Does the Bessel Function Identity J_n-1(z) + J_n+1(z) = (2n/z) J_n(z) Hold?

[bdj03001]

Complex analysis: Let J_n (z) be the Bessel function for a positive integer n of order n. Verify?

J_n-1 (z) + J_n+1 (z) = ((2n)/z) J_n (z)

 

[Orion1]

Bessel Identity...


For a Bessel function of the first kind $$J_n(z)$$

Identity confirmed:
$$J_{n-1}(z) + J_{n+1}(z) = \frac{2\,n\,J_{n}(z)}{z} \; \; \; n > 0 \; \; \; z \neq 0$$

$$\Mfunction{BesselJ}(-1 + n,z) + \Mfunction{BesselJ}(1 + n,z) = \frac{2\,n\,\Mfunction{BesselJ}(n,z)}{z} \; \; \; n > 0 \; \; \; z \neq 0$$

n = 1
Attachment 1: LHS plot
Attachment 2: RHS plot

The x-intercepts and amplitudes appear to match, therefore this is an identity.

Reference:
http://www.efunda.com/math/bessel/besselJYPlot.cfm

 

[tacman]

To verify this statement, we will use the recurrence relation for Bessel functions:

J_n+1 (z) = (2n/z) J_n (z) - J_n-1 (z)

Substituting this into the given equation, we get:

J_n-1 (z) + (2n/z) J_n (z) - J_n (z) = ((2n)/z) J_n (z)

Simplifying, we get:

J_n-1 (z) + J_n (z) = ((2n)/z) J_n (z)

which is exactly the statement that we wanted to verify. Therefore, we can conclude that the given equation is true and the statement has been verified.