`Therefore, the system of equations ()
can be written compactly as
`

`We want to find a solution of the system, i.e. a set of values
such that all the equations are simultaneously
satisfied
`

`Newton's method builds a sequence of points
`

`In one dimension, the sequence is built using Newton's formula
`

`For -dimensional problems, this formula generalizes to
`

`The place of the derivative is now taken by the Jacobian matrix
defined as
`

`Let
`

`The Newton formula () can be written as
`

`This is a system of linear equations; is a matrix,
and and are -dimensional vectors.
`

`One step of the Newton method (say, the step)
proceeds as follows:
`

- 0
- We have available, and want to compute the next iterate, ;
- 1
- Evaluate the vector function (we evaluate each component function individually);
- 2
- Evaluate the derivative matrix (each entry is a function of and need to be evaluated individually);
- 3
- Solve the system (). For this, we need to
- 3.1
- compute the LU decomposition (with pivoting) of the matrix ;
- 3.2
- apply the back-substitution algorithm to the right hand side , to obtain the solution ;

- 4
- Compute the next iterate as .

`Note that one Newton step is very expensive. We have to evaluate
functions (the entries of ) and we need to
calculate the LU decomposition of each step.
`