Solving a Nonlinear Equation using Newton-Raphson Method

It's required to solve that equation: f(x) = x.^3 - 0.165*x.^2 + 3.993*10.^-4 using Newton-Raphson Method with initial guess (x0 = 0.05) to 3 iterations and also, plot that function.

Please help me with the code (i have MATLAB R2010a) ... I want the code to be with steps and iterations and if possible calculate the error also, please

[DEBADITYA GUPTA]

suppose I need to solve f(x)=a*x.^3+b*x.^2+c using Newton-Raphson method where a,b,c are to be import from excel file or user defined, the what i need to do?

[Syed nisar Abbas]

Write a code in matlab of newton rephson method to solve cos(x)+2sin(x)+x^2 any one can answer me quickly

[Rajesh]

to solve the given equation (x^2-1)/(x-1) taking minimum values for x and draw the stem graph and step graph

[Walter Roberson]

@Rajesh

You should open your own Question, after reading http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Answer 1

[Bruno Pop-Stefanov]

The following code implements the Newton-Raphson method for your problem:

fun = @(x)x^3 - 0.165*x^2 + 3.993e-4;  

x_true = fzero(fun,[0.01 0.1],optimset("Display","iter")); 

 
 Func-count    x          f(x)             Procedure
    2             0.1    -0.0002507        initial
    3       0.0644397  -1.82743e-05        interpolation
    4       0.0617709   5.41455e-06        interpolation
    5       0.0623809  -2.96286e-08        interpolation
    6       0.0623776  -4.33628e-11        interpolation
    7       0.0623776    1.0842e-17        interpolation
    8       0.0623776             0        interpolation
 
Zero found in the interval [0.01, 0.1]

x = 0.1;
x_old = 100;

iter = 0;

while abs(x_old-x) > 1e-10 && iter <= 10 % x ~= 0  
    x_old = x;  
    x = x - (x^3 - 0.165*x^2 + 3.993e-4)/(3*x^2 - 0.33*x);  
    iter = iter + 1;  
    fprintf('Iteration %d: x=%.18f, err=%.18f\n', iter, x, x_true-x);  
    pause(1);  
end 

Iteration 1: x=0.016433333333333161, err=0.045944248180416342
Iteration 2: x=0.094298416648742320, err=-0.031920835134992817
Iteration 3: x=0.042655687778369436, err=0.019721893735380067
Iteration 4: x=0.063158887832661437, err=-0.000781306318911934
Iteration 5: x=0.062375951756182137, err=0.000001629757567366
Iteration 6: x=0.062377581507153931, err=0.000000000006595571
Iteration 7: x=0.062377581513749496, err=0.000000000000000007

You can plot the function with, for example:

x = linspace(0,0.1);  
f = x.^3 - 0.165*x.^2 + 3.993*10^-4;  
figure;  
plot(x,f,'b',x,zeros(size(x)),'r--')  
grid on 

[Hassan Mohamed]

If possible, i want the code to be with steps and iterations, and if possible calculate the error also, please. thnx so much for helping

[Bruno Pop-Stefanov]

I edited my answer to display the value of x at each iteration and the error with the true value. pause stops the loop until the you press a key.

[Hassan Mohamed]

i'm sorry but, the value of error in incorrect Relative Approximate Ture Error = (True Error / True value) * 100

[Skyler Kolb]

if your initial guess is

x = 0.05

but then late declare your initial guess as a new value

x_old = x

in your while loop, how is that going to work? Won't that defeat the purpose of having an initial guess?

[Brian Reinhard]

ajarin master

[Cesar Castro]

It did not work , May you help me, please? My F = x(1)^4+x(2)^4+2*x(1)^2*x(2)^2-4*x(1)+3 I Know the root is 1.

I do not know why, it did not work. Thanks in advance.

Answer 2

[Dhruv Bhavsar]

1. Solve the system of non-linear equations.

x^2 + y^2 = 2z

x^2 + z^2 =1/3

x^2 + y^2 + z^2 = 1

using Newton’s method having tolerance = 10^(−5) and maximum iterations upto 20

%Function NewtonRaphson_nl() is given below.
fn = @(v) [v(1)^2+v(2)^2-2*v(3) ; v(1)^2+v(3)^2-(1/3);v(1)^2+v(2)^2+v(3)^2-1];
jacob_fn = @(v) [2*v(1) 2*v(2) -2 ; 2*v(1) 0 2*v(3) ; 2*v(1) 2*v(2) 2*v(3)];
error = 10^-5 ;


v = [1 ;1 ;0.1] ;
no_itr = 20 ;

[point,no_itr,error_out]=NewtonRaphson_nl(v,fn,jacob_fn,no_itr,error)
NewtonRaphson_nl_print(v,fn,jacob_fn,no_itr,error);

# OUTPUT.

Functions Below.

function [v1 , no_itr, norm1] = NewtonRaphson_nl(v,fn,jacob_fn,no_itr,error)
    % nargin = no. of input arguments
    if nargin <5 , no_itr = 20 ; end
    if nargin <4 , error = 10^-5;no_itr = 20 ; end
    if nargin <3 ,no_itr = 20;error = 10^-5; v = [1;1;1]; end
    
    v1 = v;
    fnv1 = feval(fn,v1);
    i = 0;
    while true
        jacob_fnv1 = feval(jacob_fn,v1);
        H = jacob_fnv1\fnv1;
        v1 = v1 - H;
        fnv1 = feval(fn,v1);
        i = i + 1 ;
        norm1 = norm(fnv1);
        if i > no_itr && norm1 < error, break , end
        %if norm(fnv1) < error , break , end
    end
end

function [v1 , no_itr, norm1] = NewtonRaphson_nl_print(v,fn,jacob_fn,no_itr,error)
    v1 = v;
    fnv1 = feval(fn,v1);
    i = 0;
    fprintf('      Iteration|    x     |     y      |    z      | Error      | \n')
    while true
        norm1 = norm(fnv1);
        fprintf('%10d     |%10.4f| %10.4f | %10.4f| %10.4d |\n',i,v1(1),v1(2),v1(3),norm1)
        jacob_fnv1 = feval(jacob_fn,v1);
        H = jacob_fnv1\fnv1;
        v1 = v1 - H;
        fnv1 = feval(fn,v1);
        i = i + 1 ;
        norm1 = norm(fnv1);
        if i > no_itr && norm1 < error, break , end
        %if norm(fnv1) < error , break , end
        
    end
end

This covers answer to your question and also queries for some comments I read in this thread.

[Juan Sebastian Castro Barrero]

i need make a program in matlab to solve

F(1)=1+(x(1)^2)-(x(2)^2)+((exp(x(1)))*cos(x(2)));

F(2)=(2*(x(1))*(x(2)))+((exp(x(1)))*sin(x(2)));

starting initials (-1,4)

5 iterations.

can you help me?

[Upendra Kumar]

i need to solve 5 non linear equations with 5 unknowns in matlab so how i can write program for solving those equations

[Munish Jindal]

Got this error.

unrecognized function or variable 'NewtonRaphson_nl_print'.

[Walter Roberson]

the code is given above starting at the line

function [v1 , no_itr, norm1] = NewtonRaphson_nl_print(v,fn,jacob_fn,no_itr,error)

Answer 3

[Pourya Alinezhad]

you can use the following line of code;

x = fzero(@(x)x.^3 - 0.165*x.^2 + 3.993*10.^-4,0.05)

Answer 4

[Mohamed Hakim]

function NewtonRaphsonMethod
%Implmentaton of Newton-Raphson method to determine a solution.
%to approximate solution to x = cos(x), we let f(x) = x - cos(x)
i = 1;
p0 = 0.5*pi;            %initial conditions
N = 100;                %maximum number of iterations
error = 0.0001;         %precision required

syms 'x'
f(x) = x - cos(x);      %function we are solving
df = diff(f);           %differential of f(x)
while i <= N
    p = p0 - (f(p0)/df(p0));          %Newton-Raphson method 
    
    if (abs(p - p0)/abs(p)) < error                     %stopping criterion when difference between iterations is below tolerance
        fprintf('Solution is %f \n', double(p))
        return
    end
    
    i = i + 1;
    p0 = p;             %update p0
end
fprintf('Solution did not coverge within %d iterations at a required precision of %d \n', N, error)     %error for non-convergence within N iterations
end

[Taha Anum]

can i have a generic code which may ask user for the equation

[Vicki]

What is the role of the "syms" function here?

[Torsten]

By using x as a symbolic variable, f is a symbolic function. The "diff" command can analytically differentiate symbolic functions. Thus if you are not able to compute the derivative of x-cos(x), symbolic math and the diff command will do it for you.

Equivalently, you could just set

f = @(x) x-cos(x);
df = @(x)1+sin(x);

without the

syms x 

line.

Answer 5

[likhith krishna]

function [y, exitflag, iter_count] = newton(R, J, x0, tol, maxit)

y = x0;
iter_count = 0;
exitflag = 0;

R_val = R(y);
if norm(R_val, 2) < tol
    exitflag = 1;  
    return;
end

for k = 1:maxit-1
    J_val = J(y);

    if rcond(J_val) < 1e-12
        exitflag = -1;
        iter_count = k - 1;
        return;
    end

    dx = -J_val \ R_val;

    y = y + dx;
    iter_count = k;

    R_val = R(y);


    if norm(R_val, 2) < tol
        exitflag = 1;
        return;
    end
end

exitflag = 0;
iter_count = maxit;
end

[pinninti]