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

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

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

My textbook derives this by showing that

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

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?
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Vargo is offline
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 is offline
Nov26-12, 04:15 PM
P: 224
What math course is this? Just wondering

JJacquelin is offline
Nov27-12, 01:42 AM
P: 746

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 is offline
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.

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