# Fixed Point Iteration Proof

Given the fixed point iteration $$x_{n+1}=\frac{-x_n^2-c}{2b}$$ where $b$ and $c$ are fixed, $x_n\longrightarrow x$, what does $x$ solve? Additionally, what is the region for $(b,c)$ values where our iteration converges at a rate of $O(2^{-n})$ or better from an interval of starting values $x_0$ near $x$?

Say $P(x)=x^2+2bx+c$. It's clear that if $x_n\to x$ then $P(x)=0$. If you want to investigate under what conditions $x_n$ actually converges you might "note" that $$P(x_{n+1})=\frac1{4b^2}(P(x_n)^2+4bx_nP(x_n))$$and try to figure out when $P(x_n)$ small will imply that $P(x_{n+1})$ is even smaller.
• Also, what does it mean for $x_n\to x$? – user310867 Feb 5 '16 at 0:59
• All I was trying to imply was exactly what I said. Regarding what $x_n\to x$ means, you used that notation in your question! What did you mean by the notation? (Oh, maybe you didn't mean you didn't know what $x_n\to x$ meant, you were asking how what I wrote has anything to do with showing $x_n$ converges. If that's the question, the answer is it's not hard to show that if $P(x_n)\to0$ then $x_n$ must converge to a zero of $P$ (or possibly do some oscillating between near one zero and near the other...)) – David C. Ullrich Feb 5 '16 at 1:35