A series involving Bessel functions

[gugo82]

As some of you may know there are functions in Analysis which, although non elementary (as exponentials, powers, logarithms, ...), can be described in a very detailed way: these are usually called special functions and include Bessel functions, hypergeometic functions, Euler's [tex]$\Gamma$[/tex], the [tex]$B$[/tex] function, Riemann's [tex]$\zeta$[/tex], Legendre and Laguerre polynomials, Airy functions, the [tex]$\text{erf}$[/tex] function and many others.

Some of these functions often come up in problems of Applied Mathematics; for example, Bessel functions and Legendre polynomials are eigenfunctions of the Laplace operator with null Dirichlet boundary condition, respectively, in the [tex]$\mathbb{R}^2$[/tex] unitary disc (so that they are the normal modes of vibration of a circular membrane -or drum-) and in the [tex]$\mathbb{R}^3$[/tex] unitary ball (so that they are the normal modes of vibration of a solid ball).

Here we focus our attenction on Bessel functions of first kind, i.e. the ones usually denoted by [tex]$\text{J}_\nu (z)$[/tex].

***

Just few prerequisites: (for more see e.g. Watson, A Treatise on the Theory of Bessel functions second edition, Cambridge University Press, 1944).

Let [tex]$\nu \in \mathbb{R}$[/tex].
The following second order linear ODE:

(B) [tex]$z^2 u^{\prime \prime} (z) +z u^\prime (z)+(z^2-\nu^2) u(z)=0$[/tex]

is called Bessel equation of order [tex]$\nu$[/tex]; since the equations corresponding to [tex]$\pm \nu$[/tex] are the same, we can choose [tex]$\nu \geq 0$[/tex].

The set of solutions of (B) is a linear space of dimension [tex]$2$[/tex]; if [tex]$\nu \notin \mathbb{N}$[/tex], a base of this space is formed by the functions [tex]$\text{J}_{\pm \nu} (z)$[/tex] defined by:

[tex]$\text{J}_{\pm \nu} (z) =\sum_{k=0}^{+\infty} \frac{(-1)^k}{k!\ \Gamma (k\pm \nu +1)} \ \left( \frac{z}{2} \right)^{2k\pm \nu}$[/tex];

otherwise, if [tex]$\nu =n \in \mathbb{N}$[/tex], the linear independence of [tex]$\text{J}_{\pm n}$[/tex] fails (because [tex]$\text{J}_{-n} (z)=(-1)^n \text{J}_n (z)$[/tex]) and a base for the space of the solutions is given by the functions [tex]$\text{J}_n (z),\ \text{Y}_n (z)$[/tex], the first defined as above and the latter defined by:

[tex]$\text{Y}_n (z)=\lim_{\nu \to n} \frac{\text{J}_\nu (z) \cos \nu x -\text{J}_{-\nu} (z)}{\sin \nu x}$[/tex].

The functions [tex]$\text{J}_{\pm \nu}$[/tex] are called Bessel functions of the first kind of order [tex]$\nu$[/tex] (while the [tex]$Y_n (z)$[/tex] are called Bessel functions of the second kind, or Neumann functions, of integer order).

The series expansion of [tex]$\text{J}_\nu$[/tex] (which converges in the whole complex plane iff [tex]$\nu \geq 0$[/tex] or [tex]$\nu \text{ is a negative integer}$[/tex]) makes it clear that [tex]$0$[/tex] is a singular point iff [tex]$\nu \in ]-\infty ,0] \setminus \mathbb{Z}$[/tex]; otherwise [tex]$0$[/tex] is a continuity point iff [tex]$\nu \geq 0 \text{ or a negative integer}$[/tex] and a zero of [tex]$\text{J}_\nu$[/tex] iff [tex]$\nu >0 \text{ or a negative integer}$[/tex].
Moreover, from the series we can directly recover some interesting recurrence relations which are extremely useful in applications, as:

(1) [tex]$\left[ z^\nu\ \text{J}_\nu (z) \right]^\prime =z^\nu\ \text{J}_{\nu -1} (z)$[/tex] and [tex]$\left[ \frac{1}{z^\nu}\ \text{J}_\nu (z) \right]^\prime =-\frac{1}{z^\nu}\ \text{J}_{\nu +1} (z)$[/tex];

(2) [tex]$2\ \text{J}_\nu^\prime (z) =\text{J}_{\nu -1} (z)-\text{J}_{\nu +1} (z)$[/tex];

(3) [tex]$\frac{2\nu}{z}\ \text{J}_\nu (z) =\text{J}_{\nu -1} (z) +\text{J}_{\nu +1} (z)$[/tex],

and many many others.

***

Exercise:

Prove that the series:

(*) [tex]$\sum_{n\geq 1} \frac{1}{(2n)!!}\ \text{J}_{n-1} (z)\ z^n$[/tex]

converges for [tex]$z\in \mathbb{C}$[/tex] and evaluate its sum.

Hint: Use (3) to evaluate the partial sums of (*).

[gugo82]

Thank you, Mathematico!

[salvozungri]

Thank you gugo!! Now I'm free .
it is so interesting, a little masterpiece. For what it's worth, I congratulate you!!

[gugo82]

I'm going to split the proof in two parts:

1. Derivation of the series from one of the recurrence relations.

Using (3), one can write down the following relations:

(I) [tex]$\text{J}_1(z) = \frac{z}{2} \ \text{J}_0(z) +\frac{z}{2}\ \text{J}_2 (z)$[/tex]

(II) [tex]$\text{J}_2 (z)=\frac{z}{4} \ \text{J}_1(z) +\frac{z}{4}\ \text{J}_3 (z)$[/tex]

(III) [tex]$\text{J}_3 (z)=\frac{z}{6} \ \text{J}_2(z) +\frac{z}{6}\ \text{J}_4 (z)$[/tex]

(IV) [tex]$\text{J}_4 (z)=\frac{z}{8} \ \text{J}_3(z) +\frac{z}{8}\ \text{J}_5 (z)$[/tex]

[tex]$\ldots$[/tex]

(G) [tex]$\text{J}_n (z)=\frac{z}{2n} \ \text{J}_{n-1}(z) +\frac{z}{2n}\ \text{J}_{n+1} (z)$[/tex] (This is simply (3) with [tex]$\nu =n$[/tex])

[tex]$\ldots$[/tex]

Now plugging (II) in (I) one has:

(I.1) [tex]$\text{J}_1 (z)=\frac{z}{2} \ \text{J}_0 (z) +\frac{z^2}{8} \ \text{J}_1(z) +\frac{z^2}{8}\ \text{J}_3 (z)$[/tex];

plugging (III) in (I.1) one finds:

(I.2) [tex]$\text{J}_1 (z)=\frac{z}{2} \ \text{J}_0 (z) +\frac{z^2}{8} \ \text{J}_1(z) +\frac{z^3}{48}\ \text{J}_2 (z) +\frac{z^3}{48} \ \text{J}_4 (z)$[/tex];

with the same procedure from (IV) and (I.2) one has:

(I.3) [tex]$\text{J}_1 (z)=\frac{z}{2} \ \text{J}_0 (z) +\frac{z^2}{8} \ \text{J}_1(z) +\frac{z^3}{48}\ \text{J}_2 (z) +\frac{z^4}{384} \ \text{J}_3 (z) +\frac{z^4}{384} \ \text{J}_5 (z)$[/tex]...

It's easy to see that the denominators are [tex]$2=2!!$[/tex], [tex]$8=4!!$[/tex], [tex]$48=6!!$[/tex], [tex]$384=8!!$[/tex], hence (I.3) can be rewritten in the following manner:

(I.3) [tex]$\text{J}_1 (z) =\sum_{n=1}^4 \frac{z^n}{(2n)!!}\ \text{J}_n (z) +\frac{z^4}{8!!}\ \text{J}_5 (z)$[/tex].

Iterating the process with the aid of (G), one finds the general relation:

(I.G) [tex]$\text{J}_1 (z) =\sum_{n=1}^N \frac{z^n}{(2n)!!}\ \text{J}_n (z) +\frac{z^N}{(2N)!!}\ \text{J}_{N+1} (z)$[/tex]

which provides a convenient expression for the remainders of series (*): in fact the l.h.s. of (I.G) is the function supposed to be the sum of (*) and the r.h.s of (I.G) is in the form "[tex]$N$[/tex]-th partial sum of (*) plus a remainder term".

2. Convergence of the series via the analysis of the remainders.

Hence one wins if he can prove that [tex]$|R_N(z)|:=\frac{|z|^N}{(2N)!!}\ |\text{J}_{N+1} (z)| \to 0$[/tex] as [tex]$N\to 0$[/tex] at least on compact subsets [tex]$K\subseteq \mathbb{C}$[/tex].

Now choose [tex]$K$[/tex] compact: if [tex]$z\in K$[/tex] then [tex]$|z|^N < \rho^N$[/tex], where [tex]$\rho >0$[/tex] is s.t. [tex]$K\subseteq B(0;\rho)$[/tex]; moreover [tex]$|J_{N+1} (z)|$[/tex] is bounded from above by a constant [tex]$M_{N+1} (\rho) >0$[/tex] in [tex]$B(0;\rho)$[/tex] (by Weierstrass) and hence also in [tex]$K$[/tex]; putting these things together (and reminding that [tex]$(2N)!!=2^N\ N!$[/tex]) yelds:

(a) [tex]$|R_N(z)|\leq \frac{\rho^N}{(2N)!!}\ M_{N+1} (\rho) =\left( \frac{\rho}{2}\right)^N\ \frac{1}{N!}\ M_{N+1} (\rho)$[/tex].

But one has (using Stirling's relation [tex]$N! \approx \sqrt{2\pi N}\ (\tfrac{N}{e})^N$[/tex]):

[tex]$\left( \frac{\rho}{2}\right)^N\ \frac{1}{N!} \approx \left( \frac{\rho \ e}{2}\right)^N\ \frac{1}{\sqrt{2\pi N}\ N^N} $[/tex],

so relation [tex]$|R_N (z)|\to 0$[/tex] in [tex]$K$[/tex] holds if the sequence [tex]$M_{N+1} (\rho )$[/tex] doesn't increase "too quickly".

As Mathematico noted before, one has:

[tex]$|\text{J}_{N+1} (z)| \leq e^{\frac{|z^2|}{4}}\ \frac{|z|^N}{2}$[/tex]

for all [tex]$N$[/tex]s and [tex]$z$[/tex]s, therefore:

[tex]$M_{N+1} (\rho) \leq e^{\frac{\rho^2}{4}}\ \frac{\rho^N}{2}$[/tex]*

and it's plain that for all positive [tex]$\rho$[/tex] one has:

[tex]$\lim_N \left( \frac{\rho \ e}{2}\right)^N\ \frac{1}{\sqrt{2\pi N}\ N^N}\ e^{\frac{\rho^2}{4}}\ \frac{\rho^N}{2} = \frac{e^\frac{\rho^2}{4}}{2} \lim_N \left( \frac{\rho^2 \ e}{2}\right)^N\ \frac{1}{\sqrt{2\pi N}\ N^N}=0$[/tex].

Finally one can state that:

[tex]$\lim_N |R_N (z)|=0 \qquad \text{for $z\in K$}$[/tex]

hence series (*) converges totally on each compact subset [tex]$K\subset \mathbb{C}$[/tex].

Passing (I.G) to the limit one obviously finds:

[tex]$\text{J}_1 (z) =\sum_{n=1}^{+\infty} \frac{z^n}{(2n)!!}\ \text{J}_{n-1} (z)$[/tex]

which was the claim. $\square$

__________
* Though this estimate works, it seems too loose if [tex]$z$[/tex] is real.
I think the sequence [tex]$M_{N+1} (\rho)$[/tex] can actually be bounded from above by a constant which doesn't depend on [tex]$\rho$[/tex] for real values of [tex]$z$[/tex]... But I'm not sure.

[salvozungri]

Sorry gugo, I must give up. I don't know what to do . I used your hint, but I have problem with recursions..

[salvozungri]

[work in progress]

Ok, I want to give a try to solve this one. First of all, I observed that for all [tex]n\in\mathbb{N}, \,\Gamma(n)= (n-1)![/tex], a proof of this fact can be found here, so i decided to rewrite the Bessel function as follow:
[tex]$J_{n-1}(z) = \sum_{k=0}^\infty\frac{(-1)^k}{k! (n+k-1)!}\left(\frac{z}{2}\right)^{2k+n-1}\,\, n\in \mathbb{N}_{>0}[/tex]
because... well... I think it's much better, the Bessel functions are not so interesting in the original form .
Now I will give an upper-bound-function to [tex]|J_{n-1}(z)|[/tex]

(1.1) [tex]$|J_{n-1}(z)| =\left| \sum_{k=0}^\infty\frac{(-1)^k}{k! (n+k-1)!}\left(\frac{z}{2}\right)^{2k+n-1}\right|\le \left|\frac{z}{2}\right|^{n-1}\sum_{k=0}^\infty \frac{1}{k!(n+k-1)!}\left|\frac{z}{2}\right|^{2k}\le_{(1.2)} \left|\frac{z}{2}\right|^{n-1}e^\frac{|z|^2}{4}[/tex].

I work with our serie, in particular I will study the absolutely convergence, using (1.1):
[tex]$\sum_{n=1}^\infty\frac{1}{(2n)!!} \left| J_{n-1}(z)z^n\right|\le e^\frac{|z|^2}{4}\sum_{n=1}^\infty \left\frac{|z|^{2n -1}}{2^{n-1}(2n)!!}[/tex]

The last serie can be written as a power serie, and easily we can find the radius of convergence, in this case it's [tex]\rho=+\infty[/tex].
Finally we discovered that our serie is absolutely convergent for all [tex]z\in \mathbb{C}[/tex], then it's convergent in [tex]\mathbb{C}[/tex]

Is this ok?

For sum... I don't find the way to evaluate it , but I'm working on it
__________
(1.2): It suffices to note that [tex](n+k-1)!k!\ge k![/tex], and to remember the Taylor serie of [tex]e^x[/tex]