What can we say about $a_n = 0.9a_{n-1}+0.3\sqrt{a_{n-1}}$? Suppose we have the following recurrence relation:
$$a_0 = k$$
$$a_n = 0.9a_{n-1}+0.3\sqrt{a_{n-1}}$$
where $a_n \in \Bbb R^+$ and $n$ is a non-negative integer.
I would like to get a better understanding the behaviour of $a_n$. Given $a_0 = k$ as the starting point, it is possible to evaluate it recursively. However, is there any way to analyse it and do a proof of convergence or divergence? Is there any theoretical machinery I can use to approach this?
Edit: I've tried evaluating the sequence numerically with multiple positive $k \in \Bbb R$ as starting points, and with $1000$ iterations, it seems to converge to $9$, always. But again, how do we prove this theoretically? I'm mainly interested in proving its convergence, now that it numerically seems to converge to $9$.
A further note, because the question might arise: I've inferred this from an excel homework explaining an economic model, but this function was not explicitly given to me anywhere.
 A: We define the following recursion relation (same as you did in the comments):
$$a_0 = k$$
$$a_{n+1} = 0.9a_n+0.3\sqrt{a_n}$$As told by @lulu in the comments, we need to show if $k =0$, $\lim_{n\to\infty}a_n = 0$ (which is obvious), if $0<k<9$, $\lim_{n\to\infty}a_n = 9$, if $k=9$, $\lim_{n\to\infty}a_n = 9$ (again, obvious) and if $k>9$, $\lim_{n\to\infty}a_n = 9$. It doesn't make sense to consider $k<0$ since the square root of negative numbers is not defined in reals.
Say $0<k<9$: We proceed by induction to show $a_{n-1}<a_n<9$ for any finite $n>0$.
The base case can be easily shown for $n=1$.
We assume, for some $p>0$:
$$a_{p-1} < a_p<9$$
We need to show $a_p<a_{p+1}<9$
$$a_{p+1} = 0.9a_p+0.3\sqrt{a_p} < 0.9\times 9 + 0.3\times 3 = 9$$
Thus, $$\color{blue}{a_{p+1} < 9}$$Now, $$3>\sqrt{a_p}$$So,
$$3\sqrt{a_p} > a_p \implies 0.3\sqrt{a_p} > 0.1a_p$$
So, $$0.9a_p+0.3\sqrt{a_p} > a_p$$Thus, $$\color{blue}{a_{p+1}>a_p}$$
From the blue inequalities, we get the following:
$$\color{green}{a_p<a_{p+1}<9}$$which completes our proof.
Say $k>9$:
Again, we proceed by induction to show: $a_{n-1}>a_n>9$.
The base case can be easily shown for $n=1$.
Let us assume, $a_{q-1}>a_q>9$ for some $q>0$.We need to show $a_q>a_{q+1}>9$.
$$a_{q+1} = 0.9a_q+0.3\sqrt{a_q} > 0.9\times 9 + 0.3 \times 3 = 9$$
Thus, $$\color{blue}{a_{q+1}>9}$$We know,$$3<\sqrt{a_q}\implies 3\sqrt{a_q}<a_q\implies 0.3\sqrt{a_q}<0.1a_q\implies 0.9a_q + 0.3\sqrt{a_q}<a_q$$Thus, we have:
$$\color{blue}{a_{q+1}<a_q}$$From the blue inequalities, we get:
$$\color{green}{a_q>a_{q+1}>9}$$
which is the final nail in the coffin.
We get the following results:

*

*When $k=0$, all the $a_n$ will be $0$.

*When $0<k<9$, the sequence increases, but it is always less than $9$, so the sequence approaches $9$.

*When $k=9$, all the $a_n$ will be $9$.

*When $k>9$, the sequence decreases, but it is always greater than $9$, so the sequence approaches $9$.

We call such numbers ($0$ and $9$) fixed points since the function always seems to approach these two or when the initial value is among the two, the function remains constant.
