The irrationality exponent of pi and sine-denominator series

Source and verification

This article explains family 017 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Irrationality exponents

For an irrational real number $\alpha$, define

$$\mu (\alpha )=\sup \left\{\nu >0:\begin{array}{c}0<|\alpha -p/q|<q^{-\nu }\\ \text{infinitely often}\end{array}\right\},$$

where the fractions are reduced and $q\ge 2$. The pigeonhole principle gives $\mu (\alpha )\ge 2$. One places the $Q+1$ fractional parts of $0,\alpha ,\dots ,Q\alpha$ in $Q$ equal intervals; two lie within $1/Q$, giving $1\le q\le Q$ with $\Vert q\alpha \Vert \le 1/Q$. Irrationality forces unbounded denominators as $Q$ grows, and hence infinitely many approximations of order $q^{-2}$.

Here $\Vert t\Vert$ denotes distance to the nearest integer.

Claimed optimal exponent

The manuscript asserts $\mu (\pi )=2$. More precisely, for every $\nu >2$, all sufficiently large denominators satisfy

$$\left|\pi -\frac{p}{q}\right|\ge q^{-\nu }(p\in \mathbb{Z}).$$

The threshold depends on $\nu$ and is not effective in the argument. Fractions need not be reduced in this eventual inequality. (OpenAI, 2026)

This does not assert a fixed $c>0$ with $|\pi -p/q|\ge c/q^2$. That stronger property would be equivalent to bounded continued-fraction partial quotients.

🏷️ Interpolation and determinant size

The source argues by assuming infinitely many approximations at one fixed exponent $\nu >2$. It selects several denominators successively, with separated scales.

A jet is a finite list of Taylor coefficients. The interpolation problem prescribes jets after substitutions

$$\begin{array}{c}Y=1+t,\\ X_i=c_{ji}+u_i+\log (1+t).\end{array}$$

The variable $t$ follows a logarithmic curve; $u_i$ describes transverse displacement. Polynomial monomials share one weighted degree budget, so their exponent vectors fill a simplex.

The key interpolation theorem asserts full rank for the required jet map, uniformly in the allowed centers after the weights have been chosen. Having more coefficients than conditions is not enough: a structured matrix can have dependent rows regardless of its dimensions.

Writing $\mathrm{i}=\sqrt{-1}$, the chosen centers $c_{ji}=2\mathrm{i}j{p}_i/q_i$ approximate the periods $2\pi \mathrm{i}j$. Interpolation supplies a nonzero rational-complex determinant. Clearing denominators bounds its absolute value from below.

For the upper bound, translate to the exact periods and expand. Rows with the same transverse multi-index test the same entire functions. Repeated Taylor degrees then make a determinant term vanish. If many rows have small transverse index, the few available groups force a large saving through these collisions. If few do, large indices produce high powers of the small approximation errors instead.

The source develops the algebraic interpolation theorem separately; these notes use it as a stated mechanism without covering the excluded algebraic-geometry background. Its threshold independence from the centers is essential to avoid circular parameter choices.

🏷️ Why an approximation bound can imply series convergence

The Flint–Hills series is

$$\sum _{n\ge 1}\frac{1}{n^3{\sin }^{2}n},$$

with radians understood. Terms become large when $n$ is close to a multiple of $\pi$. A termwise approximation bound alone is too crude; one must also exploit separation between near-resonances.

Choose $2<\nu <5/2$. The claimed theorem, with finitely many small denominators absorbed into a constant, gives

$$\Vert q\pi \Vert \ge c{q}^{1-\nu }(q\ge 1).$$

For $K\le q<2K$, the points $q\pi$ modulo one are at least $d=c(2K{)}^{1-\nu }$ from zero and from one another. The latter follows by applying the same estimate to the difference of two indices.

Ordering distances on the two half-circles gives

$$\begin{array}{rl}\sum _{K\le q<2K}\frac{1}{q^3\Vert q\pi {\Vert }^2}& \le \frac{2}{K^3{d}^2}\sum _{j\ge 1}{j}^{-2}\\ & =O(K^{2\nu -5}).\end{array}$$

Because $2\nu -5<0$, summing over dyadic $K$ converges.

For each integer $n$, take the nearest nonnegative integer $q$ to $n/\pi$. There are at most four $n$‘s in each group, and for $q\ge 1$,

$$\begin{array}{c}n\ge \pi q/2,\\ |\sin n|\ge (2/\pi )\Vert q\pi \Vert .\end{array}$$

These inequalities compare the Flint–Hills series with the convergent distance series. This proves the implication from the claimed approximation theorem using only elementary spacing and sine concavity. It is not an independent proof of that theorem. (OpenAI, 2026)

🏷️ The generalized threshold

The source extends the argument to

$$\sum _{n\ge 1}\frac{1}{n^a|\sin n{|}^b},a,b>0,$$

claiming convergence exactly when $a>\max \{1,b\}$. For $a\le 1$, comparison with $\sum n^{-a}$ proves divergence. For $1<a\le b$, continued-fraction convergents $p_j/q_j$ satisfy $|\sin p_j|<q_j^{-1}$ and $p_j\asymp q_j$; the corresponding summands fail to tend to zero. The boundary cases therefore diverge.

For the convergent range, the same dyadic spacing estimate uses ${\sum }_{j\le K}{j}^{-b}$, producing respectively a power, a logarithm, or a bounded factor as $b<1$, $b=1$, or $b>1$.

🏷️ Key steps and proof outlines

Interpolation supplies a nonzero minor with controlled denominators

Assume infinitely many approximations at a fixed exponent $\nu >2$. Choose finitely many of them, put $w_i=\lceil \log q_i\rceil$ and $r_i=2\mathrm{i}{p}_i/q_i$, and write $w\alpha ={\sum }_i{w}_i{\alpha }_i$ for a multi-index $\alpha$. The matrix columns are monomials $Y^h{X}^{\alpha }$ with $w_{0}h+w\alpha \le H$. Its rows extract coefficients $t^s{u}^{\beta }$ after substituting

$$Y=1+t,X_i=j{r}_i+G_i(t)+u_i,$$

where $G_i$ is a finite Taylor polynomial for $\log (1+t)$. The row constraint is $v_{0}s+w\beta /\theta <H$, with $0<\theta <1$.

Proof outline of the rank interface. The source’s separated-weight interpolation theorem makes the jet map surjective once the successive weights are large enough and satisfy its volume inequalities. The weight thresholds do not depend on the centers $j{r}_i$, although the final threshold for $H$ may. Replacing the logarithm by $G_i$ preserves rank: the omitted tail starts above the transverse-variable weight, so the change is triangular and invertible on the finite truncated jet space. Select a nonzero square minor using all $M$ rows. This uses the source’s interpolation theorem as a stated input; the algebraic-geometric proof of that theorem is outside this article’s scope. (OpenAI, 2026)

Proof outline of the arithmetic lower bound. Multiply column $\alpha$ by ${\prod }_i{q}_i^{{\alpha }_i}$ and row $\beta$ by ${\prod }_i{q}_i^{-{\beta }_i}$. The rational centers become Gaussian integers, meaning numbers in $\mathbb{Z}[\mathrm{i}]$. Coefficients of $G_i$ have denominators dividing $L_i=\mathrm{lcm}(1,\dots ,T_i-1)$, where $T_i$ is its first omitted degree. A common entry denominator divides $D_H={\prod }_i{L}_i^{\lfloor H/w_i\rfloor }$. Multiplying every row by $D_H$ makes the determinant a nonzero Gaussian integer, whose absolute value is at least one. Undoing these operations yields

$$\frac{\log |{\Delta }_H|}{MH}\ge -(1-\bar{b})-E_{\mathrm{ar}},\bar{b}=\frac{1}{MH}\sum _{\text{rows}}w\beta \le \theta .$$

The nonnegative error $E_{\mathrm{ar}}$ records the common-denominator cost and rounding losses; the source gives its explicit bound. This is the lower bound both analytic alternatives must beat.

Translate approximate periods without losing the Taylor-tail budget

Let $\omega =2\pi \mathrm{i}$ and define the entire functions

$$f_{a,P}(z)=[u^a]P(e^z,z+u_1,\dots ,z+u_m).$$

Proof outline. Since $e^{j\omega }=1$, the original row can be expanded at $z=j\omega +\log (1+t)$. The discrepancy in the $i$th coordinate is the sum of ${ϵ}_i=j(r_i-\omega )$ and ${\tau }_i(t)=G_i(t)-\log (1+t)$. Expand powers of these two terms. The approximation hypothesis gives $|{ϵ}_i|\le 2K{q}_i^{-\nu }$. If a term uses $d_i$ copies of ${\tau }_i$, its Taylor order is at least ${\sum }_i{d}_i{T}_i$. To contribute to a row of order $s$, this must be at most $s$, which bounds ${\sum }_i{w}_i{d}_i$ by $H/F_0$ for the chosen truncation parameter $F_0$. Thus only a controlled portion of the weight can be spent on tails instead of small approximation errors. Cauchy’s coefficient estimate bounds the remaining tail coefficients. Each translated row coefficient with index $a\ge \beta$ is consequently at most $\exp [-\nu w(a-\beta )+H{E}_{\mathrm{tr}}]$.

Repeated coefficient rows force a quadratic collision saving

For a term in the translated determinant, let $n_a$ count rows with the same transverse index $a$. Expand each entire function on a disk of radius $R$ as $f_{a,P}(z)={\sum }_{d\ge 0}{c}_{a,d,P}(z/R{)}^d$. The tested points lie in a disk of radius $R/2$.

Proof outline. If two rows choose the same pair $(a,d)$, their coefficient vectors $(c_{a,d,P}{)}_P$ agree, so that determinant term vanishes. Hence within each group $a$, the selected nonnegative degrees are distinct and sum to at least $0+1+\cdots +(n_a-1)=\left(\genfrac{}{}{0ex}{}{n_a}{2}\right)$. Each degree has a geometric cost $2^{-d}$. Reserve half of that decay for the distinct-degree lower bound and use the other half to sum all degrees absolutely. Cauchy’s bounds for the coefficients and the permutation count then give

$$|\det T|\le \exp \left[-\frac{\log 2}{4}\sum _a{n}_a^2+MH(E_{\mathrm{hol}}+o(1))\right].$$

The decisive saving is quadratic in repeated group sizes. It comes from determinant antisymmetry, not from estimating the rows separately by their norms.

Every translated term pays either a collision cost or an approximation cost

Choose a cutoff $AH$ for transverse weight and a small fraction $\eta >0$. If at least $\eta M$ rows have $wa\le AH$, they occupy at most $D_A$ possible indices. Cauchy–Schwarz forces ${\sum }_a{n}_a^2\ge (\eta M{)}^2/D_A$, producing the collision saving. Otherwise more than $(1-\eta )M$ rows have $wa>AH$, so the translation coefficients supply a total saving at least $\nu MH(A(1-\eta )-\bar{b})$.

Proof outline of closure. Choose the fixed parameters so both savings exceed $MH(1-\bar{b})$ plus the arithmetic and analytic errors. The source arranges a strict margin using $A^2<\theta$ and $\nu (A-\theta )>1-\theta$, then makes $\eta$ and the error allowances small. Every row has only polynomially many translation terms in $H$; their combined contribution after normalization is $\log Q_H/H\to 0$. Choose the fixed finite dimension and separated approximation denominators first, and only then let $H\to \infty$. Both alternatives contradict the same nonzero-determinant lower bound. This excludes infinitely many approximations of exponent $\nu$; since $\nu >2$ was arbitrary, the claimed upper bound is $\mu (\pi )\le 2$.

flowchart TD
  A[Nonzero interpolation determinant] --> B[Arithmetic lower bound]
  A --> C[Translate rows to exact periods]
  C --> D[Many low-weight indices]
  C --> E[Many high-weight indices]
  D --> F[Repeated coefficient rows force degree separation]
  E --> G[High powers of small approximation errors]
  F --> H[Strictly smaller analytic upper bound]
  G --> H
  B --> I[Contradiction]
  H --> I

🏷️ Connections

Catalan’s constant shares the determinant comparison principle but does not supply an approximation exponent. Metric approximation with a shift concerns almost every real number, rather than the particular number $\pi$.

🏷️ Manuscripts and verification scope

• The irrationality exponent of π is 2 — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the exponent-two assertion; the Flint–Hills consequence lies outside that selected statement. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. The irrationality exponent of π is 2.