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Subtracting erf from one amplifies absolute input errors in the tail

Statement

For $x > 0$ , suppose the inputs to the subtraction $1 - \erf (x)$ are approximations $1 + δ_{1}$ and $\erf (x) + δ_{2}$ , where $δ_{1}, δ_{2}$ are signed errors. Its output error is exactly $δ_{1} - δ_{2}$ , so its relative error is at most $\frac{∣ δ_{1} ∣ + ∣ δ_{2} ∣}{erfc (x)} .$ For fixed nonzero input error bounds, this conditioning factor grows without bound as $x \to \infty$ , because $erfc (x) > 0$ and tends to zero. The identity $erfc (x) = e^{- x^{2}} erfcx (x)$ and a certified direct $erfcx$ enclosure avoid this particular subtraction.

Facts & Assumptions

Given: A real $x > 0$ and approximations $\hat{1} = 1 + δ_{1}$ , $\overset{}{\erf} = \erf (x) + δ_{2}$ .

[F1]

Fresnel, error, and Gauss hypergeometric series and Scaled complementary error function and its moments: $erfc (x) = 1 - \erf (x)$ and $erfcx (x) = e^{x^{2}} erfc (x)$ .

[F2]

Error-function series geometric tail bound: for real $x$ , $\erf (x) = 2 π^{- 1 / 2} \int_{0}^{x} e^{- t^{2}} d t$ .

[F3]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ : $\int_{- \infty}^{\infty} e^{- t^{2}} d t = \sqrt{π}$ .

[F4]

Soundness of finite erfcx enclosures: a successful direct $erfcx$ evaluation supplies a containing absolute enclosure.

[F5]

Certified real elementary-function evaluations enclose their ranges: every finite successful real elementary-range output encloses its exact value or range on an admissible dyadic input interval; this applies to $e^{- x^{2}}$ for exact dyadic $x$ .

Proof

1.1F1F2F3algebra

By [F1], the exact output is $erfc (x)$ . Subtracting the perturbed inputs gives $\hat{1} - \overset{}{\erf} = erfc (x) + δ_{1} - δ_{2} .$ Since [F2] and [F3] give $erfc (x) = 2 π^{- 1 / 2} \int_{x}^{\infty} e^{- t^{2}} d t > 0$ , division by the exact value and the triangle inequality yield the stated relative-error bound.

2.1F1F3F4F5step 1.1∎

The same tail integral tends to zero as $x \to \infty$ , because the nonnegative Gaussian is integrable by [F3]. If $∣ δ_{1} ∣, ∣ δ_{2} ∣ \leq η$ with fixed $η > 0$ , the choice $δ_{1} = η$ , $δ_{2} = - η$ attains error $2 η$ , so the relative error is $2 η / erfc (x)$ and is unbounded. By [F1], multiplying containing intervals for $erfcx (x)$ and $e^{- x^{2}}$ with outward rounding encloses $erfc (x)$ and avoids the subtraction. For exact dyadic $x > 0$ , [F4] and [F5] supply those factors whenever their finite evaluations succeed; this asserts no finite-cap success for arbitrary real $x$ .

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Sources

DLMF, §7.2, definitions of erf and erfc (standard reference, not scraped)
DLMF, §§7.2 and 7.12(i) (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §2 (standard reference, not scraped)