Numerical Integration matlab Help

3 visualizaciones (últimos 30 días)
mutah
mutah el 2 de Mayo de 2014
Comentada: mutah el 5 de Mayo de 2014
Hi dears
this is my code please help me and check it
close ; clear ; clc; g=4.6;
k=0.0000000000001;
m=1;n=100;
rou=3.5/sqrt(2);
p=0.00001:0.01:0.81;
c=(rou*sqrt(pi*m)/sqrt(3*g*(1+k)))*(1/(m*exp(m*k)))*(gamma(m+0.5)/gamma(m))*kummer(m+0.5,m,m*k);
f =@(r)((1-erfc(n*p*c/r)).*((2*m.*(1+k).^((m+1)./2)./k.^((m-1)./2)./exp(m.*k)).* r.^(m).*exp(-m.*(1+k).*r.^2).*besseli(m-1, 2*m.*sqrt(k.*(k+1)).*r)));
F=integral(f,0,inf);
  2 comentarios
José-Luis
José-Luis el 2 de Mayo de 2014
kummer is not a Matlab function.

Iniciar sesión para comentar.

Respuestas (1)

Roger Stafford
Roger Stafford el 2 de Mayo de 2014
To compute this integration numerically you will have to do some kind of simplification in expressing the integrand and in calculating the value of the variable 'c'.
In the expression for 'c', at one point you divide by exp(m*k) where m*k = 1e-13. Using 'double' numbers this would certainly underflow to zero, so you would be dividing by zero at this point, which produces an infinity. The third argument of 'kummer' is also 1e-13. I do not know what this produces but I would suspect it also underflows to zero. If so, the net effect would be infinity times zero for which matlab would produce a NaN.
That same kind of problem is present in calculating the values of the anonymous function 'f'. There are tiny 'k' values within 'exp' and the bessel function in its definition, which will undoubtedly create the same deleterious effect.
As a further difficulty, in the anonymous function 'f' you have included the vector 'p' which has 81 elements. That is not the proper way to define a function which is to serve as an integrand. The integrand should produce a scalar value for 'f' for each individual value of 'r'.

Categorías

Más información sobre Numerical Integration and Differentiation en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by