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This is a past prelim question,

Let $F(x)=2x+qx^2$ with $\frac{1}{2}\leq q \leq 1$ and $x_{n+1}=F(x_{n})$.

For which interval does this iteration converge ?

I tried these but did not work;

Fixed points are $0, -\frac{1}{q}$, Since $F'(0)>1$ if there is such an interval with initials from that interval converging to a fixed point this fixed point will be $-\frac{1}{q}$. I tried to use the idea; find closed interval C with $|F'|\leq\lambda <1$ on $C$ and $F(C)\subset C$. Suppose we just want to find an interval not the maximal one. Then $F' = 2+2px$, $|2+2qx|\leq \frac{1}{2}$ this will give smaller interval, then $-\frac{1}{2}\leq 2+2qx \leq \frac{1}{2}$, $-\frac{5}{4q} \leq x \leq -\frac{3}{4q} =: C$. But $F(C)$ is not a subset of $C$. I tried for smaller number $a$ for $|F'| < a$, but it did not work.

Question 1: Is there any way to find such $C$ ?

Question 2: How can we find maximal interval ?

any clues/ideas would be appreciated.

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  • $\begingroup$ Note that $F(x) < x $ on $I:=(-1/q,0)$ so starting from $x \in I$ the iteration $x_n$ converges just because it is decreasing and bounded below. The limit is $-1/q$ as it is the unique negative fixed point. Since $F'(-1/q)=0$ there is quadratic convergence. Also note that all the $F_q$ are conjugated, so it suffices to consider $q=1$. $\endgroup$
    – Pietro Majer
    Commented Nov 22, 2011 at 11:04
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    $\begingroup$ btw, this question seems more suitable for math.stackexchange.com $\endgroup$
    – Pietro Majer
    Commented Nov 22, 2011 at 11:15

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