Solving a Nonlinear Equation using Newton-Raphson Method

5,350 views (last 30 days)
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
  2 Comments
Syed nisar Abbas
Syed nisar Abbas on 5 Jul 2021
Write a code in matlab of newton rephson method to solve cos(x)+2sin(x)+x^2 any one can answer me quickly

Sign in to comment.

Accepted Answer

Bruno Pop-Stefanov
Bruno Pop-Stefanov on 25 Nov 2013
Edited: MathWorks Support Team on 27 Sep 2022
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
  6 Comments
Cesar Castro
Cesar Castro on 21 Sep 2021
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.

Sign in to comment.

More Answers (13)

Dhruv Bhavsar
Dhruv Bhavsar on 28 Aug 2020
  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.
  2 Comments
Upendra Kumar
Upendra Kumar on 13 Apr 2021
i need to solve 5 non linear equations with 5 unknowns in matlab so how i can write program for solving those equations

Sign in to comment.


Pourya Alinezhad
Pourya Alinezhad on 25 Nov 2013
you can use the following line of code;
x = fzero(@(x)x.^3 - 0.165*x.^2 + 3.993*10.^-4,0.05)

Eduardo
Eduardo on 13 Apr 2014
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.

Mohamed Hakim
Mohamed Hakim on 21 May 2021
Edited: Walter Roberson on 12 Feb 2022
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

Mark Mikofski
Mark Mikofski on 16 May 2014

Valdes Corleone
Valdes Corleone on 1 Sep 2018
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

Sayeed
Sayeed on 21 Oct 2018
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’.
  1 Comment
madhan ravi
madhan ravi on 21 Oct 2018
Please ask a separate question question because it would make this thread unrelated

Sign in to comment.


Sayeed
Sayeed on 21 Oct 2018
Thanks, I did. Can u help with that?
https://se.mathworks.com/matlabcentral/answers/425183-code-for-equation-with-ln

ali raza
ali raza on 28 Mar 2021
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

Christine Joy Africa
Christine Joy Africa on 23 Mar 2022
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

rendel deligos
rendel deligos on 28 May 2022
calculate the root of the function g(x) = (X^3 -X) sinX
  2 Comments
Walter Roberson
Walter Roberson on 28 May 2022
factor. You can see from examination that it has a root when X=-1 or X=0 or X=1 or anywhere sin(X)=0 which is all the integer multiples of π

Sign in to comment.


Fethi Ds
Fethi Ds on 29 May 2022
hello everyone.
Can you please help me to creat a program Newton Raphson for tcsc (Thyristor-Controlled Series Capacitor) to control the active power flow in transmission lines in the power system to specific values.

Jefte
Jefte on 8 Sep 2022
Newton-Raphson method to approximate the root of the nonlinear equation Compute and presets the results of five iterates for each of the initial guesses x0=1, x0=2, x0=100

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!