Effacer les filtres
Effacer les filtres

Info

Cette question est clôturée. Rouvrir pour modifier ou répondre.

working on a bisection code and im pretty lost is it right so far?

2 vues (au cours des 30 derniers jours)
adriana resendez
adriana resendez le 25 Fév 2015
Clôturé : MATLAB Answer Bot le 20 Août 2021
function [xr ea iter] = bisect(func,x1, xu, es, maxit)
if func(x1)*func(xu)>=0, error('no sign change'); end
xold = x1;
for iter=1:maxit
xr = ???
test = func(x1)*func(xr)
ea = abs((xr-xold)/xr)*100;
if test<0, xu = xr;
elseif test>0; ???;
else ea = 0;
end
  2 commentaires
John D'Errico
John D'Errico le 25 Fév 2015
Please use the code formatting button "{} Code" when you post code. Your code is unreadable as it is.
John D'Errico
John D'Errico le 25 Fév 2015
Modifié(e) : John D'Errico le 25 Fév 2015
Having fixed your code (I'll do that once) to make it readable...
Is it right? Of course not.
You have assignments like
xr = ???
and
???;
That last was a good one, though not truly valid syntax. I'll admit that when I stop writing code to take a rest, I usually add a random line like
jw56ywykmej
This way I cannot miss finding it when I resume work later on. At the very least, mlint will tell me about it.
Your code is missing end statements. Lots of other issues I could pick at. For example, why use a for loop instead of a while loop? A for loop forces the code to run for maxit iterations. What if it has converged before then? Yes, this last is a minor issue that relates to efficiency, instead of does the code work.
Essentially, you are going in a reasonable direction, but you are not there yet. So why not finish writing it? Try it out? Learn to debug your code, so learn to use the debugger.

Réponses (0)

Cette question est clôturée.

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by