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Exact finite terminal-contamination error identity

Statement

Fix $x > 0$ , integers $N \geq 1$ and $0 \leq n < N$ , and define the exact backward recurrence solution by $u_{N + 1} = 0$ , $u_{N} = 1$ , and $u_{j - 1} = 2 j u_{j} / x - u_{j + 1}$ for $j = N, N - 1, \dots, 1$ . If $D_{N + 1} (x) \neq 0$ , put $q = - J_{N + 1} (x) / D_{N + 1} (x)$ . For $s \in {0, 1}$ such that $J_{s} (x) + q D_{s} (x) \neq 0$ , let ${\hat{J}}_{n} = J_{s} (x) u_{n} / u_{s}$ . Then

${\hat{J}}_{n} - J_{n} = \frac{q (J_{s} D_{n} - J_{n} D_{s})}{J_{s} + q D_{s}} .$

This is a finite- $N$ identity. It makes no assertion that the error tends to zero as $N \to \infty$ .

Facts & Assumptions

Given: $x, N, n, s$ as above; the sequences $J, D$ ; and the terminal recurrence sequence $u$ .

[F1]

Positive discrete Wronskian for the Bessel recurrence pair: $J, D$ are linearly independent solutions of the same second-order recurrence.

[F2]

Bessel differential equations and order recurrences from the series: $J_{j + 1} = 2 j J_{j} / x - J_{j - 1}$ for $j \geq 1$ .

[F3]

Complementary solution for the integer Bessel recurrence: $D$ obeys the same recurrence.

Proof

1.1givenF1F2F3algebra

The sequence $J + q D$ obeys the recurrence by [F2] and [F3], and its value at $N + 1$ is $J_{N + 1} + q D_{N + 1} = 0$ . Its value at $N$ is nonzero: if it too were zero, backward recurrence would make it zero at every index, contradicting the linear independence in [F1] and its coefficient $1$ on $J$ . Therefore $(J + q D) / (J_{N} + q D_{N})$ has the terminal values $u_{N + 1} = 0, u_{N} = 1$ . Uniqueness of the finite backward recurrence gives $u_{j} = (J_{j} + q D_{j}) / (J_{N} + q D_{N})$ for $0 \leq j \leq N$ .

2.1givenstep 1.1algebra∎

Taking the ratio of the formulas for $j = n$ and $j = s$ yields $u_{n} / u_{s} = (J_{n} + q D_{n}) / (J_{s} + q D_{s})$ . Since the denominator is nonzero, ${\hat{J}}_{n} - J_{n} = J_{s} \frac{J_{n} + q D_{n}}{J_{s} + q D_{s}} - J_{n} = \frac{q (J_{s} D_{n} - J_{n} D_{s})}{J_{s} + q D_{s}} .$ Every equality uses this single fixed terminal index $N$ ; no limit in $N$ was taken.

Depends on

Used by

Dependency tree · two levels

7 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, §3.6(iii), Miller's Algorithm (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 4 (standard reference, not scraped)