Newton's method in Matlab
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I am trying to apply Newton's method in Matlab, and I wrote a script:
syms f(x)
f(x) = x^2-4
g = diff(f)
x_1=1 %initial point
while f(['x_' num2str(i+1)])<0.001;% tolerance
for i=1:1000 %it should be stopped when tolerance is reached
['x_' num2str(i+1)]=['x_' num2str(i)]-f(['x_' num2str(i)])/g(['x_' num2str(i)])
end
end
I am getting this error:
Error: An array for multiple LHS assignment cannot contain M_STRING.
Newton's Method formula is x_(n+1)= x_n-f(x_n)/df(x_n) that goes until f(x_n) value gets closer to zero.
1 件のコメント
John D'Errico
2019 年 1 月 26 日
You should realize that things like this:
['x_' num2str(i+1)]=['x_' num2str(i)]-f(['x_' num2str(i)])/g(['x_' num2str(i)])
are not valid MATLAB syntax, that you cannot create or access variables on the fly like that.
採用された回答
Basil C.
2019 年 1 月 26 日
編集済み: Basil C.
2019 年 1 月 26 日
You seem to have made some fudamentals errors in your code.
1. the variable 'x' cannot be used however you feel right
2. also with the for loop in your code would have to run for 1000 iteration every time which makes it inefficient so always put a condition inside it
Please use the below code....
syms f(x) x
f(x) = x^2-4;
g = diff(f);
x(1)=1 ;%initial point
for i=1:1000 %it should be stopped when tolerance is reached
x(i+1) = x(i) - f(x(i))/g(x(i));
if( abs(f(x(i+1)))<0.001) % tolerance
disp(double(x(i+1)));
break;
end
end
3 件のコメント
Sergio E. Obando
2024 年 6 月 17 日
Basil's code works for me on R2024a. For an alternate solution that does not rely on symbolic variables, you can try this:
f = @(X) X.^2 - 4;
dfdx = @(X) 2*X;
nIter = 1000;
tol = 0.001;
x = zeros(1,nIter);
x(1) = 1;
for ii = 1:nIter
x(ii+1) = x(ii) - f(x(ii))/dfdx(x(ii));
if abs(f(x(ii+1))) < tol
disp(x(ii+1));
break
end
end
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