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Introduction

The Bessel function of the first kind $J_n(x)$ ($n \in \mathbb{Z},\ x \in \mathbb{R}$) appeared early among other topics, in Celestial Mechanics, in the series expression of the true anomaly in function of the mean anomaly, used for the description of the elliptic motion as a function of time.

Such developments seemed to have found the series solution for $J_n(x)$ by favoring the use of Laurent series expansion and the generating function approach. See for instance Tisserand's Treatise on Celestial Mechanics p.206-207 here.

These developments do not use complex analysis and the residue theorem (that otherwise makes for a good proof), but despite their elegance I find them lackluster on the mathematical side, since they seem to omit 'obvious' steps.

What I would like to see, and could not achieve yet, is a robust proof along these lines, with no other tool (since these methods seemed, again, 'obvious', and convincing enough back in the day).

I looked at a good amount of literature from the 1800s, early 1900s, and then 1960s and up (after Celestial Mechanics was studied again due to the Space Race), but could not find any clear proof along these lines, cf the Notes at the the end.

Historical developments

So these proofs usually go as follows: let us consider the following function $f(x,z)$ with $x \in \mathbb{R},\ z \in \mathbb{C}$ (the generating function): $$ \tag{*} f(x,z) = e^ {\frac{x}{2}\left(z-\frac{1}{z}\right)}. $$ This function being holomorphic on $\mathbb{C}^*$, which is an annulus, it admits an expansion in Laurent series on $\mathbb{C}^*$: $$ \tag{**} f(x,z) = \sum_{n=-\infty}^{+\infty} J_n(x) z^n. $$ The $J_n(x)$ coefficients being what is to be determined (i.e. the connection to the Bessel function is unknown so far).

On the other hand, we also have that $$ f(x,z) = e^{\frac{xz}{2}} e^{\frac{-x}{2z}}= \sum_{q=0}^{+\infty} \frac{1}{q!} \left( \frac{x z}{2} \right)^q \times \sum_{p=0}^{+\infty} \frac{(-1)^p}{p!} \left( \frac{x}{2z} \right)^p, $$ and these two series being absolutely convergent, their product is also convergent and equal to the following double sum: $$ \tag{1} f(x,z) = \sum_{q=0}^{+\infty}\sum_{p=0}^{+\infty} \left( \frac{x}{2} \right)^{q+p}\frac{(-1)^{p}}{q!p!}z^{q-p}. $$

The specific problems and my attempt

This is where for me the problems arise. If one follows Tisserand for instance, he sets $q=p+n$ and says without further explanation that one directly obtains by identification the following usual expression for $J_n(x)$ for $n \geq 0$: $$ \tag{2} J_n(x) = \sum_{p=0}^{+\infty}\frac{(-1)^p}{p!(p+n)!}\left( \frac{x}{2} \right)^{2p+n}. $$ The case for $n<0$ yielding an identical expression is solved by realizing that $J_{-n}(x) = (-1)^n J_n(x)$, which is easily seen by doing $z \to -\frac{1}{z}$ in $(*)$ and seeing that $(*)$ does not change, and then comparing the two expressions using $(**)$.

From my point of view, to be able to say directly by identification that we have $(2)$ can only come from the fact that $(1)$ should be able to be recast as $$ \tag{3} f(x,z) = \sum_{n=-\infty}^{+\infty}\sum_{p=0}^{+\infty} \epsilon(n) \left( \frac{x}{2} \right)^{2p+n}\frac{(-1)^{p}}{(p+n)!p!}z^n $$ with $\epsilon(n) = 1$ if $n \geq 0$, and $(-1)^n$ otherwise.

So, since $(1) \to (2)$ did not seem obvious at all to me, I did the actual calculation by changing the indices in the double sum $(1)$, this yields: $$ \tag{4} f(x,z) = \sum_{n=-p}^{+\infty}\sum_{p=0}^{+\infty} \left( \frac{x}{2} \right)^{2p+n}\frac{(-1)^{p}}{(p+n)!p!}z^n $$ and this is for me where the real problem is because:

  1. The first sum starts at $-p$ instead of $-\infty$
  2. There is no $\epsilon(n)$ function anywhere

So this is where I am, I cannot find a nice change of indices, or a nice way to reorder $(4)$ to get $(3)$. Now, I am unsure of the actual rules to switch infinite double sums, which may or may not be one of the cruxes to what my problem boils down to...but even by assuming everything is fine with switching I cannot find a way.

Notes

  • I know that Watson in his reference treaty on the subject Treatise on the theory of Bessel functions (p. 14-16, see here) solves the problem by looking at each term yielding a $z^n$ or $z^{-n}$ and collects all of them to yield $(2)$. This is pretty convincing, but nevertheless, he somewhat masks the double sums by doing so, and I feel that there is a way of solving the problem more robustly as I described above, by some nice reordering or change of indices.
  • There are also a number of proofs in earlier works (that don't even use the summation sign), that claim that by 'looking' at the sums and extracting the good terms, you can 'see' that you have $(2)$, but that does not cut it for me.
  • Maybe there is no way exactly as I described, but I of course welcome any other convincing solution not hiding anything, and at the same time not using complex analysis, or other kinds of tools.

N.B.: I posted this question some months ago on Math.SE which got one comment that did not really help, but I feel like this is a more appropriate place given the historical context at play, so I hope to get more answers here.

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  • $\begingroup$ This does not look like a history question to me, so much as a mathematics question with a historical motivation. I can see why Math SE might not have cared for it, because of the heavy personal-taste constraints you put on the mathematics problem. $\endgroup$ – kimchi lover Jan 6 at 15:35
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    $\begingroup$ It's a fair point, but I am asking here if anybody knows of any proof along the lines of what I describe (Tisserand's approach) in the abundant and old literature on the subject that I most likely have missed, especially since I focused when I searched on literature on the Celestial Mechanics side, and not in any other field that I may not know much about (and Bessel functions arise in many places), so I feel like this is relevant for the HS&M.SE. Maybe I should replace a tag with reference-request and rephrase a little? $\endgroup$ – viiv Jan 6 at 16:16

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