integral(exp(double integral)) ?===>integral of riemann
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Marwen Tarhouni
le 20 Juil 2019
Commenté : Marwen Tarhouni
le 21 Juil 2019
Hi,
i have an equation to 3 unknown (x,y theta)
i try this code
a =2.2;
b =0:10:100;
r = 50;
for i = 1:length(b)
bin=100;
resultat=0;
for k=1:bin
y=k*r/bin;
eq1= @(x,theta) exp(-b(i).*(sqrt(x.^2.*(cos(theta)).^2+(x.*sin(theta)-y).^2).*x)) ;
In2=integral2(eq1,0,r,0,2*pi);
resultat=resultat+ exp(-a* (1-In2))*2*y/(r^2);
end
end
==== >result does not work correctly
2 commentaires
John D'Errico
le 20 Juil 2019
Without even looking more carefully at the numbers, as soon as I see this start:
exp(-3.7154e+05*d(i).*(
I will predict your problem is in the form of exponential underflows. The result will be numerical garbage.
Réponse acceptée
Sulaymon Eshkabilov
le 20 Juil 2019
Hi,
In your code, the variable d is assigned instead of b on line 2: d= 0:10:100; When this is fixed then everything works ok.
You can also start your simulation with the command: clearvars
Good luck
Plus de réponses (0)
Voir également
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!