Matlab Numerical integral improvement

6 Ansichten (letzte 30 Tage)
Shan  Chu
Shan Chu am 4 Feb. 2017
Bearbeitet: Karan Gill am 17 Okt. 2017
Hi, I have the integral below:
F_A_I=@(x) besselj(1,x.*3.5).*besselj(1,x.*0.5);
A=integral(F_A_I,0,Inf,'RelTol',1e-6,'AbsTol',1e-12,'ArrayValued',true);
But Matlab said:
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.7e+00. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
while Mathematica can give the answer straightforward A=0.0205664
Could you please help me to improve my code. Thanks
  1 Kommentar
Niels
Niels am 4 Feb. 2017
probably a definition gap in your function, integral might converge to inf, in these cases matlab displays -> Reached the limit on the maximum number of intervals in use.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Karan Gill
Karan Gill am 13 Feb. 2017
Bearbeitet: Karan Gill am 17 Okt. 2017
Updated answer for R2017b. Use int and convert the symbolic solution to floating point.
>> syms x
f = int(besselj(1, x/2)*besselj(1, (7*x)/2),x,0,inf)
f =
-(4*(100*ellipticE(1/49) - 99*ellipticK(1/49)))/(21*pi)
>> f_dbl = double(ans)
f_dbl =
0.0022
>> f_vpa = vpa(f)
f_vpa =
0.0022054352588140668793354496265733
OLD ANSWER from 13-Feb-2017
The convert to double using "double".
  2 Kommentare
Walter Roberson
Walter Roberson am 14 Feb. 2017
Bearbeitet: Walter Roberson am 14 Feb. 2017
vpaintegral() with up to 10000 MaxFunctionCalls complains it cannot reach required precision.
The ratio oscillates a lot.
Karan Gill
Karan Gill am 25 Sep. 2017
Updated answer with solution starting R2017b.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

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

Start Hunting!

Translated by