Diagonal reduction in a two-variable Newton iteration ($x_{n+1}=y_{n+1}=2x_n^3/(3x_n^2-1)$)

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 $3x_n^2\ne 1$. Around $(1,1)$ its error factors exactly asThus 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 $3/(2\sqrt{2})$. Around $(-1,-1)$, replacing $x$ by $-x$ gives the corresponding conclusion.