For what range of values will the fixed point iteration converge?

Question

For what range of values of $c$ will the fixed point iteration $x_{n+1} = x_n + c{x_n}^2 - 9$ converge, and for what particular value of $c$ will it converge much faster?

Answer 1

Perhaps the computation is easier if you multiply the recursion by $c$, $$ cx_{n+1}=(cx_n)+(cx_n)^2+9c=(cx_n+\tfrac12)^2+(9c-\tfrac14) $$ and consider $y_n=cx_n+\tfrac12$, so that $$ y_{n+1}=y_n^2+9c+\tfrac14 $$ which now has the form of the Mandelbrot iteration with all the methods to explore its convergence.

Answer 2

For a moment, suppose the sequence converges $x_n \to x$. Then, $$ x = x + cx^2-9 \implies x^2 = 9/c $$

Now, we have local attraction to this fixed point if $\left|\frac{dx_{n+1}}{dx_n}\right| < 1$ in a neighborhood of the fixed point. Let us consider $c>0$. Thus, for $x=3/\sqrt{c}$, $$ \frac{dx_{n+1}}{dx_n} = 1+2cx = 1+6\sqrt{c}. $$ Note that this will never be locally attracting, so we "must" approach $-3/\sqrt{c}$. In this case, $$ \frac{dx_{n+1}}{dx_n} = 1+2cx = 1-6\sqrt{c} \implies 6\sqrt{c} < 2 \implies c < 1/9. $$

Of course, local stability analyses don't give global asymptotic stability, but it's a start. To summarize, this partial analysis yields that you converge to $-3/\sqrt{c}$ for sufficiently nice $x_0$ and $c<1/9$.