Example 1.  Use fixed point iteration to find a solution to the nonlinear system  
        [Graphics:Images/NewtonSystemMod_gr_75.gif]   

Solution 1.

First, enter the coordinate functions [Graphics:../Images/NewtonSystemMod_gr_76.gif], and vector function [Graphics:../Images/NewtonSystemMod_gr_77.gif] into Mathematica.

[Graphics:../Images/NewtonSystemMod_gr_78.gif]



[Graphics:../Images/NewtonSystemMod_gr_79.gif]

Second, graph the curves [Graphics:../Images/NewtonSystemMod_gr_80.gif] using Mathematica.  
The points of intersection are the solutions we seek.  

[Graphics:../Images/NewtonSystemMod_gr_81.gif]


[Graphics:../Images/NewtonSystemMod_gr_82.gif]

 

 

How many points of intersection are there ?  
Use fixed point iteration to find a numerical approximation to the solution near [Graphics:../Images/NewtonSystemMod_gr_83.gif].  

[Graphics:../Images/NewtonSystemMod_gr_84.gif]



[Graphics:../Images/NewtonSystemMod_gr_85.gif]

[Graphics:../Images/NewtonSystemMod_gr_86.gif]

[Graphics:../Images/NewtonSystemMod_gr_87.gif]

[Graphics:../Images/NewtonSystemMod_gr_88.gif]

[Graphics:../Images/NewtonSystemMod_gr_89.gif]

[Graphics:../Images/NewtonSystemMod_gr_90.gif]

[Graphics:../Images/NewtonSystemMod_gr_91.gif]

[Graphics:../Images/NewtonSystemMod_gr_92.gif]

[Graphics:../Images/NewtonSystemMod_gr_93.gif]

[Graphics:../Images/NewtonSystemMod_gr_94.gif]

[Graphics:../Images/NewtonSystemMod_gr_95.gif]

[Graphics:../Images/NewtonSystemMod_gr_96.gif]

A fixed point satisfies the equation  [Graphics:../Images/NewtonSystemMod_gr_97.gif].  Our last approximation is stored in  [Graphics:../Images/NewtonSystemMod_gr_98.gif],  check it out.  

[Graphics:../Images/NewtonSystemMod_gr_99.gif]


[Graphics:../Images/NewtonSystemMod_gr_100.gif]

[Graphics:../Images/NewtonSystemMod_gr_101.gif]

Do you think that iteration produced the solution ? Why ?  

Accuracy is determined by the tolerance and number of iterations.  How accurate was the solution "really"?

[Graphics:../Images/NewtonSystemMod_gr_102.gif]


[Graphics:../Images/NewtonSystemMod_gr_103.gif]

[Graphics:../Images/NewtonSystemMod_gr_104.gif]

Since our tolerance was only  [Graphics:../Images/NewtonSystemMod_gr_105.gif],  the accuracy is what we should expect.

Use fixed point iteration to attempt finding a numerical approximation to the solution near [Graphics:../Images/NewtonSystemMod_gr_106.gif].  
But this time attempt only 10 iterations !  

[Graphics:../Images/NewtonSystemMod_gr_107.gif]



[Graphics:../Images/NewtonSystemMod_gr_108.gif]

[Graphics:../Images/NewtonSystemMod_gr_109.gif]

[Graphics:../Images/NewtonSystemMod_gr_110.gif]

[Graphics:../Images/NewtonSystemMod_gr_111.gif]

[Graphics:../Images/NewtonSystemMod_gr_112.gif]

[Graphics:../Images/NewtonSystemMod_gr_113.gif]

[Graphics:../Images/NewtonSystemMod_gr_114.gif]

[Graphics:../Images/NewtonSystemMod_gr_115.gif]

[Graphics:../Images/NewtonSystemMod_gr_116.gif]

[Graphics:../Images/NewtonSystemMod_gr_117.gif]

[Graphics:../Images/NewtonSystemMod_gr_118.gif]

[Graphics:../Images/NewtonSystemMod_gr_119.gif]

A fixed point satisfies the equation  [Graphics:../Images/NewtonSystemMod_gr_120.gif].  Our last approximation is stored in  [Graphics:../Images/NewtonSystemMod_gr_121.gif],  check it out.  

[Graphics:../Images/NewtonSystemMod_gr_122.gif]


[Graphics:../Images/NewtonSystemMod_gr_123.gif]

[Graphics:../Images/NewtonSystemMod_gr_124.gif]

Did iteration find the desired solution near [Graphics:../Images/NewtonSystemMod_gr_125.gif]?  Why ?  

Use Mathematica to find that solution  near [Graphics:../Images/NewtonSystemMod_gr_126.gif].  
This time use the procedure  NSolve  (if you use Solve you might be unhappy).

[Graphics:../Images/NewtonSystemMod_gr_127.gif]



[Graphics:../Images/NewtonSystemMod_gr_128.gif]

[Graphics:../Images/NewtonSystemMod_gr_129.gif]

[Graphics:../Images/NewtonSystemMod_gr_130.gif]

[Graphics:../Images/NewtonSystemMod_gr_131.gif]

 

 

Where did all those complex number come from ?  Thought we were looking at curves in the plane ?

Try solving the first equation for x then substituting one of the solutions in for x is the second equation.  

[Graphics:../Images/NewtonSystemMod_gr_132.gif]


[Graphics:../Images/NewtonSystemMod_gr_133.gif]

[Graphics:../Images/NewtonSystemMod_gr_134.gif]

[Graphics:../Images/NewtonSystemMod_gr_135.gif]

[Graphics:../Images/NewtonSystemMod_gr_136.gif]

[Graphics:../Images/NewtonSystemMod_gr_137.gif]

[Graphics:../Images/NewtonSystemMod_gr_138.gif]

[Graphics:../Images/NewtonSystemMod_gr_139.gif]

[Graphics:../Images/NewtonSystemMod_gr_140.gif]

[Graphics:../Images/NewtonSystemMod_gr_141.gif]

So we can solve for y, in a single polynomial equation of degree 12  which is known to have 12 roots, some of which are complex.

[Graphics:../Images/NewtonSystemMod_gr_142.gif]

[Graphics:../Images/NewtonSystemMod_gr_143.gif]

 

 

For our course, we don't want the complex solutions because we usually work with real numbers and our iteration schemes produces more real numbers.
However, you should be convinced that the second solution we seek is

{x -> 1.5560974843220556, y -> -0.5756507696188449}  

Do you think we should have started closer to the solution and that fixed point iteration should then work ?   
If so, try the following iteration.  

[Graphics:../Images/NewtonSystemMod_gr_144.gif]



[Graphics:../Images/NewtonSystemMod_gr_145.gif]

[Graphics:../Images/NewtonSystemMod_gr_146.gif]

[Graphics:../Images/NewtonSystemMod_gr_147.gif]

[Graphics:../Images/NewtonSystemMod_gr_148.gif]

[Graphics:../Images/NewtonSystemMod_gr_149.gif]

[Graphics:../Images/NewtonSystemMod_gr_150.gif]

[Graphics:../Images/NewtonSystemMod_gr_151.gif]

[Graphics:../Images/NewtonSystemMod_gr_152.gif]

[Graphics:../Images/NewtonSystemMod_gr_153.gif]

[Graphics:../Images/NewtonSystemMod_gr_154.gif]

[Graphics:../Images/NewtonSystemMod_gr_155.gif]

[Graphics:../Images/NewtonSystemMod_gr_156.gif]

[Graphics:../Images/NewtonSystemMod_gr_157.gif]

[Graphics:../Images/NewtonSystemMod_gr_158.gif]

[Graphics:../Images/NewtonSystemMod_gr_159.gif]

[Graphics:../Images/NewtonSystemMod_gr_160.gif]

[Graphics:../Images/NewtonSystemMod_gr_161.gif]

[Graphics:../Images/NewtonSystemMod_gr_162.gif]

[Graphics:../Images/NewtonSystemMod_gr_163.gif]

[Graphics:../Images/NewtonSystemMod_gr_164.gif]

[Graphics:../Images/NewtonSystemMod_gr_165.gif]

[Graphics:../Images/NewtonSystemMod_gr_166.gif]

Is iteration converging to the desired solution {1.5560974843220556, -0.5756507696188449}  ?  

Our goal is to obtain this second solution.  

[Graphics:../Images/NewtonSystemMod_gr_167.gif]


[Graphics:../Images/NewtonSystemMod_gr_168.gif]

[Graphics:../Images/NewtonSystemMod_gr_169.gif]

Theoretically, if we know the fixed point and look at the partial derivatives of the equations then the row sum should be small.

Case (i)  Two dimensions.  If [Graphics:../Images/NewtonSystemMod_gr_170.gif] is sufficiently close to [Graphics:../Images/NewtonSystemMod_gr_171.gif] and if  

    
[Graphics:../Images/NewtonSystemMod_gr_172.gif],  

    
[Graphics:../Images/NewtonSystemMod_gr_173.gif],  

then fixed point iteration will converge to the fixed point [Graphics:../Images/NewtonSystemMod_gr_174.gif].  

Consider the first inequality for our problem.

[Graphics:../Images/NewtonSystemMod_gr_175.gif]



[Graphics:../Images/NewtonSystemMod_gr_176.gif]

[Graphics:../Images/NewtonSystemMod_gr_177.gif]

[Graphics:../Images/NewtonSystemMod_gr_178.gif]

[Graphics:../Images/NewtonSystemMod_gr_179.gif]
[Graphics:../Images/NewtonSystemMod_gr_180.gif]

[Graphics:../Images/NewtonSystemMod_gr_181.gif]

[Graphics:../Images/NewtonSystemMod_gr_182.gif]

[Graphics:../Images/NewtonSystemMod_gr_183.gif]
[Graphics:../Images/NewtonSystemMod_gr_184.gif]

[Graphics:../Images/NewtonSystemMod_gr_185.gif]

The sum of the absolute values of the partial derivatives is too big, so we should not have expected that iteration would converge.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004