MATLAB Answers

Please help to solve the following equations by the Newton–Raphson method

29 views (last 30 days)
Seahawkgo
Seahawkgo on 11 Jan 2021
Answered: Divija Aleti on 25 Jan 2021
Hi All,
Please help to solve the following equations by the Newton–Raphson method:
fx1 = (2*x1^2) + (x2^2) -10;
fx2 = (x1^2) - (x2^2) + (x1*x2) - 4;
Start with an initial guess of x1 = 1 and x2 = 1., y =0
My code as below:
clc;clear;close all;
syms x1 x2 fx1 fx2 y1 y2;
fx1 = (2*x1^2) + (x2^2) -10;
fx2 = (x1^2) - (x2^2) + (x1*x2) - 4;
Fx = [fx1;fx2]
J11 = diff(fx1,x1);
J12 = diff(fx1,x2);
J21 = diff(fx2,x1);
J22 = diff(fx2,x2);
JJ = [J11 J12 ; J21 J22]
JJ1=[J11; J12]
JJ2=[J21 ; J22]
y1 = 0;
y2 = 0;
YY=[y1;y2]
x1(1) = 1;
x2(1) = 1;
X = [1;1];
epsilon = 0.001;
n = 10;
func1 = inline(Fx(1))
func2 = inline(Fx(2))
M = diff(sym(Fx(1)))
N = diff(sym(Fx(2)))
DF1 = inline(M)
DF2 = inline(N)
for i=1:4
X(i+1)=X(i)+(YY-func1(X(i)))
end
Thanks in advance
  1 Comment
Walter Roberson
Walter Roberson on 11 Jan 2021
Do not use inline(). Use matlabFunction() to convert symbolic expressions into anonymous functions.

Sign in to comment.

Answers (1)

Divija Aleti
Divija Aleti on 25 Jan 2021
Hi,
Have a look at the following code which solves the above mentioned equations by Newton-Raphson Method :
clc;clear;close all;
syms fx1(x1,x2) fx2(x1,x2)
fx1(x1,x2) = (2*x1^2) + (x2^2) -10;
fx2(x1,x2) = (x1^2) - (x2^2) + (x1*x2) - 4;
Fx = [fx1;fx2]
J11 = diff(fx1,x1);
J12 = diff(fx1,x2);
J21 = diff(fx2,x1);
J22 = diff(fx2,x2);
JJ = [J11 J12 ; J21 J22];
epsilon = 0.001;
x1_o = 1;
x2_o = 1;
x = [x1_o];
y = [x2_o];
for i=1:10
h = det([-fx1(x(i),y(i)) J12(x(i),y(i)); -fx2(x(i),y(i)) J22(x(i),y(i))])/det(JJ(x(i),y(i)));
k = det([J11(x(i),y(i)) -fx1(x(i),y(i)); J21(x(i),y(i)) -fx2(x(i),y(i))])/det(JJ(x(i),y(i)));
x(i+1) = x(i) + h;
y(i+1) = y(i) + k;
if abs(x(i+1)-x(i)) <= epsilon && abs(y(i+1)-y(i)) <= epsilon
break
end
end
% The two roots are :
x_1 = x(i+1)
x_2 = y(i+1)
For additional information on 'syms', refer to the following link:
Regards,
Divija

Community Treasure Hunt

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

Start Hunting!

Translated by