Help with newton-raphson

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

[Torsten]

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]

i have tried to follow codes from the web and also previous worksheets and this is what was recommended

function c=newtonRaphson(f,c0,n)
    syms x;
    df=diff(f,x);
    c(1)=c0;
    for i=1:n
        c(i+1)=c(i)-(f(c(i))/df(c(i)))
    end
    display (c);
end

[Ethan Cole]

however i always get an error about Index exceeds the number of array elements. Index must not exceed 1. and am unsure why this is or how to fix this

[Torsten]

Then show us the code together with the error message and maybe someone will be able to help.

[Sam Chak]

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]

i am using aprogramme called matlab grader so a screenshot of the code is alli can give

Answer 1

[Torsten]

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

[Ethan Cole]

through matlab grader i cant use the inline function,which is where i am getting stuck.the code needs to be able to run a variety of functions

[Torsten]

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]

i willcheck if it throws an error now, if youlook aboveat the screenshot i posted the programme gives a function and variables for the newton-raphson

[Torsten]

yes, you did not initialize c:

c = zeros(n+1,1)

[Ethan Cole]

at what point in the code should that go. aologies im extremely new to all this

[Torsten]

At the point where I did it in my code.

[Ethan Cole]

i have added it into my code but its still throwing an error.ive captured the question,my code and error if that will help

[Torsten]

After the line

df = diff(f,x)

include the line

df = matlabFunction(df);

in your code.

[Ethan Cole]

thank you

[Sam Chak]

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]

linking from the previous question, any chance you can help me convert it correctly to satisfy for all

[Ethan Cole]

[Torsten]

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

Answer 2

[Sam Chak]

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.

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;

[Ethan Cole]

Ive seen a similar codelikethis online,the only downside is that i have tostay within certain inputs that being f-function, c0- intitial and n-number of steps to be computed

[Sam Chak]

No worries. I understand that the assignment requires you to input the number of iterations/steps. @Torsten will surely fix it. But I wonder if you put n = 1e9, will the MATLAB Grader going to execute the 1 billion iterations? 😅

Answer 3

[John]

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