Newton's method in Matlab
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FURKAN CEVAHIR
am 26 Jan. 2019
Kommentiert: Sergio E. Obando
am 17 Jun. 2024
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 Kommentar
John D'Errico
am 26 Jan. 2019
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.
Akzeptierte Antwort
Basil C.
am 26 Jan. 2019
Bearbeitet: Basil C.
am 26 Jan. 2019
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 Kommentare
Sergio E. Obando
am 17 Jun. 2024
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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