don't know the problem in my while loop

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hamza kharbouch
hamza kharbouch 2022 年 1 月 1 日
回答済み: hamza kharbouch 2022 年 1 月 1 日
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);

採用された回答

the cyclist
the cyclist 2022 年 1 月 1 日
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 件のコメント
hamza kharbouch
hamza kharbouch 2022 年 1 月 1 日
編集済み: hamza kharbouch 2022 年 1 月 1 日
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 2022 年 1 月 1 日
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);

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その他の回答 (1 件)

hamza kharbouch
hamza kharbouch 2022 年 1 月 1 日
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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