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Adjacent initial Bessel values cannot both vanish

Statement

For every real $x > 0$ , $J_{0} (x)$ and $J_{1} (x)$ are not both zero.

Facts & Assumptions

Given: A real number $x > 0$ and the integer-order Bessel functions from Integer-order Bessel series.

[F1]

Bessel differential equations and order recurrences from the series: $J_{0}$ is entire, satisfies $z^{2} J_{0}^{''} + z J_{0}^{'} + z^{2} J_{0} = 0$ , has $J_{0} (0) = 1$ , and $J_{0}^{'} = - J_{1}$ .

[F2]

Identity theorem for holomorphic functions: two holomorphic functions on a complex domain that agree on a set with an interior accumulation point agree throughout that domain.

Proof

1.1givenF1algebra

Suppose $J_{0} (x) = J_{1} (x) = 0$ . By [F1], $J_{0} (x) = J_{0}^{'} (x) = 0$ . Expand the entire function about $x$ , $J_{0} (x + w) = \sum_{m \geq 0} a_{m} w^{m}$ . The coefficient of $w^{m}$ in the differential equation in [F1] is $x^{2} (m + 2) (m + 1) a_{m + 2} + x (m + 1) (2 m + 1) a_{m + 1} + (m^{2} + x^{2}) a_{m} + 2 x a_{m - 1} + a_{m - 2} = 0,$ with $a_{- 1} = a_{- 2} = 0$ . Since $a_{0} = a_{1} = 0$ and $x > 0$ , induction gives $a_{m} = 0$ for all $m$ , so $J_{0}$ vanishes on a disk about $x$ .

2.1F1F2step 1.1∎

Apply [F2] to $J_{0}$ and the zero function on the connected domain $C$ . It follows that $J_{0}$ is identically zero, contradicting $J_{0} (0) = 1$ from [F1]. Thus $J_{0} (x)$ and $J_{1} (x)$ cannot both vanish.

Depends on

Used by

Positive discrete Wronskian for the Bessel recurrence pair Lemma

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

DLMF, §§10.2(i)–(ii) (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 4 (standard reference, not scraped)