Implementing Newton Iteration Method for Solving Nonlinear Equations Using MATLAB

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The known nonlinear equations are as follows:
3 * x1-cos (X2 * X3)-1/2 = 0

X1 ^ 2-81*(X2 + 0.1) ^ 2 + sin (X3) + 1.06 = 0

Exp (-X1 * x2) + 20 * X3 + (10 * pi-3)/3 = 0

The accuracy of the solution must reach 0.00001.

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First, create the function fun

The storage equations program saves fun. m to the working path as follows:

Function f = fun (X );

% Defines the nonlinear equations as follows:

% Variable X1 X2 X3

% Function F1 F2 F3

Syms X1 X2 X3

F1 = 3 * x1-cos (X2 * X3)-1/2;

F2 = x1 ^ 2-81*(X2 + 0.1) ^ 2 + sin (X3) + 1.06;

F3 = exp (-X1 * x2) + 20 * X3 + (10 * pi-3)/3;

F = [F1 F2 F3];

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Create Function dfun

Save dfun. m to the working path of the yake matrix used to find the equations:

Function df = dfun (X );

% The yakebi matrix used to solve the equations is stored in dfun.

F = fun (X );

DF = [diff (F, 'x1'); diff (F, 'x2 '); diff (F, 'x3')];

DF = conj (DF ');

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The programming Newton method solves the nonlinear equations and saves Newton. m to the working path:

Function x = Newton (x0, EPS, N );

Con = 0;

% X0 indicates the initial iteration value. EPS indicates the accuracy. N indicates the maximum number of iterations. Con indicates whether the result is converged.

For I = 1: N;

F = Subs (fun (x0), {'x1' X2 'x3'}, {x0 (1) x0 (2) x0 (3 )});

DF = Subs (dfun (x0), {'x1 ''X2 ''x3'}, {x0 (1) x0 (2) x0 (3 )});

* = X0-f/DF;

For j = 1: length (x0 );

Il (I, j) = x (j );

End

If norm (x-x0) <EPS

Con = 1;

Break;

End

X0 = X;

End

 

In the following example, the TXT file name is iteration.txt.

Fidgefopen('iteration.txt ', 'w ');

Fprintf (FID, 'iteration ');

For j = 1: length (x0)

Fprintf (FID, 'x % d', J );

End

For j = 1: I

Fprintf (FID, '\ n % 6D', J );

For k = 1: length (x0)

Fprintf (FID, '% 10.6f', il (j, k ));

End

End

If con = 1

Fprintf (FID, '\ n calculation result convergence! ');

End

If con = 0

Fprintf (FID, '\ n too many iterations may not converge! ');

End

Fclose (FID );

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Run the program

Enter the following content in MATLAB:

Newton ([0.1 0.1-0.1], 0.00001, 20)

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Output result

Ans =

 

0.5000 0.0000-0.5236

 

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View iteration process in Iteration

 

Iteration X1 X2 X3

1 0.490718 0.031238-0.519661

2 0.509011 0.003498-0.521634

3 0.500928 0.000756-0.523391

4 0.500227 0.000076-0.523550

5 0.500019 0.000018-0.523594

6 0.500005 0.000002-0.523598

7 0.500000 0.000000-0.523599

Computation result convergence!