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An arbitrarily small seed perturbation flips a later Bessel sign

Statement

At $x = 1$ , every $J_{n} (1)$ is positive and tends to zero as $n \to \infty$ , whereas the complementary recurrence solution $D_{n} (1)$ defined in Complementary solution for the integer Bessel recurrence is positive and unbounded for $n \geq 1$ . Hence, for every $ε > 0$ , the perturbed seed pair ${\tilde{J}}_{0} = J_{0} (1) - ε D_{0} (1), {\tilde{J}}_{1} = J_{1} (1) - ε D_{1} (1)$ can be arbitrarily close to the exact seeds, yet its forward recurrence solution ${\tilde{J}}_{n} = J_{n} (1) - ε D_{n} (1)$ is negative for some index where the exact $J_{n} (1)$ is positive. This is a recurrence-sensitivity example, not a claim that every floating-point implementation follows this particular error direction.

Facts & Assumptions

Given: The Bessel series at $x = 1$ , the order recurrence, and the complementary solution $D$ .

[F1]

Integer-order Bessel series: at $x = 1$ , $J_{n} (1) = \sum_{k \geq 0} (- 1)^{k} (1 / 2)^{2 k + n} / (k! (k + n)!)$ .

[F2]

Convergence and geometric tails for integer-order Bessel series: these series converge absolutely, and their successive absolute term ratio is $1 / [4 (k + 1) (k + n + 1)]$ at $x = 1$ .

[F3]

Bessel differential equations and order recurrences from the series: $J_{n + 1} (1) = 2 n J_{n} (1) - J_{n - 1} (1)$ for $n \geq 1$ .

[F4]

Complementary solution for the integer Bessel recurrence: $D_{0} = - J_{1}$ , $D_{1} = J_{0}$ , and $D_{n + 1} = 2 n D_{n} - D_{n - 1}$ at $x = 1$ .

Proof

1.1F1F2algebra

Let $a_{k} = (1 / 2)^{2 k + n} / (k! (k + n)!)$ . By [F2], the positive terms $a_{k}$ decrease strictly to zero. Grouping the alternating series in consecutive pairs gives $J_{n} (1) = (a_{0} - a_{1}) + (a_{2} - a_{3}) + \dots > 0$ . Also $(k + n)! \geq n!$ , so $0 < J_{n} (1) \leq \sum_{k \geq 0} a_{k} \leq \frac{2^{- n}}{n!} \sum_{k \geq 0} \frac{(1 / 4)^{k}}{k!} = \frac{2^{- n} e^{1 / 4}}{n!} ⟶ 0.$ In particular, $J_{0} (1) \geq 1 - 1 / 4 = 3 / 4$ and $J_{1} (1) \geq 1 / 2 - 1 / 16 = 7 / 16$ .

2.1F1F3F4step 1.1induction

By [F4], $D_{0} (1) = - J_{1} (1) < 0$ and $D_{1} (1) = J_{0} (1) > 0$ ; hence $D_{2} = 2 D_{1} - D_{0} > D_{1} > 0$ . If $n \geq 2$ and $D_{n} \geq D_{n - 1} > 0$ , then $D_{n + 1} = 2 n D_{n} - D_{n - 1} \geq (2 n - 1) D_{n} \geq 3 D_{n}$ . Induction gives $D_{n} \geq 3^{n - 2} D_{2}$ for $n \geq 2$ , so $D_{n} \to \infty$ .

3.1F1F3F4step 1.1step 2.1algebra∎

The exact sequence $J_{n} - ε D_{n}$ obeys the same linear recurrence by [F3] and [F4], so it is the forward-recurrence solution from the displayed perturbed seeds. Since $J_{n} \to 0$ and $D_{n} \to \infty$ by steps 1.1 and 2.1, some finite $n$ satisfies $ε D_{n} > J_{n}$ . At that index ${\tilde{J}}_{n} < 0 < J_{n}$ . The seed differences have magnitudes $ε ∣ D_{0} ∣$ and $ε ∣ D_{1} ∣$ , which tend to zero with $ε$ .

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Sources

DLMF, §10.6(i), recurrence relations (standard reference, not scraped)
DLMF, §§3.6(iii) and 10.6(i) (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 4 (standard reference, not scraped)