mestavk wrote: $(x_{n})_{-\infty<n<\infty}$ is a sequence of real numbers which satisfies $x_{n+1}=\frac{x_{n}^2+10}{7}$ for every $n \in \mathbb{Z}$. If there exist a real upperbound for this sequence, find all the values $x_{0}$ can take. If $|x|>5$ it won't have an upperbound, in the other cases it has. But if $x_0$ is the biggest one, we let count? Then the ans is $x_0\in [-5,5]$

NO. For one thing, $x_0$ can never take negative values, since $x_0 = \frac {x_{-1}^2 + 10} {7} > 0$. Define $f\colon \mathbb{R} \to \mathbb{R}$ by $f(x) = \frac {x^2 + 10} {7}$. Then $x_{n+1} = f(x_n) > 0$ for all $n\in \mathbb{Z}$; this means all terms of the sequence are positive. Now, $f(x) = x$ writes as $(x-2)(x-5) = 0$, so the graph of $f$ meets the first bisector $y=x$ at $x=2$ and $x=5$. The behaviour of such sequences is well known and documented; for a term of value greater than $5$, the sequence is thereon increasing to $+\infty$, thus is unbounded; for a term of value equal to $5$, the sequence is thereon constant (equal to $5$); for a term of value in the interval $(2,5)$, the sequence is thereon decreasing to $2$; for a term of value equal to $2$, the sequence is thereon constant (equal to $2$); for a term of value in the interval $(0,2)$, the sequence is thereon increasing to $2$. Therefore, when the sequence is bounded, all its terms are to be found in the interval $(0,5)$. No term $x_k$ can be strictly less than $2$, since then all (infinitely many) terms of lesser indices will also be less than $2$, and so the term $x_k$ will be as close to $2$ as wanted (iterating $g(x) = \sqrt{7x-10}$ with a seed in $[10/7,2)$ eventually leads to a term of value less than $10/7$, impossible). Value $2$ is possible, only when the sequence is constant (equal to $2$); similarly, value $5$ is possible, only when the sequence is constant (equal to $5$). Any value for $x_k$ in the interval $(2,5)$ is possible, since then for $n>k$ we have valid values for $x_n = f^{n-k}(x_k)$, while for $n<k$ we have valid values for $x_n = g^{k-n}(x_k)$, where $g\colon (2,5) \to (2,5)$ given by $g(x) = \sqrt{7x-10}$ is the inverse of the restriction of $f$ to the interval $(2,5)$. Thus the answer is that the range of possible values for $x_0$, in fact for any term $x_k$, is the interval $[2,5]$.

We claim the answer is $x_0\in[2,5]$, and any such $x_0$ works. In what follows, we use the fact without proof that if $(y_n)_{n\ge 1}$ is a monotonically increasing sequence that admits an upper bound, then $\lim_{n\to\infty}y_n$ is well-defined, and finite. We begin by observing that $x_n>0$ for all $n$; and \[ x_{n+1} - x_n = \frac{\bigl(x_n-2\bigr)\bigl(x_n-5\bigr)}{10}. \]In particular, $x_{n+1}\ge x_n$ if $x_n\notin [2,5]$; and $x_{n+1}<x_n$ if $x_n\in [2,5]$. Step 1: $x_0>5$ is impossible. Suppose there is such a sequence with $x_0>5$. A straightforward induction then implies $x_k>5$ for all $k\ge 1$. In particular, $(x_n)_{n\ge 0}$ is monotonically increasing. Assume now that it admits a (finite) upper bound. Then it has a limit, denote by $L$. Passing to limits in the recursion equation, we find $L=\frac{L^2+10}{7}$, that is $L\in\{2,5\}$. Note, however, that as $x_k>x_0>5$ for all $k$, the sequence remains bounded away from $5$, hence $L\ne 5$. Thus, no such sequence exists for $x_0>5$. Step 2: $x_0<2$ is impossible. Once again, suppose the contrary; and define $y_k=x_{-k}$; and observe that $y_k<2$ for all $k$ by induction. In particular, $y_k\in(0,2)$ for all $k$. Now $y_k$ is a monotonically decreasing sequence that is bounded from below. Hence, it admits a limit, again denoted by $L$; and once again the recursion yields $L\in\{2,5\}$. Notice, however, that $y_k$ remains bounded away from $2$, thus $L\ne 2$. Again, a contradiction. Step 3: Any $x_0\in[2,5]$ works. If $x_0\in[2,5]$, we first obtain $x_k\in[2,5]$ for all $k\in\mathbb{Z}$. It suffices to check the sequence is also well-defined. To that end, the forward recursion is trivial, hence we restrict our attention to the backward recursion: set $y_k\triangleq x_{-k}$ and consider $y_{n+1}=\sqrt{7y_n-10}$. As $y_0\in[2,5]$, all terms of this sequence are clearly well-defined, completing the proof.