= Solution
By the <diagonal reduction in a two-variable Newton iteration>, $y_n=x_n$ for $n\ge1$. Put $e_n=x_n-1$. Direct factorization gives the exact <Newton map in optimization> error identity
$$
e_{n+1}=\frac{2x_n^3-(3x_n^2-1)}{3x_n^2-1}=e_n^2\frac{2x_n+1}{3x_n^2-1}.
$$
The assumed convergence $x_n\to1$ makes the final factor tend to $3/2$, so it is bounded for all sufficiently large $n$. Hence $|e_{n+1}|\le C|e_n|^2$. For the <Euclidean norm> error $E_n=\|(x_n,y_n)-(1,1)\|=\sqrt2|e_n|$, this becomes \b[$E_{n+1}\le(C/\sqrt2)E_n^2$], proving <quadratic convergence>. Unless the exact limit is reached in finitely many steps, $E_{n+1}/E_n^2\to3/(2\sqrt2)$.
Back to article page
