triple integral of parametrized function

5 vues (au cours des 30 derniers jours)
Bill Francois
Bill Francois le 6 Mai 2015
Hi
I would like to numerically compute the integral of a parametrized function to use it as a function of the parameter. Is this possible? (I know it works with a simple integral, but in the Help folder of integral3, they assign a value to the parameter BEFORE computing the integral so they do not get a function of the parameter in the end).
My function is the following one:
fun1 = @(k,e,x,y,z)((e.*psf(x,y,z)).^k).*exp(-e.*psf(x,y,z))/factorial(k)
Where psf is a function of x, y , z: @(x,y,z)exp(-2*(x.^2+y.^2)-2*z.^2) (a 3D Gaussian)
I would like to get the triple integral of fun1 between the limits -100 and 100 for x,yamd z (for example; the best woul be for me to be able to tune the limits of the integral) and use it as a function of k (which is a natural integer), to plot it for various values of e.
Thanks in advance
Bill

Réponse acceptée

Mike Hosea
Mike Hosea le 7 Mai 2015
Something like this?
psf = @(x,y,z)exp(-2*(x.^2+y.^2)-2*z.^2);
fun1 = @(k,e,x,y,z)((e.*psf(x,y,z)).^k).*exp(-e.*psf(x,y,z))/factorial(k);
% Make a function that takes a scalar k and a scalar e and returns the
% integral. Can use -100,100 limits (faster), or expressions involving k.
% Can use different tolerances.
scalar_k_scalar_e_fun = @(k,e)integral3(@(x,y,z)fun1(k,e,x,y,z),-inf,inf,-inf,inf,-inf,inf,'Abstol',1e-4,'RelTol',1e-3);
% Make the latter function work with an array input for e, to facilitate
% plotting.
scalar_k_array_e_fun = @(k,e)arrayfun(@(e)scalar_k_scalar_e_fun(k,e),e);
e = 0:0.1:1;
k = 3:5;
q = zeros(length(k),length(e));
for i = 1:length(k)
q(i,:) = scalar_k_array_e_fun(k(i),e);
end
hold on
for i = 1:length(k)
plot(e,q(i,:));
end
hold off

Plus de réponses (1)

Walter Roberson
Walter Roberson le 6 Mai 2015
No, you cannot do that numerically. There is a possibility that you could do it symbolically, but it would be common that no closed-form integral existed.
At the time you do numeric integration, all variables must be assigned particular values, with the particular x, y, z to integrate at being the only free variables. There is no way to produce a formula out of numeric integration. And that's what you seem to be wanting to do, produce a formula that has k as a free variable. If you want a formula output, then you need symbolic integration.
  1 commentaire
Bill Francois
Bill Francois le 7 Mai 2015
Thanks for the answer however, I do not want a general formula of the function defined by the integral. What I would like would be to is to be able to compute the value of the integral, numerically, for various values of the parameter, and plot them. How can i do that?
The best would be for me to have a function of k, e and l, (l being the integration limit), that returns the numerical value of the integral when I give it the values of k, e, and l. Is there a way I can do this?
It is possible to do it with a single integral but I find no way of doing it with a triple integral.

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