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Signed finite remainder for the erfcx asymptotic expansion

Statement

For real $x > 0$ and integer $N \geq 1$ , put

$(1 / 2)_{0} = 1, (1 / 2)_{k} = \prod_{j = 0}^{k - 1} (j + 1 / 2) .$

$S_{N} (x) = \frac{1}{x \sqrt{π}} \sum_{k = 0}^{N - 1} (- 1)^{k} \frac{(1 / 2)_{k}}{x^{2 k}} .$

Then

$erfcx (x) - S_{N} (x) = (- 1)^{N} R_{N} (x), 0 \leq R_{N} (x) \leq \frac{1}{x \sqrt{π}} \frac{(2 N)!}{4^{N} N! x^{2 N}} .$

This is a finite signed remainder bound, not a convergence assertion for the asymptotic series.

Facts & Assumptions

Given: $x > 0$ , integer $N \geq 1$ , and the definitions of erfcx and $M_{k} (x)$ in Scaled complementary error function and its moments.

[F1]

Moment recurrence for the scaled complementary error function: $erfcx (x) = 2 M_{0} (x) / \sqrt{π}$ .

[F2]

Change of variable in an improper integral: a valid increasing $C^{1}$ substitution preserves convergence and value of an improper integral.

[F3]

Taylor's Lagrange remainder follows from repeated Rolle steps: the Taylor remainder after degree $N - 1$ of a $C^{N}$ function is $f^{(N)} (ξ) t^{N} / N!$ for some $ξ$ between the expansion point and $t$ .

[F4]

Gamma at the positive integers: $Γ (N + 1) = N!$ for integer $N \geq 0$ .

[F5]

Euler's Gamma function on the right half-plane and Euler's Gamma integral converges locally uniformly on the right half-plane: for $a > 0$ , $Γ (a) = \int_{0}^{\infty} s^{a - 1} e^{- s} d s < \infty$ .

[L9]

Continuity and derivatives of positive-base real powers, Sum, product, quotient, and chain rules from difference quotients: the displayed real functions are smooth on their stated domains, and their derivatives follow by the product, power and chain rules; iterate these rules for the finite higher derivatives used below.

Proof

1.1givenF1F2algebra

In the integral for $M_{0}$ , set $s = t^{2} + 2 x t$ . Its derivative $2 (t + x) > 0$ for $t \geq 0$ , so [F2] applies. Then $d s = 2 (t + x) d t$ , $t + x = \sqrt{x^{2} + s}$ , and $erfcx (x) = \frac{1}{\sqrt{π}} \int_{0}^{\infty} \frac{e^{- s}}{\sqrt{x^{2} + s}} d s = \frac{1}{x \sqrt{π}} \int_{0}^{\infty} e^{- s} {(1 + \frac{s}{x^{2}})}^{- 1 / 2} d s .$

1.2F3L9algebra

Direct differentiation gives $f^{(N)} (u) = (- 1)^{N} (1 / 2)_{N} (1 + u)^{- N - 1 / 2}$ . Apply [F3] to $f (t) = (1 + t)^{- 1 / 2}$ at $0$ to obtain $f (t) = \sum_{k = 0}^{N - 1} (- 1)^{k} \frac{(1 / 2)_{k}}{k!} t^{k} + (- 1)^{N} ρ_{N} (t), 0 \leq ρ_{N} (t) \leq \frac{(1 / 2)_{N}}{N!} t^{N} (t \geq 0),$ because $∣ f^{(N)} (u) ∣ = (1 / 2)_{N} (1 + u)^{- N - 1 / 2} \leq (1 / 2)_{N}$ for $u \geq 0$ .

2.1F4F5step 1.1step 1.2algebra∎

Substitute $t = s / x^{2}$ in step 1.2 and integrate against $e^{- s} / (x \sqrt{π})$ . By the Gamma integral and [F4], $\int_{0}^{\infty} e^{- s} s^{N} d s = Γ (N + 1) = N!$ by [F5] and [F4]. The signed remainder is nonnegative after removing $(- 1)^{N}$ and is at most $(1 / 2)_{N} / (x^{2 N + 1} \sqrt{π})$ . The identity $(1 / 2)_{N} = (2 N)! / (4^{N} N!)$ follows by separating even and odd factors in $(2 N)!$ . This gives the stated bound.

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)