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I'm trying to make sense of Hankel's expansion for the first Bessel function $J_0$. According to NIST, it is 10.17(i) here. Using their notation, we have $\nu=0.$ Hence the linked formula becomes:

Let

$$a_k = \frac{(-1^k)1^2*3^2* . . . *(2k-1)^2}{8^kk!}$$

and

$$\omega = z - \pi/4.$$

Then

$$J_0(z) \approx \sqrt{\frac{2}{\pi z}}\left(\cos(\omega)\sum_{k=0}^{\infty}{(-1)^k \frac{a_{2k}}{z^{2k}} - \sin(\omega)\sum_{k=0}^{\infty}{(-1)^k \frac{a_{2k+1}}{z^{2k+1}}}}\right)$$

But that makes no sense, as the series on the right both diverge! The problem is that the absolute value of the numerator of $a_k$ is greater than $(k!)^2$. Therefore, the absolute value of $a_k$ is greater than $\frac{k!}{8^k}$; hence for large $k$, it dominates the $z^{(2k+1)}$ in the denominator.

Can anyone rectify the apparent contradiction? (NIST goofed, maybe?)

William Jockusch
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1 Answers1

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$$\frac{(2k-1)!!^2}{8^k k!}=\frac{(2k)!^2}{8^k k! (2k)!!^2}=\frac{(2k)!^2}{32^k k!^3}=\frac{k!}{32^k}\binom{2k}{k}^2 $$ behaves like $\frac{k!}{2^k}$, hence you are right, the series involved in your last expression are not analytic functions of $\frac{1}{z}$. The issue here lies in the meaning of $\approx$. For instance, a continuous but not analytic function of $\frac{1}{n}$ also appears in Stirling's approximation

$$ \log n! \approx n\log\frac{n}{e}+\frac{1}{2}\log(\pi n)+\color{red}{\sum_{k\geq 2}\frac{(-1)^k B_k}{k(k-1)n^{k-1}}} $$ So, here it is the solution of the mistery: the notation $\approx$ means that for any $m\geq 2$ we have

$$ \log n! \approx n\log\frac{n}{e}+\frac{1}{2}\log(\pi n)+{\sum_{k=2}^m\frac{(-1)^k B_k}{k(k-1)n^{k-1}}}+O\left(\frac{1}{n^m}\right)\quad\text{as }n\to +\infty $$ but no further uniformity (about the coefficient hidden in the Landau notation $O$) is granted.

On the other hand, $J_0(z)$ is a non-constant entire function, hence the point at infinity of the Riemann sphere has to be an essential singularity of $J_0(z)$, and that apparently pathological behaviour is not pathological at all.

Jack D'Aurizio
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