**Problem**: Find n-Dimensional Zero of a Vector-Valued Function of a Vector Argument**f**(**x**),**f**(**z**) =**0**,where

**x**= [x_{i}]_{n×1},**z**= [z_{i}]_{n×1}and**f**(**x**) = [f_{i}(**x**)]_{n×1}(Vectors are indicated by boldfaced variables and functions).**Taylor Approximation**: Letting the (k+1)th iteration be such that**x**^{(k+1)}__=__**x**^{(k)}+**delta_x**^{(k)}__~__**z**, so that**0**=**f**(**x**^{(k+1)}) =**f**(**x**^{(k)}+**delta_x**^{(k)})__~__**f**(**x**^{(k)}) + J^{(k)}***delta_x**^{(k)},assuming the change |

**delta_x**^{(k)}| is sufficiently small, where**x**^{(k)}is the kth Newton approximation,**f**^{(k)}=**f**(**x**^{(k)}) is the kth value of the n-dim. function**f**,- J
^{(k)}= ([f_{i,j}]_{n×n})^{(k)}is the kth Jacobian function, - and f
_{i,j}is the jth partial derivative (with respect to x_{j}) of the ith component of**f**.

**Conversion to Linear Algebra Problem**:J

^{(k)}***delta_x**^{(k)}= -**f**^{(k)},can be converted to the Augmented Matrix:

["A" | "

**b**" ] = [ J^{(k)}| -**f**^{(k)}].which can be solved by Forward Gaussian Elimination to get an approximation for

**delta_x**^{(k)}which can be used to get the new Newton n-dimensional iterate,**x**^{(k+1)}=**x**^{(k)}+**delta_x**^{(k)}.**Numerical Example**:*

*Intersection*of a circle of radius 2 and center (0,0) with a exponential function function y = 1 - e^{x}. So let the component functions be f_{1}(x_{1}, x_{2}) and f_{2}(x_{1}, x_{2}) or alternately f(x,y) and g(x,y) without the subscripts.*

*Vector Function with n=2*:[ f(x,y) ] = [ x

^{2}+ y^{2}- 4 ]

[ g(x,y) ] = [ e^{x}+ y - 1 ]*

*2×2 Jacobian J*:[ J

_{11}J_{11}] = [ 2x 2y ]

[ J_{21}J_{22}] = [ e^{x}1]*

*Starting Guess (MATLAB Notation)*:[x; y]

^{(0)}= [0; -2];*

*0th Interate Interation: Initial Augmented Matrix using Forward Gaussian Elimination and Virtual Full Pivoting (*: The maximal full pivot is found in the element (1,2) of J**pr**= row pivot vector and**pc**= column pivot vector)^{(1)}:[J | -

**f**||**pr**|**pc**]^{(0)}=[0 -4 | 0 || 1 | 1] -> [ 0 -4.000 | 0 || 1 | 2 ]

[1 +1 | 2 || 2 | 2] -> [ 1 (-0.25)_{m}| 2 || 2 | 1 ]The multiplier (1/(-4))

_{m}= (-0.25)_{m}is calculated and is not really need at this step, except to do the elimination: J_{22}-> J_{22}- (-0.25)*J_{12}= 0, since the two zeros in the second row due not cause a change in J_{21}and "J_{23}" = - g^{(0)}= +2Backsubstitution is trivial due to zeros in the pivot (1st) row, yielding

**delta_x**^{(0)}= [2; 0] =>**x**^{(1)}=**x**^{(0)}+**delta_x**^{(0)}= [0; -2] + [2; 0] = [2; -2];for the first iterate answer.

*

*1st Iterate Elimination*: The first pivot is found in element (2,1) of J^{(1)}after substiting**x**^{(1)}= [x^{(1)}; y^{(1)}] into the augmented matrix function [J(x,y) | -[f(x,y); g(x,y)] ]:[J | -

**f**||**pr**|**pc**]^{(1)}=(4r)[4.000 -4.000 | -4.000 || 1 | 1] -> [ (0.5413)

_{m}-4.541 | -1.624 || 1 | 2 ]

[7.389 +1.000 | -4.389 || 2 | 2] -> [ 7.389 +1.000 | -4.389 || 2 | 1 ]The elimination step used was J

_{1,j}= J_{1,j}- (0.5413)*J_{2,j}for j=1, 3, where J_{1,3}is the element of the RHS (Right Hand Side) in the first row.Backsubstitution yields

**delta_x**^{(0)}=(4r) [-0.6424; 0.3576]; =>**x**^{(2)}=**x**^{(1)}+**delta_x**^{(1)}=(4r) [2; -2] + [-0.6424; 0.3576] =(4r) [1.358; -1.642];for the second iterate answer.

*

*2nd Iterate Elimination*: The first pivot is found in element (2,1), again, of J^{(2)}, after substituting**x**^{(2)}= [x^{(2)}; y^{(2)}] into the augmented matrix function:[J | -

**f**||**pr**|**pc**]^{(2)}=(4r)[2.716 -3.284 | -0.5903|| 1 | 1] -> [ (0.6986)

_{m}-3.983 | +0.3302 || 2 | 1 ]

[3.888 +1.000 | -1.246 || 2 | 2] -> [ 3.888 +1.000 | -1.246 || 1 | 2 ]The elimination step used was J

_{1,j}= J_{1,j}- (0.6986)*J_{2,j}for j=1, 3, where J_{1,3}is the element of the RHS (Right Hand Side) in the first row.Backsubstitution yields

**delta_x**^{(1)}=(4r) [-0.2992; -0.0.08290] =>**x**^{(2)}=**x**^{(1)}+**delta_x**^{(1)}=(4r) [1.358; -1.642] + [-0.2992; -0.0.08290] =(4r) [1.059; -1.725];Two norm measures of the closeness of the iterates that might be used for stopping criteria are the infinity (inf or maximal) norms:

||

**delta_x**^{(1)}|| = ||**x**^{(2)}-**x**^{(1)}|| =(4r) 0.2992;||

**f**^{(1)}|| = ||[f^{(1)}; g^{(1)}]|| =(4r) 1.246.For more terms, this procedure can be implemented in MATLAB.

*Remarks:
*

- These calculation were done in 4 digit rounding exam
precision as indicated by the modified equal sign "=(4r)".
- Also, this webpage was done in pure HTML, so the spacing and symbols
will not be as good as typeset graphics webapage or a text.
- See also Heath's short and different Section 5.6.2 on N-Dimensional Newton's Method. In addition, Gerald and Wheatley in the Section 2.12 on Systems of Nonlinear Equations treats the same example somewhat differently.

**Email Comments or Questions to Professor Hanson
**