Find the values of $\lambda$ for which the iterative method $x_{n+1} = x_n + \lambda (x_n^2 - 3)$ converges to $\sqrt{3}$.

Question

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.

Answer 1

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.

Answer 2

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 $$x_{n}= \begin{pmatrix}1 & x_0\end{pmatrix}P^{-1}D^nP\begin{pmatrix}1\\x_0\end{pmatrix}$$