= Diagonal reduction in a two-variable Newton iteration
{title2=$x_{n+1}=y_{n+1}=2x_n^3/(3x_n^2-1)$}
For $f(x,y)=x^4-4xy+2y^2+4$, the <Newton step> equations imply $y_{n+1}=x_{n+1}$. Solving the remaining equation gives the displayed scalar recurrence, defined when $3x_n^2\ne1$. Around $(1,1)$ its error factors exactly as
$$
x_{n+1}-1=(x_n-1)^2\frac{2x_n+1}{3x_n^2-1}.
$$
Thus any well-defined sequence converging to $(1,1)$ has <quadratic convergence>, or reaches that point exactly in finitely many steps. The scalar asymptotic error constant is $3/2$; the <Euclidean norm> error constant on the diagonal is $3/(2\sqrt2)$. Around $(-1,-1)$, replacing $x$ by $-x$ gives the corresponding conclusion.
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