Systems of nonlinear equations (intro)
General · Mathematics
Study notes
Q: One Newton step for x²+y²=4, xy=1 from (1.8, 0.6). F = (x²+y²-4, xy-1). J = [[2x,2y],[y,x]]. At (1.8,0.6): F = (3.24+0.36-4, 1.08-1) = (-0.4, 0.08). J = [[3.6,1.2],[0.6,1.8]]. det = 6.48-0.72 = 5.76. J⁻¹F = (1/5.76)[1.8(-0.4)-1.2(0.08), -0.6(-0.4)+3.6(0.08)]. = (1/5.76)(-0.816, 0.528) = (-0.1417, 0.0917). x_new = (1.8+0.1417, 0.6-0.0917) = (1.9417, 0.5083)! (True: (1.9319, 0.5176) - close!)