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Error-function series geometric tail bound

Statement

For the kth term ek of the erf⁡ series in Fresnel, error, and Gauss hypergeometric series,

∣ek+1∣∣ek∣=∣z∣2k+12k+12k+3≤∣z∣2k+1.

The displayed quotient is asserted when ek≠0. At z=0, every ek=0, so the series and every tail are zero; the corresponding division-free inequality ∣ek+1∣≤∣z∣2∣ek∣/(k+1) holds there as well.

Thus, if ∣z∣2/(K+2)=q<1, the tail after terms 0,…,K is bounded by ∣eK+1∣/(1−q). The series converges locally uniformly. For every piecewise-C1 path γ:[0,1]→C with γ(0)=0 and γ(1)=z, it equals 2π∫γe−t2 dt, where the complex line integral is defined by The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral. In particular, the straight-segment integral has this value, and the integral is path-independent.

Facts & Assumptions

Given: The erf⁡ series in Fresnel, error, and Gauss hypergeometric series.

[F1]

Fresnel, error, and Gauss hypergeometric series: ek=(2/π)(−1)kz2k+1/(k!(2k+1)).

[F2]

The complex exponential by its power series and The complex exponential series converges absolutely for every complex argument: exp⁡w=∑k≥0wk/k! with absolute convergence for every complex w.

[F5]

The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral: for a continuous integrand on a rectifiable path, the complex line integral is defined by its four componentwise real Riemann--Stieltjes integrals.

[F6]

For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals: on a piecewise-C1 path, the line integral equals the sum of the parameter integrals of f(γ(s))γ′(s) over its smooth pieces.

Proof

1.1F1algebra

For z≠0, every term in [F1] is nonzero, so dividing successive absolute terms gives the stated ratio. At z=0 all terms vanish, and the stated division-free inequality and zero tail follow directly without division. For k≥K+1 it is at most q<1, so the tail is dominated by ∣eK+1∣∑j≥0qj=∣eK+1∣/(1−q). On every disk the ratio is eventually below 1, proving local uniform convergence.

2.1F1F2F3step 1.1algebra

The series in [F1] has infinite radius by step 1.1, so [F3] gives erf⁡′(z)=2π∑k=0∞(−1)kz2kk!=2πexp⁡(−z2), using [F2] and the definition of the exponential series. Thus erf is an entire primitive of the stated integrand and erf⁡(0)=0.

3.1F4F5F6step 2.1algebra∎

Let γ:[0,1]→C be any piecewise-C1 path with γ(0)=0 and γ(1)=z. On a smooth piece, put ηh=γ(s+h)−γ(s). Complex differentiability gives erf⁡(γ(s)+ηh)−erf⁡(γ(s))=erf⁡′(γ(s))ηh+ηhϵ(ηh) with ϵ(η)→0 as η→0; set ϵ(0)=0. Since ηh/h→γ′(s), division by h makes the remainder tend to zero and proves (erf⁡∘γ)′(s)=erf⁡′(γ(s))γ′(s). At a piece endpoint the same limit with one-sided increments gives the relative derivative, and continuity of γ′ extends this derivative continuously to the closed piece. By [F4], the real and imaginary parts of this derivative integrate to the endpoint differences. By [F6], these parameter integrals sum to the line integral defined in [F5]. Hence ∫γ2πe−t2 dt=erf⁡(z)−erf⁡(0)=erf⁡(z). This proves the segment identity and path independence without interchanging a series limit with a line integral.

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Cited to discharge well-definedness by Fresnel, error, and Gauss hypergeometric series.

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