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 个评论
Torsten
2022-3-31
There are so many MATLAB Newton codes in the net.
If you don't know where to start, you should choose one of them and try to understand it.
Ethan Cole
2022-3-31
Ethan Cole
2022-3-31
Torsten
2022-3-31
Then show us the code together with the error message and maybe someone will be able to help.
I see... @Ethan Cole, you have tried writing many codes for the Newton–Raphson Method in solving root-finding problem 
.
. 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:
,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?
Ethan Cole
2022-3-31
编辑：Cris LaPierre
2022-3-31
i am using aprogramme called matlab grader so a screenshot of the code is alli can give
采纳的回答
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 个评论
Ethan Cole
2022-3-31
Torsten
2022-3-31
But you were told to define f as an inline function. What is MATLAB grader ? Does the code throw an error when executed with usual MATLAB ?
Ethan Cole
2022-3-31
yes, you did not initialize c:
c = zeros(n+1,1)
Ethan Cole
2022-3-31
Torsten
2022-3-31
At the point where I did it in my code.
Ethan Cole
2022-3-31
Torsten
2022-3-31
After the line
df = diff(f,x)
include the line
df = matlabFunction(df);
in your code.
Ethan Cole
2022-3-31
Sam Chak
2022-3-31
Wait @Ethan Cole. You need to test various nonlinear functions. Don't submit first. The assignment doesn't say that you cannot do have hybrid approach to check the accuracy of the solution. Some nonlinear function may require more iterations to converge to the solution.
Ethan Cole
2022-3-31
Ethan Cole
2022-3-31
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
更多回答（2 个）
Sam Chak
2022-3-31
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.
is satisfied.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;
John
2023-7-31
0 个投票
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
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