why doesnt matlab calculate this divisions

function deneme()

syms x y
f = @(x,y) (((1/5) * (exp(-2 * x) - 6 * sin(x * y))) - 0.4325 )

g = @(x,y) (((1/5) * (x^2 * y + 6 * cos(x))) - 0.0643 )

J = jacobian([((1/5) * (exp(-2 * x) - 6 * sin(x * y))) - 0.4325 , ((1/5) * (x^2 * y + 6 * cos(x))) - 0.0643],[x , y])

A = inv(J)
H(x,y) = det(J)

C(x,y) = A(1,1)
D(x,y) = A(1,2)
E(x,y) = A(2,1)
F(x,y) = A(2,2)
error = 100;
tc = 10^(-20);
i = 0;
x0 = 1;
y0 = 1;
x(1) = x0
y(1) = y0

while (error > tc)
  
  if H(x(i+1),y(i+1)) == 0
      fprintf('System cannot be solved')
  end
  x(i+2) = eval(x(i+1) -  C(x(i+1),y(i+1)) * feval(f,x(i+1),y(i+1)) + D(x(i+1),y(i+1)) * feval(g,x(i+1),y(i+1)));
  
  y(i+2) = eval(y(i+1) -  E(x(i+1),y(i+1)) * feval(f,x(i+1),y(i+1)) + F(x(i+1),y(i+1)) * feval(g,x(i+1),y(i+1)));
 
  e1 = abs((x(i+2)-x(i+1))/x(i+2)) * 100;
  e2 = abs((y(i+2)-y(i+1))/y(i+2)) * 100;
  error = eval(max(e1,e2))
  i = i+1;
end
disp(error)
fprintf('After %d iterations the approx root is %f %f',i,x(i+1),y(i+1))
end

Answer 1

[Stephan]

Hi,

x,y are symbolic variables. Matlab shows them this way, due to accurracy. Insert a type cast to make them numbers:

x = double(x);
y = double(y);

Best regards

Stephan

[hadoune oussama]

it doesn't work!

[hadoune oussama]

here is a pic of the results!

Answer 2

[Walter Roberson]

In new enough versions you can control this with sympref

https://www.mathworks.com/help/symbolic/sympref.html#mw_1464af98-5a29-4ac0-b0a7-3d101ea82566

[hadoune oussama]

it does really work!! thank you