I would like to see why below recursive scheme is convergent :
$x_{n+1}= \sqrt{\alpha\cdot x_n+\beta}$
Here, $\alpha>0$ and $\beta\in\mathbb{R}\setminus\left\{0\right\}$.
I tried something like:
$x_1=\sqrt{\alpha\cdot x_0+\beta}$
$x_2=\sqrt{\alpha\underbrace{\left(\sqrt{\alpha\cdot x_0+\beta}\right)}_{x_1} +\beta}$
$x_3= \sqrt{\alpha\underbrace{\left(\sqrt{\alpha\left(\sqrt{\alpha\cdot x_0+\beta}\right) +\beta}\right)}_{x_2}+\beta}$
$\cdots$
$x_{n+1}=\sqrt{\alpha\cdot x_n+\beta}$
Note 1 :
Also, I managed to boil down $x_2$ as:
$x_2=\sqrt{\alpha\cdot\left(\sqrt{\alpha\cdot x_0+\beta}\right)+\beta}\\$
or,
$x_2=\sqrt{\sqrt{\alpha^2\left(\alpha\cdot x_0+\beta\right)}+\beta}\\$
and further simplified as:
$x_2=\sqrt{\sqrt{\left(\alpha^3\cdot x_0+\alpha^2\beta\right)}+\beta}\\$
Note 2 :
And, with the same approach I did on $x_2$, also I simplified $x_3$ further as:
$x_3= \sqrt{\alpha\cdot \sqrt{\sqrt{\left(\alpha^3\cdot x_0+\alpha^2\beta\right)}+\beta}+\beta}$
or further simplified as:
$x_3= \sqrt{ \sqrt{\sqrt{\alpha^4\left(\alpha^3\cdot x_0+\alpha^2\beta\right)}+\beta}+\beta}$
that could be even more simplified by introducing the $\alpha^4$ inside the $\left(\cdot\right)$ as:
$x_3= \sqrt{ \sqrt{\sqrt{\left(\alpha^7\cdot x_0+\alpha^6\beta\right)}+\beta}+\beta}$
Note 3:
Based on (Note 1) and (Note 2), I could write my recursive scheme in a more compact ways as:
$x_1=\sqrt{\alpha\cdot x_0+\beta}$
$x_2=\sqrt{\sqrt{\alpha^3\cdot x_0+\alpha^2\beta}+\beta}\\$
$x_3= \sqrt{ \sqrt{\sqrt{\alpha^7\cdot x_0+\alpha^6\beta}+\beta}+\beta}$
$x_4= \sqrt{ \sqrt{ \sqrt{ \sqrt{ \alpha^{13}x_0+\alpha^{12}\beta }+\beta }+\beta }+\beta }$
$\cdots$
Now from here, I try to rewrite $x_1$, $x_2$, $x_3$ and $x_4$ such that they look to have a repeative term:
$x_1=\sqrt{\alpha^0\alpha\cdot x_0+\alpha^0\beta}$
$x_2=\sqrt{\sqrt{\alpha^2\alpha\cdot x_0+\alpha^2\beta}+\beta}\\$
$x_3= \sqrt{ \sqrt{\sqrt{\alpha^6\alpha\cdot x_0+\alpha^6\beta}+\beta}+\beta}$
$x_4= \sqrt{ \sqrt{ \sqrt{ \sqrt{ \alpha^{12}\alpha\cdot x_0+\alpha^{12}\beta }+\beta }+\beta }+\beta }$
and by factoring $\alpha^n$ out, inside the most inner square root we have:
$x_1=\sqrt{\alpha^{\overbrace{0}^{\left(1\cdot 0\right)}}\left(\alpha\cdot x_0+\beta\right)}$
$x_2=\sqrt{\sqrt{\alpha^{\overbrace{2}^{\left(2\cdot 1\right)}}\left(\alpha\cdot x_0+\beta\right)}+\beta}\\$
$x_3= \sqrt{ \sqrt{\sqrt{\alpha^{\overbrace{6}^{\left(3\cdot 2\right)}}\left(\alpha\cdot x_0+\beta\right)}+\beta}+\beta}$
$x_4= \sqrt{ \sqrt{ \sqrt{ \sqrt{ \alpha^{\overbrace{12}^{\left(4\cdot 3\right)}}\left(\alpha\cdot x_0+\beta\right) }+\beta }+\beta }+\beta }$
This, finally leads us to write a general iterative scheme as:
$x_{n}= \sqrt{ \cdots \sqrt{ \sqrt{ \sqrt{ \alpha^{n\left(n-1\right)}\left(\alpha\cdot x_0+\beta\right) }+\beta }+\beta \cdots }+\beta }$
Here, number of square roots is $n$.
Now, I need to see why the concluded iterative scheme is convergent.
Such a sequence is always monotone and bounded. Let for example $\alpha >0$ and $\beta >0$. Let $\ell$ the unique solution to $\sqrt{\alpha x+\beta }=x$. If $x_0<\ell$, then $x_n$ will be increasing and upper-bounded by $\ell$ (prove by induction). If $x_0>\ell$, then the sequence will be decreasing and lower-bounded by $\ell$ (prove it by induction). If $x_0=\ell$, then it will be stationary. Try to adapt this when $\alpha <0$ and $\beta >0$ or when $\alpha <0$ and $\beta <0$ or when $\alpha >0$ and $\beta <0$.