Use Newton’s method with line searches to solve f(x) = 0, where f : ^{2 2} is defined as:

Use seed value = (1, 1). Do not use scaling.

Details of this solution are available as an Excel spreadsheet.

The Jacobian of f is:

Based upon this, we apply Newton’s method with line searches. Results are indicated below.

k
1	(1.0000, 1.0000) (2.0000, 0.3333) (1.9619, 0.5525) (1.7507, 1.0072) (1.7478, 0.9233) (1.7476, 0.9148) (1.7476, 0.9147)	(–1.0000, –4.0000) (–0.9259, 3.9259) (–0.6817, 3.2208) (0.3427, –0.4232) (0.0288, –0.0368) (0.0003, –0.0004) (0.0000, 0.0000)
2
3
4
5
6
7

A line search was used to obtain . Newton’s method originally determined = (1.7515, 1.7636), which failed criterion [2.152]. Applying a line search, yielded * = 0.1533. This corresponded to value x* = (1.9619, 0.5525). We set = x*. This did satisfy criterion [2.152]. No other line searches were required.