integral(exp(double integral)) ?===>integral of riemann
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Marwen Tarhouni
am 20 Jul. 2019
Kommentiert: Marwen Tarhouni
am 21 Jul. 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 Kommentare
John D'Errico
am 20 Jul. 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.
Akzeptierte Antwort
Sulaymon Eshkabilov
am 20 Jul. 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
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