Alphabeta Math
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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 30 results · all verified · 25 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Stable special function recurrences and certified evaluation

1 · Prerequisites

2 · Summary

Objective

This page defines finite, domain-specific enclosures for integer-order Bessel and modified Bessel functions, Airy functions, Fresnel integrals, the error function and a Gauss hypergeometric series, Gamma and a continued log-Gamma branch, and the scaled complementary error function. The evaluation contract separates exact mathematical error bounds from the finite interval operations that propagate them.

Each evaluator has explicit input domains, finite work and precision caps, absolute-radius success tests, and inconclusive, domain, or range outcomes. Relative certification is reported only when the returned enclosure excludes zero. Power-series branches use finite geometric tails; the erfcx expansion uses a signed finite remainder; the log-Gamma product uses an explicit tail; and the large-argument branch uses a locally proved sectorial constant.

For integer Bessel order, the page proves the series recurrences and integral representations, a composite midpoint certificate, and a finite Miller terminal-contamination identity from two exact recurrence solutions and their positive discrete Wronskian. The Miller certificate is evaluated only when all normalization and denominator intervals exclude zero. It makes no claim that increasing the terminal index forces convergence. The modified Bessel output is the scaled value e−xIn(x) on x≥0.

The spherical branch evaluates jℓ, outgoing hℓ(1), and their first derivatives on positive-real point interfaces or dyadic input intervals. It supplies both unscaled values and explicit scaled values; each scaled derivative is the scaled derivative of the original function. Exact finite Laurent expressions and a rational Taylor remainder certify the interval propagation. Absolute success, optional zero-excluding relative flags, and finite-cap, domain, and range outcomes follow the common result contract. The represented point need not be dyadic: a caller may propagate certified parameter intervals to arguments such as kR or k2+V0R, provided each refinement encloses the exact positive argument and terminates. The exact scattering statements retain their full real-parameter domains; numerical certification separately requires those input interfaces.

The Airy series uses Bi⁡(0)=1/(31/6Γ(2/3)); all initial values are read against DLMF §9.2(ii). The Gauss-series branch is restricted to ∣z∣<1 with c not a nonpositive integer. The log-Gamma branch uses the principal-log slit plane C∖(−∞,0].

The required quadrature pages provide surrounding library context. The Bessel midpoint certificates are proved locally with their exact second-derivative bounds. Outward interval, rectangle, and elementary range operations are supplied by the listed interval-arithmetic page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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Airy initial-value series

Definition

For y(z)=∑k≥0ckzk, set c2=0 and ck+3=ck/((k+3)(k+2)) for k≥0. Define Ai⁡ by this series with

c0=132/3Γ(2/3),c1=−131/3Γ(1/3),

and define Bi⁡ by the same recurrence with

c0=131/6Γ(2/3),c1=31/6Γ(1/3).

Here Γ is Euler's integral on the right half-plane from Euler's Gamma function on the right half-plane. The standard Airy initial values are those in DLMF §9.2(ii); in particular, the power of 3 in Bi⁡(0) is in the denominator.

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Certified special-function enclosure result

Definition

A certified enclosure for a real value is a nonempty closed interval B=[a,b] with finite dyadic endpoints; for a complex value it is a nonempty closed rectangle B=[a1,b1]+i[a2,b2] with finite dyadic endpoints. Its radius is half its length in the real case and half its Euclidean diagonal in the complex case; d(B,0) is the Euclidean distance from B to 0. An evaluator item declares whether its input is an exact dyadic point or a dyadic interval/rectangle enclosure; a successful range result contains the exact function image of every admissible point in an input enclosure. Constants not exactly dyadic must be supplied with certified enclosing intervals or rectangles.

The interval backend uses the endpoint operations in Outward-rounded interval, rectangle and ball operations; the soundness of finite arithmetic expression DAGs follows from Outward-rounded operations are inclusion-isotone and enclose DAG evaluations. Real elementary function ranges use Certified real elementary-function evaluations enclose their ranges, and complex exponential and principal-logarithm ranges use Certified complex exponential and principal logarithm enclose their ranges. An operation that returns a non-finite status supplies no finite enclosure claim.

A finite-budget evaluator receives a branch input enclosure wholly within the stated domain, absolute tolerance εa>0, optional relative tolerance εr>0, finite integer work and precision caps, and the fixed format and truncation parameters required by the backend. Its result has one of four statuses:

  • success: the returned enclosure contains the exact requested value and its radius is at most εa. If 0∉B and relative tolerance was requested, the evaluator may also report relative_certified: true only when rad⁡(B)/d(B,0)≤εr.
  • inconclusive: the procedure exhausted a finite cap before its requested enclosure test passed.
  • domain: the input is outside the stated branch domain, violates a stated parameter condition, or lies on a branch cut excluded by that branch.
  • range: a certified finite enclosure cannot be represented in the requested output format; the procedure returns no unsound finite interval.

If a success enclosure contains 0, it still supplies the stated absolute bound, but it carries no relative-accuracy claim. The work and precision caps do not imply that success occurs for every input.

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Fresnel, error, and Gauss hypergeometric series

Definition

For z∈C, define the Fresnel functions and error function by

C(z)=∑k=0∞(−1)k(π/2)2kz4k+1(2k)!(4k+1),S(z)=∑k=0∞(−1)k(π/2)2k+1z4k+3(2k+1)!(4k+3),erf⁡(z)=2π∑k=0∞(−1)kz2k+1k!(2k+1),erfc⁡(z)=1−erf⁡(z).

The following local lemmas prove convergence and show that C,S,erf⁡ equal the standard entire integral functions with their usual normalizations.

For a,b,c,z∈C with c∉{0,−1,−2,…} and ∣z∣<1, define

(a)0=1,(a)k=∏j=0k−1(a+j),2F1(a,b;c;z)=∑k=0∞(a)k(b)k(c)kk!zk.

Only this Gauss-series branch of 2F1 is defined on this page; no value outside ∣z∣<1 or across a continuation cut is implicit.

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Integer-order Bessel series

Definition

For an integer n≥0 and z∈C, define

Jn(z)=∑k=0∞(−1)k(z/2)2k+nk!(k+n)!,In(z)=∑k=0∞(z/2)2k+nk!(k+n)!.

These are the integer-order Bessel and modified Bessel functions used on this page. For real x≥0, the scaled modified Bessel value means Insc(x)=e−xIn(x). This notation does not assert a finite evaluation method or a truncation bound; those are proved separately.

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Log-Gamma branch on the right half-plane

Definition

On Re⁡z>0, define

ℓ+(z)=−Log⁡z−γEz+∑n=1∞(zn−Log⁡(1+zn)).

Here Log⁡ is the principal logarithm from Complex logarithms, the principal logarithm, and principal and multivalued complex powers, and γE denotes Euler's constant γ from The Euler-Mascheroni constant. Both z and 1+z/n lie in the principal-logarithm domain. The local uniform convergence of this series and the identity eℓ+(z)=Γ(z) are established by Weierstrass log-Gamma series converges to a Gamma branch ↗.

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Finite harmonic enclosure for Euler's constant

Statement

For each integer M≥1, put HM=∑k=1M1/k. Euler's constant γ satisfies 0≤HM−log⁡M−γ≤1M. Thus [HM−log⁡M−1/M, HM−log⁡M] is a finite enclosure for γ.

Facts & Assumptions

Given: An integer M≥1.

[F1]

The Euler–Mascheroni constant and the harmonic asymptotic: γn=Hn−log⁡n decreases to γ.

[F2]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: log⁡(n+1)−log⁡n=∫nn+1dt/t for n≥1.

Proof

1.1F1F2algebra

For each n≥M, [F1] and [F2] give γn−γn+1=∫nn+1dtt−1n+1=∫nn+1(1t−1n+1)dt. The integrand is nonnegative and at most 1/n−1/(n+1) on this interval. Therefore 0≤γn−γn+1≤1n−1n+1.

2.1F1step 1.1algebra∎

Summing step 1.1 for n=M,…,K telescopes to 0≤γM−γK+1≤1M−1K+1. Let K→∞ and use γK+1→γ from [F1]. Since γM=HM−log⁡M, the claimed two-sided bound follows.

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Gamma functional equation away from the poles

Statement

For every z∈C∖{0,−1,−2,…}, Γ(z+1)=zΓ(z), where Γ is the meromorphic continuation of Euler's Gamma function.

Facts & Assumptions

Given: Euler's Gamma integral on Re⁡z>0 and z∈C∖{0,−1,−2,…}.

[F2]

Euler's Gamma function is holomorphic on the right half-plane: this integral defines a holomorphic function on the right half-plane.

[F3]

The Gamma functional equation: if Re⁡w>0, then Γ(w+1)=wΓ(w).

[F4]

Dedekind completeness implies the Archimedean property: for every real r, some integer m≥1 satisfies r>−m.

Proof

1.1givenF1F2algebraconstruct

For each integer m≥0, let Vm={w∈C:Re⁡w>−m−1} and define Gm(w)=Γ(w+m+1)/(w(w+1)⋯(w+m)) as a meromorphic function on Vm. Its numerator is holomorphic on this half-plane by [F1] and [F2], and its denominator is a polynomial.

2.1givenF3F4step 1.1algebra

For every w∈Vm, Re⁡(w+m+1)>0. Applying [F3] at w+m+1 and cancelling the common meromorphic factor w+m+1 gives Gm+1(w)=Gm(w) on the overlap Vm⊂Vm+1. By [F4], the union of the Vm is C, so these compatible meromorphic formulas define a meromorphic continuation of Euler's Gamma function on the whole plane.

3.1givenF4step 2.1algebrachoose∎

Choose m≥1 large enough that z∈Vm; then z+1∈Vm−1. Because z is not a nonpositive integer, the denominator factors below are nonzero. The glued formulas give Γ(z)=Γ(z+m+1)/(z(z+1)⋯(z+m)) and Γ(z+1)=Γ(z+m+1)/((z+1)⋯(z+m)). Multiplying the first identity by z yields the second.

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Spherical Bessel and outgoing Hankel function conventions

Definition

For real z>0, let Dz:=z−1ddz and define, for integers ℓ≥0, jℓ(z):=(−z)ℓDzℓ(sin⁡zz),hℓ(1)(z):=(−z)ℓDzℓ(−ieizz). The definition fixes the outgoing eiz convention. The functions solve the radial Helmholtz equation and have the recurrence, origin, Wronskian, and infinity behavior by Spherical Bessel and outgoing Hankel recurrences solve the radial Helmholtz equation ↗.

These symbolic definitions assert no stable or certified numerical evaluation algorithm for large ℓ or z.

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Complementary solution for the integer Bessel recurrence

Definition

Fix real x>0. Let D0(x)=−J1(x) and D1(x)=J0(x), and for every integer j≥1 define

Dj+1(x)=2jxDj(x)−Dj−1(x).

Thus D is the second sequence obtained from the same order recurrence as Jj(x). It is a local auxiliary solution, not the Bessel function of the second kind.

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Scaled complementary error function and its moments

Definition

For real x>0, define erfcx⁡(x)=ex2erfc⁡(x). For each integer k≥0, define the finite positive moment

Mk(x)=∫0∞tke−t2−2xt dt.

The moments are auxiliary quantities for a finite-depth interval enclosure of erfcx⁡. They are not the modified Bessel functions Ik.

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Log-Gamma branch on the slit plane

Definition

Let Ω=C∖(−∞,0]. For every z∈Ω, an admissible integer m≥0 with Re⁡(z+m)>0 exists by the Archimedean property Dedekind completeness implies the Archimedean property. Choose any such m and set

ℓ(z)=ℓ+(z+m)−∑j=0m−1Log⁡(z+j),

where the sum is empty if m=0 and ℓ+ and Log⁡ are defined in Log-Gamma branch on the right half-plane. The next lemma proves that this value is independent of the admissible m and satisfies eℓ(z)=Γ(z). This is a continued branch of log-Gamma, not a separate principal Log⁡(Γ(z)).

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Airy coefficient recurrence and geometric tail bounds

Statement

The coefficient recurrence in Airy initial-value series makes both series solve y′′=zy. In a residue class k≡r(mod3), successive nonzero terms have absolute ratio

∣z∣3(k+3)(k+2).

If the ratio of each term after a chosen first omitted term is bounded by q<1, that class's tail is at most the first omitted term's absolute value divided by 1−q. The series and their first two derivative series converge locally uniformly on C.

Facts & Assumptions

Given: The coefficient recurrence, c2=0, and initial coefficients from Airy initial-value series.

[F1]

Airy initial-value series: ck+3=ck/((k+3)(k+2)) and the four initial values specify Ai⁡ and Bi⁡.

[F2]

Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term: a complex power series may be differentiated term by term strictly inside its radius of convergence.

Proof

1.1F1algebra

Multiplying the coefficient relation in [F1] by zk+3 shows that the coefficient of zk+1 in y′′ is ck, which is the coefficient of zk+1 in zy. At degree 0, the coefficients agree because c2=0; at degree 1, they agree by the recurrence at k=0. Thus the formal series satisfy y′′=zy.

1.2F1algebra

For Tk=ckzk with ck≠0, ∣Tk+3∣/∣Tk∣=∣z∣3/((k+3)(k+2)). This tends to zero and decreases in k; once all subsequent ratios are at most q<1, summing the geometric majorant gives ∑j≥0∣Tk0+3j∣≤∣Tk0∣/(1−q).

2.1F2step 1.1step 1.2algebra∎

On every disk ∣z∣≤R, the ratios of each residue-class term and of its first two formal derivative series tend to zero uniformly as k→∞. Applying the geometric bound from step 1.2 to each class gives locally uniform convergence of the series and differentiated series. The series have infinite radius by these bounds, so [F2] permits termwise differentiation; step 1.1 then proves the Airy equation for the sums.

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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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Fresnel series geometric tail bounds

Statement

For the terms ck of C and sk of S in Fresnel, error, and Gauss hypergeometric series,

∣ck+1∣∣ck∣≤π2∣z∣4/4(2k+1)(2k+2),∣sk+1∣∣sk∣≤π2∣z∣4/4(2k+2)(2k+3).

Each displayed quotient is asserted only when its denominator is nonzero. At z=0, every ck=sk=0, so both series and every tail are zero. The corresponding division-free inequalities obtained by multiplying by ∣ck∣ or ∣sk∣ hold for every z.

Once the applicable bound is at most q<1 for all terms after a first omitted term, the absolute tail is bounded by that term divided by 1−q. Both series converge locally uniformly on C. For every piecewise-C1 path γ:[0,1]→C with γ(0)=0 and γ(1)=z, they equal C(z)=∫γcos⁡(πt2/2) dt,S(z)=∫γsin⁡(πt2/2) 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, these integrals are path-independent and agree with the straight-segment integrals.

Facts & Assumptions

Given: The series definitions of C,S in Fresnel, error, and Gauss hypergeometric series.

[F1]

Fresnel, error, and Gauss hypergeometric series: ck=(−1)k(π/2)2kz4k+1/((2k)!(4k+1)) and sk=(−1)k(π/2)2k+1z4k+3/((2k+1)!(4k+3)).

[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

If z=0, all terms in [F1] vanish, and their division-free bounds and zero tails hold directly. For z≠0, every term is nonzero, and taking ratios in [F1] gives ∣ck+1∣∣ck∣=π2∣z∣444k+1(2k+1)(2k+2)(4k+5),∣sk+1∣∣sk∣=π2∣z∣444k+3(2k+2)(2k+3)(4k+7). The omitted factors are each less than 1, yielding the displayed bounds.

2.1step 1.1algebra

The denominators in the bounds grow quadratically in k, so the ratios tend to zero; summing the resulting geometric majorant gives the stated tail bound and uniform convergence on every disk.

3.1F1F2F3step 2.1algebra

The infinite-radius power series in [F1] may be differentiated termwise by [F3]. The resulting derivatives are C′(z)=∑k=0∞(−1)k(π/2)2kz4k(2k)!=cos⁡(πz2/2),S′(z)=∑k=0∞(−1)k(π/2)2k+1z4k+2(2k+1)!=sin⁡(πz2/2), where the last equalities use [F2].

4.1F4F5F6step 3.1algebra∎

Both C and S are entire primitives of their displayed integrands. For any piecewise-C1 path γ:[0,1]→C from 0 to z, put ηh=γ(s+h)−γ(s) on each smooth piece. Complex differentiability gives C(γ(s)+ηh)−C(γ(s))=C′(γ(s))ηh+ηhϵC(ηh) with ϵC(η)→0 as η→0; set ϵC(0)=0. Since ηh/h→γ′(s), division by h makes the remainder tend to zero and proves (C∘γ)′(s)=C′(γ(s))γ′(s). The same argument gives (S∘γ)′(s)=S′(γ(s))γ′(s). At each piece endpoint the same limit with one-sided increments gives the relative derivatives; they extend continuously to the closed piece. By [F4], the real and imaginary parts integrate to endpoint differences. By [F6], the parameter integrals sum to the complex line integrals defined in [F5]. Since C(0)=S(0)=0, the line integrals equal C(z) and S(z). Hence they are path-independent and agree with the straight-segment integrals.

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Gauss hypergeometric series geometric tail bound

Statement

Let a,b,c,z∈C, ∣z∣<1, and c∉{0,−1,−2,…}. If z=0, then h0=1 and hk=0 for k≥1, so the series equals 1 and every tail after h0 is zero. Suppose instead that z≠0 and neither a nor b is a nonpositive integer. Then every term hk=(a)k(b)kzk/((c)kk!) is nonzero, and for any integer N>∣c∣,

∣hk+1∣∣hk∣=∣z∣∣a+k∣∣b+k∣∣c+k∣(k+1)≤qN:=∣z∣(1+∣a∣/N)(1+∣b∣/N)1−∣c∣/N(k≥N).

For sufficiently large N, qN<1; then the tail starting at hN is at most ∣hN∣/(1−qN). If a or b is a nonpositive integer, the series terminates and its tail after the last nonzero term is zero.

Facts & Assumptions

Given: The Gauss series in Fresnel, error, and Gauss hypergeometric series and its parameter restrictions.

[F1]

Fresnel, error, and Gauss hypergeometric series: hk=(a)k(b)kzk/((c)kk!) and (a)k+1=(a+k)(a)k (and likewise for b,c).

Proof

1.1givenF1algebra

In the nonterminating case with z≠0, every term is nonzero. The Pochhammer product gives hk+1/hk=z(a+k)(b+k)/((c+k)(k+1)). For k≥N>∣c∣, ∣a+k∣≤k(1+∣a∣/N), ∣b+k∣≤k(1+∣b∣/N), ∣c+k∣≥k(1−∣c∣/N), and k+1≥k. These inequalities yield the stated qN.

2.1givenstep 1.1algebra∎

For fixed nonterminating parameters, qN→∣z∣<1, so a finite N has qN<1; summing the geometric majorant proves the stated tail bound. On a compact subset with ∣z∣≤r<1, replacing ∣z∣ by r gives a uniform ratio bound below 1 for all sufficiently large N, proving local uniform convergence. If z=0, the tail is exactly zero by the Statement. If a or b is a nonpositive integer, its Pochhammer product becomes zero after finitely many terms, so the series terminates and its tail after the last nonzero term is zero.

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Convergence and geometric tails for integer-order Bessel series

Statement

For n≥0, z≠0, and either series in Integer-order Bessel series, the absolute term ratio is

rk=∣z∣24(k+1)(k+n+1).

At z=0, the order-zero series have only the constant term a0=1 and positive orders have only zero terms; in every case the tail after terms k=0,…,K is zero. The series for Jn and In converge absolutely and locally uniformly in z∈C. If K≥0 and q=∣z∣2/[4(K+2)(K+n+2)]<1, then the tail after terms k=0,…,K has absolute value at most ∣aK+1∣/(1−q), where ak is the corresponding series term. The first two formal termwise derivative series also converge locally uniformly.

Facts & Assumptions

Given: An integer n≥0, z∈C, and the two power series in Integer-order Bessel series.

[F1]

Integer-order Bessel series: the terms are akJ=(−1)k(z/2)2k+n/(k!(k+n)!) and akI=(z/2)2k+n/(k!(k+n)!).

Proof

1.1givenF1algebra

For z≠0, every term is nonzero, and cancellation of the factorials gives ∣ak+1∣/∣ak∣=∣z∣2/[4(k+1)(k+n+1)]. This ratio decreases to 0. At z=0, the series have only the possible order-zero constant term, so every tail after k=0 is zero without using a quotient.

2.1step 1.1algebra

For z≠0 and k≥K+1 the ratios are at most q, so ∑k=K+1∞∣ak∣≤∣aK+1∣∑j=0∞qj=∣aK+1∣/(1−q). At z=0 the same tail bound holds because the tail is zero. On each disk ∣z∣≤R, put qR=R2/[4(K+2)(K+n+2)]. For sufficiently large K, qR<1 and the nonzero-argument tail is bounded uniformly by sup⁡∣z∣≤R∣aK+1(z)∣/(1−qR); at z=0 it is zero. Thus the series converge uniformly on every disk.

3.1step 1.1algebra∎

For derivative order d∈{1,2}, put m=2k+n. When m≥d, the absolute value of the formal derivative term on ∣z∣≤R is at most bk=m(m−1)⋯(m−d+1)Rm−d/(2mk!(k+n)!). For R>0 the ratio bk+1/bk equals R2/[4(k+1)(k+n+1)] times a factor tending to 1, so it tends to 0. The ratio test makes ∑bk finite; the Weierstrass test then gives uniform convergence of each formal derivative series on the disk. The finitely many terms with m<d have zero derivative, and a disk of radius R=0 is immediate.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Weierstrass log-Gamma series converges to a Gamma branch

Statement

For every z with Re⁡z>0, the series defining ℓ+(z) in Log-Gamma branch on the right half-plane converges locally uniformly, defines a holomorphic function, and satisfies eℓ+(z)=Γ(z) for Euler's Gamma function on that half-plane.

Facts & Assumptions

Given: Re⁡z>0 and the definitions in Log-Gamma branch on the right half-plane.

[F1]

Euler's Beta function on the right half-planes: B(p,q)=∫01tp−1(1−t)q−1 dt(Re⁡p,Re⁡q>0).

[F2]

The principal logarithm is the normalised holomorphic branch on the slit plane: Log⁡′(w)=1/w and exp⁡(Log⁡w)=w on the slit plane.

[F3]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a) and A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuously differentiable real function integrates its derivative to its endpoint difference, and a continuous derivative on a compact interval is integrable; apply componentwise to complex-valued functions of a real parameter.

[F7]

exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0: for u>0, ∣uz−1∣=uRe⁡z−1 under the real-log power convention.

[F8]

Euler's Gamma function on the right half-plane and Euler's Gamma integral converges locally uniformly on the right half-plane: for Re⁡z>0, Γ(z)=∫0∞uz−1e−u du, and the corresponding positive real majorant is integrable.

[F9]

The Euler-Mascheroni constant and The Euler–Mascheroni constant and the harmonic asymptotic: if HN=∑n=1N1/n, then HN−log⁡N→γ.

[F10]

The exponential function is strictly increasing, The exponential function is smooth and (exp⁡)′=exp⁡, and Mean value theorem on a closed real interval: the real exponential is increasing, has derivative ex, and its mean-value estimate applies on compact intervals.

[F12]

Change of variable in an improper integral: monotone affine changes of variable preserve the corresponding proper and improper integrals; apply componentwise to the complex Beta integrand.

Proof

1.1givenF2F3F6F11algebra

For w=z/n with Re⁡z>0, the path 1+tw lies in the principal-logarithm domain. Its derivative integrand is continuous on [0,1] and is integrable by [F3]. By [F2], [F3], and the real-parameter chain rule, Log⁡(1+w)=∫01w1+tw dt,w−Log⁡(1+w)=∫01tw21+tw dt. Since ∣1+tw∣2=1+2tRe⁡w+t2∣w∣2≥1, the summand has absolute value at most ∣z∣2/(2n2). Summing this bound is uniform on every compact subset of the right half-plane. Each summand is holomorphic there, so the series converges locally uniformly and defines a holomorphic function by [F11].

1.2F1F3F4F6algebrainduction

For N≥1, the Beta integrals satisfy B(z,1)=1z,B(z,N+1)=Nz+NB(z,N). For the base formula, apply [F3] componentwise on [ε,1] to the primitive tz/z and let ε↓0; the lower boundary tends to 0 since ∣εz∣=εRe⁡z. For N≥1, differentiate g(t)=tz(1−t)N on [ε,1]. By [F6], g′(t)=ztz−1(1−t)N−Ntz(1−t)N−1. The derivative is continuous and integrable on this compact interval by [F3]. Applying the componentwise fundamental theorem and then letting ε↓0 gives zB(z,N+1)=NB(z+1,N); both boundary terms vanish. The integrand identity tz−1(1−t)N−1=tz−1(1−t)N+tz(1−t)N−1 and [F4] give B(z,N)=B(z,N+1)+B(z+1,N). Combining these two equalities yields (z+N)B(z,N+1)=NB(z,N), proving the recurrence. Iterating it from B(z,1)=1/z gives B(z,N+1)=N!z(z+1)⋯(z+N). All denominators are nonzero because Re⁡(z+j)>0 for j≥0.

1.3F5F9F10F12algebra

Fix z with a:=Re⁡z>0, and define KN(u)=(1−u/N)N for 0≤u≤N and KN(u)=0 for u>N. For 0≤u<N, [F5] and [F12] give log⁡(1−u/N)=−∫0u/Nds1−s≤−uN, so 0≤KN(u)≤e−u by exponential monotonicity and the addition law; the same bound holds at u=N. On a fixed interval [0,R], for N>2R, Nlog⁡(1−u/N)=−u−δN(u),0≤δN(u)=N∫0u/Ns1−s ds≤R2N. The mean value theorem in [F10] and et≤1 for t≤0 give ∣KN(u)−e−u∣≤R2/N uniformly on [0,R].

2.1F1F5F12step 1.2algebra

Define AN(z):=N! Nzz(z+1)⋯(z+N),Nz:=exp⁡(zlog⁡N). By step 1.2 and the substitution u=Nt in [F12], AN(z)=NzB(z,N+1)=∫0Nuz−1(1−uN)Ndu. The real logarithm laws in [F5] preserve the stated power convention under this substitution.

2.2F4F7F8step 1.3algebra

By [F7], ∣uz−1∣=ua−1. For each real or imaginary component, the integral difference on [0,R] is bounded, using [F4], by R2N∫0Rua−1 du=Ra+2aN. On [R,∞), KN(u)≤e−u bounds the difference in each component by 2∫R∞ua−1e−u du, which tends to 0 as R→∞ by [F8]. First choosing R large and then N large proves AN(z)→Γ(z).

3.1F2F6F8F9step 1.1step 1.2step 2.1step 2.2algebra∎

Let ℓN(z) be the defining partial sum through N and put PN(z):=zexp⁡(γEz)∏n=1N(1+zn)e−z/n. The principal-log identity exp⁡(Log⁡w)=w, the exponential addition law in [F6], and the formula from step 1.2 give exp⁡(ℓN(z))=PN(z)−1,PN(z)AN(z)=exp⁡ ⁣(z(γE+log⁡N−HN)). By [F9] the second expression tends to 1. By step 1.1, ℓN(z)→ℓ+(z); continuity of the complex exponential gives PN(z)→exp⁡(−ℓ+(z)). Step 2.2 gives AN(z)→Γ(z), so exp⁡(−ℓ+(z))Γ(z)=1. Multiplying by exp⁡(ℓ+(z)) and using the addition law proves eℓ+(z)=Γ(z).

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Spherical Bessel and outgoing Hankel recurrences solve the radial Helmholtz equation

Statement

For the functions of Spherical Bessel and outgoing Hankel function conventions, define Lℓy:=y′′+2zy′+(1−ℓ(ℓ+1)z2)y(z>0). Then Lℓjℓ=Lℓhℓ(1)=0, and for y=j,h(1), yℓ+1(z)=ℓzyℓ(z)−yℓ′(z),yℓ′(z)=yℓ−1(z)−ℓ+1zyℓ(z)(ℓ≥1). The regular-origin and outgoing-infinity normalizations are jℓ(z)=zℓ(2ℓ+1)!!+O(zℓ+2),hℓ(1)(z)=−i(2ℓ−1)!!z−ℓ−1(1+O(z))(z↓0), jℓ(z)=sin⁡(z−ℓπ/2)z+O(z−2),hℓ(1)(z)=(−i)ℓ+1eizz+O(z−2)(z→∞), where (−1)!!:=1. Their Wronskian is jℓ(z)(hℓ(1))′(z)−jℓ′(z)hℓ(1)(z)=iz2.

Facts & Assumptions

Given: The exact finite-derivative definitions of jℓ and hℓ(1) from Spherical Bessel and outgoing Hankel function conventions, their power-series meanings, and ordinary differentiation rules on z>0.

[F1]

Spherical Bessel and outgoing Hankel function conventions: for z>0 and integer ℓ≥0, jℓ=(−z)ℓ(z−1d/dz)ℓ(sin⁡z/z) and hℓ(1)=(−z)ℓ(z−1d/dz)ℓ(−ieiz/z).

[F2]

Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0 and The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c): real-valued differentiable functions satisfy the product, quotient, and chain rules at limit points of their domains. For complex-valued functions of the real variable, define (u+iv)′=u′+iv′; expansion into real and imaginary components then gives the same product rule. Every finite derivative used below exists on (0,∞): sine and cosine derivatives alternate by [F3], the derivative of eiz is ieiz by [F3], and derivatives of z−1 follow from the real quotient and chain rules.

[F3]

The derivatives of sine and cosine are cosine and minus sine gives (sin⁡z)′=cos⁡z and (cos⁡z)′=−sin⁡z; Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ gives eiz=cos⁡z+isin⁡z. Componentwise differentiation therefore gives (eiz)′=ieiz for real z.

[F5]

The sine and cosine power series converge absolutely for every real argument: the defining factorial series for sine and cosine converge absolutely at every real argument.

[F6]

A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant: a real function continuous on an interval and differentiable with derivative zero at every interior point is constant on that interval.

[F7]

In the displayed normalizations, (2m+1)!!=∏r=0m(2r+1) and (2m−1)!!=∏r=1m(2r−1) for m≥0, with the second product empty and equal to 1 at m=0. For z↓0, q(z)=O(za) means ∣q(z)∣≤Cza for some C and all sufficiently small z>0; for z→∞, q(z)=O(z−a) means ∣q(z)∣≤Cz−a for all sufficiently large z. A Wronskian here means W[y,v]=yv′−y′v.

[F8]

Trigonometric identities, Machin's formula and reduced Taylor tails: the power-series sine and cosine satisfy the addition identities, sin⁡2z+cos⁡2z=1, and sin⁡(π/2)=1, cos⁡(π/2)=0; it defines π=4arctan⁡(1). Together with [F3], this gives ∣sin⁡z∣,∣cos⁡z∣,∣eiz∣≤1 for real z.

[F9]

The factorial n! and the falling factorial nk‾, defined by recursion in N defines 0!=1, the factorial recursion n!=n(n−1)!, and proves every factorial is nonzero; Canonical naturals are positive and strictly increasing gives positivity of the natural factors in their real embedding. Regrouping the even and odd factors in (2ℓ+1)! gives (2ℓ+1)!=2ℓℓ!(2ℓ+1)!!.

Proof

Given: The functions and conventions in the Statement.

1.1F1F2F3algebra

Write f0(z)=sin⁡z/z and g0(z)=−ieiz/z. By [F2]–[F3], f0′=cos⁡z/z−sin⁡z/z2 and f0′′=−sin⁡z/z−2cos⁡z/z2+2sin⁡z/z3, so L0f0=0. Also g0′=eiz(1/z+i/z2) and g0′′=eiz(i/z−2/z2−2i/z3), so L0g0=0. For any function with the required finite derivatives on (0,∞), put yℓ=(−z)ℓDℓf with D=z−1d/dz. Since (Dℓf)′=zDℓ+1f, differentiating the product gives yℓ′=ℓyℓ/z−yℓ+1, hence yℓ+1=ℓyℓ/z−yℓ′. This recurrence applies to both functions in [F1].

1.2F1F4F5F9algebra

The sine series gives sin⁡z/z=∑m≥0(−1)mz2m/(2m+1)!. After r applications of D, the formal series is ∑m≥r(−1)m2rm!z2m−2r/((m−r)!(2m+1)!). For z≠0 its successive nonzero absolute-term ratio is ∣z∣2(m+1)/((m+1−r)(2m+2)(2m+3))→0; at z=0 only its constant term remains. Thus each such series has infinite radius by [F4]. Termwise differentiation and division by z give the next series: the m=r term is constant and differentiates to zero, so no negative power appears. The operation at r=ℓ kills m<ℓ; its first remaining coefficient is 2ℓℓ!/(2ℓ+1)!=1/(2ℓ+1)!! by [F9]. The later powers are zℓ+2,zℓ+4,…; their coefficient series is bounded on ∣z∣≤1 by compact uniform convergence in [F4], so their sum is O(zℓ+2). This proves the stated origin expansion for jℓ.

2.1F2step 1.1algebra

Put λ=ℓ(ℓ+1) and μ=(ℓ+1)(ℓ+2). Direct differentiation gives Lℓ+1(ℓy/z−y′)=−y′′′+(ℓ−2)y′′/z+(−1+μ/z2)y′+(ℓ/z−ℓμ/z3)y=ℓLℓy/z−(Lℓy)′. Thus if Lℓy=0, the recurrence in step 1.1 gives Lℓ+1yℓ+1=0. Starting from step 1.1 proves both radial equations by induction. For ℓ≥1, differentiate yℓ=(ℓ−1)yℓ−1/z−yℓ−1′ and substitute Lℓ−1yℓ−1=0; this yields yℓ′=(ℓ+1)yℓ−1′/z+(1−(ℓ2−1)/z2)yℓ−1. Substituting yℓ−1′=(ℓ−1)yℓ−1/z−yℓ from step 1.1 and collecting terms gives yℓ′=yℓ−1−(ℓ+1)yℓ/z.

2.2F2F3F7F8step 1.1algebrainduction

Write hℓ(1)=Cℓeiz, starting with C0=−i/z. Applying the recurrence of step 1.1 and the product rule in [F2]–[F3] gives Cℓ+1=ℓCℓ/z−Cℓ′−iCℓ. Induction shows Cℓ is a finite Laurent polynomial with exponents −1,…,−ℓ−1. If its most singular term is aℓz−ℓ−1, the coefficient of z−ℓ−2 in Cℓ+1 is (2ℓ+1)aℓ; the term −iCℓ is less singular. Since a0=−i, this gives aℓ=−i(2ℓ−1)!!. Also eiz=1+O(z) as z↓0 by [F3] and differentiability at 0, so the outgoing origin expansion follows. For infinity, induct on ℓ using step 1.1. The base terms are j0=sin⁡z/z and h0(1)=−ieiz/z. If jℓ=sin⁡θ/z+Rℓ, θ=z−ℓπ/2, with Rℓ a finite sum of z−ksin⁡z and z−kcos⁡z terms for k≥2, then Rℓ′=O(z−2) by [F3],[F7],[F8], and ℓjℓ/z=O(z−2); hence jℓ+1=−cos⁡θ/z+O(z−2)=sin⁡(z−(ℓ+1)π/2)/z+O(z−2) by the addition identities [F8]. For hℓ(1), the coefficient recurrence gives a finite form eiz(bℓ/z+∑k=2ℓ+1bℓkz−k). Its leading coefficient obeys bℓ+1=−ibℓ because the terms ℓCℓ/z and Cℓ′ have powers at most z−2; thus bℓ=(−i)ℓ+1. The remainder and its derivative are O(z−2) by [F3],[F7],[F8], so the recurrence gives hℓ+1(1)=(−i)ℓ+2eiz/z+O(z−2). Differentiating the leading terms and finite remainders also gives jℓ′=cos⁡(z−ℓπ/2)/z+O(z−2) and (hℓ(1))′=i(−i)ℓ+1eiz/z+O(z−2).

3.1F2F3F6F8step 2.1step 2.2algebra∎

For two solutions y,v of Lℓy=0, differentiation and substitution of their equations gives (yv′−y′v)′=−2(yv′−y′v)/z, hence (z2W[y,v])′=0. Applying [F6] separately to the real and imaginary parts on (0,∞) shows z2W[jℓ,hℓ(1)] is a constant complex number. With θ=z−ℓπ/2, the expansions and derivative expansions of step 2.2 give z2W[jℓ,hℓ(1)]⟶(−i)ℓ+1eiz(isin⁡θ−cos⁡θ)=i. Indeed, Euler's formula and the addition identities give eiz(cos⁡θ−isin⁡θ)=ei(z−θ)=eiℓπ/2; repeated use of sin⁡(π/2)=1 and cos⁡(π/2)=0 gives eiℓπ/2=iℓ. Since isin⁡θ−cos⁡θ=−(cos⁡θ−isin⁡θ) and (−i)ℓ+1iℓ=−i, the displayed limit is i. The constant is therefore i, which proves the Wronskian formula.

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Scaled spherical Bessel evaluation on positive real arguments

Definition

Let ℓ≥0 be an integer and z>0 real. Put cℓ=(2ℓ+1)!!=∏r=0ℓ(2r+1),dℓ=(2ℓ−1)!!=∏r=1ℓ(2r−1), where the second product is empty and equals 1 when ℓ=0. Using the outgoing convention of Spherical Bessel and outgoing Hankel function conventions, define Jℓsc(z)=cℓz−ℓjℓ(z),DJℓsc(z)=cℓz−ℓjℓ′(z), Hℓsc(z)=dℓ−1zℓ+1e−izhℓ(1)(z),DHℓsc(z)=dℓ−1zℓ+1e−iz(hℓ(1))′(z). The symbols DJℓsc and DHℓsc mean the scaled derivatives of the original functions, not derivatives of the scaled functions. The available targets are jℓ, jℓ′, hℓ(1), (hℓ(1))′, and these four scaled quantities; every derivative is with respect to z.

An input is either (i) a closed dyadic interval I=[a,b]⊂(0,∞), requesting an enclosure of the whole target image on I, or (ii) a positive real z supplied by an interface that, on each integer request n≥1, terminates and returns a closed dyadic interval In containing z with width at most 2−n. The intervals need not be nested. This is a representation hypothesis on a supplied input, not a claim that every positive real has such a representation. An exact positive dyadic point is the constant-interface case In=[z,z]. A represented point need not be dyadic: arguments such as kR or k2+V0R are admitted when a caller constructs a terminating interface by rigorous interval propagation from represented parameters. Each returned box must contain the exact composed argument; a midpoint or approximate point alone is not an admissible interface. The caller's construction must retain positivity and prove termination of every requested refinement.

For each target, outputs and optional relative tests use Certified special-function enclosure result: a real enclosure is an interval, a complex enclosure is a rectangle, and its radius is half its length or half its Euclidean diagonal, respectively. Each target has an exact positive rational absolute tolerance εa; an exact positive rational relative tolerance εr may also be requested. A relative flag may be true only when the returned enclosure excludes zero and rad⁡(B)/d(B,0)≤εr. No relative flag is available when the enclosure contains zero.

Intermediate interval endpoints may be exact rationals. Final endpoints are rounded outward to multiples of 2−p for an integer p≥0. The mathematical procedure has no fixed exponent limit. A concrete bounded implementation must return range or inconclusive rather than claim a finite result it cannot represent. Finite caps are specified by the evaluating algorithm. A fixed nondegenerate input interval requests its whole image and is not promised arbitrarily small radius; eventual absolute success is asserted separately for represented-point inputs.

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Finite certified power-series evaluation for Fresnel, erf, and Gauss functions

Definition

Inputs are a target label T∈{C,S,erf⁡,2F1}, an exact dyadic complex argument z, and, for the Gauss-series target, exact dyadic complex parameters a,b,c. Also input positive exact rational absolute tolerance εa, optional positive exact rational relative tolerance εr, a fixed exponent range (emin⁡,emax⁡), and finite positive integer caps Smax⁡,Nmax⁡,pmax⁡,W, with pmax⁡≥2. The stage cap Smax⁡ bounds retries, Nmax⁡ bounds series terms, pmax⁡ bounds dyadic precision, and W bounds this procedure's exact interval operations, term constructions, and elementary-DAG node operations. Every nested elementary-range call shares this counter and is interrupted with inconclusive before its next operation would exceed the remaining budget. At stage q, use precision p=max⁡(2,min⁡(q,pmax⁡)) and truncation index q for every elementary-range subcall.

Reject an invalid target, nonpositive tolerance, invalid cap, invalid exponent range emin⁡>emax⁡, or a Gauss input with ∣z∣2≥1 or c∈{0,−1,−2,…} by returning domain. The Fresnel and error-function branches accept every finite complex z. Outputs are a finite interval or rectangle with status success, or a status inconclusive, domain, or range as in Certified special-function enclosure result.

All rational and interval arithmetic uses Outward-rounded interval, rectangle and ball operations. Embed real coefficient intervals X as X+i[0,0] in complex arithmetic; typed elementary DAG nodes never use an implicit real-to-complex conversion. If a required intermediate operation or output endpoint overflows its selected format, return range. If an elementary-range subcall returns a non-finite status, skip the current candidate; if no later stage succeeds, return inconclusive. The same candidate-skip rule applies to every other non-finite arithmetic status, including split-domain or entire when a rounded divisor contains zero; never use such a status as a finite interval. Charge each exact arithmetic operation, coefficient term, rounding, and elementary-range call against W; before an operation that would exceed W, return inconclusive.

At each stage, form Pq, a certified dyadic interval containing π, only when T∈{C,S,erf⁡}. For r∈{1/5,1/239}, form Aq(r)=∑j=0q−1(−1)jr2j+1/(2j+1) and Eq(r)=r2q+1/((2q+1)(1−r2)) by exact rational arithmetic. The arctangent series and geometric-tail argument in Range-reduced exponential, logarithm and arctangent series have certified tails gives arctan⁡r∈[Aq(r)−Eq(r),Aq(r)+Eq(r)]. Combine these intervals as 16arctan⁡(1/5)−4arctan⁡(1/239), then round the resulting endpoints outward to the selected format at precision p. The Machin identity in Trigonometric identities, Machin's formula and reduced Taylor tails proves π∈Pq. If this conversion is non-finite, skip this stage for the selected Fresnel or erf target. For the Gauss-series target, omit Pq and all square-root calls.

Only when T=erf⁡, call Certified range evaluation for elementary real and complex functions on the one-node real DAG Pq with the same exponent range and precision p, and truncation index q. If it returns a finite interval Rq, continue only if inf⁡Rq>0; otherwise skip this stage. This call encloses π by Certified real elementary-function evaluations enclose their ranges. The C,S branches use Pq directly and do not call square root. For a complex rectangle B=X+iY, write M(B)=max⁡(∣inf⁡X∣,∣sup⁡X∣)+max⁡(∣inf⁡Y∣,∣sup⁡Y∣); this rational quantity bounds the modulus of every point in B.

At each stage q=1,…,min⁡(Smax⁡,Nmax⁡,pmax⁡), form the following candidate for the selected target.

  • For C, sum its defining terms c0,…,cq−1 from Fresnel, error, and Gauss hypergeometric series by interval arithmetic, using Pq/2 for π/2. Let cq be the first omitted term and set rC=(sup⁡Pq)2M(z)44(2q+1)(2q+2). If rC<1, enlarge the partial-sum rectangle in both coordinates by M(cq)/(1−rC); otherwise skip this stage.
  • For S, sum its first q defining terms using Pq/2, let sq be the first omitted term, and set rS=(sup⁡Pq)2M(z)44(2q+2)(2q+3). If rS<1, enlarge the partial-sum rectangle in both coordinates by M(sq)/(1−rS); otherwise skip this stage.
  • For erf, use the interval 2/Rq from the single square-root call above for 2/π, sum the first q defining terms, and let eq be the first omitted term. Set rE=M(z)2/(q+1). If rE<1, enlarge the partial-sum rectangle in both coordinates by M(eq)/(1−rE); otherwise skip this stage.
  • For the Gauss series, compute its terms by h0=1 and hk+1=hk z (a+k)(b+k)(c+k)(k+1) using outward interval operations. The exact c+k is nonzero by the domain check, but its rounded enclosure may contain zero; in that case skip this candidate before division. If a or b is a nonpositive integer, let L be one more than the smaller absolute value of those nonpositive integer parameters. If L≤q, sum exactly through hL−1 and use zero tail. Otherwise put A=∣ℜa∣+∣ℑa∣,B=∣ℜb∣+∣ℑb∣,C0=∣ℜc∣+∣ℑc∣,Z0=∣ℜz∣+∣ℑz∣, If q≤C0, skip this stage before forming a quotient. Otherwise set rH=Z0(1+A/q)(1+B/q)1−C0/q. If rH<1, sum h0,…,hq−1, let hq be the first omitted term, and enlarge the partial-sum rectangle in both coordinates by M(hq)/(1−rH); otherwise skip this stage. The geometric bound in Gauss hypergeometric series geometric tail bound applies because A,B,C0,Z0 dominate the corresponding complex moduli.

Round each candidate's final endpoints outward to precision p. Return success at the first stage where its radius is at most εa. If relative tolerance was requested, set relative_certified to true exactly when zero is excluded and the contract's relative-radius inequality holds; failure of that relative test does not prevent absolute success. If no stage produces a passing candidate, return inconclusive. The finite stage loop and the finite elementary-range calls define the procedure completely.

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The Airy Ai function has a negative real zero

Statement

The real Airy function Ai⁡ has at least one zero on the negative real axis.

Facts & Assumptions

Proof

1.1F1F2F3F4algebra

By [F3], Γ(2/3)>0, so [F2] gives Ai⁡(0)>0; in particular, Ai⁡ is not the zero function. Put u(t)=Ai⁡(−t). By [F1] and the chain rule in [F4], u′′(t)+tu(t)=0 for real t.

2.1givenF4step 1.1algebra

Suppose u has no zero on [1,1+π]. Continuity gives a constant sign; replace u by −u if necessary so that u>0 there. Set W(t)=u′(t)sin⁡(t−1)−u(t)cos⁡(t−1). The product rule and sine and cosine derivatives in [F4], together with the equation in step 1.1, give W′(t)=(1−t)u(t)sin⁡(t−1)≤0 on this interval.

3.1step 2.1F4contradiction∎

But W(1)=−u(1)<0 and W(1+π)=u(1+π)>0, contradicting that W is nonincreasing. Therefore u has a zero in [1,1+π], and Ai⁡ has a zero in [−(1+π),−1].

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Integral representations and a range bound for integer Bessel functions

Statement

For integer n≥0 and real x, the series in Integer-order Bessel series satisfy

Jn(x)=1π∫0πcos⁡(nθ−xsin⁡θ) dθ,In(x)=1π∫0πexcos⁡θcos⁡(nθ) dθ.

Consequently, for x≥0, 0≤e−xIn(x)≤1.

Facts & Assumptions

Given: Integer n≥0, real x, and the series definitions of Integer-order Bessel series.

[F1]

Integer-order Bessel series: Jn and In are the absolutely convergent power series with the displayed factorial coefficients.

[F2]

Convergence and geometric tails for integer-order Bessel series: the defining Jn and In series converge absolutely and locally uniformly; their coefficients are the factorial coefficients used in the generating-function calculation below.

[F3]

The complex exponential by its power series and The complex exponential series converges absolutely for every complex argument: the exponential is its absolutely convergent factorial power series for every complex argument.

[F4]

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums: the Cauchy product of two absolutely convergent complex series converges to the product of their sums. Every absolutely convergent complex series converges, and rearrangements preserve its sum permits regrouping the resulting absolutely convergent terms by their Laurent exponent.

[F7]

Change of variable in an improper integral: the proper affine substitution θ↦2π−θ preserves the integral after reversing the endpoints.

Proof

1.1F1F3F4algebraF2

For fixed z and ∣t∣=1, both exponential series in exp⁡ ⁣(z2(t−t−1))=(∑p≥0(z/2)ptpp!)(∑q≥0(−z/2)qt−qq!) have absolute tails bounded independently of t, by [F3]. Their product therefore has an absolutely and uniformly convergent double series on the unit circle; grouping its terms by the Laurent exponent p−q is legitimate by [F4] and the absolute bound. For n≥0, the coefficient of tn is ∑q≥0(−1)q(z/2)2q+n(q+n)!q!=Jn(z) by [F1]. Replacing the minus sign in the first exponential by a plus sign gives the coefficient ∑q≥0(z/2)2q+n(q+n)!q!=In(z).

2.1F5F6step 1.1algebra

For every integer m, direct integration of the real and imaginary parts gives ∫02πeimθ dθ={2π,m=0,0,m≠0. For m≠0, the sine and cosine antiderivatives and their 2π-periodicity from [F5] give zero. Integrating each finite Laurent partial sum against e−inθ therefore extracts exactly its tn coefficient. The uniform convergence in step 1.1 permits passage to the full integral by [F6], applied componentwise. Consequently each coefficient is 12π∫02πG(eiθ)e−inθ dθ, where G is the corresponding generating function.

3.1F5F7step 2.1algebra

For real x, Euler's formula in [F5] gives GJ(eiθ)=exp⁡(ixsin⁡θ) and GI(eiθ)=exp⁡(xcos⁡θ). In the first coefficient integral, the real part is cos⁡(xsin⁡θ−nθ); the substitution θ↦2π−θ on [π,2π], justified by [F7], shows that the imaginary parts cancel and the real parts on the two half intervals agree. Thus the coefficient is π−1∫0πcos⁡(nθ−xsin⁡θ) dθ. For the second integral, cos⁡(2π−θ)=cos⁡θ and the sine Fourier terms cancel under the same substitution, giving π−1∫0πexcos⁡θcos⁡(nθ) dθ. This proves both representations.

4.1F1step 3.1algebra∎

For x≥0, every term in the series for In(x) is nonnegative, so In(x)≥0. The integral gives e−xIn(x)≤π−1∫0πe−x(1−cos⁡θ)dθ≤1, since 1−cos⁡θ≥0.

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Binet integral identity for the right-half-plane log-Gamma branch

Statement

For Re⁡z>0, let ℓ+(z) be the branch in Log-Gamma branch on the right half-plane. Define Φ(u)=∫0u(⌊v⌋−v+1/2) dv. Then the absolutely convergent integral satisfies ℓ+(z)=(z−12)Log⁡z−z+12log⁡(2π)+∫0∞Φ(u)(u+z)2 du. On each unit interval, Φ(n+t)=t(1−t)/2 for 0≤t≤1; hence 0≤Φ(u)≤1/8.

Facts & Assumptions

Given: Re⁡z>0 and an integer N≥2 tending to infinity in the proof.

[F1]

Log-Gamma branch on the right half-plane: ℓ+ is defined by −Log⁡z−γEz+∑k=1∞(zk−Log⁡(1+z/k)).

[F2]

Weierstrass log-Gamma series converges to a Gamma branch: the defining series converges locally uniformly on the right half-plane.

[F3]

The Euler-Mascheroni constant and The Euler–Mascheroni constant and the harmonic asymptotic: γE=γ=lim⁡n→∞(Hn−log⁡n).

[F4]

Stirling's formula for factorials: n!∼2πn(n/e)n as n→∞ through positive integers.

[F5]
[F6]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion and The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): continuous functions on compact intervals are integrable, and an integrable derivative integrates to its endpoint difference; the real statement applies to real and imaginary parts of a complex-valued path.

[F8]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: the real logarithm is continuous on (0,∞) and satisfies log⁡′(x)=1/x.

[F9]

The natural logarithm as the inverse of the exponential function and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: writing e=exp⁡(1) gives log⁡e=1, and log⁡(xy)=log⁡x+log⁡y for positive x,y.

Proof

1.1givenF7algebra

Define f(u)=⌊u⌋−u+1/2. For u=n+t with integer n≥0 and 0≤t<1, f(u)=1/2−t. Each full unit interval has integral zero, and its partial integral is t(1−t)/2. Therefore Φ(n+t)=t(1−t)/2, Φ(n)=0, and 0≤Φ≤1/8.

2.1F5F6F7step 1.1algebra

On [n,n+1], Φ′=f and [F7] gives ∫nn+1f(u)u+z du=[Φ(u)u+z]nn+1+∫nn+1Φ(u)(u+z)2 du=∫nn+1Φ(u)(u+z)2 du, since Φ(n)=Φ(n+1)=0. Summing over n<N gives IN(z):=∫0Nf(u)u+z du=∫0NΦ(u)(u+z)2 du. Because ∣u+z∣≥u+Re⁡z and Φ≤1/8, the right-hand side converges absolutely as N→∞.

2.2givenF5F6F9step 1.1algebra

On each [n,n+1], write f(u)=n+12−u=(n+z+12)−(u+z). By [F5] the function (n+z+12)Log⁡(u+z)−u has derivative f(u)/(u+z); its derivative is continuous and integrable on the compact interval by [F6]. Applying [F6] and summing the resulting endpoint differences yields IN(z)=(N+z−12)Log⁡(N+z)−(z+12)Log⁡z−N−log⁡((N−1)!)−zHN−1+∑k=1N−1(zk−Log⁡(1+z/k)). The coefficient of each internal Log⁡(k+z) in the endpoint sum is −1. For k≥1, Log⁡(k+z)=log⁡k+Log⁡(1+z/k) because k>0 is real and multiplication by it preserves the principal argument; [F9] gives ∑k=1N−1log⁡k=log⁡((N−1)!). These identities give the displayed form.

3.1F1F2F3step 2.2algebra

Add −Log⁡z−γEz to the finite sum in step 2.2. By [F1]--[F3], the resulting left side converges to ℓ+(z), while ℓ+(z)=IN(z)−(N+z−12)Log⁡(N+z)+(z−12)Log⁡z+N+log⁡((N−1)!)+z(HN−1−γE)+o(1).

4.1F3F4F5F6F8F9step 3.1algebra

Taking logarithms in [F4] is valid by continuity [F8]. The laws [F9] and e=exp⁡(1) give log⁡(n!)=(n+12)log⁡n−n+12log⁡(2π)+o(1). At n=N−1, since log⁡(1−1/N)=−1/N+O(N−2), this becomes log⁡((N−1)!)=(N−12)log⁡N−N+12log⁡(2π)+o(1). The logarithm estimate follows by integrating 1/(1−t) on [0,1/N]; the difference from 1/N is bounded by 1/(N2(1−1/N)). Also [F3] gives HN−1−γE=log⁡N+o(1). Finally, for fixed z and N>2∣z∣, Log⁡(N+z)=log⁡N+z/N+Oz(N−2), since N>0 and 1+z/N,N+z lie in the right half-plane, so Log⁡(N+z)=log⁡N+Log⁡(1+z/N). By [F5] and [F6], Log⁡(1+w)−w=−∫01tw2/(1+tw) dt for w=z/N; here ∣1+tw∣≥1 when Re⁡z>0. Therefore (N+z−12)Log⁡(N+z)=(N+z−12)log⁡N+z+o(1).

5.1step 2.1step 3.1step 4.1algebra∎

Substitute these asymptotics into step 3.1. The coefficients of log⁡N cancel, and N cancels the −N in the factorial term, leaving ℓ+(z)=lim⁡N→∞IN(z)+(z−12)Log⁡z−z+12log⁡(2π). By step 2.1, the limit is the absolutely convergent integral of Φ(u)/(u+z)2. This is the claimed identity.

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Moment recurrence for the scaled complementary error function

Statement

For x>0 and integer k≥0, the moments in Scaled complementary error function and its moments are finite and positive. They satisfy

2xM0+2M1=1,kMk−1=2xMk+2Mk+1(k≥1).

With rk=Mk+1/Mk, it follows that

rk=k+12(x+rk+1),0<rk<k+12x,erfcx⁡(x)=1π(x+r0).

Facts & Assumptions

Given: Real x>0, integer k≥0, and the moment and scaled-error definitions in Scaled complementary error function and its moments.

[F1]

Scaled complementary error function and its moments: Mk(x)=∫0∞tke−t2−2xt dt and erfcx⁡(x)=ex2erfc⁡(x).

[F2]

Fresnel, error, and Gauss hypergeometric series: erfc⁡(x)=1−erf⁡(x).

[F3]

Error-function series geometric tail bound: for real x, erf⁡(x)=2π−1/2∫0xe−u2 du.

[F4]

The Gaussian integral ∫−∞∞e−x2 dx=π: ∫−∞∞e−u2 du=π.

[F5]

If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v: on compact intervals, if its hypotheses hold, ∫uv′=uv∣∂−∫u′v. In particular the derivatives used below are integrable on each compact interval because they are continuous.

[F6]

Change of variable in an improper integral: an admissible increasing C1 substitution preserves convergence and value of an improper integral.

[F7]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: each continuous real derivative on a compact truncation interval is Riemann integrable; apply to real and imaginary parts for the complex-valued integration-by-parts formulas.

[F8]

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, ∫0∞sa−1e−sds=Γ(a)<∞.

[L9]

The exponential function is smooth and (exp⁡)′=exp⁡, 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

1.1givenF1F6F7F8algebra

The integrands in [F1] are positive. Near 0 they are bounded by tk. For t≥1, e−2xt≤1; the substitution s=t2 and [F6] bound the tail by 12∫1∞s(k−1)/2e−sds, which is finite by the Gamma integral convergence in [F8]. The integrands are continuous on every compact truncation interval, so their proper integrals there exist by [F7]. Thus each Mk is finite; it is strictly positive because its integrand is positive on [1,2].

2.1F1F5F7L9step 1.1algebra

Apply [F5] on [ε,R] to w(t)=e−t2−2xt and to tkw(t). These functions and their derivatives are continuous on the compact interval, so all derivatives required by [F5] are integrable by [F7]; the complex-valued integration-by-parts statement is applied componentwise. The boundary terms tend to 0 at R→∞; for k≥1 they also tend to 0 at ε↓0, while for k=0 the lower boundary tends to 1. For the infinite endpoint, choose an integer m>k/2; the exponential series gives et2/2≥(t2/2)m/m!, so tke−t2−2xt≤2mm!tk−2me−t2/2→0. Passing to the limits in the integration-by-parts identities gives 2xM0+2M1=1 and, for k≥1, kMk−1=2xMk+2Mk+1.

3.1step 2.1algebra

Divide the recurrence with index k+1 by 2Mk>0. Since rk=Mk+1/Mk and rk+1=Mk+2/Mk+1, this gives (k+1)/2=xrk+rkrk+1 and hence rk=(k+1)/(2(x+rk+1)). Positivity gives 0<rk<(k+1)/(2x).

4.1F1F2F3F4F6step 2.1algebra∎

The integrand in [F4] is even, so ∫0∞e−u2du=π/2. Subtract the integral identity in [F3] and use the definition in [F2] to get erfc⁡(x)=2π∫x∞e−u2 du. The increasing substitution u=t+x from [F6] then gives erfcx⁡(x)=2π∫0∞e−t2−2xt dt=2M0π. The first identity in step 2.1 gives 2M0(x+r0)=1; substitution proves erfcx⁡(x)=1/(π(x+r0)).

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Bessel differential equations and order recurrences from the series

Statement

For z≠0, the functions in Integer-order Bessel series satisfy

z2Jn′′+zJn′+(z2−n2)Jn=0,z2In′′+zIn′−(z2+n2)In=0.

For real or complex z≠0 and integer n≥1, they also satisfy

Jn−1(z)+Jn+1(z)=2nzJn(z),In−1(z)−In+1(z)=2nzIn(z).

In addition, J0′(z)=−J1(z).

At z=0, the series limits are J0(0)=I0(0)=1 and Jn(0)=In(0)=0 for n≥1.

Facts & Assumptions

Given: The series definitions of Integer-order Bessel series and their compact-uniform convergence and termwise differentiability from Convergence and geometric tails for integer-order Bessel series.

[F1]

Integer-order Bessel series: the coefficient of degree 2k+n is (−1)k/(22k+nk!(k+n)!) for Jn and 1/(22k+nk!(k+n)!) for In.

[F2]

Convergence and geometric tails for integer-order Bessel series: the series and their first two derivative series converge uniformly on compact disks.

[F3]

Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term: each defining complex power series may be differentiated term by term strictly inside its radius of convergence.

Proof

1.1F1F2F3algebra

The convergence of the first two formal derivative series in [F2], together with the infinite radius supplied by the ratio bound in Convergence and geometric tails for integer-order Bessel series, lets [F3] justify each termwise differentiation used below. For a monomial z2k+n, (z2D2+zD−n2)z2k+n=4k(k+n)z2k+n. The coefficient ratio in [F1] is ak/ak−1=∓1/[4k(k+n)] for k≥1, with the minus sign for Jn and plus sign for In. Thus the coefficient of each power cancels in the respective differential equation.

1.2F1F2F3algebra

For Jn−1+Jn+1, reindex the Jn+1 series so both terms have power z2k+n−1. For k≥1 their combined coefficient is (−1)kn/[22k+n−1k!(k+n)!], and for k=0 it is 1/[2n−1(n−1)!]. These are exactly the coefficients of (2n/z)Jn. For In−1−In+1 the same reindexing gives the same positive combined coefficient, proving the second recurrence.

1.3F1F2F3algebra

Differentiating the J0 series termwise by [F3] and reindexing k=j+1 gives J0′(z)=∑j=0∞(−1)j+1z2j+122j+1j!(j+1)!=−J1(z).

2.1F1F2F3∎

Setting z=0 in the convergent series leaves only the k=0,n=0 term; every term has a positive power of z when n≥1. Hence the stated limits hold.

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Explicit tail bound for the log-Gamma product series

Statement

For Re⁡z>0 and integer N>2∣z∣, the tail in the series for ℓ+(z) after its first N summands satisfies

∣∑n=N+1∞(zn−Log⁡(1+zn))∣≤∣z∣22N(1−∣z∣/(N+1)).

Facts & Assumptions

Given: Re⁡z>0 and an integer N>2∣z∣.

[F1]

Log-Gamma branch on the right half-plane: the summands are z/n−Log⁡(1+z/n).

[F2]
[F3]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): the derivative of a continuously differentiable complex-valued path integrates to its endpoint difference, by applying the real theorem to each component.

Proof

1.1givenF1F2F3F4algebra

Put w=z/n, so Re⁡w>0. The integrand in the path formula is continuous and integrable by [F4]. By [F2], [F3], and the chain rule, w−Log⁡(1+w)=∫01tw21+tw dt. Since ∣1+tw∣2=1+2tRe⁡w+t2∣w∣2≥1, this summand has modulus at most ∣z∣2/(2n2).

2.1givenstep 1.1algebra∎

The decreasing function t↦t−2 satisfies ∑n=N+1∞n−2≤∫N∞t−2dt=1/N. Summing step 1.1 gives ∣z∣2/(2N). Since 0<1−∣z∣/(N+1)<1 under N>2∣z∣, this is no larger than the stated bound.

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Shifted log-Gamma branch is independent of the shift

Statement

For every z∈Ω=C∖(−∞,0], the value ℓ(z) in Log-Gamma branch on the slit plane is independent of the integer m≥0 with Re⁡(z+m)>0, is holomorphic on Ω, and satisfies eℓ(z)=Γ(z).

Facts & Assumptions

Given: z∈Ω and the branch definitions Log-Gamma branch on the right half-plane and Log-Gamma branch on the slit plane.

[F1]

Weierstrass log-Gamma series converges to a Gamma branch: eℓ+(w)=Γ(w) and ℓ+ is holomorphic for Re⁡w>0.

[F2]

The Gamma functional equation: if Re⁡w>0, then Γ(w+1)=wΓ(w).

[F3]

Gamma functional equation away from the poles: the same functional equation holds when w is not a nonpositive integer.

[F6]

A differentiable map on a connected open Euclidean set has zero derivative exactly when it is constant: a holomorphic function with zero derivative on a connected open domain is constant.

[F7]

exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential and The complex exponential by its power series: exp⁡(u+v)=exp⁡(u)exp⁡(v) and exp⁡(0)=1. In particular, exp⁡(−v) is the reciprocal of exp⁡(v).

[F9]

Real and imaginary parts, complex conjugation, and modulus, Real and complex modulus product and triangle inequalities, and Every path-connected space is connected, and every path component lies inside a component: the right half-plane is open, since if x=Re⁡w0>0 and ∣w−w0∣<x, then Re⁡w≥x−∣Re⁡(w−w0)∣≥x−∣w−w0∣>0. For w0,w1 in it, γ(t)=(1−t)w0+tw1 has positive real part and ∣γ(t)−γ(s)∣=∣t−s∣∣w1−w0∣, which proves continuity by the metric definition. The half-plane contains 1 and is open and connected.

[F10]

Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with ∂zˉf=0, or with the Cauchy–Riemann equations: under C≅R2, a complex-differentiable function is totally differentiable and its real derivative is multiplication by its complex derivative; in particular, H′=0 implies DH=0.

Proof

1.1F1F2F4F5F6F7F8F9F10algebra

For Re⁡w>0, the function H(w)=ℓ+(w+1)−ℓ+(w)−Log⁡w is holomorphic by [F1]. The functional equation [F2], branch identity [F1], and exp⁡(Log⁡w)=w from [F4] give exp⁡(ℓ+(w+1))=exp⁡(ℓ+(w)+Log⁡w). By [F7], exp⁡(H(w))=1. Differentiating this identity by [F5] gives exp⁡(H(w))H′(w)=0, hence H′(w)=0. By [F10], the real derivative of H vanishes; the right half-plane is nonempty, open, and connected by [F9], so [F6] makes H constant there. For positive real w, all three terms are real by the defining series for ℓ+ and the real logarithm, so [F8] and exp⁡(H(w))=1 imply H(w)=log⁡1=0. Therefore ℓ+(w+1)=ℓ+(w)+Log⁡w throughout the right half-plane.

2.1givenstep 1.1algebra

If m2>m1 are both admissible shifts, then w=z+m1 has positive real part, as do w,w+1,…,w+(m2−m1−1). Iterating step 1.1 gives ℓ+(z+m2)−∑j=0m2−1Log⁡(z+j)=ℓ+(z+m1)−∑j=0m1−1Log⁡(z+j), so the definition is independent of m. Locally m can be fixed, making the formula holomorphic; independence makes these local formulas agree.

3.1F1F3step 2.1algebra∎

Exponentiating the formula with any admissible m gives eℓ(z)=Γ(z+m)/∏j=0m−1(z+j) by [F1]. None of the factors vanishes because z∈Ω. Iterating [F3] downward therefore identifies this quotient with Γ(z).

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Finite exact-interval evaluation of scaled spherical Bessel values

Definition

Inputs are those of Scaled spherical Bessel evaluation on positive real arguments, a nonempty finite list of valid targets, finite exact positive rational absolute tolerances, optional exact positive rational relative tolerances, and finite integer caps W,Q,Mmax⁡,pmax⁡≥1. W is the maximum exact-arithmetic work count, Q the maximum stage and represented-point interface index, Mmax⁡ the maximum Taylor degree, and pmax⁡ the maximum final dyadic precision. A Gaussian rational is a+ib with a,b∈Q. A Laurent polynomial is a finite sum ∑kakzk with integer exponents and Gaussian-rational coefficients; its formal derivative uses (zk)′=kzk−1. An expression graph below is a finite directed acyclic graph of exact rational arithmetic operations.

Factorials are those of The factorial n! and the falling factorial nk‾, defined by recursion in N; they are nonzero by that definition and Canonical naturals are positive and strictly increasing, so every displayed factorial denominator is defined. For a rational x, ⌈x⌉ denotes the least integer greater than or equal to x, computed by exact integer division.

Reject an invalid order, target, tolerance, cap, or interval input not contained in (0,∞) with domain. For a represented point, an interval returned by the input interface may meet zero; such a stage is skipped and a later interface index is tried.

Construct Laurent polynomials by A0=0,B0=z−1,C0=−iz−1, Ar+1=rz−1Ar−Ar′−Br,Br+1=rz−1Br−Br′+Ar,Cr+1=rz−1Cr−Cr′−iCr for r=0,…,ℓ−1. With s and c denoting the sine and cosine input enclosures, form the unscaled outputs u=Aℓc+Bℓs,v=(Aℓ′+Bℓ)c+(Bℓ′−Aℓ)s,w=Cℓ(c+is),t=(Cℓ′+iCℓ)(c+is). Form the scaled outputs cℓz−ℓu,cℓz−ℓv,dℓ−1zℓ+1Cℓ,dℓ−1zℓ+1(Cℓ′+iCℓ). Collect like Laurent powers exactly; in particular cancel negative powers in the scaled hℓ(1) value when they cancel algebraically. A remaining z−1 in the scaled derivative is evaluated only on positive input intervals.

For real intervals use exact rational endpoint hulls: [a,b]+[c,d]=[a+c,b+d], [a,b][c,d]=[min⁡(ac,ad,bc,bd),max⁡(ac,ad,bc,bd)], and [a,b]−1=[1/b,1/a] when 0<a≤b. For complex rectangles, apply these real interval operations to the real and imaginary components of (x+iy)(u+iv)=(xu−yv)+i(xv+yu). Thus interval arithmetic encloses every pointwise sum and product; division is only by a positive power of the positive argument or by a positive integer constant. Compute powers, factorials, and rational endpoints by finite exact arithmetic.

At a positive current box I=[a,b], choose M≥0 with M+2≥2b and form the rational interval polynomial and rational tail TM(I)=∑m=0M(iI)mm!,EM=2bM+1(M+1)!. The interval power is repeated complex-rectangle multiplication. Enclose cos⁡(I) by Re⁡TM(I)+[−EM,EM] and sin⁡(I) by Im⁡TM(I)+[−EM,EM]. Evaluate each requested output graph with these leaves and z∈I, and round each final endpoint outward to the grid 2−pZ.

Use the following deterministic finite search. Charge the Laurent recurrence, graph construction, and coefficient collection to the work counter as they are performed. For stages q=1,…,Q, first take the fixed interval unchanged, or call the represented-point interface at index q. Skip a represented-point stage unless its returned interval has a>0. Set p=q and M=max⁡(q,⌈2b⌉). If p>pmax⁡ or M>Mmax⁡, stop with inconclusive. Evaluate all requested graphs and the displayed TM,EM using the work counter: each exact rational or Gaussian-rational arithmetic operation or comparison, each Taylor term, each endpoint-rounding operation, and each interface call costs one unit. If the next operation would exceed W, stop with inconclusive before performing it. If the output format cannot represent a certified endpoint, return range with no finite claim. If every requested target has radius at most its absolute tolerance, return success; for each optional relative tolerance, set its flag true exactly when the enclosure excludes zero and the relative test in the contract passes. A failed relative test does not prevent absolute success. If no stage succeeds before a cap or the search ends, return inconclusive. No floating-point recurrence is used.

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Composite midpoint enclosure for the integer Bessel integral

Statement

Let n≥0 be an integer, x∈R, m≥1, and h=π/m. Put ϕ(θ)=cos⁡(nθ−xsin⁡θ) and Qm=h∑j=0m−1ϕ((j+1/2)h). Then

∣Jn(x)−Qmπ∣≤h224((n+∣x∣)2+∣x∣).

This is the local composite-midpoint certificate for the Bessel integral; it does not certify floating-point summation unless each operation is enclosed.

Facts & Assumptions

Given: Integer n≥0, real x, and positive integer m.

[F1]

Integral representations and a range bound for integer Bessel functions: Jn(x)=π−1∫0πϕ(θ) dθ.

[F2]

Taylor's Lagrange remainder follows from repeated Rolle steps: for a C2 function, the first-order Taylor remainder at c is f′′(ξ)u2/2 for some point ξ between c and c+u.

[L9]

Sum, product, quotient, and chain rules from difference quotients, The derivatives of sine and cosine are cosine and minus sine: 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

1.1givenL9algebra

Write g(θ)=nθ−xsin⁡θ, so ∣g′(θ)∣≤n+∣x∣ and ∣g′′(θ)∣≤∣x∣. Differentiating ϕ=cos⁡g gives ∣ϕ′′(θ)∣≤∣g′(θ)∣2+∣g′′(θ)∣≤(n+∣x∣)2+∣x∣=:M.

2.1step 1.1F2algebra

On a panel centered at c with half-width h/2, [F2] gives ∣ϕ(c+u)−ϕ(c)−uϕ′(c)∣≤Mu2/2. Integrating over −h/2≤u≤h/2 cancels the odd term and bounds the panel error by Mh3/24.

3.1F1step 2.1algebra∎

Summing the m panel errors gives ∣∫0πϕ−Qm∣≤mMh3/24=πMh2/24. Divide by π and use [F1] to obtain the stated bound.

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Adjacent initial Bessel values cannot both vanish

Statement

For every real x>0, J0(x) and J1(x) are not both zero.

Facts & Assumptions

Given: A real number x>0 and the integer-order Bessel functions from Integer-order Bessel series.

[F1]

Bessel differential equations and order recurrences from the series: J0 is entire, satisfies z2J0′′+zJ0′+z2J0=0, has J0(0)=1, and J0′=−J1.

[F2]

Identity theorem for holomorphic functions: two holomorphic functions on a complex domain that agree on a set with an interior accumulation point agree throughout that domain.

Proof

1.1givenF1algebra

Suppose J0(x)=J1(x)=0. By [F1], J0(x)=J0′(x)=0. Expand the entire function about x, J0(x+w)=∑m≥0amwm. The coefficient of wm in the differential equation in [F1] is x2(m+2)(m+1)am+2+x(m+1)(2m+1)am+1+(m2+x2)am+2xam−1+am−2=0, with a−1=a−2=0. Since a0=a1=0 and x>0, induction gives am=0 for all m, so J0 vanishes on a disk about x.

2.1F1F2step 1.1∎

Apply [F2] to J0 and the zero function on the connected domain C. It follows that J0 is identically zero, contradicting J0(0)=1 from [F1]. Thus J0(x) and J1(x) cannot both vanish.

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Signed finite remainder for the erfcx asymptotic expansion

Statement

For real x>0 and integer N≥1, put

(1/2)0=1,(1/2)k=∏j=0k−1(j+1/2).

SN(x)=1xπ∑k=0N−1(−1)k(1/2)kx2k.

Then

erfcx⁡(x)−SN(x)=(−1)NRN(x),0≤RN(x)≤1xπ(2N)!4NN!x2N.

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

Facts & Assumptions

Given: x>0, integer N≥1, and the definitions of erfcx and Mk(x) in Scaled complementary error function and its moments.

[F1]

Moment recurrence for the scaled complementary error function: erfcx⁡(x)=2M0(x)/π.

[F2]

Change of variable in an improper integral: a valid increasing C1 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 CN function is f(N)(ξ)tN/N! for some ξ between the expansion point and t.

[F4]

Gamma at the positive integers: Γ(N+1)=N! for integer N≥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)=∫0∞sa−1e−s ds<∞.

[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 M0, set s=t2+2xt. Its derivative 2(t+x)>0 for t≥0, so [F2] applies. Then ds=2(t+x)dt, t+x=x2+s, and erfcx⁡(x)=1π∫0∞e−sx2+s ds=1xπ∫0∞e−s(1+sx2)−1/2ds.

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)=∑k=0N−1(−1)k(1/2)kk!tk+(−1)NρN(t),0≤ρN(t)≤(1/2)NN!tN(t≥0), because ∣f(N)(u)∣=(1/2)N(1+u)−N−1/2≤(1/2)N for u≥0.

2.1F4F5step 1.1step 1.2algebra∎

Substitute t=s/x2 in step 1.2 and integrate against e−s/(xπ). By the Gamma integral and [F4], ∫0∞e−ssNds=Γ(N+1)=N! by [F5] and [F4]. The signed remainder is nonnegative after removing (−1)N and is at most (1/2)N/(x2N+1π). The identity (1/2)N=(2N)!/(4NN!) follows by separating even and odd factors in (2N)!. This gives the stated bound.

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Finite continued-fraction enclosure for erfcx

Statement

Fix real x>0 and integer N≥0. Set IN=[0,(N+1)/(2x)], and for k=N−1,…,0 define

Ik=[k+12(x+sup⁡Ik+1),k+12(x+inf⁡Ik+1)].

Then rk∈Ik for every 0≤k≤N, where rk is the moment ratio from Moment recurrence for the scaled complementary error function. Therefore

erfcx⁡(x)∈[1π(x+sup⁡I0),1π(x+inf⁡I0)].

Each finite depth gives a valid enclosure; no convergence or shrinking-width claim as N grows is made.

Facts & Assumptions

Given: x>0, integer N≥0, and the intervals Ik defined above.

[F1]

Moment recurrence for the scaled complementary error function: rk=(k+1)/(2(x+rk+1)) and 0<rN<(N+1)/(2x).

[F2]

Moment recurrence for the scaled complementary error function: erfcx⁡(x)=1/(π(x+r0)).

Proof

1.1givenF1algebrainduction

The upper bound in [F1] puts rN in IN. The map fk(t)=(k+1)/(2(x+t)) is positive and decreasing for t≥0. Hence, if rk+1∈Ik+1, then fk(sup⁡Ik+1)≤fk(rk+1)≤fk(inf⁡Ik+1); by [F1] the middle term is rk. Backward induction gives rk∈Ik down to k=0.

2.1step 1.1F2algebra∎

The function t↦1/(π(x+t)) is positive and decreasing on t≥0. Applying it to the interval I0 and using [F2] gives the stated erfcx enclosure. Every step used a finite number of inclusions, so the enclosure is valid at this N regardless of whether its width decreases with larger N.

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Composite midpoint enclosure for the scaled modified Bessel integral

Statement

Let n≥0 be an integer, x≥0, m≥1, h=π/m, and f(θ)=e−x(1−cos⁡θ)cos⁡(nθ),Qm=h∑j=0m−1f((j+1/2)h). Then ∣e−xIn(x)−Qmπ∣≤h224((x+n)2+x). The bound certifies the exact midpoint integral. Floating-point summation is certified only if its operations are enclosed separately.

Facts & Assumptions

Given: Integer n≥0, real x≥0, and integer m≥1.

[F1]

Integral representations and a range bound for integer Bessel functions: In(x)=1π∫0πexcos⁡θcos⁡(nθ) dθ,0≤e−xIn(x)≤1.

[F2]

Taylor's Lagrange remainder follows from repeated Rolle steps: for a C2 function, the first-order Taylor remainder at c is f′′(ξ)u2/2 for some point between c and c+u.

[L9]

The exponential function is smooth and (exp⁡)′=exp⁡, Sum, product, quotient, and chain rules from difference quotients, The derivatives of sine and cosine are cosine and minus sine: 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

1.1givenL9algebra

Put g(θ)=e−x(1−cos⁡θ). Since 1−cos⁡θ≥0, we have 0<g≤1 on [0,π]. Differentiation gives g′=−xsin⁡θ g and g′′=(−xcos⁡θ+x2sin⁡2θ)g. Thus ∣g′∣≤x and ∣g′′∣≤x+x2. For f=gcos⁡(nθ), ∣f′′∣≤∣g′′∣+2n∣g′∣+n2∣g∣≤x+x2+2nx+n2=(x+n)2+x.

2.1step 1.1F2algebra

On each panel centered at c, [F2] bounds ∣f(c+u)−f(c)−uf′(c)∣ by Mu2/2, where M=(x+n)2+x by step 1.1. Integrating over −h/2≤u≤h/2 cancels the odd linear term and bounds that panel's error by Mh3/24.

3.1F1step 2.1algebra∎

Summing the m panel errors gives ∣∫0πf−Qm∣≤mMh3/24=πMh2/24. By [F1], e−xIn(x)=π−1∫0πf(θ) dθ; dividing by π proves the claim.

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Explicit sectorial remainder bound for the right-half-plane log-Gamma branch

Statement

Fix 0<δ≤π/2 and z∈C with Re⁡z>0. For the branch ℓ+ from Log-Gamma branch on the right half-plane, Binet integral identity for the right-half-plane log-Gamma branch gives ℓ+(z)=(z−12)Log⁡z−z+12log⁡(2π)+R(z),R(z)=∫0∞Φ(u)(u+z)2 du, where Φ(u)=∫0u(⌊v⌋−v+1/2) dv. Then ∣R(z)∣≤18sin⁡2(δ/2)∣z∣. For δ=π/2, the bound is 1/(4∣z∣). The relative error after exponentiating the three main terms is at most exp⁡(∣R(z)∣)−1.

Facts & Assumptions

Given: 0<δ≤π/2 and Re⁡z>0.

[F1]

Binet integral identity for the right-half-plane log-Gamma branch: the exact branch identity and its remainder representation hold, with Φ(n+t)=t(1−t)/2 on each unit interval.

[F2]

Real and imaginary parts, complex conjugation, and modulus: for z=x+iy, ∣z∣2=x2+y2.

[F3]

Signs, monotonicity intervals, and ranges of sine and cosine and Trigonometric identities, Machin's formula and reduced Taylor tails: cosine decreases on [0,π] and cos⁡(π/2)=0, hence cos⁡δ≥0 for 0<δ≤π/2.

[F4]

Trigonometric identities, Machin's formula and reduced Taylor tails: the cosine addition identity and sin⁡2x+cos⁡2x=1 give sin⁡2(δ/2)=(1−cos⁡δ)/2 and cos⁡2(δ/2)=(1+cos⁡δ)/2.

Proof

1.1F1algebra

The formula in [F1] defines Φ(n+t)=t(1−t)/2 for integer n≥0 and 0≤t≤1. Hence 0≤Φ(u)≤1/8 for every u≥0.

1.2givenF2F3F4algebra

Write z=x+iy, with x=Re⁡z>0, and put r=∣z∣. By [F2], ∣u+z∣2=(u+x)2+y2=u2+r2+2ux≥u2+r2≥u2+r2−2urcos⁡δ, since u≥0 and cos⁡δ≥0 by [F3]. The difference between the last expression and sin⁡2(δ/2)(u+r)2 is cos⁡2(δ/2)(u−r)2≥0 by [F4]. Thus ∣u+z∣−2≤sin⁡−2(δ/2)(u+r)−2.

2.1F1step 1.1step 1.2algebra∎

Applying steps 1.1 and 1.2 to the remainder in [F1] gives ∣R(z)∣≤18sin⁡2(δ/2)∫0∞du(u+r)2=18sin⁡2(δ/2)r. For any complex w, ∣ew−1∣≤e∣w∣−1 by the exponential series; apply this with w=R(z).

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Soundness of the finite Fresnel, erf, and Gauss-series enclosures

Statement

If Finite certified power-series evaluation for Fresnel, erf, and Gauss functions returns success for C,S,erf⁡, or a Gauss-series value on its stated domain, its returned interval or rectangle contains that exact function value and passes the requested absolute-radius test. A relative certificate is valid only when the enclosure excludes zero and the contract's relative-radius test passes. Every execution with finite caps terminates. The theorem makes no claim of success for every valid input.

Facts & Assumptions

Given: A successful return of the procedure in Finite certified power-series evaluation for Fresnel, erf, and Gauss functions with a sound outward interval backend.

[A1]
[F1]

Finite certified power-series evaluation for Fresnel, erf, and Gauss functions: a successful series branch encloses its finite partial sum and enlarges it by the applicable geometric tail bound before applying the radius test.

[F2]

Fresnel, error, and Gauss hypergeometric series: the four functions on the stated branches are represented by the specified series.

[F3]

Fresnel series geometric tail bounds: the Fresnel tail bound contains the exact omitted series tail when the tested ratio is below 1.

[F4]

Error-function series geometric tail bound: the erf⁡ tail bound contains the exact omitted series tail when the tested ratio is below 1.

[F5]

Gauss hypergeometric series geometric tail bound: the Gauss tail bound contains the exact omitted series tail when N>∣c∣ and qN<1; terminating parameter cases have zero tail after their last nonzero term.

[F6]

Certified special-function enclosure result: the radius and relative tests specify exactly which claims accompany success.

Proof

1.1F1F2F3F4F5

In each nonterminating success branch, the exact value is the sum of the partial series and its omitted tail by the series definitions in the algorithm contract. The relevant lemma among [F3], [F4], and [F5] bounds the modulus of that tail by the radius used to enlarge the partial-sum enclosure. In a terminating Gauss branch [F5] makes the tail exactly zero.

2.1A1A2F1step 1.1

The outward interval backend encloses the exact arithmetic partial sums by [A1]. Every elementary-range subcall used for constants or interval images is sound by [A2]. Adding the symmetric tail enlargement therefore gives an interval or rectangle containing the exact full value.

3.1A1A2F1F6step 2.1

The procedure reports success only after the containing enclosure passes the absolute-radius test; its relative flag is set only under the stated zero-exclusion and relative-radius conditions. These are exactly the certificate claims in [F6]. If no candidate passes, the procedure returns inconclusive; no universal success is asserted.

4.1givenA2step 3.1algebra∎

The outer stage count is finite, each partial sum has at most the capped number of terms, and each elementary-range call has a finite DAG, format, and truncation index. Thus every capped execution terminates.

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Soundness and eventual absolute precision for spherical Bessel enclosures

Statement

Every success output of Finite exact-interval evaluation of scaled spherical Bessel values encloses each requested scaled or unscaled value, or the entire requested image for an interval input, and satisfies its absolute-radius test. Its relative flags satisfy Certified special-function enclosure result. Every capped execution terminates. For each fixed order, positive represented real point, finite nonempty target list, and positive rational absolute tolerances, some finite work, degree, interface-index, and precision caps permit absolute success. The same holds for an exact positive dyadic point. For a nonzero requested target, any positive rational relative tolerance can also eventually be certified by rerunning with sufficiently small positive rational absolute tolerances and sufficiently large finite caps; no relative certification is promised at a zero. The theorem gives no uniform floating-point condition number, bit-complexity bound, success guarantee for fixed small caps, or scattering uniqueness result.

Facts & Assumptions

Given: An integer ℓ≥0, a nonempty finite list of targets, positive rational absolute tolerances, optional positive rational relative tolerances, and either a positive dyadic interval or a positive real point with the terminating interval interface specified in Scaled spherical Bessel evaluation on positive real arguments.

[F1]

Spherical Bessel and outgoing Hankel function conventions defines jℓ=(−z)ℓ(z−1d/dz)ℓ(sin⁡z/z) and hℓ(1)=(−z)ℓ(z−1d/dz)ℓ(−ieiz/z) for integers ℓ≥0 and real z>0.

[F2]

Sine and cosine defined by their real power series and The complex exponential by its power series define sine, cosine, and exponential by their factorial-denominator power series. The complex exponential series converges absolutely for every complex argument states that the exponential series is absolutely convergent for every complex argument. The factorial n! and the falling factorial nk‾, defined by recursion in N defines 0!=1, n!=(n−1)!n for n≥1, and proves n!≠0; Canonical naturals are positive and strictly increasing states each positive natural has positive real embedding.

[F3]

Scaled spherical Bessel evaluation on positive real arguments fixes the positive-real point/interval modes, all eight targets, their scaling conventions, and the absolute/relative meanings. In particular, a scaled derivative is the scaled derivative of the original function.

[F4]

Finite exact-interval evaluation of scaled spherical Bessel values uses exact rational Laurent expressions, interval operations, the Taylor polynomial TM, the error bound EM, final outward rounding, and finite caps as stated in its Definition.

[F6]

The derivatives of sine and cosine are cosine and minus sine gives (sin⁡z)′=cos⁡z and (cos⁡z)′=−sin⁡z. Euler's formula: exp⁡(iθ)=cos⁡θ+isin⁡θ for every real θ gives eiz=cos⁡z+isin⁡z for real z, so componentwise differentiation gives (eiz)′=ieiz.

[F7]

The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums: if two complex series converge absolutely, their Cauchy product converges absolutely and its sum is the product of their sums.

[F8]
[F9]

Certified special-function enclosure result: a successful real enclosure is an interval and a complex enclosure a rectangle; their radii are half the length and half the Euclidean diagonal, respectively. success requires the exact requested image to be enclosed and the radius to be at most the absolute tolerance. A relative flag can be true only when zero is excluded and rad⁡(B)/d(B,0)≤εr; inconclusive, domain, and range make no finite-value claim.

[F10]

For ∣r∣<1 the sequence rk is null, and for ∣r∣>1 the sequence ∣r∣k diverges to +∞: if ∣r∣<1, then rq→0 as the natural index q→∞; in particular 2−q=(1/2)q→0.

Proof

Given: The order, targets, tolerances, and input mode in the Statement.

1.1F1F4F5F6algebrainduction

For yℓ=(−z)ℓDℓf, differentiation as in [F1] gives yℓ+1=ℓyℓ/z−yℓ′. Starting from j0=sin⁡z/z=A0cos⁡z+B0sin⁡z with A0=0,B0=z−1, the derivative formula (Aℓcos⁡z+Bℓsin⁡z)′=(Aℓ′+Bℓ)cos⁡z+(Bℓ′−Aℓ)sin⁡z gives exactly the algorithm's Aℓ+1,Bℓ+1. Starting from h0(1)=−ieiz/z=C0eiz with C0=−i/z, the componentwise product rule [F5]–[F6] gives (Cℓeiz)′=(Cℓ′+iCℓ)eiz and exactly the algorithm's recurrence for Cℓ+1. Euler's identity [F6] identifies the algorithm's c+is with eiz. Thus u,v,w,t represent jℓ,jℓ′,hℓ(1),(hℓ(1))′, respectively.

1.2F2F7algebra

The two series for e−iz and eiz are absolutely convergent by [F2], so [F7] identifies their product with the sum of the Cauchy-product coefficients. For total degree n, its coefficient is znSn, where Sn=∑r=0n(−i)rin−r/(r!(n−r)!). For 1≤r≤n, combining over the denominator r!(n+1−r)! and using (n+1−r)!=(n+1−r)(n−r)! and r!=r(r−1)! gives 1/(r!(n+1−r)!)=(1/(n+1))(1/(r!(n−r)!)+1/((r−1)!(n+1−r)!)). For r=0 or r=n+1, omit the summand whose index is outside 0,…,n; the remaining summand equals the left side by 0!=1. Splitting the finite sum for Sn+1 with this identity and reindexing its second part gives Sn+1=(iSn−iSn)/(n+1)=0 for every n≥0, while S0=1. The factorials and n+1 are nonzero by [F2], so these divisions are defined. Hence the product series has constant coefficient 1 and every positive-degree coefficient 0, so e−izeiz=1.

1.3F2F4F6algebra

Fix a positive input box I=[a,b] and choose M with M+2≥2b. For every z∈I, the first omitted absolute exponential-series term is at most bM+1/(M+1)!. Each subsequent ratio is at most b/(M+2)≤1/2, so the sum of the omitted terms is at most 2bM+1/(M+1)!=EM. Thus the real and imaginary parts of TM(z), enlarged by [−EM,EM], contain cos⁡z and sin⁡z. Exact rational interval addition and multiplication enclose pointwise addition and multiplication by their endpoint-hull formulas; inversion of a positive interval [a,b] is [1/b,1/a]. Evaluation over the finite graph and outward rounding therefore preserve containment of every graph value on I.

1.4givenF4choose

Each cap bounds its corresponding loop or finite expansion. The Laurent graph for fixed ℓ is finite; each rational operation and each terminating represented-point interface call takes finite work, and every cap exhaustion has an explicit status return. Therefore every capped execution terminates, including one that never obtains a positive point box or never passes its requested tests.

1.5givenF2F3F4F8F10choosealgebra

Fix a positive represented point z0. By [F10], choose a sufficiently large index q0 that 2−q0<z0/4. Write Iq0=[a0,b0]; the interface width bound gives a0>0. For all q≥q0, Iq⊂[a0/2,b0+1]=:[a,B]⊂(0,∞), because Iq contains z0, has width at most 2−q0, and z0−2−q0>a0/2 while z0+2−q0<b0+1. Thus M=q eventually and EM≤2Bq+1/(q+1)!→0, since the successive ratios are eventually at most 1/2. Put δq=width⁡(Iq). For m≥1, positivity of Iq=[aq,bq] makes repeated complex-rectangle multiplication of iIq exact along one coordinate: its mth power is im[aqm,bqm] up to the sign absorbed by the interval endpoints. Therefore width⁡(Iqm)=bqm−aqm≤mBm−1δq by factoring the difference of powers. Each degree-m term in TM(Iq) has component width at most δqBm−1/(m−1)!, so each component of the polynomial interval has width at most δq∑m=1MBm−1/(m−1)!. The terms of the last series are nonzero because B>0, and their successive ratio is B/m→0; [F8] makes its sum finite, say CB. With the Taylor tail added, the sine and cosine enclosure widths tend to zero.

2.1F1F3F4step 1.1step 1.2algebrainduction

Multiplying the unscaled expressions of step 1.1 by the scales in Scaled spherical Bessel evaluation on positive real arguments gives the four scaled identities. For Hℓsc and DHℓsc, cancel e−izeiz by step 1.2; the derivative scale acts on the original hℓ(1) derivative as defined, not on Hℓsc. Starting with C0=−iz−1, if Cℓ has exponents −1,…,−ℓ−1, then ℓz−1Cℓ and Cℓ′ have exponents at least −ℓ−2, while −iCℓ has exponents at least −ℓ−1; induction gives only exponents −1,…,−ℓ−1 in Cℓ. Thus zℓ+1Cℓ is a polynomial and zℓ+1(Cℓ′+iCℓ) has at worst exponent −1, which is well-defined on the positive input domain.

2.2F4F10step 1.5algebrainduction

The remaining expression graph is fixed and finite. On [a,B], each reciprocal has width δq/(aqbq)≤δq/a2 and is bounded. Sums of bounded intervals with widths tending to zero also have widths tending to zero; for bounded real intervals X,Y, width⁡(XY)≤sup⁡∣X∣width⁡(Y)+sup⁡∣Y∣width⁡(X). Applying these bounds to the real component operations in each complex rectangle proves by induction that every output rectangle or interval has width tending to zero. By [F10], the outward rounding increment 21−q also tends to zero, so all output radii tend to zero. Thus every positive rational absolute tolerance is passed at some finite stage, and the finite work through that stage gives finite sufficient caps. Constant interfaces prove the exact positive-dyadic case as well.

3.1F4F9step 1.1step 2.1step 1.3algebrainduction

By steps 1.1 and 2.1 the graph roots are the exact requested targets. Step 1.3 encloses the sine and cosine leaves for every admissible argument; induction through the finite graph, using the interval formulas in [F4], encloses every sum, product, reciprocal, and requested output value. Outward endpoint rounding preserves these containments. The success-radius and relative-flag tests are precisely the contract in [F9], so every finite success is sound for point or whole-interval input.

4.1F9step 3.1step 2.2algebra∎

If a requested target has value v≠0, any sound enclosure B of step 3.1 has diameter 2rad⁡(B) and contains v, so d(B,0)≥∣v∣−2rad⁡(B). Eventually d(B,0)≥∣v∣/2>0 and rad⁡(B)/d(B,0)→0, giving any requested positive rational relative tolerance along the refined sequence. The algorithm may already have stopped with absolute success before that relative test passes. To obtain the relative flag, rerun with successively smaller positive rational absolute tolerances and increasing finite caps: step 2.2 supplies a successful finite capped run for each such tolerance, and its returned radius is bounded by that tolerance. For a finite list of nonzero requested targets, sufficiently small absolute tolerances make all requested relative flags true. If v=0, every sound enclosure contains zero and no relative flag is available. A fixed interval input encloses its whole image, which may have positive diameter, so the point-mode limiting argument asserts no arbitrary-radius guarantee there.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite certified evaluation of the scaled complementary error function

Definition

Input an exact dyadic real x, positive rational absolute tolerance εa, optional positive rational relative tolerance εr, a fixed exponent range (emin⁡,emax⁡), and finite positive integer caps Smax⁡,Dmax⁡,Nmax⁡,pmax⁡,W, with pmax⁡≥2. Here Smax⁡ bounds stages, Dmax⁡ the continued-fraction depth, Nmax⁡ the asymptotic truncation, pmax⁡ the dyadic precision, and W the exact interval operations, terms, and subroutine calls performed by this procedure. Reject nonpositive tolerances, invalid caps, emin⁡>emax⁡, or x≤0 with domain. At stage q, use D=min⁡(q,Dmax⁡), N=min⁡(q,Nmax⁡), precision p=max⁡(2,min⁡(q,pmax⁡)), and truncation index q for elementary-range calls. Outputs and status meanings are those of Certified special-function enclosure result.

All rational and interval arithmetic uses Outward-rounded interval, rectangle and ball operations. Charge every exact arithmetic operation, range-evaluator node operation, term, rounding, and loop update to the shared counter W. Interrupt any nested elementary-range call with inconclusive before its next operation would exceed the remaining budget. Return inconclusive before any local operation would exceed W. A final endpoint overflow returns range. A non-finite required arithmetic operation or elementary-range subcall makes its candidate unavailable; if neither candidate can be formed, the current stage is inconclusive and later stages may still be tried.

For each stage q=1,…,Smax⁡, compute Pq, a finite dyadic interval containing π, as follows. For r∈{1/5,1/239}, form Aq(r)=∑j=0q−1(−1)jr2j+1/(2j+1) and Eq(r)=r2q+1/((2q+1)(1−r2)) exactly; the arctangent-tail lemma gives arctan⁡r∈[Aq(r)−Eq(r),Aq(r)+Eq(r)]. Combine these intervals by π=16arctan⁡(1/5)−4arctan⁡(1/239) and round outward to the selected format at precision p. The Machin identity is supplied by Trigonometric identities, Machin's formula and reduced Taylor tails; the arctangent series and its geometric-tail argument are in Range-reduced exponential, logarithm and arctangent series have certified tails. If the interval conversion is non-finite, skip the stage. Call Certified range evaluation for elementary real and complex functions on the real one-node DAG Pq, using exponent range (emin⁡,emax⁡), precision p, and truncation q. If this call is not finite or its output Rq does not have positive lower endpoint, skip this stage. Otherwise Rq encloses π by Certified real elementary-function evaluations enclose their ranges.

Form the continued-fraction candidate as follows. Set ID=[0,(D+1)/(2x)]. For k=D−1,…,0, set Ik=[k+12(x+sup⁡Ik+1),k+12(x+inf⁡Ik+1)] by outward interval operations. Apply the decreasing map r↦1/(π(x+r)) to I0, evaluating π with Rq, to obtain a candidate interval Cq. The finite-depth containment lemma Finite continued-fraction enclosure for erfcx makes Cq a valid enclosure for every D≥0.

If N≥1, also form the asymptotic candidate. Evaluate SN(x)=1xπ∑k=0N−1(−1)k(1/2)kx2k,BN(x)=1xπ(2N)!4NN!x2N with Rq and exact factorial/rational interval arithmetic. Write the computed sum interval as [s−,s+] and take a nonnegative outward upper endpoint bN for the remainder bound BN. Set Aq=[s−,s++bN] when N is even and Aq=[s−−bN,s+] when N is odd, rounding the final endpoints outward. The signed remainder lemma Signed finite remainder for the erfcx asymptotic expansion makes Aq a valid enclosure.

If both candidates are finite, intersect them when the intersection is nonempty; if it is empty, discard the stage. If only one is finite, use that candidate. Round the selected final endpoints outward to precision p. Return success at the first stage whose radius is at most εa. Set relative_certified to true exactly when relative tolerance was requested, the interval excludes zero, and the contract's relative-radius test passes; a failed relative test does not prevent absolute success. If no stage passes before the finite caps, return inconclusive.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite certified evaluation of Gamma and a continued log-Gamma branch

Definition

Inputs are a finite dyadic complex rectangle Z, a target label Γ or ℓ, positive exact rational absolute tolerance εa, optional positive exact rational relative tolerance εr, a fixed dyadic format precision p0≥2 and exponent range (emin⁡,emax⁡), fixed integer elementary truncation index K≥1, finite positive integer work cap W, and integer precision cap P≥p0. The target ℓ is the continued branch in Log-Gamma branch on the slit plane. Reject an invalid target, empty or non-finite rectangle, invalid cap, tolerance, emin⁡>emax⁡, or Z⊄C∖(−∞,0] by returning domain. The exact rectangle test is: either the imaginary interval omits zero, or the real interval has strictly positive lower endpoint. Reject a selected shift m>W with inconclusive. For stages q=1,…,P, use precision p=max⁡(p0,q), harmonic cutoff M=min⁡(q,W), product cutoff N=min⁡(q,W), and the fixed elementary truncation index K for every range call. Output statuses are those of Certified special-function enclosure result.

Use Outward-rounded interval, rectangle and ball operations for interval arithmetic. Count every exact rational or interval operation, series term, elementary-DAG node operation, rounding, and loop update against the shared counter W. If the procedure is called by another evaluator, charge these same primitive operations directly to that caller's shared counter, so the caller can debit the actual child work; the local cap W remains in force. Before any operation would exceed either remaining cap, return inconclusive. Run nested elementary-range calls under that counter and interrupt a subcall before its next operation would exceed the remaining budget. A non-finite range result makes the current candidate unavailable. Any other non-finite arithmetic result also makes its candidate unavailable; no partial result is retained. Final output overflow returns range.

Choose m=max⁡ ⁣(0,⌊−inf⁡ℜZ⌋+1), which is the least integer m≥0 for which inf⁡ℜZ+m>0. Compute this exactly from the rational endpoint; if m>W, return inconclusive before forming the shift. Put WZ=Z+m. Before every elementary-range call, round each rational interval input outward to the stage dyadic format; a conversion overflow makes that candidate unavailable. A rounded Log input must still avoid the cut; otherwise its candidate is unavailable, rather than declaring the valid original input outside the domain. Embed a real interval X explicitly as X+i[0,0] whenever it participates in a complex expression; real and complex DAG nodes retain their declared types. All elementary-range calls below use Certified range evaluation for elementary real and complex functions with the selected format and fixed truncation K; its soundness for finite real and complex results is given by Certified real elementary-function evaluations enclose their ranges and Certified complex exponential and principal logarithm enclose their ranges.

For stage q=1,…,P, form a dyadic Pq containing π: for r∈{1/5,1/239}, compute exactly Aq(r)=∑k=0q−1(−1)kr2k+1/(2k+1) and Eq(r)=r2q+1/((2q+1)(1−r2)), combine arctan⁡r∈[Aq(r)−Eq(r),Aq(r)+Eq(r)] by the Machin identity, and round outward to the selected format at precision p. The arctangent series and its geometric-tail argument are in Range-reduced exponential, logarithm and arctangent series have certified tails, and the Machin identity is in Trigonometric identities, Machin's formula and reduced Taylor tails. If the conversion is non-finite or inf⁡Pq≤0, the Stirling candidate is unavailable for this stage; still attempt the product candidate. Let aW=inf⁡ℜWZ>0 and RW=max⁡(∣inf⁡ℜWZ∣,∣sup⁡ℜWZ∣)+max⁡(∣inf⁡ℑWZ∣,∣sup⁡ℑWZ∣); RW is an exact rational upper bound for ∣w∣ on WZ.

For this stage, call the elementary-range algorithm on the principal-log node for each Z+j, j=0,…,m−1. Translation by these real integers preserves avoidance of the cut for an input rectangle wholly in the slit plane; if the exact branch-domain check fails, return domain. Add the finite log rectangles to form the shift sum LZ. If any shift-log call is non-finite, discard this whole stage, since both candidates require that sum. When m=0, set LZ=0.

Construct the product candidate when N>2RW. Call the elementary-range algorithm for the real node log⁡M to get an enclosure GM=[g−,g+] of log⁡M, and form the outward-rounded Euler-constant interval [HM−g+−1/M, HM−g−], where HM=∑k=1M1/k exactly. Call the complex principal-log node on WZ and on each 1+WZ/k for 1≤k≤N. All these rectangles lie in the principal-log domain since ℜWZ>0. Form QN=−Log⁡(WZ)−γEWZ+∑k=1N(WZk−Log⁡(1+WZ/k)) by interval arithmetic, and enlarge each coordinate by an outward upper bound for TN=RW22N(1−RW/(N+1)). The tail lemma Explicit tail bound for the log-Gamma product series proves that this is an enclosure of ℓ+(WZ), uniformly on the whole rectangle. Subtract LZ to get the shifted product candidate for ℓ(Z). If N≤2RW or a required call is non-finite, this candidate is unavailable at this stage.

Construct the Stirling candidate only when Pq is finite and has positive lower endpoint. Call the complex principal-log node on WZ, and the real logarithm node on 2Pq. Form SZ=(WZ−12)Log⁡(WZ)−WZ+12log⁡(2π)−LZ by interval arithmetic and enlarge each coordinate by an outward upper bound for 1/(4aW). The sectorial remainder lemma Explicit sectorial remainder bound for the right-half-plane log-Gamma branch proves this is an enclosure of ℓ(Z). If a required call is non-finite, this candidate is unavailable at this stage.

If both log candidates are available, intersect them when the intersection is nonempty; if empty, discard the stage. If only one is available, select it. For target ℓ, use this rectangle as the output candidate. For target Γ, call the complex exponential node on the selected rectangle and use its finite output rectangle. Return success at the first stage whose candidate radius is at most εa. Set relative_certified to true exactly when relative tolerance was requested, zero is excluded, and the contract's relative-radius test passes; a failed relative test does not prevent absolute success. If no stage succeeds before its caps, return inconclusive.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Positive discrete Wronskian for the Bessel recurrence pair

Statement

Fix real x>0, let Jj(x) be as in Integer-order Bessel series, and let Dj(x) be as in Complementary solution for the integer Bessel recurrence. Define Wj=JjDj+1−Jj+1Dj. Then

Wj=J0(x)2+J1(x)2>0(j≥0).

In particular, the two recurrence solutions J and D are linearly independent.

Facts & Assumptions

Given: A real x>0 and the two sequences in the statement.

[F1]

Bessel differential equations and order recurrences from the series: Jj+1=2jJj/x−Jj−1 for j≥1.

[F2]

Complementary solution for the integer Bessel recurrence: Dj+1=2jDj/x−Dj−1, with D0=−J1 and D1=J0.

[F3]

Adjacent initial Bessel values cannot both vanish: J0(x) and J1(x) are not both zero.

Proof

1.1F1F2algebra

Using [F1] and [F2], for j≥1, Wj=Jj(2jxDj−Dj−1)−(2jxJj−Jj−1)Dj=Jj−1Dj−JjDj−1=Wj−1. Thus Wj=W0 for every j≥0.

2.1F2F3step 1.1algebra∎

By [F2], W0=J0D1−J1D0=J02+J12, which is strictly positive by [F3]. A linear dependence of J,D would make every 2×2 determinant Wj zero, so they are independent.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Exact finite terminal-contamination error identity

Statement

Fix x>0, integers N≥1 and 0≤n<N, and define the exact backward recurrence solution by uN+1=0, uN=1, and uj−1=2juj/x−uj+1 for j=N,N−1,…,1. If DN+1(x)≠0, put q=−JN+1(x)/DN+1(x). For s∈{0,1} such that Js(x)+qDs(x)≠0, let J^n=Js(x)un/us. Then

J^n−Jn=q(JsDn−JnDs)Js+qDs.

This is a finite-N identity. It makes no assertion that the error tends to zero as N→∞.

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: Jj+1=2jJj/x−Jj−1 for j≥1.

Proof

1.1givenF1F2F3algebra

The sequence J+qD obeys the recurrence by [F2] and [F3], and its value at N+1 is JN+1+qDN+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+qD)/(JN+qDN) has the terminal values uN+1=0,uN=1. Uniqueness of the finite backward recurrence gives uj=(Jj+qDj)/(JN+qDN) for 0≤j≤N.

2.1givenstep 1.1algebra∎

Taking the ratio of the formulas for j=n and j=s yields un/us=(Jn+qDn)/(Js+qDs). Since the denominator is nonzero, J^n−Jn=JsJn+qDnJs+qDs−Jn=q(JsDn−JnDs)Js+qDs. Every equality uses this single fixed terminal index N; no limit in N was taken.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Soundness of finite erfcx enclosures

Statement

If Finite certified evaluation of the scaled complementary error function returns success for x>0, the returned interval contains erfcx⁡(x) and passes the requested absolute-radius test. A relative certificate is valid only if zero is outside the interval and the required relative-radius test passes. Every execution with finite caps terminates. The theorem makes no success claim for every x or finite cap.

Facts & Assumptions

Given: A successful return of Finite certified evaluation of the scaled complementary error function using a sound outward interval backend.

[A1]

Certified range evaluation for elementary real and complex functions supplies the finite square-root range call for π; soundness of that real output is given by Certified real elementary-function evaluations enclose their ranges.

[F1]

Finite continued-fraction enclosure for erfcx: each finite-depth continued- fraction interval contains the exact value.

[F2]

Signed finite remainder for the erfcx asymptotic expansion: the signed remainder interval formed from SN and BN contains the exact value for every N≥1.

[F3]

Certified special-function enclosure result: success requires a containing interval with the stated radius, and relative certification requires zero exclusion.

[F4]

Finite certified evaluation of the scaled complementary error function: the procedure returns only a continued-fraction or signed-asymptotic candidate, or a non-success status.

Proof

1.1A1A2F1F2F4

A continued-fraction candidate contains erfcx⁡(x) by [F1]. A signed-asymptotic candidate contains it by [F2]. The interval evaluation of each candidate is sound by [A1] and [A2]. If the procedure intersects both candidates, their common exact value lies in the intersection; otherwise the selected candidate remains an enclosure.

2.1A2F4step 1.1

Thus every interval eligible for a success return contains the exact value. The finite arithmetic and intersection preserve inclusion by [A2] and the algorithm definition.

3.1F3F4step 2.1

The procedure reports success only after the absolute-radius test in [F3] and sets a relative flag only when the interval excludes zero and the relative-radius test passes. No width decrease or universal success is asserted.

4.1givenA1step 2.1step 3.1algebra∎

The stage loop is finite; every continued-fraction recurrence, asymptotic sum, and elementary-range call has finite depth, terms, format, and truncation index. Every branch has an explicit return or proceeds to the next bounded stage. Therefore every capped execution terminates.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Soundness of finite Gamma and log-Gamma enclosures

Statement

For any finite dyadic rectangle Z⊂C∖(−∞,0], whenever Finite certified evaluation of Gamma and a continued log-Gamma branch returns success, its returned enclosure contains the range of Γ or the continued log-Gamma branch on Z and passes the stated absolute-radius test. Any reported relative certificate obeys the zero-exclusion condition. No assertion guarantees success for every valid input or finite cap. Every execution with finite caps terminates.

Facts & Assumptions

Given: An invocation of Finite certified evaluation of Gamma and a continued log-Gamma branch with its finite backend parameters. For enclosure assertions assume a successful return on the admissible rectangle Z; termination also covers unsuccessful returns.

[F1]

Finite certified evaluation of Gamma and a continued log-Gamma branch defines the least admissible shift, finite stage cutoffs, actual outward range calls, coordinate enlargements, candidate intersection, final tests and shared work accounting. This is a procedure specification; its correctness is proved below.

[F2]

Certified special-function enclosure result defines the absolute-radius and optional zero-excluding relative-radius tests and the four statuses.

[F3]

Shifted log-Gamma branch is independent of the shift: on the slit plane, for an admissible shift m, ℓ(z)=ℓ+(z+m)−∑j=0m−1Log⁡(z+j),exp⁡(ℓ(z))=Γ(z).

[F4]

Weierstrass log-Gamma series converges to a Gamma branch: the series defining ℓ+ converges to the stated Gamma branch on the right half-plane. Explicit tail bound for the log-Gamma product series bounds the tail after N>2∣w∣ terms by ∣w∣2/(2N(1−∣w∣/(N+1))).

[F5]

Finite harmonic enclosure for Euler's constant: HM−log⁡M−1/M≤γE≤HM−log⁡M for M≥1.

[F6]

Explicit sectorial remainder bound for the right-half-plane log-Gamma branch: on the right half-plane, ℓ+(w)=(w−12)Log⁡w−w+12log⁡(2π)+R(w),∣R(w)∣≤14∣w∣.

[F7]

Trigonometric identities, Machin's formula and reduced Taylor tails gives the Machin identity; Range-reduced exponential, logarithm and arctangent series have certified tails supplies the arctangent series and its geometric-tail argument used to enclose π.

Proof

1.1givenA1F1F3algebra

For an admissible Z, the prescribed integer shift has aW=inf⁡ℜ(Z+m)>0. Translation by a nonnegative real integer preserves cut avoidance: either the imaginary interval omits zero, or the real lower endpoint was already positive. Thus each exact shift-log input lies in its domain. Its outward-rounded input still contains it; the procedure uses it only when its domain is certified. By [A1] the finite shift-log sum therefore encloses the corresponding exact sum for every z∈Z. An unavailable shift sum discards the stage and makes no finite claim.

2.1A1F1F3F4F5step 1.1algebra

Write GM=[g−,g+] for a finite enclosure of log⁡M. By [F5], the outward interval [HM−g+−1/M,HM−g−] contains the same real γE used by the branch. Finite outward evaluation of the product partial sum contains its exact value for every w∈WZ=Z+m by [A1]. The exact rational majorant RW bounds ∣w∣ uniformly. When N>2RW, [F4] bounds each omitted tail by TN=RW2/(2N(1−RW/(N+1))), since the numerator increases and the positive denominator decreases when ∣w∣ is replaced by RW. The disk of radius TN lies in the coordinate rectangle used to enlarge the partial sum. Subtracting the containing shift-log sum from step 1.1 and using [F3] therefore encloses the full range ℓ(Z).

2.2A1F1F3F6F7step 1.1algebra

The finite Machin construction encloses π by [F7]. Its positive interval and the real logarithm call enclose log⁡(2π) by [A1]. Therefore the finite Stirling main-term rectangle contains the exact main term on WZ. For every such w, ∣w∣≥ℜw≥aW>0, so [F6] gives ∣R(w)∣≤1/(4aW). The prescribed outward coordinate enlargement contains this remainder disk. After subtracting the shift-log sum, [F3] again identifies every enclosed target as the same continued branch ℓ(z).

3.1A1F1F3step 2.1step 2.2algebra

Either finite log candidate thus contains ℓ(Z); their coordinatewise intersection, when used, also contains that entire image. For a log-Gamma target the selected rectangle is therefore sound. For a Gamma target, [A1] encloses its complex exponential, and [F3] identifies the pointwise exponential with Γ(z) for every z∈Z. This proves full input-range containment for both targets.

4.1F1F2step 3.1

The algorithm tests the actual final outward rectangle's radius before reporting success, and sets a relative flag only when the rectangle excludes zero and passes the relative-radius test. By [F2] these arithmetic tests give exactly the stated absolute and reported relative guarantees. Failed or non-finite candidates are never used for a finite success claim; no finite-cap success guarantee is asserted.

5.1givenF1A1algebra∎

The stage count is at most P, each shift and sum has a finite integer cutoff, and each elementary call has a finite DAG, format and truncation index. Moreover [F1] checks the shared counter before each local or nested primitive operation and loop update, interrupting before either its own or its caller's remaining work budget is exceeded. Consequently every capped invocation returns a status after finitely many charged operations, whether or not any candidate succeeds.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite certified Airy power-series evaluation

Definition

Inputs are an exact dyadic complex number z, a target label Ai⁡ or Bi⁡, positive rational absolute tolerance εa, optional positive rational relative tolerance εr, a fixed exponent range (emin⁡,emax⁡), and finite positive integer caps Smax⁡,Kmax⁡,pmax⁡,W, with pmax⁡≥2. Reject an invalid target, nonpositive tolerance, invalid cap, or emin⁡>emax⁡ with domain. The caps bound stages, terms per residue class, final dyadic precision, and this procedure's exact interval operations and subroutine calls. For stage q, use K=q, precision p=max⁡(2,min⁡(q,pmax⁡)), and elementary-range truncation q; run stages q=1,…,min⁡(Smax⁡,Kmax⁡,pmax⁡).

All arithmetic uses Outward-rounded interval, rectangle and ball operations. Charge every exact arithmetic operation, range-evaluator node operation, term, rounding, and loop update to the shared counter W. Return inconclusive before any local operation would exceed W; interrupt a nested subroutine before its next operation would exceed the remaining budget. When nested, charge every Gamma-child operation directly to the same parent counter. If a final endpoint overflows, return range. A non-finite subroutine result makes the current stage unavailable; later stages may still be tried. Other statuses are those of Certified special-function enclosure result.

At stage q, outward-round the rational points 1/3 and 2/3 to dyadic intervals Gr at precision p. For each r∈{1/3,2/3}, charge the child-call operation to the parent counter. If no parent work unit remains, return inconclusive; otherwise pass the positive remaining budget as the child's work cap and call Finite certified evaluation of Gamma and a continued log-Gamma branch with target Γ on the complex rectangle Gr+i[0,0], absolute tolerance 2−q, no relative tolerance, the same exponent range, fixed elementary truncation q, base format precision p, and precision cap p. Debit every operation performed by the Gamma child to the shared counter. A success output rectangle encloses Γ(r), since r∈Gr; if either call is not successful or its rectangle contains zero, skip this stage. A failed rational-to-dyadic conversion also makes the stage unavailable.

For s∈{1/6,1/3,2/3}, outward-round the exact rational s to a dyadic interval Ss,p, then call Certified range evaluation for elementary real and complex functions on the real expression DAG exp⁡(Ss,plog⁡3), with the same exponent range, precision p, and truncation q. These calls enclose 3s by Certified real elementary-function evaluations enclose their ranges. If a required call is not finite, skip this stage. Embed each real power interval X as the complex rectangle X+i[0,0], then combine it with the Gamma rectangles by outward complex arithmetic to enclose the two initial coefficients (a0,a1)=(132/3Γ(2/3),−131/3Γ(1/3)) for Ai, or (b0,b1)=(131/6Γ(2/3),31/6Γ(1/3)) for Bi. If any divisor rectangle contains zero or an arithmetic operation is non-finite, skip this stage.

Put R=∣ℜz∣+∣ℑz∣, an exact rational upper bound for ∣z∣. For each residue r=0,1, initialize tr,0=crzr using the selected pair of initial coefficient intervals, and compute tr,j+1=tr,jz3/((k+3)(k+2)), where k=r+3j. Sum tr,0,…,tr,q−1 and compute the first omitted term tr,q. Put kr=r+3q and ρr=R3(kr+3)(kr+2). If either ρr≥1, skip the stage. Otherwise enlarge the sum of both partial rectangles in each coordinate by M(t0,q)1−ρ0+M(t1,q)1−ρ1, where M(X+iY)=max⁡(∣inf⁡X∣,∣sup⁡X∣)+max⁡(∣inf⁡Y∣,∣sup⁡Y∣). The c2=0 residue class vanishes identically. The ratio and tail lemma Airy coefficient recurrence and geometric tail bounds therefore makes the enlarged rectangle a containing enclosure of the selected Airy value.

Round the final rectangle outward to precision p. Return success at the first stage whose radius is at most εa. Set relative_certified to true exactly when relative tolerance was requested, zero is excluded, and the contract's relative-radius test passes; a failed relative test does not prevent absolute success. If no stage succeeds before its caps, return inconclusive. The stage loop and all subroutine calls have finite caps, so the procedure has no unbounded retry.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite certified evaluation of integer-order Bessel functions

Definition

Inputs are an integer order n≥0, a target J or scaled I, an exact dyadic real argument x, positive rational absolute tolerance εa, optional positive rational relative tolerance εr, a fixed exponent range (emin⁡,emax⁡), and finite positive integer caps Smax⁡,Kmax⁡,Mmax⁡,Nmax⁡,pmax⁡,W, with pmax⁡≥2. The caps bound stages, series terms, midpoint panels, Miller terminal index, dyadic precision, and exact interval operations and subroutine calls. Run stages q=1,…,min⁡(Smax⁡,Kmax⁡,Mmax⁡,pmax⁡), using precision p=max⁡(2,min⁡(q,pmax⁡)) and elementary-range truncation q. Reject n<0, invalid targets, nonpositive tolerances, invalid caps, emin⁡>emax⁡, or x<0 for the scaled I target with domain. The scaled target is always e−xIn(x).

Use exact rational endpoint formulas for finite sums, recurrences, mesh construction, and error bounds, with final endpoints rounded outward to precision p. Charge every exact arithmetic operation, range-evaluator node operation, term, rounding, and loop update to the shared counter W; interrupt a nested elementary-range call with inconclusive before its next operation would exceed the remaining budget. Factorials use The factorial n! and the falling factorial nk‾, defined by recursion in N. Call Certified range evaluation for elementary real and complex functions for the real sine, cosine, and exponential ranges in the midpoint branches. Before each call, round its rational input interval outward to the fixed dyadic format. Its finite outputs are sound by Certified real elementary-function evaluations enclose their ranges. Count each exact operation, term, rounding, and subroutine call against W; if the next operation would exceed W, return inconclusive. A non-finite elementary call makes that branch unavailable. Final endpoint overflow returns range.

Before forming any stage constant, if the target is scaled I and x=0, use precision p=pmax⁡, form [1,1] when n=0 and [0,0] when n≥1, and round its endpoints outward to that format. If either conversion overflows, return range. Otherwise apply the ordinary absolute-radius test to the rounded candidate and return success only when it passes; return inconclusive if it fails. The internal exact candidate has radius zero, but outward underflow may widen its stored enclosure. Set a successful output's relative flag by the usual contract test, so a zero value receives no relative certificate. This branch does not call the elementary-range procedure.

At stage q, compute an outward dyadic interval Pq containing π from the Machin identity in Trigonometric identities, Machin's formula and reduced Taylor tails and the arctangent series and geometric-tail argument in Range-reduced exponential, logarithm and arctangent series have certified tails: for r=1/5,1/239, use the q-term sum ∑k=0q−1(−1)kr2k+1/(2k+1) and tail bound r2q+1/((2q+1)(1−r2)), combine by 16arctan⁡(1/5)−4arctan⁡(1/239), and round outward to the selected format at precision p. If this conversion is non-finite or inf⁡Pq≤0, midpoint candidates are unavailable at this stage; the independent power-series or Miller branches may still be attempted.

For a requested Jn(x), first form a power-series candidate if possible. For any order r≥0, define the finite subroutine SeriesJ⁡(r): start a0=(x/2)r/(0! r!) and ak+1=−akx2/(4(k+1)(k+r+1)); sum a0,…,aq−1, and compute aq. Put ρJ=x24(q+1)(q+r+1). If ρJ<1, return the interval partial sum enlarged by ∣aq∣/(1−ρJ); otherwise mark this series candidate unavailable. The tail lemma Convergence and geometric tails for integer-order Bessel series proves each returned interval contains Jr(x).

For each requested target, also form its composite-midpoint candidate with m=q panels, h=π/m, and nodes θj=(j+12)h, j=0,…,m−1, using Pq as the interval for π. For Jn, call the elementary-range algorithm for sin⁡(θj), form nθj−xsin⁡(θj), and call it for the cosine of that interval. Let Qm be the interval sum h∑jcos⁡(nθj−xsin⁡θj) and EJ=hmax⁡224((n+∣x∣)2+∣x∣),hmax⁡=sup⁡(Pq)/m. The candidate is Qm/Pq+[−EJ,EJ]. For the scaled I target with x>0, call the elementary algorithm for cos⁡(θj), form −x(1−cos⁡(θj)), call it for the real exponential, and also call it for cos⁡(nθj). Let Qm be h times the sum of the resulting product intervals and set EI=hmax⁡224((x+n)2+x). The candidate is Qm/Pq+[−EI,EI]. The exact midpoint error lemmas Composite midpoint enclosure for the integer Bessel integral and Composite midpoint enclosure for the scaled modified Bessel integral prove these inclusions.

For a Jn(x) Miller candidate, only try this branch when x>0 and N=n+q≤Nmax⁡. Obtain SeriesJ⁡(r) for r=0,1,n,N+1; if any is unavailable, skip the branch. Set D0=−J1, D1=J0, and compute Dj+1=2jDj/x−Dj−1 for j=1,…,N by interval arithmetic. Set uN+1=0,uN=1, and compute uj−1=2juj/x−uj+1 for j=N,…,1. Require the intervals for DN+1, us, Js, and Js+qMDs below to exclude zero. Form qM=−JN+1/DN+1. In the fixed order s=0,1, whenever the tests pass, form J^n=Jsunus,EM=qM(JsDn−JnDs)Js+qMDs,CM=J^n−EM. The exact finite terminal-contamination identity Exact finite terminal-contamination error identity proves that each CM contains Jn(x). The Casoratian lemma Positive discrete Wronskian for the Bessel recurrence pair supplies the independent recurrence pair and positive invariant required by that identity.

Collect every available sound candidate for the requested target. If more than one is available, intersect them; if the intersection is empty, discard this stage. Round the selected candidate outward to precision p. Return success at the first stage whose radius is at most εa. Set relative_certified to true exactly when relative tolerance was requested, zero is excluded, and the contract's relative-radius test passes; failure of that relative test does not prevent absolute success. If no stage succeeds before the finite caps, return inconclusive. No branch uses a limit in the Miller terminal index.

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Soundness of finite Airy enclosures

Statement

Whenever Finite certified Airy power-series evaluation returns success, its interval or rectangle contains the requested Ai⁡(z) or Bi⁡(z) and passes the requested absolute-radius test. Any relative certificate is valid only when the enclosure excludes zero. No success guarantee is made for every input or finite cap. Every execution with finite caps terminates.

Facts & Assumptions

Given: A successful return of Finite certified Airy power-series evaluation using a sound outward interval backend.

[A1]

Finite certified evaluation of Gamma and a continued log-Gamma branch supplies each finite Gamma call; Soundness of finite Gamma and log-Gamma enclosures says every successful call encloses the Gamma range on its dyadic input rectangle.

[A2]

Certified range evaluation for elementary real and complex functions supplies the finite real range calls for powers of 3; their finite outputs are sound by Certified real elementary-function evaluations enclose their ranges.

[F1]

Certified special-function enclosure result: the success radius test and zero-exclusion rule have the stated absolute and relative meanings.

[F2]

Airy coefficient recurrence and geometric tail bounds: each nonzero Airy residue-class tail is bounded by its first omitted term divided by 1−q when all later ratios are at most q<1.

[F3]

Finite certified Airy power-series evaluation: the procedure encloses the two partial sums, adds the two residue-class tail bounds, and uses certified Gamma initial constants.

Proof

1.1A1A2A3F3

The initial coefficients used by the procedure enclose the exact Airy initial values by [F3], with the Gamma calls sound by [A1] and the powers of 3 sound by [A2]. Arithmetic on these intervals is sound by [A3]. The series definition and recurrence determine all later coefficients, so the resulting partial sums enclose their exact finite series sums.

2.1A3F2F3step 1.1

By [F2], each nonzero omitted class tail lies within its computed absolute tail bound; the third class is zero because its initial coefficient is zero and the recurrence preserves zero. Therefore the exact Airy value lies in the partial-sum enclosure enlarged by the sum of the two tail bounds.

3.1A1A2F1F3step 2.1

The algorithm reports success only when that containing rectangle passes the requested radius test, and its relative flag obeys [F1]. If its finite caps do not allow that test, the result is inconclusive; no claim of universal success is made.

4.1givenA1A2step 3.1algebra∎

The stage loop is finite; each Gamma and elementary-range subcall has finite caps and truncation index, and the residue-class sums have at most the capped number of terms. Thus every capped execution terminates.

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Soundness of finite integer-order Bessel enclosures

Statement

Whenever Finite certified evaluation of integer-order Bessel functions returns success, its interval contains the requested Jn(x) or scaled modified Bessel value e−xIn(x) and passes the requested absolute-radius test. A relative certificate is valid only when zero is outside the interval and the relative test passes. The theorem makes no success guarantee for every input or finite cap. Every execution with finite caps terminates.

Facts & Assumptions

Given: A successful return of Finite certified evaluation of integer-order Bessel functions with a sound outward interval backend.

[A1]

Certified range evaluation for elementary real and complex functions supplies the finite real sine, cosine, and exponential ranges used by the midpoint branches; every finite range is sound by Certified real elementary-function evaluations enclose their ranges.

[F1]

Finite certified evaluation of integer-order Bessel functions: series, midpoint, and finite Miller candidates are returned only after their respective tail or terminal-error enclosures are applied; its exact scaled-I input x=0 branch returns the corresponding defining-series value. Candidate intersections are taken only when nonempty.

[F2]

Integer-order Bessel series and Real elementary functions and their domains: at x=0, the defining series gives I0(0)=1, In(0)=0 for n≥1, and e0=1.

[F3]

Certified special-function enclosure result: success requires exact inclusion and the stated radius; relative certification requires zero exclusion.

[F4]

Convergence and geometric tails for integer-order Bessel series: the truncated Jn series tail is enclosed by the geometric bound when its ratio test passes.

[F5]

Exact finite terminal-contamination error identity: the finite normalized backward recurrence error is exactly the displayed terminal-contamination expression whenever its denominators are nonzero.

[F6]

Composite midpoint enclosure for the integer Bessel integral: the midpoint Jn enclosure contains the exact cosine integral.

[F7]

Composite midpoint enclosure for the scaled modified Bessel integral: the midpoint enclosure contains the exact scaled In integral.

[F8]

Trigonometric identities, Machin's formula and reduced Taylor tails supplies the Machin identity; the arctangent series and geometric-tail argument in Range-reduced exponential, logarithm and arctangent series have certified tails give the finite bounds used to enclose π for the midpoint mesh.

[F9]

Complementary solution for the integer Bessel recurrence: D0=−J1, D1=J0, and Dj+1=2jDj/x−Dj−1 for x>0 and j≥1.

Proof

1.1A1A2F1F2F4F6F7F8

If the target is scaled I and x=0, the defining series [F2] gives e−xIn(x)=1 for n=0 and 0 for n≥1, so the exact internal singleton contains the target; its final outward conversion also contains it by [A2], and the algorithm tests the actual rounded radius. Otherwise, the power-series candidate contains Jn(x) by [F4]. The two midpoint candidates contain their integrals by [F6] and [F7]; their finite sine, cosine, and exponential ranges are sound by [A1], their mesh constant is enclosed by [F8], and their arithmetic interval operations are sound by [A2].

1.2A2F1F4F5F9algebrainduction

For a Miller candidate, [F1] requires x>0 and N=n+q>n, and obtains J0,J1,Jn,JN+1 from successful SeriesJ⁡ calls. Their intervals contain these exact values by [F4] and [A2]. Starting from D0=−J1,D1=J0, induction through the finite recurrence [F9] and [A2] encloses every Dj required by the error formula, including Dn,Ds,DN+1. The prescribed terminal singletons and backward recurrence similarly enclose every exact uj. The procedure divides only after the interval for DN+1 excludes zero, so the resulting interval contains qM=−JN+1/DN+1. It also requires the intervals for us, Js and Js+qMDs to exclude zero. Thus the exact denominators in [F5] are nonzero, and [A2] encloses both J^n=Jsun/us and its exact error EM from [F5]. Their interval difference contains J^n−EM=Jn. Repeated occurrences of the same enclosed values may widen this difference but preserve containment; unavailable calls or failed guards discard the candidate.

2.1F1step 1.1step 1.2

Every candidate used by the procedure therefore contains the same exact requested value. Any nonempty intersection of candidates also contains that value.

3.1A1F1F3step 2.1

The algorithm reports success only after the containing interval passes the absolute-radius test and, when reported, the zero-excluding relative test in [F3]. Otherwise it returns an inconclusive or range status under finite caps. Thus the success claims are sound and do not rely on a limit as the Miller terminal index grows.

4.1givenA1step 3.1algebra∎

The stage and candidate loops are finite, every series and recurrence uses at most its stated finite index, and each elementary-range call has a finite DAG, format, and truncation index. Hence every capped execution terminates.

5 · Examples, counterexamples and false statements

None yet.

Sources