Wronskian of Bessel Functions of non-integral order v, -v


by mjordan2nd
Tags: bessel, functions, nonintegral, order, wronskian
mjordan2nd
mjordan2nd is offline
#1
Nov23-12, 06:41 PM
P: 118
My textbook states

[tex]
J_v(x) J'_{-v}(x) - J'_v(x) J_{-v}(x) = -\frac{2 \sin v \pi}{\pi x}
[/tex]

My textbook derives this by showing that

[tex]
J_v(x) J'_{-v}(x) - J'_v(x) J_{-v}(x) = \frac{C}{x}
[/tex]

where C is a constant. C is then ascertained by taking x to be very small and using only the first order of the power series expansion for Bessel functions. Does this mean that this computation for C is inexact? It seems that there should be some error terms in there from higher powers of x, or am I missing something?

By the way, I'm using Arfken/Weber and N.N. Lebedev as my guide here.

Thanks for any help.

Edit: Perhaps this would have been better in the differential equations section?
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
Vargo
Vargo is offline
#2
Nov26-12, 12:07 PM
P: 350
No it doesn't sound like an approximation.

If you have two power series that are equal for all x (in some interval), then their coefficients have to be equal at all orders. Strictly speaking these aren't power series since there is a term of order -1 but that doesn't change the fact that the coefficients on the right side have to match the coefficients on the left side. On the right side, C is the lowest order coefficient. So if you calculate the lowest order coefficient on the left side, it has to be equal to C.

The fact that the right side has only the one term C/x means that all the higher order terms cancel each other out on the left side.
piercebeatz
piercebeatz is offline
#3
Nov26-12, 04:15 PM
P: 224
What math course is this? Just wondering

JJacquelin
JJacquelin is offline
#4
Nov27-12, 01:42 AM
P: 744

Wronskian of Bessel Functions of non-integral order v, -v


[tex]J_v(x) J'_{-v}(x) - J'_v(x) J_{-v}(x) = -\frac{2 \sin v \pi}{\pi x} [/tex]
is not an approximate. It is an identity for any variable x and any order v

Considering a constant order v, then [tex] C=-\frac{2 \sin v \pi}{\pi} [/tex] is constant. hence [tex] J_v(x) J'_{-v}(x) - J'_v(x) J_{-v}(x) = \frac{C}{x} [/tex]
mjordan2nd
mjordan2nd is offline
#5
Dec3-12, 08:47 PM
P: 118
I think I understand! Thank you for your explanations.

Pierce: This is my mathematical methods for physicists course.


Register to reply

Related Discussions
integral of spherical bessel function (first kind), first order Calculus 2
while finding a second solution with wronskian for bessel... Differential Equations 0
Integral of Bessel functions combination? Calculus 1
Integral representation of Bessel functions of the second kind Differential Equations 0
integral of first order (first kind) bessel function Differential Equations 2