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

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Discussion Overview

The discussion revolves around the relationship between the Bessel functions of positive and negative orders, specifically how to compute Bessel[-v,x] given Bessel[v,x] in the context of numerical computations using Fortran. The scope includes theoretical aspects of Bessel functions and their properties.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks assistance in finding the value of Bessel[-v,x] given Bessel[v,x], indicating a lack of understanding of their relationship.
  • Another participant states that the Bessel function satisfies a specific differential equation and suggests that the sign of the order may not matter due to the squaring in the equation. They provide the relationship J_{-n}(x) = (-1)^nJ_n(x).
  • A subsequent reply reiterates the relationship for integer orders but notes that complications arise when the order is not an integer.
  • Further, it is mentioned that for non-integer orders, the relationship can be expressed as J_{-\nu} = cos(νπ)J_ν - sin(νπ)Y_ν, referencing Numerical Recipes.
  • One participant acknowledges the cleverness of another's contribution without further elaboration.

Areas of Agreement / Disagreement

Participants agree on the relationship for integer orders but express uncertainty and difficulty regarding non-integer orders. The discussion remains unresolved regarding the general case.

Contextual Notes

The discussion highlights the complexity of the relationship between Bessel functions for non-integer orders and references specific mathematical expressions without resolving the implications of these relationships.

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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