# Non linear recurrence

Let

$$a_{n+1}=3a_{n}^2(1-a_{n})$$

Where $$a_0$$ is any complex number

For $$f(x)=3x^2(1-x)$$ I found fixed point $$0,\frac{1}{2}(1\pm \frac{i}{\sqrt3})$$

Setting $$a_0$$ equal to one fixed point we get $$a_n$$ constant for all $$n$$.

Is it possible to find a formula for $$a_n$$ ?

There is a general technique for this situation. Let $$\, f(x) := 3x^2(1-x).\,$$ Now at the fixed point $$\,x=0\,$$ we have $$\, f(x) = 3x^2 + O(x^3).\,$$ We want a function $$\,g(x)\,$$ such that $$\, g(x^2) = f(g(x)).\,$$ Given an initial ansatz of $$\, g(x) = O(x)\,$$ we use the previous equation to solve for the power series coefficients one at a time with the result $$g(x) = \frac13 x + \frac1{18} x^2 + \frac{11}{216} x^3 + \frac{7}{324} x^4 + \frac{389}{10368} x^5 + O(x^6).$$ Now if $$\,a_0 = g(z)\,$$ then $$\, a_n = g(z^{2^n})\,$$ which is quadratic convergence to the fixed point $$0.$$ It is unlikely that there is a closed form for $$\,g(x).\,$$ A similar result holds for the other fixed points.