Not Enough Input Arguments: Newtons Method/Jacobian Matrix

4 visualizaciones (últimos 30 días)
meman4567
meman4567 el 9 de Mzo. de 2023
Editada: Jan el 9 de Mzo. de 2023
Hey all, I am getting a "not enough input arguments" in the following code in the "gradfun" line, which is the jacobian matrix of the three functions listed in the vector "fun". Can anyone help with this?
tol = 10^-9;
x0 = [-1 1 1];
x = x0(1);
y = x0(2);
z = x0(3);
fun = @(x,y,z) [x^2-y-sin(z)+1, x+1+sin(10*y)-y, (1-x)*z-y];
gradfun = @(x,y,z) [2*x, -1, -cos(z); 1, 10*cos(10*y)-1, 0;-z,-1,1-x];
newton(fun,gradfun,x0,tol)
The newton function, which i from a textbook of mine, is as follows:
function [root,numits] = newton(fun,gradfun,x0,tol)
% Solve fun(x)=0 using Newton's method given the function and its gradient
% gradfun starting from the initial guess x0.
x0 = x0(:); % this will force x0 to be a column vector
xold = x0+1; % this needs to be ~= x0 so that we enter the while loop
xnew = x0;
numits = 0;
n = length(x0);
%data_x=[xnew(1)];
%data_y=[xnew(2)];
while norm(xnew-xold)>tol
gradfxk = gradfun(xnew);
fxk = fun(xnew);
fxk = fxk(:); % this will force fxk to be a column vector
[a,b]=size(fxk);
if a~=n || b~=1
error('function has wrong dimension, expecting %d x 1, but got %d x %d',n, a, b)
end
[a,b]=size(gradfxk);
if a~=n || b~=n
error('gradient has wrong dimension, expecting %d x %d, but got %d x %d',n, n, a, b)
end
xold = xnew;
xnew = xold - (gradfxk)^{-1} * fxk, but implement as a linear solve
xnew = xold - gradfxk \ fxk;
numits = numits+1;
plot(xnew(1), xnew(2),'-o','MarkerEdgeColor','black');
%xlim([-5 5])
%ylim([-80 10])
pause(0.1)
if (numits>=100)
root = xnew;
fprintf('current step:\n')
disp(xnew)
error('no convergence after %d iterations', numits);
end
end
root = xnew;
root
end
  1 comentario
Jan
Jan el 9 de Mzo. de 2023
Editada: Jan el 9 de Mzo. de 2023
Please post a copy of the complete message instead of paraphrasing some parts of it. Which line is "in the "gradfun" line"?
Please use the tools to format code. This improves the readability of the code.

Iniciar sesión para comentar.

Respuestas (1)

Jan
Jan el 9 de Mzo. de 2023
Editada: Jan el 9 de Mzo. de 2023
You define gradfun with 3 inputs:
gradfun = @(x,y,z) ...
In the code you call it with 1 input:
gradfxk = gradfun(xnew);
The same happens in the next line with fun():
fxk = fun(xnew);
Either call them as
gradfxk = gradfun(xnew(1), xnew(2), xnew(3));
fxk = fun(xnew(1), xnew(2), xnew(3));
Or define the functions as:
fun = @(x) [x(1)^2-x(2)-sin(x(3))+1, x(1)+1+sin(10*x(2))-x(2), (1-x(1))*x(3)-x(2)];
gradfun = @(x) [2*x(1), -1, -cos(x(3)); 1, 10*cos(10*x(2))-1, 0;-x(3),-1,1-x(1)];

Categorías

Más información sobre Software Development Tools en Help Center y File Exchange.

Etiquetas

Productos

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by