TITLE: Is there a closed form formula for the recursive sequence: $x_n = x_{n-1} + \alpha\sqrt{x_{n-1}}$
QUESTION [9 upvotes]: I saw this link for closed form formula for a recursive sequence (How to derive a closed form of a simple recursion?)
However, what if my formula is:
$$x_n = x_{n-1} + \alpha\sqrt{x_{n-1}} \quad \text{ and } \quad x_0 = \beta \quad \text{ and }\quad\alpha,\beta> 0$$
Is there a closed form solution to determine the value of $x_n$ for a given $n$?
Bonus: If there is a closed form solution, is there an inverse? That is, if one is given a value $y$, can one deduce the closest $n$ such that $x_n$ is the closest to $y$ compared to any other possible $x_n$ value?
I couldn't figure out the mathemical formula for either question. Since we're CS engineers, we know that we could create a lookup table for the first 1 million values of $n$ and that should work for us. (It's arguably a reasonable solution, since we need to compute $x_n$ for 100 million values of $n$ (so obviously, $n$ will be the same thousands of times and so precomputing is not such a bad idea.) But still, it would "cleaner" if there were a closed form analytical solution to the above recursive sequence that we could use/consider.

REPLY [3 votes]: Some notes on the asymptotic behavior of the sequence in questions.

You are iterating the function $f(x) = x + \alpha \sqrt{x}$. For positive $\alpha$, this will obviously produce a growing sequence, and it is not hard to see that it will grow to infinity.
If you define $g(x) = \sqrt{x^2 + \alpha x}$, you can verify that $f(x^2) = g(x)^2$. 
Defining $y_n = \sqrt{x_n}$, you can see that $y_n = g(y_{n-1})$.

Consider the expansion (valid for large $x$)
$$g(x) = \sqrt{x^2 + \alpha x} = x \sqrt{1 + \frac{\alpha }{x}} = x \left( 1 + \frac{\alpha}{2x} - \frac{\alpha^2}{8x^2} + o(x^{-2})\right)$$
$$= x + \frac{\alpha}{2} - \frac{\alpha^2}{8x} + o(x^{-1})$$
We can thus infer that, for large $n$
$$y_n \approx \frac{\alpha n}{2}$$
Feeding this back into the asymptotic expansion, we can show further that:
$$y_n \approx \frac{\alpha}{4}(2n - \log n)$$
Deriving extra terms in this expansion is possible, but not necessarily enlightening.
One can then infer that, for large $n$,
$$x_n \approx \frac{\alpha^2}{16}\left(2n - \log n\right)^2$$
This does not really touch on the dependence on $\beta$, which is harder to broach by studying asymptotics.