Convergence of recurrence sequence $a_n=a_{n-1}+1/2(a-a_{n-1})^2$, where $|a|\leq 1$ and $a_0=0$ The recurrence sequence $a_n=a_{n-1}+1/2(a-a_{n-1})^2$, where $|a|\leq 1$ and $a_0=0$ is clearly increasing sequence. How to prove it is positive bounded above? What is the limit of the sequence? Any hints how to proceed please.
 A: We can see that the sequence is always non-decreasing by studying the first difference:
$$a_n-a_{n-1}=0.5(a-a_{n-1})^2>0$$
$$(a-a_{n-1})^2>0\implies a_{n-1}<a\,\vee\,a_{n-1}>a $$
$$a_n-a_{n-1}=0.5(a-a_{n-1})^2=0$$
$$(a-a_{n-1})^2=0\implies a_{n-1}=a $$
$$\implies a_n\geq a_{n-1}\,\forall n \in \mathbb{N}$$
If we assume that the sequence has a limit $L$, then
$$L-L=0.5(a-L)^2\implies L=a$$
Now let us see under which conditions this sequence is bounded above by $a$:
$$a_n=a_{n-1}+0.5(a-a_{n-1})^2< a$$
$$a_{n-1}+0.5a^2+0.5a_{n-1}^2-aa_{n-1}< a$$
$$a_{n-1}+0.5a^2+0.5a_{n-1}^2-aa_{n-1}< a$$
$$0.5a_{n-1}^2+a_{n-1}(1-a)+(0.5a^2-a)< 0$$
$$a^*_{n-1}=-(1-a)\pm \sqrt{(1-a)^2-2(0.5a^2-a)}=\\
=-(1-a)\pm\sqrt{1+a^2-2a-a^2+2a}=a-1\pm1$$
$$a^*_{n-1}\in(a-2,a)$$
and under which conditions $a$ is reached:
$$a_n=a_{n-1}+0.5(a-a_{n-1})^2= a \implies a^*_{n-1}\in \{a-2,a\}$$
Indeed by setting $a_0<a-2$ or $a_0>a$, the sequence shoots up and diverges. So if we set $a_0=0$ and $a\in [0,2]$, then the sequence converges to $a$ because it will be always strictly increasing towards $a$ or reaches $a$ in finite time.
A: The limit of the sequence, if it exists, is a fixed point of $g(x)=x + \frac 12(a-x)^2$. This means that the only possible limit is $\lim a_n = a$. If you apply the fixed point theorem on the interval $[0,a]$, you'll get the result (in spite of having $g'(a)=1$).
