Help with newton-raphson
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Construct a MATLAB function called newtonRaphson that implements n steps of the
Newton-Raphson method to find the root of some function f(z).
Your function should receive the following three inputs (in this order): an inline function
f; an initial guess cO; and the number of steps to be computed, n.
Your function should return a vector called c with n + 1 entries containing a vector of all
the approximations computed. The first entry in c should contain the initial guess; the
remaining n entries should contain the approximations that your function has computed
in order.
The definition of your function should look like this:
function c=newtonRaphson(f,c0,n)
Im unsure where to start as every type of code i try to use ends up coming back with an error. please help
6 Commenti
Sam Chak
il 31 Mar 2022
Modificato: Sam Chak
il 31 Mar 2022
I see... @Ethan Cole, you have tried writing many codes for the Newton–Raphson Method in solving root-finding problem
.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/947924/image.png)
Most likely you are a little rigid (restricted) and overwhelmed by the technical terms in the instructions, such as to have an inline function f, the number of steps to be computed, n, and then the solver shall return a vector called c with n + 1 entries. If you look at the Newton–Raphson Iteration Formula in mathematics book:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/947929/image.png)
it definitely does not tell you what an inline function is, what steps are, and what a vector c is. That's why you don't know where to begin.
Can you put up the latest version of your code here?
Risposta accettata
Torsten
il 31 Mar 2022
Modificato: Torsten
il 31 Mar 2022
f = inline('x.^2-3','x');
c0 = 2;
n = 10;
c = newtonRaphson(f,c0,n)
function c = newtonRaphson(f,c0,n)
df = @(x) (f(x+1.0e-6)-f(x))*1e6;
c = zeros(n+1,1);
c(1) = c0;
for i = 1:n
c(i+1) = c(i)-(f(c(i))/df(c(i)))
end
end
13 Commenti
Torsten
il 31 Mar 2022
Modificato: Torsten
il 31 Mar 2022
Isn't this exactly Sam Chak's code ?
function c = newtonRaphson2(f,c0,epsilon)
syms x
df = matlabFunction(diff(f,x));
error = 2*epsilon;
itermax = 30;
cold = c0;
iter = 0;
flag = 0;
while error > epsilon
iter = iter + 1;
cnew = cold - f(cold)/df(cold);
error = abs(cnew-cold)/max(1,abs(cold));
if iter > itermax
disp('Iteration limit exceeded.');
flag = 1;
break
end
cold = cnew;
end
if flag == 0
c = cnew;
else
c = c0;
end
end
Più risposte (2)
Sam Chak
il 31 Mar 2022
Hi @Ethan Cole
The method presented by @Torsten satisfies your requirements in your assignment. Here is an alternative. The main difference is the termination condition, where the program does not execute a fixed number of iterations, but the interation will stop once the condition
is satisfied.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/948009/image.png)
format long g
f = @(x) x^2 - 3; % to find the square root of 3
epsilon = 1e-6;
x0 = 1;
[c, Iterations] = NewtonRaphson(f, x0, epsilon)
function [x, Iter] = NewtonRaphson(f, x0, epsilon)
Iter = 0;
df = @(x) (f(x+1.0e-6)-f(x))*1e6;
x1 = x0 - f(x0)/df(x0);
while abs(f(x1)) > epsilon
x0 = x1;
x1 = x0 - f(x0)/df(x0);
Iter = Iter + 1;
end
x = x1;
2 Commenti
John
il 31 Lug 2023
function [p, PN] = Newton_371(p0,N,tol,f,fp)
p=p0;
PN(1)=p0;
for n= 1:N
p=p-f(p)/fp(p);
PN(n+1) =p;
if abs(p-PN(n)) <=tol
break
end
end
end
0 Commenti
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