Detailed explanation of the implementation method of iterative method of nonlinear equations in matlab

Detailed description of matlab implementation of nonlinear equation system iteration method

Newton iteration method:

function[x0,n]=newton(fx,dfx,x0,tol,N)

% Newton iteration method

% The first parameter fx is the desired function expression about variable x.

% The second parameter dfx is the first derivative of fx.

% x0 is the initial value of iteration.

% tol is the iteration error limit.

% N Maximum number of iterations.

x=x0;f0=eval(fx);df0=eval(dfx);

n=0;

disp('[ n xn xn 1 delta ]');

while n

x1=x0-f0/df0;

x=x1;f1=eval(fx);df1=eval(dfx);

delta=abs(x0-x1);

% X=[n,x0,x1,delta];

disp(X); % is used to display intermediate results

if delta

fprintf('Iterative calculation successful')

return

else

n=n 1;

x0=x1;f0=f1;df0=df1;

end

end

if n==N 1

fprintf('Iteration calculation failed')

end

The other two can be modified slightly on this basis.

MATLAB program to use Newton iteration to solve nonlinear equations

Give you a complete version:

% Newton’s method for solving nonlinear equations

function main()

clc; clear all;

f = @(x)log(x sin(x)); % test function

df = @(x)(1 cos(x))/(x sin(x)); % derivative function

x0 = 0.1; % iteration initial value

x = TestNewton(f, df, x0) % Newton’s method solution

function x = TestNewton(fname, dfname, x0, e, N)

% Purpose: Newton iteration method to solve nonlinear equation f(x)=0

% fname and dfname respectively represent the M function handle or embedded function expression of f(x) and its derivative function

% x0 is the iteration initial value, e is the accuracy (default value 1e-7)

% x returns a numerical solution and displays the calculation process. Set the upper limit of the number of iterations N to prevent divergence (default 500 times)

% Input parameters

if nargin

N = 500;

end

if nargin

e = 1e-7;

end

x = x0; % initial value

x0 = x 2*e; % float

k = 0; % number of steps

fprintf('x[%d]= .9f\n', k, x) % print information

while abs(x0-x)>e & k

k = k 1; % record the number of steps

x0 = x; % update x(k)

x = x0 - feval(fname,x0)/feval(dfname,x0); % update x(k 1)

fprintf('x[%d]= .9f\n', k, x) % print information

end

if k == N

fprintf('The maximum number of iterations has been reached'); % End of iteration

end

result:

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