Aug 15 at 7:20

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