Editing Newton's method (section)

==== Example ====
For example, the following set of equations needs to be solved for vector of points <math>\ [\ x_1, x_2\ ]\ ,</math> given the vector of known values <math>\ [\ 2, 3\ ] ~.</math>{{refn | This example is similar to one in reference,<ref name=":3" /> pages 451 and 452, but simplified to two equations instead of three.}}

<math>
\begin{array}{lcr} 
5\ x_1^2 + x_1\ x_2^2 + \sin^2( 2\ x_2 ) &= \quad 2  \\ 
e^{ 2\ x_1 - x_2 } + 4\ x_2              &= \quad 3 
\end{array}</math>

the function vector, <math>\ F (X_k)\ ,</math> and Jacobian Matrix, <math>\ J(X_k)\ </math> for iteration k, and the vector of known values, <math>\ Y\ ,</math> are defined below.

<math>\begin{align}
~ & F(X_k) ~ = ~
\begin{bmatrix} 
\begin{align}
~ & f_{1}(X_{k}) \\ 
~ & f_{2}(X_{k}) 
\end{align}
\end{bmatrix} ~ = ~
\begin{bmatrix} 
\begin{align}
~ & 5\ x_{1}^2 + x_{1}\ x^2_{2} + \sin^2( 2\ x_{2} ) \\ 
~ & e^{ 2\ x_{1}-x_{2} } + 4\ x_{2} 
\end{align}
\end{bmatrix}_k
\\
~ & J(X_k) = 
\begin{bmatrix} 
~ \frac{\ \partial{ f_{1}(X) }\ }{ \partial{x_{1}} }\ , & ~ \frac{\ \partial{ f_{1}(X) }\ }{ \partial{x_{2}} } ~\\
~ \frac{\ \partial{ f_{2}(X) }\ }{ \partial{x_{1}} }\ , & ~ \frac{\ \partial{ f_{2}(X) }\ }{ \partial{x_{2}} } ~
\end{bmatrix}_k
~ = ~
\begin{bmatrix} 
\begin{align}
~ & 10\ x_{1} + x^2_{2}\ ,  & & 2\ x_1\ x_2+4\ \sin( 2\ x_{2} )\ \cos( 2\ x_{2} ) \\ 
~ & 2\ e^{ 2\ x_{1} - x_{2} }\ ,  & &-e^{ 2\ x_{1} - x_{2}} + 4
\end{align}
\end{bmatrix}_k
\\
~ & Y = 
 \begin{bmatrix}~ 2 ~\\~ 3 ~\end{bmatrix}

\end{align} </math>

Note that <math>\ F(X_k)\ </math> could have been rewritten to absorb <math>\ Y\ ,</math> and thus eliminate  <math>Y</math> from the equations. The equation to solve for each iteration are

<math>\begin{align}
\begin{bmatrix}
\begin{align}
~ & ~ 10\ x_{1} + x^2_{2 }\ , & & 2 x_1 x_2 + 4\ \sin( 2\ x_{2} )\ \cos( 2\ x_{2} ) ~\\
~ & ~ 2\ e^{ 2\ x_{1} - x_{2} }\ ,  & & -e^{ 2\ x_{1} - x_{2} } + 4 ~
\end{align}
\end{bmatrix}_k

\begin{bmatrix}
~ c_{1} ~\\
~ c_{2} ~
\end{bmatrix}_{k+1}
=
\begin{bmatrix}
~ 5\ x_{1}^2 + x_{1}\ x^2_{2} + \sin^2( 2\ x_{2} ) - 2 ~\\
~ e^{ 2\ x_{1} - x_{2} } + 4\ x_{2} - 3 ~
\end{bmatrix}_k

\end{align}</math>

and

<math> X_{ k+1 } ~=~ X_k - C_{ k+1 } </math>

The iterations should be repeated until <math>\ \Bigg[ \sum_{i=1}^{i=2} \Bigl| f(x_i)_k - (y_i)_k \Bigr|\Bigg] < E\ ,</math> where <math>\ E\ </math> is a value acceptably small enough to meet application requirements.

If vector <math>\ X_0\ </math> is initially chosen to be <math>\ \begin{bmatrix}~ 1 ~&~ 1 ~\end{bmatrix}\ ,</math> that is, <math>\ x_1 = 1\ ,</math> and <math>\ x_2=1\ ,</math> and <math>\ E\ ,</math> is chosen to be 1.{{10^|−3}}, then the example converges after four iterations to a value of <math>\ X_4 = \left[~ 0.567297,\ -0.309442 ~\right] ~.</math>