Bessel generating function PDF

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MATH7402: The generating function for Bessel functions The Bessel functions Jn (x) have a generating function that can be used in a similar way to that for the Legendre Polynomials. It turns out that n=∞ X

Jn (x)tn = exp

n=−∞



x 2

  1 t− . t

(1)

This can be derived by considering solutions of Helmholtz’ equation for G(x) in Cartesian and in polar coordinates. ∇2 G + G = 0,

r2 Grr + rGr + Gθθ + r 2 G = 0.

Gxx + Gyy + G = 0,

One solution is G = exp(iy), corresponding to a plane wave. In polar coordinates, this is G = exp(ir sin θ). We have seen that the solution in polar coordinates can be found by separating out the angular dependence exp(inθ) leaving the radial dependence satisfying Bessel’s equation. The plane wave is regular at the origin so we do not need the Yn solutions. In the next chapters where we come back to separation of variables, we will see that we should be able to write G as follows ∞ X G = exp(ir sin θ) = An Jn (r) exp(inθ ) (2) n=−∞

for some constants An . If we write t = exp(iθ) so that sin θ = (t − t−1 )/2, and replace r by x then we have    n=∞ X 1 x t− = An Jn (x)tn . (3) exp 2 t n=−∞ To show that the An are all equal to one, we will use our knowledge of the expansions of Jn (x) for small x, obtained from the series solutions shown in the lectures. In particular, we can obtain from the series d n Jn (0) = 2−n . dxn We start by isolating a particular Jm from the sum (2) by effectively using the orthogonality properties of cos(nθ) and sin(nθ) which lie behind the idea of Fourier Series. We multiply (2) by exp(−imθ) and integrate Z

2π

exp(ir sin θ − imθ) dθ =

0

∞ X

n=−∞

An Jn (r)

Z

2π

exp(inθ − imθ) dθ = 2πAm Jm (r )

0

Now differentiate m times with respect to r and put r = 0 to find Z 2π d m Jm im (sinm θ) exp(ir sin θ) exp(−imθ ) dθ = 2πAm (r), dr m 0 Z 2π =⇒ im sinm θ exp(−imθ)dθ = 2πAm /2m . 0

Using the exponential form for sin θ gives Z Z 2π m 2m 2π 1  1 1 Am = 2π = 1, (1 − exp(−2imθ ))m dθ = exp iθ − exp( −iθ) exp( −imθ) dθ = 2π 2π 0 2m 2π 0 as the only non-zero contribution to the integral comes from integrating the first term in the binomial expansion of the integrand. This proves our result....