What is Newton's method when there is a lot of input?

I am trying to estimate 3parameters of mixed weibull distribution. And I think the Nweton method for it.

But, error happend

'Error: exit :There are too many output arguments.

Error: etimation_weibull> accelerated_newton (line 30)

if ~ exit ('n_iter','var')

Error: etimation_weibull (line 21)

[x, x_err, x_h] = accelerated_newton (func, dfunc, x0, conv_tol, n_iter);

I think that the error is related to the part where the @ function is used. Is it not possible if there are multiple inputs?

Also, is there any other method for estimating the three parameters of the mixed Weibull distribution?

%% Estimation 3parameters in weibull distribution
load('weibul.mat')

Error using load
Unable to find file or directory 'traj.mat'.

X = Euclidean_distance;  % input data

%% Define first-order and second-order derivatives of Weibull distribution

a = 0.1;
b = 0.1;
c = 0.1;

func = @(x,a,b,c) 1-exp((x-c)/a).^b;
dfunc = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-(x-c)/a).^b;

x0 = Euclidean_distance(1);
conv_tol = 1e-10;
n_iter = 30;


[x, x_err, x_h] = accelerated_newton(func,dfunc,x0,conv_tol,n_iter); %● error happend


%% newton-methods function
function [x, x_err, x_h] = accelerated_newton(func,dfunc,x0,conv_tol,n_iter)

          
         if ~isa(func, 'function_handle') || ~isa(dfunc,'function_handle')
             error('function must be a handle (add"@" before the function name')
         end

           if ~exit('n_iter', 'var')
               n_iter = 30;
           end

          if ~exist('conv_tol', 'var')
              conv_tol = 1e-15;
          end

          x = x0; 
          x_h = nan(n_iter + 1, 1);
          x_h(1) = x;

          for k=1:n_iter
              f = func(x);
              df = dfunc(x);
              delta = f./df;

              if k==1
                  df_23  = dfunc(x - 2.*delta./3);
                  FF = 0.5 + sqrt(max(0 ,0.75.*df_23 - 0.5));
              else
                  d2f = (4.*df + 2.*df_old - 6.*(f - f_old)./(x - x_old))./(x-x_old);
                  A2_hat = -.5.*d2f./df;
                  FF = 0.5 + sqrt(max(0, 0.25 - A2_hat.*delta));
              end
              x_old = x;
              x = x_old - delta./FF;
              x_h(k+1) = x;
              x_err = abs(x_old - x);
              if ((x_err < conv_tol) && (abs(f) - conv_tol))
                  x_h(k+2:n_iter+1) = [];
                  break;
              end

              df_old = df;
              f_old = f;
              if k == n_iter
                  disp('The function did not converge')
              end
          end
end

Answer 1

[Lokesh]

Hi,

There are a few errors in the code that lead to the errors "There are too many output arguments" and "Not enough input arguments."

The first error is due to the typo 'exit', which should be 'exist'. Here is the corrected code:

if ~exist('n_iter', 'var')
    n_iter = 30;
end

Additionally, the functions func and dfunc are defined to expect four input variables (x, a, b, c), but they are being called with only a single input variable x, which leads to the error "Not enough input arguments."