$ x_{n + 1} = \sqrt {c – \sqrt {c + x_n}}$ Find $ c$ such that $ x_0$ in $ (0, c)$, $ (x_n)$ has a finite limit $ \lim x_n$ when $ n\to + \infty$

Question

Given a real number $ c > 0$, a sequence $ (x_n)$ of real numbers is defined by $ x_{n + 1} = \sqrt {c – \sqrt {c + x_n}}$ for $ n \ge 0$. Find all values of $ c$ such that for each initial value $ x_0$ in $ (0, c)$, the sequence $ (x_n)$ is defined for all $ n$ and has a finite limit $ \lim x_n$ when $ n\to + \infty$.
For $f(1)$ to be defined, then $c – \sqrt {c + x_0} \geq 0 \Rightarrow c^2-c \geq x_0$
But $x_0 \in (0;c) \Rightarrow x_0<c$
Thus $c^2-c>c \Rightarrow c(c-2)>0$.
If $c<0 \Rightarrow c – \sqrt {c + x_n} < 0$, which is a contradiction.
Therefore $c \geq 2$.
I think we will induct with $x_{n+1}$ to have $ (x_n)$ is defined for all $ n$ by finding the range of $x_n$.
$\textbf{Definition.}$
The function $f : D \rightarrow D$ is called Lipschitz function on $D$ if there is exist $q \in \mathbb{R}, 0 < q < 1$ such that $|f(x) − f(y)| ≤ q|x − y| \Leftrightarrow |f'(x)| \leq q$ for all $x$ and $y$ in $D$.
$\textbf{Theorem.}$
If $f(x)$ is Lipschitz function on $D$, then the sequence $\{x_n\}$ defined by $x_0 = a \in D, x_{n+1} = f(x_n)$ converges (has a finite limit $ \lim x_n$ when $ n\to + \infty$).
Then we can prove $|{f'(x)}| \leq q < 1$ for all $x \in (0;?)$ to have $(x_n)$ has a finite limit $ \lim x_n$ when $ n\to + \infty$.

Answer 1

We suppose that $c\geqslant 2$. Let $\displaystyle f(x)=\sqrt{c-\sqrt{c+x}}$

• f is strictly decreasing on $[0,c^2-c]\ $ and $\ f([0,c^2-c])=[0,f(0)]$.

• $\forall x \in [0,f(0)] \ , \ |f'(x)| = \dfrac{1}{4f(x)\sqrt{x+c}} \leqslant \dfrac{1}{4\sqrt{c}} \dfrac{1}{f\circ f(0)}$

• $\displaystyle f(0)=\sqrt{c-\sqrt{c}}\leqslant \sqrt{c} $

• $\displaystyle f\circ f(0) \geqslant f(\sqrt{c})=\sqrt{c-\sqrt{c+\sqrt{c}}} = \sqrt{c} \sqrt{1-\sqrt{\dfrac{1}{c}+\dfrac{1}{c\sqrt{c}}}}\geqslant \sqrt{2}\sqrt{1-\sqrt{\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}}}$

• So: $\ \forall x \in [0,f(0] \ , \ |f'(x)| \leqslant \dfrac{1}{8}\ \dfrac{1}{ \sqrt{1-\sqrt{\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}}}} < 1$

Then, for all $a\in [0,f(0)]$, the sequence $\{x_n\}$ defined by $\ x_0 = a\ , \ x_{n+1}=f(x_n)\ $ converges.

For all $a \in (0,c)$ , the sequence $\{x_n\}$ defined by $\ x_0= a \ , \ x_{n+1}=f(x_n) \ $ verifies $\ x_1=f(x_0) \in [0,f(0)]\ $ and, therefore, converges.