ExerciseNonlinear systemsadvanced

Exercise: Newton on a 2×2 system with iteration table

Full solution of $e^x e^y+x\cos y=0$ , $x+y=1$ with Newton from $[2,-1]$ : table of iterates, residual and increment norms, and ACOC settling at 2.

Setup

Solve the system $e^xe^y+x\cos y=0$ , $x+y=1$ . The derivation builds $F$ and the Jacobian. Use initial estimate $x^{(0)}=[2,-1]^T$ , tolerance $10^{-20}$ (extended-precision arithmetic) and at most 40 iterations.

Iterations

iter	$x^{(k)}$	$\\|F(x^{(k+1)})\\|$	$\\|x^{(k+1)}-x^{(k)}\\|$	ACOC
1	[5.3247, −4.3247]	0.7051	4.7018	n/a
2	[5.1697, −4.1697]	0.0486	0.2191	n/a
3	[5.1573, −4.1573]	0.0003	0.0176	0.8229
4	[5.1572, −4.1572]	1.7091·10⁻⁸	0.0001	1.9620
5	[5.1572, −4.1572]	4.3406·10⁻¹⁷	6.2690·10⁻⁹	1.9989
6	[5.1572, −4.1572]	2.7997·10⁻³⁴	1.5921·10⁻¹⁷	2.0000

Newton iterations from

x^{(0)}=[2,-1]

; the solution is

\alpha\approx[5.15723,\,-4.15723]