Help with newton-raphson

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Ethan Cole 2022-3-31

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1685284-help-with-newton-raphson

回答： John 2023-7-31

采纳的回答： Torsten

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

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Sam Chak 2022-3-31

编辑：Sam Chak 2022-3-31

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

Screenshot 2022-03-31 at 14.10.19.png

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

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采纳的回答

Torsten 2022-3-31

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https://ww2.mathworks.cn/matlabcentral/answers/1685284-help-with-newton-raphson#answer_931404

编辑：Torsten 2022-3-31

在 MATLAB Online 中打开

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

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Ethan Cole 2022-3-31

Torsten 2022-3-31

编辑：Torsten 2022-3-31

在 MATLAB Online 中打开

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

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更多回答（2 个）

Sam Chak 2022-3-31

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https://ww2.mathworks.cn/matlabcentral/answers/1685284-help-with-newton-raphson#answer_931429

在 MATLAB Online 中打开

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;

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Ethan Cole 2022-3-31

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 2022-3-31

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? 😅