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?

Answer 1

[Bruno Pop-Stefanov]

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

x = 0.05;
x_old = 100;
x_true = 0.0623776;
iter = 0;
while abs(x_old-x) > 10^-3 && x ~= 0
    x_old = x;
    x = x - (x^3 - 0.165*x^2 + 3.993*10^-4)/(3*x^2 - 0.33*x);
    iter = iter + 1;
    fprintf('Iteration %d: x=%.20f, err=%.20f\n', iter, x, x_true-x);
    pause;
end

You can plot the function with, for example:

x = -10:0.01:10;
f = x.^3 - 0.165*x.^2 + 3.993*10^-4;
figure;
plot(f)
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?

Answer 2

[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 3

[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?

Answer 4

[Eduardo]

Hello,

I am trying to make a program using the Newton-Raphson method for complex equations, but I am in a infinite loop.

If I change the expression,

 x = x - (x^3 - 0.165*x^2 + 3.993*10^-4)/(3*x^2 - 0.33*x);

For this:

 x = x - (sqrt(-x))/(-0.5.*(1./sqrt(-x)));

The algorithm does not work.

how can I use a complex equation in this case ??

Thank you very much.

Answer 5

[Mark Mikofski]

use newtonraphson

Answer 6

[Valdes Corleone]

Hi ! I want to solved an système of three equations non-linear with three variables ,discretised with finit differential and solved with newton raphson. Please can I have a code matlab

Answer 7

[Sayeed]

How can i Code it?

u=0.23 y=0.3 v=0.00001

u/u’=5*log(u’*y/v)-3.05

u’=?

I need to find the value of u’.

[madhan ravi]

Please ask a separate question question because it would make this thread unrelated

Answer 8

[Sayeed]

Thanks, I did. Can u help with that?

https://se.mathworks.com/matlabcentral/answers/425183-code-for-equation-with-ln