Integration In Matlab

5 visualizaciones (últimos 30 días)
Pranjal Pathak
Pranjal Pathak el 23 de Mzo. de 2012
Thanks to andrei bobrov for your answer! Yes R1=R2. Your coding seems to work, but the value(answer) is a huge one to know the exact value. Can we command in MATLAB to give the calculated result only upto 4 decimal place.
syms r R1=sqrt(3)*(2*r.^2-1) R2=sqrt(3)*(2*r.^2-1) b = .7; S=2*pi*imag(int(exp(1i*b*R1)*r,r,0,1)*int(R2*exp(-1i*b*R1)*r,r,0,1)) out = vpa(S);
  2 comentarios
Jan
Jan el 23 de Mzo. de 2012
Please post comments to answers in the corresponding thread.
Alexander
Alexander el 23 de Mzo. de 2012
Also, if something works for you, could you please accept the answer? So others know that the matter is resolved.

Iniciar sesión para comentar.

Respuesta aceptada

Alexander
Alexander el 23 de Mzo. de 2012
Just use double:
syms r
R1=sqrt(3)*(2*r.^2-1)
R2=sqrt(3)*(2*r.^2-1)
b = .7;
S=2*pi*imag(int(exp(1i*b*R1)*r,r,0,1)*int(R2*exp(-1i*b*R1)*r,r,0,1))
out = vpa(S);
double(out)
ans =
-0.7308
Or if you don't need vpa you can do out = double(S) directly.

Más respuestas (0)

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by