For what $a$ the sequence $2x_{n+1}=x^2_n+a$ has finite limit? Let $(x_n)$ be the sequence defined recursively by the following formular $$x_0=0, \ 2x_{n+1}=x^2_n+a$$
Find $a$ such that the above sequence has finite limit?
With all of my effort, there is still one case left, say, $a&lt;-3$, in which I don't even know how to prove or disprove the existence of the limit.
It's evident that if the sequence has a limit, this limit must satisfy the following equation:
$$2t=t^2 +a$$
and then if the sequence converge, it should converge to one of the 2 fixed point $p_1=1 +\sqrt{1-a}$,$p_2=1 -\sqrt{1-a}$.
Let's denote function $f(x)=\frac{1}{2}(x^2+a)$ with $a&lt;-3$  (and so $x_{n-1}=f(x_n)$), we have
$$f'(p_1)=1 +\sqrt{1-a} &gt;1$$
$$f'(p_1)=1 -\sqrt{1-a}&lt;-1$$
then $|f'(p_{1,2})|&gt;1$, according to this post, the both fixed points $p_1$ and $p_2$ are unstable. Because $x_0 \notin \{\pm p_{1,2}\}$ (4 values $p_1,p2,-p_1,-p_2$), then the sequence will never be able to hit the fixed points and so doesn't converge.
Conclusion: the sequence $x_n$ doesn't converge for $a&lt;-3$.
PS: why these conditions $x_0 \notin \{\pm p_{1,2}\}$  are enough to conclude the divergence of the sequence? Let's prove by contractiction. Suppose $k$ is the smallest value that $x_k \in \{\pm p_{1,2}\}$. Then we have $f(x_{k-1})=f(p)$ with $p \in \{\pm p_{1,2}\}$. But this implies that $\frac{1}{2}(x_{k-1}^2 +a)= \frac{1}{2}(p^2+a)$ or $x_{k-1} \in \{\pm p_{1,2}\}$, hence $k-1$ which is smaller than $k$ is also satisfy the assumption(=&gt; contradiction).
You could say $2x=x^2+a$ and then you get
the discriminante $D=4-4a$.
Now for $a &lt; 1$ you get 2 real solutions.
For $a = 1$ you get 1 real solution and for $a &gt; 1$ you get none.
Moreover you have to show that the sequence is monotonously falling and limited for certain intervalls.