don't know the problem in my while loop

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hamza kharbouch
hamza kharbouch el 1 de En. de 2022
Respondida: hamza kharbouch el 1 de En. de 2022
there is also another error idk about of :
Array indices must be positive integers or logical values.
Error in sym/subsref (line 890)
R_tilde = builtin('subsref',L_tilde,Idx);
iteration=0;i=1;
x=zeros(100,1);
syms f(x);
f(x) = input('donner une function : f(x)= ');
e = input('donner une la valuer de erreur : e = ');
x(1) = input('donner la valeur de depart : x1 = ');
while ( e < abs(x(i+1)-x(i)) )
iteration = iteration +1 ;
x(i+1)=x(i)-f(x(i))/diff(f(x(i)));
i=i+1;
end
disp('la value de x apartir de e est', x(i));
disp(iteration);

Respuesta aceptada

the cyclist
the cyclist el 1 de En. de 2022
You initialize the value of i to be zero.
Then you try
x(i)
which is equivalent to
x(0)
which means you are trying to access the zeroth element of x. But MATLAB indexing begins at 1, not 0, so you get that error.
  8 comentarios
hamza kharbouch
hamza kharbouch el 1 de En. de 2022
Editada: hamza kharbouch el 1 de En. de 2022
yea the purpose of it was to calculate the root of any function u choose as f(x) depending on the value of the start x1 and the error e of how close is enough
i choosed
f(x)=x^3-10 for example
e=0.001
x1=3
for an example . and it didn't work there . i want it to work atleast in the possible cases and i think a polynomial function usually is
the cyclist
the cyclist el 1 de En. de 2022
OK. Is the following code equivalent to your example, and give the error you see?
I removed the input functions (because they won't work here), and I also used an anonymous function instead of the symbolic function (because I don't have that toolbox).
iteration=0;
i=1;
x=zeros(100,1);
% syms f(x);
% f(x) = input('donner une function : f(x)= ');
% e = input('donner une la valuer de erreur : e = ');
% x(1) = input('donner la valeur de depart : x1 = ');
f = @(x) x.^3 - 10;
e = 0.001;
x(1) = 3;
while ( e < abs(x(i+1)-x(i)) )
iteration = iteration +1 ;
x(i+1)=x(i)-f(x(i))/diff(f(x(i)));
i=i+1;
end
Error using /
Matrix dimensions must agree.
disp('la value de x apartir de e est', x(i));
disp(iteration);

Iniciar sesión para comentar.

Más respuestas (1)

hamza kharbouch
hamza kharbouch el 1 de En. de 2022
okey i did minor other simplifications and it's working now
thanks
i=1;
iteration=0;
f = @(x) cos(x);
df = @(x) -sin(x);
e = 0.00001;
x(1) = 3;
while ( e < abs(x(i+1)-x(i)) )
iteration = iteration +1 ;
x(i+1)=x(i)-f(x(i))/df(x(i));
i=i+1;
end
disp(iteration);
disp(x(i));

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