How to Find Bessel[-v,x] Given Bessel[v,x] in Fortran?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
xylai
Messages
58
Reaction score
0
I am working on some numerical works. I use the computer language: Fortran language.
Here I have a problem about the Bessel functon.

Now I know the value of Bessel[v,x], where v is positive and real.
I want to know the value of Bessel[-v,x].

I don't know their relation. Can you help me?
Thanks!
 
Physics news on Phys.org
The Bessel function satisfies the differential equation,

[tex]x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - \nu^2)y = 0[/tex]

We can see here that the sign of the order wold seem to be irrelevant because we take its square. However, the relationship is

[tex]J_{-n}(x) = (-1)^nJ_n(x)[/tex]
 
Born2bwire said:
The Bessel function satisfies the differential equation,

[tex]x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - \nu^2)y = 0[/tex]

We can see here that the sign of the order wold seem to be irrelevant because we take its square. However, the relationship is

[tex]J_{-n}(x) = (-1)^nJ_n(x)[/tex]

As far as I know, when n is integer, you are right: [tex]J_{-n}(x) = (-1)^nJ_n(x)[/tex].
But when n is not integer, it becomes very difficult.
 
xylai said:
As far as I know, when n is integer, you are right: [tex]J_{-n}(x) = (-1)^nJ_n(x)[/tex].
But when n is not integer, it becomes very difficult.

[tex]J_{-\nu} =\cos (\nu\pi)J_\nu - \sin(\nu\pi)Y_\nu[/tex]

via Numerical Recipes.