Why won't my code run? Matlab just says its busy when i run it?

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me 2015-9-16

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评论： Image Analyst 2015-9-19

when i put this in the command window I matlab just keeps saying its busy and doesn't give me any results. bisection(@(x)(cos(x)-x),0,1,1e-6)

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me 2015-9-17

在 MATLAB Online 中打开

function [xs, fs, n] = bisection(f, a, b, tol)
% This function is for finding a single isolated root of f over the interval [a,b]
% When do you get an error   
fa = feval(f, a);
fb = feval(f,b);
if sign(fa) * sign(fb) == 1 
    xs = nan;    
    fs = nan;
    n = nan;
    fprintf('error, check your inputss...');
    return
elseif a >= b; 
    xs = nan; 
    fs = nan; 
    n = nan; 
    fprintf('error, check your inputss...');
    return
end 
n=0; % initialize itteration
x0 = (a + b)/2;
f0 = feval(f, x0);
xf = x0;
while (abs(x0 - xf)) >= tol*(1+abs(x0))||abs(f0)>=tol;    
    xf = x0;
    x0 = (a + b)/2;
    n = n + 1;
    if (fa) * (f0) < 0
        a = x0;        
    else 
         b = x0;        
    end
end
xs = xo;
fs = f0;
% Printing final results
fprintf('\n Bisection Method For Functions W/ Multiple Roots \n Homework 3 Problem 1')
fprintf('\n\n xs = %f produces  \n fs = %f \n n = %i iterations\n', xs, fs, n-1);
end
%bisection(@(x)(cos(x)-x),0,1,1e-6)

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Answer 1

Titus Edelhofer 2015-9-17

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在 MATLAB Online 中打开

Hi,

I did not try the code, but I would strongly recommend to have some limit on the number of iterations, something like

it = 0;
itMax = 25;
while it <= itMax && (abs(x0 - xf)) >= tol*(1+abs(x0))||abs(f0)>=tol
   % do what you want to do
   it = it + 1;
end
% test no of iterations:
if it > itMax
  warning('Maximum number of iterations reached. Result may be wrong or inaccurate.')
end

For debugging use the debugger: put a breakpoint inside your loop and go step by step an watch the variables to see what's happening.

Titus

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Titus Edelhofer 2015-9-17

Sorry, I did not see that you already count iterations. Use your variable n instead of "my" variable it...

Image Analyst 2015-9-19

An excellent suggestion that I make a lot. NEVER have a while statement without a failsafe - a check on the iteration count so that you don't get into an infinite loop in case your main condition is never attained.

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Answer 2

James Tursa 2015-9-17

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https://ww2.mathworks.cn/matlabcentral/answers/243470-why-won-t-my-code-run-matlab-just-says-its-busy-when-i-run-it#answer_192682

编辑：James Tursa 2015-9-19

在 MATLAB Online 中打开

The f(0) looks like a typo in this line:

if sign(f(a)) * sign(f(0)) < 0

As for the rest of the code, nobody is going to type this in by hand to test it. We cannot copy code from a picture. In the future please post code as text highlighted with the { } code button ... do not post code as a picture.

EDIT: 9/18/2015

Try this (changes noted with <--)

function [xs, fs, n] = bisection(f, a, b, tol)
% This function is for finding a single isolated root of f over the interval [a,b]
% When do you get an error   
fa = feval(f, a);
fb = feval(f,b);
if sign(fa) * sign(fb) == 1 
    xs = nan;    
    fs = nan;
    n = nan;
    fprintf('error, check your inputss...');
    return
elseif a >= b; 
    xs = nan; 
    fs = nan; 
    n = nan; 
    fprintf('error, check your inputss...');
    return
end 
n=0; % initialize itteration
x0 = (a + b)/2;
f0 = feval(f, x0);
xf = x0;
while (abs(x0 - xf)) >= tol*(1+abs(x0))||abs(f0)>=tol;    
    xf = x0;
    x0 = (a + b)/2;
    f0 = feval(f, x0); % <-- added this line
    n = n + 1;
    if (fa) * (f0) > 0  % <-- changed < to >
        a = x0;        
        fa = f0;  % <-- added this line
    else 
         b = x0;
    end
end
xs = x0; % <-- changed xo to x0
fs = f0;
% Printing final results
fprintf('\n Bisection Method For Functions W/ Multiple Roots \n Homework 3 Problem 1')
fprintf('\n\n xs = %f produces  \n fs = %f \n n = %i iterations\n', xs, fs, n-1);
end
%bisection(@(x)(cos(x)-x),0,1,1e-6)