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Warning: Reached the maximum number of function evaluations (10000). The result passes the global error test.

8 Ansichten (letzte 30 Tage)
theta= xlsread('KomoriPro1.xlsx',1,'A:A');
phi= xlsread('KomoriPro1.xlsx',1,'D:D');
theta_= xlsread('KomoriPro1.xlsx',1,'G:G');
phi_= xlsread('KomoriPro1.xlsx',1,'K:K');
theta= theta*pi/180;
theta_= theta_*pi/180;
phi=phi*pi/180;
phi_=phi_*pi/180;
a = zeros(19^4,1);% matrix for chi (130321x1)
J = zeros(19^4,1);
I = zeros(19^4,1);
count = 1;
for i= 1:length(theta)
for j= 1: length(phi)
for k= 1: length(theta_)
for m=1:length(phi_)
chi= acos(sin(theta(i))*sin(theta_(k))*cos(phi(j))*cos(phi_(m))+sin(theta(i))*sin(theta_(k))*sin(phi(j))*sin(phi_(m))+ cos(theta(i))*cos(theta_(k)));
a(count) = chi;
if theta(i) == (1.571)
O = integral2(@(theta,phi) asin(theta)*dirac(pi/2) , 0,3.14,0,3.14);
J(count) = integral2(@(theta_,phi_) O*sin(chi)*sin(theta_),0,3.14,0,3.14);
I(count) = integral2(@(theta,phi) J(count)*O*sin(theta),0,3.14,0,3.14);
else
O = integral2(@(theta,phi) asin(theta) , 0,3.14,0,3.14);
J(count) = integral2(@(theta_,phi_) O*sin(chi)*sin(theta_),0,3.14,0,3.14);
I(count) = integral2(@(theta,phi) J(count)*O*sin(theta),0,3.14,0,3.14);
end
count = count + 1;
end
end
end
end

Antworten (1)

Anmol Dhiman
Anmol Dhiman am 7 Dez. 2020
Hi BalPartap,
Regards,
Anmol Dhiman
  1 Kommentar
Eugene Benilov
Eugene Benilov am 11 Mai 2023
Unfortunately, this isn't a "similar question". Anmol Dhiman's link deals with the case where the global error test is FAILED, whereas the actual question is about the case where the error test is PASSED.
I've got a similar situation, where the test is passed, and I'd like to know if I can trust the result of the computation.
I might add that, in my problem, the integrand involves a Heaviside step-function: a well-written function should be able to handle it... but I'm not sure whether the one I use (quad2d) does.

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