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.

Note however that there's an error above, inserted to force you to actually work the formulas out for yourself... something very much like the above is correct, and should allow you to prove convergence in a certain region.

  • $\begingroup$ Don't quite gather what you're trying to imply here. Your second paragraph specifically. $\endgroup$ – user310867 Feb 5 '16 at 0:53
  • $\begingroup$ Also, what does it mean for $x_n\to x$? $\endgroup$ – user310867 Feb 5 '16 at 0:59
  • $\begingroup$ 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...)) $\endgroup$ – David C. Ullrich Feb 5 '16 at 1:35
  • $\begingroup$ Still a little hazy on this whole concept. Can you provide some more details? $\endgroup$ – user310867 Feb 5 '16 at 1:44

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