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Seahawks el 11 de En. de 2021
Respondida: Divija Aleti el 25 de En. de 2021
Hi All,
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
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Walter Roberson el 11 de En. de 2021
Do not use inline(). Use matlabFunction() to convert symbolic expressions into anonymous functions.

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### Respuestas (1)

Divija Aleti el 25 de En. de 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)
Regards,
Divija
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