When did this Sequence v converge? Determine all $c>0$ such that the recursive sequence $\{a_n\}$ defined by setting $\\$
$a_1=\frac{c}{2}$   and   $a_{n+1}=\frac{c+a_n^2}{2}$ converges.
What I Did $\\$
If $\{a_n\}$ to $p$,  then  $p=\frac{c+p^2}{2}$
$\\$
Checking the discriminant we get $c\leq 1$.
$\\$In the case when $c \leq 1$
We get two roots for equation $\frac{2- \sqrt{4-4c}}{2}=r_1$ and $\frac{2+ \sqrt{4-4c}}{2}=r_2$
I was checking where does $a_1$ lies?
Is $a_1 \in \left( -\infty,r_1 \right)$ in this case if $\{a_n\}$ increases the it converges to $r_1$.
And similar for other two cases.
But while checking the conditions on $c$ if it lies in $\left( -\infty,r_1 \right)$ or  $\left( r_1,r_2 \right)$ or $\left( r_2,\infty \right)$. 
Surprisingly I am getting c=0 in all cases..
Does this mean the sequence never converges??
Or 
I am wrong? If so correct me
 A: Ok. Let's start by studying the evolution of $a_n$.
$2(a_{n+1}-a_n)=a_n^2-2a_n+c$
Studying the function $x^2-2x+c$ tell us that this equation has roots for $c\leq 1$.
Case 1: $c>1$
Can the sequence have a limit? Well, you know that this is not the case (since it would mean $l=\frac{l^2+c}{2}$, wich is not possible)
Case 2: $c=1$
Then, three subcases...
$a_n=1$, then $a_{n+1}=1$... BUT $a_1=\frac 12$, so this is not possible.
$a_n > 1$, $a_{n+1}>a_n>1$. The sequence cannot converge to $1$.
$a_n<1$, $a_{n+1}>a_n$ AND $a_{n+1}<\frac{1+1^2}{2}=1$;  $a_n$ is stricly increasing, bounded, thus converges to $1$ (the limit if $c=1$).
Since $a_1= \frac 12$, you are from the very beginning in the case 3...
Case 3: c<1
You go on with the same kind of reasoning as in the case 2, by being careful in the monotony of the sequence, the fact that it stays, or not, in the same intervals you chose for your cases etc. You will find obviously your two potential limits $r_1$ and $r_2$, with $r_1=1-\sqrt{1-c}$ and $r_2=1+\sqrt{1-c}$
