Quadratic first order recurrence relation, is there a solution? I am trying to find a solution for $a_n$, writing $a_n$ as a function of $n$,
according to the following recurrence relation:
$$a_n=3.9*a_{n-1}(1-a_{n-1}) ;   a_0 = \frac{1}{2}$$.
I have tried expanding, substitution, and looking at the ratio.
By doing linear substitution such as:
$$a_n = \frac{1}{2} + b_n$$.
a simpler form is found:
$$b_n = \frac{1.9}{4}-3.9b_{n-1}^2; b_0 = 0$$
I tried to expand the expression and hit a wall.
I also ran it in Wolframalpha, giving no solution, with this initial condition its easy to see that $0 < a_n < 1$ for all $n$.
My question is, is there a solution? if so, how do I find it, and if there isn't, can it be proven?
Thanks!
 A: You hit a wall because you expect a simple explicit solution in terms of elementary functions. This does not exist.
Using a computer you might want to first investigate what happens with lower values of the coefficients. Start with $a_n=(5/2)a_{n-1}(1-a_{n-1})$. This converges nicely to a limit of $3/5$ where $a_n/a_{n-1}$ would reach unity, no problem.
Now try $a_n=(10/3)a_{n-1}(1-a_{n-1})$. You might suppose this will equally hit a limit of $7/10$ where $a_n/a_{n-1}$ would again reach unity, but you don't actually get there. The fixed point destabilizes and instead you get an oscillation between two different limits. You would have to set up and solve a quadratic equation to define these sub-sequence limits.
Keep increasing the coefficients and you get more sub-sequence limits and more complex oscillations. By the time the coefficient gets to $3.9$ the oscillations have become so complex that they look random; even with thousands of terms you cannot clearly discern a set of sub-sequence limits at all. The limiting behavior, let alone the individual terms, is clearly nonelementary.
