Newton's Method Help
20 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Kaeden Beaty
am 12 Feb. 2021
Kommentiert: Alan Stevens
am 12 Feb. 2021
I have a code that works for Newton's method, but I am trying to make it into an algorithm that takes an imput from the interval [a, b] instead of x0. In other words I guess the initial point is calculated inside of the cody using x0=(a+b)/2. Another thing I would like to do is insert an if statement into the main loop that breaks out of the loop if the x is outside of [a, b]. This is the original that works, but uses an input of x0.
function x=mynewton(f,df,x0,tol,maxiter)
x=x0;
xold=x
for i=1:maxiter
x=x-f(x)/df(x)
err(1)=abs(x-xold)
xold=x
if err(1)<tol
break
end
end
And then this is what I tried to do but there is an error
function x=mynewton2(f,df,x0,tol,maxiter)
x=(a+b)/2;
xold=x
for i=1:maxiter
x=x-f(x)/df(x)
err(1)=abs(x-xold)
xold=x
if err(1)<tol
break
end
if x<a
break
end
if x>b
break
end
end
end
I am really sorry if this is a stupid question, I am just very new to MATLAB and am not the best with technology. Any help would be much appreciated!
0 Kommentare
Akzeptierte Antwort
Alan Stevens
am 12 Feb. 2021
Assuming you make
x0 = [a, b];
before you pass it to the function, then replace
x = (a+b)/2;
by
x = (x0(1) + x0(2))/2;
within mynewton2.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!