My code won't run. It says there is an error in line 30.

조회 수: 3 (최근 30일)
Oliver Mendoza
Oliver Mendoza 2020년 2월 20일
답변: Alex Mcaulley 2020년 2월 20일
I am trying to make this program run. When I attempt to run it, it says that there is an error in line 30. I don't know what the issue could be.
  댓글 수: 1
Jon
Jon 2020년 2월 20일
편집: Jon 2020년 2월 20일
What does it say the error is? Please cut and paste the entire error message. Also, please post the code that you use to test/call your function that can be run to produce the error.

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답변 (1개)

Alex Mcaulley
Alex Mcaulley 2020년 2월 20일
There is an error in your calls to matlabFunction:
function [] = Newton(f, x1, tol, N)
%
% Newton uses Newton’s method to approximate a root of f(x) = 0.
%
% [r, n] = Newton(f, x1, tol, N), where
%
% f is a symbolic function representing f(x),
% x1 is the initial point,
% tol is the scalar tolerance for convergence (default 1e-4),
% N is the maximum number of iterations (default 20),
%
% r is the approximate root of f(x) = 0,
% n is the number of iterations required for convergence.
%
if nargin < 4 || isempty(N), N = 20; end
if nargin < 3 || isempty(tol), tol = 1e-4; end
% Find f' and convert to MATLAB function for evaluation
% syms x
fp = matlabFunction(diff((f))); %Changed
% Convert f to MATLAB function for evaluation
% syms x
f = matlabFunction(f); %Changed
x = zeros(1,N+1); % Pre-allocate
x(1) = x1;
for n = 1:N
if fp(x(n)) == 0
r ='failure';
return
end
x(n+1) = x(n) - (f(x(n))/fp(x(n)));
if abs(x(n+1) - x(n)) < tol
r = x(n+1);
return
end
end
end

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