Find $x_0\in \mathbb{Q}$ s.t. $(x_n)_{n}$ is convergent. Let $(x_n)_{n\geq 0}$ a sequence of real numbers given by the relation $2 x_{n+1}=2x_n^2-5x_n+3$, for every $n\geq 0$. Find $x_0\in \mathbb{Q}$ s.t. $(x_n)_{n}$ is convergent.
An easy remark is that the limit of the sequence is $3$ or $\dfrac{1}{2}$, but now I am stuck.
 A: Both fixed points are locally unstable.
If you take a small perturbation around $1/2$, let say $x_n = 1/2+\varepsilon$, then $x_{n+1} = 1/2 -3\varepsilon/2 + O(\varepsilon^2)$, so that $x_{n+1}$ is now less close to $1/2$ than $x_n$. In the same way if $x_n = 3+\varepsilon$, then $x_{n+1}=3+7\varepsilon/2+O(\varepsilon^2)$.
Thus, the only way to converge is that at some point you'll have $x_n=1/2$ or $x_n=3$. Obviously $x_0=1/2$ and $x_0=3$ are possible solutions.
However they're not the only ones. If for instance you have $x_0 = 2$ then $x_1=1/2$ and the sequence will then stay at $1/2$ for all $n\geq 1$. So to find all the possible $x_0$ you should go backwards starting from $1/2$ and $3$.
Let's start from $3$. You have that $x_0=-1/2$ leads to $x_1 = 3$. Then you look for these points which can give you $x_1=-1/2$. Actually you're lucky because there's none.
Then we have to study $1/2$. $x_0=2$ leads to $1/2$. How can we arrive at $x_1 = 2$ (so that $x_2=1/2$)? Here things are more complicated. You've to solve a 2nd order equation and get that $x_0 = \frac{1}{4}(5\pm\sqrt{33})$ are both possible starting points.
Now you have that you can never reach $\frac{1}{4}(5-\sqrt{33})$, but it is possible to arrive at $\frac{1}{4}(5+\sqrt{33})$, which is attained as $x_1$ by two values of $x_0$, one positive and the other negative. The negative one will be to small, so can't be reached, but the positive one can be reached and so you need to go on. There will be infinite possible $x_0$ which you can find in this way.
Now you're interested in the $x_0$ in $\mathbb{Q}$ only. Among the ones that we've found, just $-1/2,1/2,3$ and $2$ are in $\mathbb{Q}$. Is there any more possible $x_0$ in $\mathbb{Q}$? the answer is no. Indeed you should be able, starting from such an $x_0$, to reach one of the two irrational $\frac{1}{4}(5+\sqrt{33})$. But, starting from a rational $x_0\in\mathbb{Q}$, then $x_n$ will be rational for any $n\geq 0$, so that none of those two points can be reached.
To summarize, the $x_0\in\mathbb{Q}$ which lead to a convergent sequence are $\{-1/2;1/2; 2;3\}$.
