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

AI Thread Summary
The discussion focuses on finding the value of the Bessel function Bessel[-v,x] given Bessel[v,x] in Fortran. It highlights the relationship between Bessel functions of negative and positive orders, specifically that J_{-n}(x) = (-1)^nJ_n(x) for integer n. For non-integer values, the relationship becomes more complex, with the formula J_{-\nu} = cos(νπ)Jν - sin(νπ)Yν provided as a reference. The Bessel function's differential equation is also mentioned, emphasizing that the sign of the order does not affect the equation's structure. Understanding these relationships is crucial for accurate numerical computations involving Bessel functions.
xylai
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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!
 
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The Bessel function satisfies the differential equation,

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

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

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

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

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

J_{-n}(x) = (-1)^nJ_n(x)

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

J_{-\nu} =\cos (\nu\pi)J_\nu - \sin(\nu\pi)Y_\nu

via Numerical Recipes.
 
You are very clever.
 
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