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Least-term truncation of the erfcx asymptotic expansion

Statement

For $x \geq 1$ , let $N \geq 1$ be an integer nearest to $x^{2}$ , so $N - \frac{1}{2} \leq x^{2} \leq N + \frac{1}{2}$ . In the expansion from Signed finite remainder for the erfcx asymptotic expansion, the magnitudes of the terms are $a_{k} (x) = \frac{(1 / 2)_{k}}{x^{2 k + 1} \sqrt{π}} .$ The term $a_{N} (x)$ is minimal (ties are possible), and the signed remainder after $N$ terms is bounded by exactly $a_{N} (x)$ . For example, at $x = 3$ the least-term truncation is $N = 9$ . This describes optimal truncation by term size; it does not assert convergence of the asymptotic series.

Facts & Assumptions

Given: Real $x \geq 1$ and integer $N \geq 1$ with $N - \frac{1}{2} \leq x^{2} \leq N + \frac{1}{2}$ .

[F1]

Signed finite remainder for the erfcx asymptotic expansion: after $N$ terms, $erfcx (x) - S_{N} (x) = (- 1)^{N} R_{N} (x)$ with $0 \leq R_{N} (x) \leq (1 / 2)_{N} / (x^{2 N + 1} \sqrt{π})$ .

Proof

1.1givenalgebra

The positive term ratio is $\frac{a_{k + 1} (x)}{a_{k} (x)} = \frac{k + 1 / 2}{x^{2}} .$ For $0 \leq k < N$ , this ratio is at most $(N - 1 / 2) / x^{2} \leq 1$ , so the terms do not increase up to $a_{N}$ . For $k \geq N$ , the ratio is at least $(N + 1 / 2) / x^{2} \geq 1$ , so terms do not decrease after $a_{N}$ . Hence $a_{N}$ is a least term.

2.1F1step 1.1algebra∎

The upper bound in [F1] equals $(1 / 2)_{N} / (x^{2 N + 1} \sqrt{π}) = a_{N} (x)$ , and its sign is fixed by $(- 1)^{N}$ . Thus the first omitted least-sized term also bounds the signed asymptotic remainder. At $x = 3$ , $N = 9$ satisfies $8.5 \leq 9 \leq 9.5$ ; the same finite certificate applies.

Depends on

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Sources

DLMF, §7.12(i), equation 7.12.1 (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§3 and 5 (standard reference, not scraped)