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Let $\phi(x) = x + \lambda (x^2 - 3)$, with $\phi'(x) = 1 + 2\lambda x$. As the title says, I want to find the values of $\lambda$ such that the sequence $x_{n+1} = x_n + \lambda (x_n^2 - 3)$ converges to $\sqrt{3}$.

Now my thought process so far goes like this:

$$ |\phi'(x)| < 1 \Leftrightarrow -1 < \lambda x < 0 $$

which means

\begin{align*} x \in ( -\frac{1}{\lambda}, 0),\ \text{for}\ 0 < \lambda \\ x \in ( 0, -\frac{1}{\lambda}),\ \text{for}\ \lambda < 0 \end{align*}

Also, $\phi'(x) = 0 \Leftrightarrow x = -\frac{1}{2\lambda}$ and by computing

\begin{align*} \phi(0) &= -3\lambda \\ \phi(-\frac{1}{\lambda}) &= -3\lambda \\ \phi(-\frac{1}{2\lambda}) &= -3\lambda - \frac{1}{4\lambda} \end{align*}

we have that $\phi[-\frac{1}{\lambda}, 0] = [-3\lambda, -3\lambda - \frac{1}{4\lambda}]$.

And now I must satisfy the condition $\phi[-\frac{1}{\lambda}, 0] \subset [-\frac{1}{\lambda}, 0] $ (for $\lambda > 0$), so $\lambda$ must satisfy both

$$ -\frac{1}{\lambda} < \phi(0) < 0 \Leftrightarrow \lambda \in (-\frac{1}{\sqrt{3}}, 0)\ \cup (0, \frac{1}{\sqrt{3}}) $$

and

$$ -\frac{1}{\lambda} < \phi(-\frac{1}{2\lambda}) < 0 \Leftrightarrow \lambda \in (-\frac{1}{2}, 0)\ \cup (0, \frac{1}{2}) $$

which finally means that for $\lambda \in (-\frac{1}{2}, 0)\ \cup (0, \frac{1}{2})$ the sequence converges for initial values $x_0$ with $x_0 \in ( -\frac{1}{\lambda}, 0)$.

Now, is my thought process correct? Also, playing around with an algorithm I implemented for this method, for $\lambda = -0.25$ and $x_0 = 5.5$ I see that it converges. But that is not covered by my thought since for $\lambda = -0.25$ it should converge only for $x_0 \in (0, 4)$. What is going on here?

Any help would be appreciated.

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    $\begingroup$ For a fixed point $x_0$ to be an attracting fixed point it must meet the following $2$ criteria, 1) to be a fixed point, aka $x_0=\phi(x_0)$ and 2) $|\phi'(x_0)|<1$. In your case $x_0=\sqrt{3}$, why are looking for $|\phi'(x)|<1$? $\endgroup$
    – rtybase
    Nov 11 '19 at 22:29
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    $\begingroup$ Probably because I'm not sure about what I'm doing and don't have enough experience on the field...! Thanks for your answer. $\endgroup$
    – Da Mike
    Nov 11 '19 at 22:49
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    $\begingroup$ In the last 2 hours, this is the 2nd post asking about recurrences, attracting fixed points and committing the same mistake (this is the other post) :) ... weird. $\endgroup$
    – rtybase
    Nov 11 '19 at 23:09
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I find matrices a great tool for linear and quadratic recurrence relations. Let $$M=\begin{pmatrix}-3\lambda&\frac12\\\frac12&\lambda\end{pmatrix}$$ Note that $$x_{n+1}= \begin{pmatrix}1 & x_n\end{pmatrix}M\begin{pmatrix}1\\x_n\end{pmatrix}$$ Now, take $M=P^{-1}DP$ where $D$ is a diagonal matrix to get $$x_{n}= \begin{pmatrix}1 & x_0\end{pmatrix}P^{-1}D^nP\begin{pmatrix}1\\x_0\end{pmatrix}$$

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  • $\begingroup$ I was not sure. If this is true, much better. $\endgroup$
    – ajotatxe
    Nov 11 '19 at 22:54
  • $\begingroup$ @alex.jordan I find a bit surprising that we get the same result for $x_{n+1}=x_n+\lambda(x_n^2-3)$ as for $x_{n+1}=x_n+\lambda(-3x_n^2+1)$. $\endgroup$
    – ajotatxe
    Nov 11 '19 at 23:02
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    $\begingroup$ Actually, can you explain more how $x_{n+1}= \begin{pmatrix}1 & x_n\end{pmatrix}M\begin{pmatrix}1\\x_n\end{pmatrix}$ leads to $x_{n}= \begin{pmatrix}1 & x_0\end{pmatrix}M^n\begin{pmatrix}1\\x_0\end{pmatrix}$? $\endgroup$ Nov 12 '19 at 3:36
  • $\begingroup$ @ajotatxe I don't get what you are doing. First, how did you come up with $M$. Then how did you get to the formula for $x_n$. Also, how does this solve my problem? $\endgroup$
    – Da Mike
    Nov 12 '19 at 17:45
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If you can ensure that $|ϕ'(\sqrt3)|<1$, then there exists some small interval around the fixed point where the iteration will converge towards this fixed point. The smaller that number is, the larger is this interval of attraction.

As we know $\frac32<\sqrt3<2$, a more direct strategy is to ensure that $|ϕ'(x)|<1\iff -1<1+2λx<1$ on this interval, meaning that $$ 0>λ>-\min_{x\in[1.5,2]}\frac1x=-\frac12. $$ So for instance $λ=-\frac13$ is a valid choice for this interval.

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  • $\begingroup$ Isn't it $\min_{x\in[1.5,2]}\frac1x= \frac12$? $\endgroup$
    – Da Mike
    Nov 12 '19 at 17:51
  • $\begingroup$ @DaMike : Yes, thanks. On-the-fly constructions sometimes run away. $\endgroup$ Nov 12 '19 at 17:54

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