# Not Enough Input Arguments: Newtons Method/Jacobian Matrix

6 ビュー (過去 30 日間)
meman4567 2023 年 3 月 9 日

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];
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
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
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 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Jan 2023 年 3 月 9 日

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.

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### 回答 (1 件)

Jan 2023 年 3 月 9 日

You define gradfun with 3 inputs:
gradfun = @(x,y,z) ...
In the code you call it with 1 input:
The same happens in the next line with fun():
fxk = fun(xnew);
Either call them as
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)];

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