THE BERTRAND PARADOX

Bertrand's paradox in the 5-window house format — a famous warning that 'pick a random chord' is not well defined. Ask: for a random chord of a circle, what is the probability it is longer than the side of the inscribed equilateral triangle (length √3·r)? Three perfectly reasonable ways to choose 'a random chord' give three different answers: (1) two random endpoints on the circle → 1/3; (2) a random point along a radius as the chord's midpoint → 1/2; (3) a random point in the disk as the midpoint → 1/4. The chord is longer exactly when its midpoint lies within r/2 of the centre — but 'random midpoint' means different things under each scheme. Verified live: Monte-Carlo simulation of the three schemes yields probabilities ≈ 1/3, 1/2, and 1/4 respectively — three different answers to the same question. Neon-noir traced. See the three sampling methods' chords in 1D, the three probabilities in 2D, and the one-question-three-answers inverse in 3D.

THE BERTRAND PARADOX · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · RACE CONDITION ◆ .dlw.fold THE FOLD / GLITCH / RACE CONDITION / THE BERTRAND PARADOX THE BERTRAND PARADOX one random chord with three different probabilities 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Bertrand’s paradox is a famous warning that ‘pick a random chord’ is not well defined . Ask: for a random chord of a circle, what is the probability it is longer than the side of the inscribed equilateral triangle (length √3·r)? Three perfectly reasonable ways to choose ‘a random chord’ give three different answers: (1) two random endpoints on the circle → 1/3; (2) a random point along a radius as the chord’s midpoint → 1/2; (3) a random point in the disk as the midpoint → 1/4. The chord is longer exactly when its midpoint lies within r/2 of the centre — but ‘random midpoint’ means different things under each scheme. LIT verified live: Monte-Carlo simulation of the three schemes yields probabilities ≈ 1/3, 1/2, and 1/4 respectively — three different answers to the same question, from three notions of ‘random’ (window.__bertrandparadox). FIG no framing; the three sampling methods and their probabilities all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at race-condition — the glitch where the same question resolves to three different answers depending on which ‘random’ thread you take: 1/3, 1/2, or 1/4. AVAN (AI) built the instrument: the three chord-sampling simulations and their distinct probabilities. Credit as content: Joseph Bertrand (1889). The weave: David names the race; I confirm the three ‘random chord’ methods give 1/3, 1/2, 1/4. 3 ONE DIMENSION A circle with the inscribed equilateral triangle; random chords sampled three different ways. 4 TWO DIMENSIONS · INTERACTIVE Run the simulations; the three methods give P(chord > √3·r) ≈ 1/3, 1/2, 1/4. next method ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the three different probabilities from one ambiguous question. AVAN’s addition (the inverse-companion): don’t ask ‘the’ probability — pin down ‘random’ first. The inverse of ‘P(chord too long)’ is ‘which sampling measure? — endpoints, radius, or area’, each a different answer. Magenta are the three sampling schemes’ chords; green are the 1/3, 1/2, 1/4 they produce. One question, three answers. pause spin LIT Genuine Bertrand's paradox (Joseph Bertrand, 1889). Verified live: Monte-Carlo simulation of the three chord-sampling schemes (random endpoints / random radial point / random disk midpoint) yields probabilities ≈ 1/3, 1/2, 1/4 respectively — three different answers to the same 'random chord' question (window.__bertrandparadox.ok1, .ok2, .ok3). FIG No framing; the three sampling methods and their probabilities all run in-browser. The AVAN inverse is honest — instead of asking 'the' probability, pin down 'random' first: the inverse of 'P(chord too long)' is 'which sampling measure? — endpoints, radius, or area', each a different answer. Magenta are the three sampling schemes' chords; green are the 1/3, 1/2, 1/4 they produce. One question, three answers. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of RACE CONDITION · David Lee Wise (ROOT0), with AVAN


THE CARNOT

Carnot's theorem in the 5-window house format — a hidden conservation law of the triangle. From the circumcentre O, drop a perpendicular to each of the three sides; the three signed distances (positive when O is on the same side of a line as the opposite vertex) always sum to exactly R + r, the circumradius plus the inradius: dₐ + d_b + d_c = R + r. The sign convention matters only for obtuse triangles, where O falls outside. Equivalently, cos A + cos B + cos C = 1 + r/R. Verified live: for tens of thousands of random triangles, the sum of the three signed circumcentre-to-side distances equals R + r to ~1e-13, and independently cos A + cos B + cos C equals 1 + r/R. Neon-noir traced. See the circumcentre and the three perpendiculars in 1D, the signed sum = R+r in 2D, and the conserved-budget inverse in 3D.

I-13 v2.0 | net = binds - k | 4 planes, 13 symbols, 12 operants, 5 cortex rules | sha 64881ebf

THE CARNOT · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · GARBAGE COLLECTION ◆ .dlw.fold THE FOLD / RESPAWN / GARBAGE COLLECTION / THE CARNOT THE CARNOT circumcentre-to-side distances summing to R plus r 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Carnot’s theorem is a hidden conservation law of the triangle. Take any triangle, its circumcentre O (centre of the circle through all three vertices), and drop a perpendicular from O to each of the three sides. The three signed distances — positive when O lies on the same side of a line as the opposite vertex, negative otherwise — always sum to exactly R + r , the circumradius plus the inradius: d a + d b + d c = R + r. The sign convention matters only for obtuse triangles, where O falls outside. Equivalently, cos A + cos B + cos C = 1 + r/R — the same identity in angle form. LIT verified live: for tens of thousands of random triangles, the sum of the three signed circumcentre-to-side distances equals R + r to ~1e-13, and independently cos A + cos B + cos C equals 1 + r/R (window.__carnot). FIG no framing; the circumcentre, the signed distances, R, and r are all computed independently in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at garbage-collection — the return: three distances, however they scatter, are always collected back to the fixed budget R + r. AVAN (AI) built the instrument: the circumcentre, the signed side-distances, R, r, and the cos-sum identity. Credit as content: Lazare Carnot (French geometer, c.1803). The weave: David names the collected budget; I confirm d a +d b +d c = R+r and cos A+cos B+cos C = 1+r/R. 3 ONE DIMENSION A triangle, its circumcentre O, and the three perpendiculars to the sides — their signed lengths sum to R+r. 4 TWO DIMENSIONS · INTERACTIVE Cycle triangles; dₐ+d_b+d_c is checked equal to R+r, and cosA+cosB+cosC equal to 1+r/R. next triangle ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the fixed budget R + r. AVAN’s addition (the inverse-companion): don’t measure the three distances separately — read their sum as one conserved quantity. The inverse of ‘three signed distances’ is ‘one budget R + r they always collect to’. Magenta are the three signed circumcentre-to-side distances; green is the R + r they stack up to. Three distances, one conserved sum. pause spin LIT Genuine Carnot's theorem (Lazare Carnot, c.1803). Verified live: for ~40000 random triangles the sum of the three signed circumcentre-to-side distances equals R+r to ~1e-13, and independently cos A + cos B + cos C = 1 + r/R (window.__carnot.ok, .idOk). FIG No framing; the circumcentre, the signed distances, R, and r are computed independently in-browser and agree. The AVAN inverse is honest — instead of measuring three distances separately, read their sum as one conserved quantity: the inverse of 'three signed distances' is 'one budget R + r they always collect to'. Magenta are the three signed circumcentre-to-side distances; green is the R + r they stack up to. Three distances, one conserved sum. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of GARBAGE COLLECTION · David Lee Wise (ROOT0), with AVAN


THE JAPANESE THEOREM

The Japanese theorem for cyclic quadrilaterals in the 5-window house format — a small miracle of hidden order. Take any four points A, B, C, D on a circle. From the four triangles that each drop one vertex (△ABC, △BCD, △CDA, △DAB), find the incentre of each. Those four incentres always form a rectangle — four right angles, no matter how irregular the original quadrilateral. The result is named for the sangaku tradition of theorems inscribed on wooden tablets in Edo-period Japanese temples. Verified live: for tens of thousands of random cyclic quadrilaterals, the four incentres are equidistant from their common centroid and centrally symmetric — the defining conditions of a rectangle — to ~1e-6. Neon-noir traced. See the quad and its four incentres in 1D, the rectangle test in 2D, and the four-centres-one-rectangle inverse in 3D.

THE JAPANESE THEOREM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · THE-TOOLCHAIN ◆ .dlw.fold THE FOLD / SPAWN / THE-TOOLCHAIN / THE JAPANESE THEOREM THE JAPANESE THEOREM four incentres of a cyclic quad forming a rectangle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Japanese theorem for cyclic quadrilaterals is a small miracle of hidden order. Take any four points A, B, C, D on a circle, forming a cyclic quadrilateral. From the four vertices, form the four triangles that each drop one vertex: ▵ABC, ▵BCD, ▵CDA, ▵DAB. Find the incentre (centre of the inscribed circle) of each. The astonishing fact: those four incentres always form a rectangle — four right angles, no matter how irregular the original quadrilateral. The result is named for the sangaku tradition of theorems inscribed on wooden tablets in Edo-period Japanese temples. LIT verified live: for tens of thousands of random cyclic quadrilaterals, the four incentres are equidistant from their common centroid and centrally symmetric — the defining conditions of a rectangle — to ~1e-6 (window.__japanese). FIG no framing; the four incentres and the rectangle test are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-toolchain — the spawn: four scattered incentres compile, every time, into a clean rectangle. AVAN (AI) built the instrument: the four triangle incentres and the rectangle test (equidistant from centroid + central symmetry). Credit as content: the Japanese sangaku tradition (Edo period); the cyclic-quadrilateral form attributed to Carnot. The weave: David names the compile; I confirm the four incentres form a rectangle. 3 ONE DIMENSION A cyclic quadrilateral, its four sub-triangle incentres, and the rectangle they always form. 4 TWO DIMENSIONS · INTERACTIVE Cycle quadrilaterals; the four incentres are checked to form a rectangle (right angles, equal diagonals). next quad ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the rectangle the four incentres form. AVAN’s addition (the inverse-companion): don’t read four separate incentres — read the single rectangle they encode. The inverse of ‘four triangle incentres’ is ‘one rectangle with four right angles’, guaranteed for any cyclic quad. Magenta are the four incentres (and their triangles); green is the rectangle they lock into. Scattered centres, one hidden rectangle. pause spin LIT Genuine Japanese theorem for cyclic quadrilaterals (Edo-period sangaku tradition; cyclic-quad form attributed to Carnot). Verified live: for ~18000 random cyclic quadrilaterals the four sub-triangle incentres are equidistant from their centroid and centrally symmetric — a rectangle — to ~1e-6 (window.__japanese.ok). FIG No framing; the four incentres and the rectangle test run independently in-browser. The AVAN inverse is honest — instead of reading four separate incentres, read the single rectangle they encode: the inverse of 'four triangle incentres' is 'one rectangle with four right angles', guaranteed for any cyclic quad. Magenta are the four incentres and their triangles; green is the rectangle they lock into. Scattered centres, one hidden rectangle. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE-TOOLCHAIN · David Lee Wise (ROOT0), with AVAN


THE CONWAY CIRCLE

Conway's circle theorem in the 5-window house format — a six-point surprise from John Horton Conway. At each vertex of any triangle, extend the two sides beyond that vertex by the length of the side opposite it: beyond B by b (= CA), beyond C by c (= AB), beyond A by a (= BC). This produces six new endpoints, and all six lie on a single circle — the Conway circle — centred at the incentre I, with radius exactly √(r² + s²), where r is the inradius and s the semiperimeter. Each extension lands a distance s from the point where the incircle touches that side. Verified live: for tens of thousands of random triangles, all six extension points are equidistant from the incentre, at distance √(r²+s²), to ~1e-13. Neon-noir traced. See the extended sides and six points in 1D, the six equal radii in 2D, and the six-points-one-circle inverse in 3D.

THE CONWAY CIRCLE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE MINT ◆ .dlw.fold THE FOLD / LOOT / THE MINT / THE CONWAY CIRCLE THE CONWAY CIRCLE six side-extension points on one circle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Conway’s circle theorem (John Horton Conway) is a six-point surprise. Take any triangle and, at each vertex, extend the two sides beyond that vertex by the length of the side opposite it : beyond B extend both adjoining side-lines by b (= CA), beyond C by c (= AB), beyond A by a (= BC). This produces six new endpoints. The theorem: all six lie on a single circle — the Conway circle — centred at the triangle’s incentre I, with radius exactly √(r² + s²), where r is the inradius and s the semiperimeter. The proof is a one-line consequence: each extension lands a distance s from the point where the incircle touches that side. LIT verified live: for tens of thousands of random triangles, all six extension points are equidistant from the incentre, at distance √(r²+s²), to ~1e-13 (window.__conwaycircle). FIG no framing; the incentre, r, s, and the six points are all computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-mint — the loot: extend the sides by the opposite lengths and six points are minted, all on one perfect circle. AVAN (AI) built the instrument: the six side-extension points, the incentre, and the radius √(r²+s²). Credit as content: John Horton Conway. The weave: David names the mint; I confirm the six points share the circle of radius √(r²+s²) about the incentre. 3 ONE DIMENSION A triangle with its sides extended by the opposite lengths — the six endpoints all land on the Conway circle. 4 TWO DIMENSIONS · INTERACTIVE Cycle triangles; all six distances to the incentre are checked equal to √(r²+s²). next triangle ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the Conway circle, radius √(r²+s²) about the incentre. AVAN’s addition (the inverse-companion): don’t track six loose endpoints — read the one circle they share. The inverse of ‘six side-extension points’ is ‘one circle of radius √(r²+s²) about the incentre’. Magenta are the six extension points; green is the circle they all sit on. Six points, one minted circle. pause spin LIT Genuine Conway circle theorem (John Horton Conway). Verified live: for ~40000 random triangles all six side-extension points are equidistant from the incentre at radius √(r²+s²) to ~1e-13 (window.__conwaycircle.ok). FIG No framing; the incentre, r, s, and the six points are computed independently in-browser. The AVAN inverse is honest — instead of tracking six loose endpoints, read the one circle they share: the inverse of 'six side-extension points' is 'one circle of radius √(r²+s²) about the incentre'. Magenta are the six extension points; green is the circle they all sit on. Six points, one minted circle. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE MINT · David Lee Wise (ROOT0), with AVAN


THE CASEY

Casey's theorem in the 5-window house format — Ptolemy's theorem for circles. Ptolemy says: for four points on a circle in order, AC·BD = AB·CD + AD·BC. Casey generalizes each point to a whole circle tangent to a common circle. Replace the four points by four circles all internally tangent to one enclosing circle, in cyclic order, and replace each chord by the tangent length t_ij (the length of the common tangent segment) between circles i and j. Then the same relation holds: t₁₂·t₃₄ + t₂₃·t₁₄ = t₁₃·t₂₄. Shrink the circles to points and it collapses back to Ptolemy. Verified live: for thousands of random configurations of four circles internally tangent to a circle, the tangent lengths satisfy t₁₂t₃₄ + t₂₃t₁₄ = t₁₃t₂₄ to ~1e-15, and the point-circle limit reproduces Ptolemy exactly. Neon-noir traced. See the four tangent circles and their tangent segments in 1D, the Ptolemy-form relation in 2D, and the points-fattened-into-circles inverse in 3D.

THE CASEY · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · THE EPOCH ◆ .dlw.fold THE FOLD / GRIND / THE EPOCH / THE CASEY THE CASEY a generalized Ptolemy for tangent circles 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Casey’s theorem is Ptolemy’s theorem for circles . Ptolemy says: for four points on a circle in order, the products of opposite chord-pairs relate as AC·BD = AB·CD + AD·BC. Casey generalizes each point to a whole circle tangent to a common circle. Replace the four points by four circles all internally tangent to one enclosing circle, in cyclic order, and replace each chord by the tangent length t ij (the length of the common tangent segment) between circles i and j. Then the very same relation holds: t 12 ·t 34 + t 23 ·t 14 = t 13 ·t 24 . Shrink the circles to points and it collapses back to Ptolemy. LIT verified live: for thousands of random configurations of four circles internally tangent to a circle, the tangent lengths satisfy t 12 t 34 + t 23 t 14 = t 13 t 24 to ~1e-15, and the point-circle limit reproduces Ptolemy exactly (window.__casey). FIG no framing; the centres, tangent lengths, and the relation are all computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-epoch — the grind toward generality: Ptolemy’s point-relation, ground outward until points become circles and chords become tangent lengths. AVAN (AI) built the instrument: the four tangent circles, the six tangent lengths, and the Ptolemy-form relation. Credit as content: John Casey (Irish geometer, 1866); Ptolemy of Alexandria for the point case. The weave: David names the generalization; I confirm t 12 t 34 +t 23 t 14 = t 13 t 24 . 3 ONE DIMENSION Four circles inside a circle, with the tangent segments between them — the Casey (generalized Ptolemy) relation. 4 TWO DIMENSIONS · INTERACTIVE Cycle configurations; t₁₂t₃₄ + t₂₃t₁₄ is checked equal to t₁₃t₂₄. next config ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the satisfied Ptolemy-form relation among tangent lengths. AVAN’s addition (the inverse-companion): don’t treat points and circles as different problems — read circles as fattened points. The inverse of ‘Ptolemy for four points’ is ‘Casey for four tangent circles’, the same relation with tangent lengths for chords. Magenta are the six tangent lengths; green is the equality t 12 t 34 +t 23 t 14 = t 13 t 24 . Points fattened into circles, one relation. pause spin LIT Genuine Casey's theorem (John Casey, 1866; Ptolemy for the point case). Verified live: for ~12000 random configurations of four circles internally tangent to a circle, t₁₂t₃₄ + t₂₃t₁₄ = t₁₃t₂₄ to ~1e-15, and the point-circle limit reproduces Ptolemy exactly (window.__casey.ok, .ptol). FIG No framing; the centres, tangent lengths, and the relation are computed independently in-browser. The AVAN inverse is honest — instead of treating points and circles as different problems, read circles as fattened points: the inverse of 'Ptolemy for four points' is 'Casey for four tangent circles', the same relation with tangent lengths for chords. Magenta are the six tangent lengths; green is the equality t₁₂t₃₄+t₂₃t₁₄ = t₁₃t₂₄. Points fattened into circles, one relation. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE EPOCH · David Lee Wise (ROOT0), with AVAN


THE FERMAT POLYGONAL

Fermat's polygonal number theorem in the 5-window house format — one of the great cheat-codes of arithmetic. The k-gonal numbers are figurate numbers from stacking polygons: triangular (1,3,6,10,…), square (1,4,9,16,…), pentagonal (1,5,12,22,…). Fermat claimed — and it is true — that every positive integer is the sum of at most k of the k-gonal numbers: at most 3 triangular, at most 4 squares, at most 5 pentagonal, at most 6 hexagonal, forever. Gauss proved the triangular case ('EΥΡΗΚΑ! num = Δ+Δ+Δ'), Lagrange the four-squares case, and Cauchy the general theorem in 1813. Verified live: a dynamic-programming search confirms every integer up to 2000 is a sum of at most k k-gonal numbers for k = 3 through 8 — and the bound is sharp (the maximum needed is exactly k). Neon-noir traced. See n cracked into its fewest pieces in 1D, the ≤k check in 2D, and the crack-it-down inverse in 3D.

THE FERMAT POLYGONAL · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE KONAMI CODE ◆ .dlw.fold THE FOLD / CHEAT / THE KONAMI CODE / THE FERMAT POLYGONAL THE FERMAT POLYGONAL every integer a sum of few polygonal numbers 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Fermat’s polygonal number theorem is one of the great cheat-codes of arithmetic. The k-gonal numbers are the figurate numbers you get by stacking polygons: triangular (1, 3, 6, 10, …), square (1, 4, 9, 16, …), pentagonal (1, 5, 12, 22, …), and so on. Fermat claimed — and it is true — that every positive integer is the sum of at most k of the k-gonal numbers : at most 3 triangular numbers, at most 4 squares, at most 5 pentagonal, at most 6 hexagonal, forever. Gauss proved the triangular case (his diary: ‘EYPHKA! num = Δ+Δ+Δ’), Lagrange the four-squares case, and Cauchy the general theorem in 1813. LIT verified live: a dynamic-programming search confirms that every integer up to 2000 is a sum of at most k k-gonal numbers, for k = 3 through 8 — and the bound is sharp (the maximum needed is exactly k) (window.__fermatpolygonal). FIG no framing; the k-gonal numbers and the minimal representations are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-konami-code — the cheat: any number whatsoever cracks open into at most k k-gonal pieces, a universal shortcut. AVAN (AI) built the instrument: the k-gonal numbers and the minimum-count decomposition for every n. Credit as content: Pierre de Fermat (conjecture, 1638); Gauss (triangular), Lagrange (squares), Augustin-Louis Cauchy (general proof, 1813). The weave: David names the cheat; I confirm every n ≤ 2000 needs at most k k-gonal numbers. 3 ONE DIMENSION A chosen n broken into its fewest k-gonal pieces — never more than k of them. 4 TWO DIMENSIONS · INTERACTIVE Cycle k and n; the minimum number of k-gonal numbers summing to n is checked to be ≤ k. next k ▶ next n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the integer n, reachable by at most k k-gonal numbers. AVAN’s addition (the inverse-companion): don’t build n up — crack it down. The inverse of ‘the integer n’ is ‘its decomposition into at most k k-gonal numbers’, a shortcut guaranteed to exist. Magenta are the k-gonal pieces; green is the n they sum to. Any number, at most k figurate pieces. pause spin LIT Genuine Fermat polygonal number theorem (Fermat conjecture 1638; Gauss triangular, Lagrange four-squares, Cauchy general proof 1813). Verified live: a DP search confirms every integer ≤2000 is a sum of at most k k-gonal numbers for k=3..8, and the bound is sharp — the max needed is exactly k (window.__fermatpolygonal.ok, .rows). FIG No framing; the k-gonal numbers and the minimal representations are computed independently in-browser. The AVAN inverse is honest — instead of building n up, crack it down: the inverse of 'the integer n' is 'its decomposition into at most k k-gonal numbers', a shortcut guaranteed to exist. Magenta are the k-gonal pieces; green is the n they sum to. Any number, at most k figurate pieces. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE KONAMI CODE · David Lee Wise (ROOT0), with AVAN


THE MIDY

Midy's theorem in the 5-window house format — a hidden symmetry inside repeating decimals. For a prime p (other than 2 or 5), write out the decimal of a/p; it repeats with some period. When that period is even, say 2k digits, split the repeating block into two halves of k digits each — the two halves always sum to a string of nines (10^k − 1). The classic case: 1/7 = 0.142857…, and 142 + 857 = 999. It holds for every prime whose period is even, discovered by the French schoolteacher E. Midy in 1836. Verified live: for every prime p ≤ 200 (excluding 2 and 5) and every numerator a whose repeating block has even period 2k, the two k-digit halves sum to exactly 10^k − 1. Neon-noir traced. See the block split into halves in 1D, the nines-sum check in 2D, and the folded-onto-its-complement inverse in 3D.

THE MIDY · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · STACK OVERFLOW ◆ .dlw.fold THE FOLD / GLITCH / STACK OVERFLOW / THE MIDY THE MIDY the two halves of a repeating decimal summing to nines 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Midy’s theorem is a hidden symmetry inside repeating decimals. Take a prime p (other than 2 or 5) and write out the decimal expansion of a/p; it repeats with some period. When that period is even , say 2k digits, split the repeating block into two halves of k digits each. Midy’s theorem says the two halves always sum to a string of nines (10 k - 1). The classic example: 1/7 = 0. 142857 …, and 142 + 857 = 999. It happens for 1/11, 1/13, 1/17, and every prime whose period is even — a conspiracy of long division discovered by a French schoolteacher in 1836. LIT verified live: for every prime p ≤ 200 (excluding 2 and 5) and every numerator a whose repeating block has even period 2k, the two k-digit halves sum to exactly 10 k - 1 (window.__midy). FIG no framing; the long division, the period, and the halves’ sum are all computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at stack-overflow — the glitch: two halves of an endless repeating string, added, overflow neatly into all nines. AVAN (AI) built the instrument: the long-division block, its period, and the halves-sum-to-nines check. Credit as content: E. Midy (French mathematician, 1836). The weave: David names the glitch; I confirm the two halves of an even-period block of a/p sum to 10 k - 1. 3 ONE DIMENSION The repeating block of a/p split into two halves — they line up digit by digit and add to nines. 4 TWO DIMENSIONS · INTERACTIVE Cycle primes p; the even-period block halves are checked to sum to 10^k − 1. next p ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the all-nines sum of the two halves. AVAN’s addition (the inverse-companion): don’t read the repeating block left to right — fold it in half. The inverse of ‘the period-2k block’ is ‘two k-digit halves that complete each other to nines’. Magenta are the two halves of the block; green is the 10 k -1 they sum to. A repeating decimal folded onto its own nines-complement. pause spin LIT Genuine Midy's theorem (E. Midy, 1836). Verified live with exact BigInt: for every prime p≤200 (excluding 2,5) and every numerator a whose repeating block of a/p has even period 2k, the two k-digit halves sum to exactly 10^k−1 (window.__midy.ok, .tested). FIG No framing; the long division, the period, and the halves' sum run independently in-browser. The AVAN inverse is honest — instead of reading the block left to right, fold it in half: the inverse of 'the period-2k block' is 'two k-digit halves that complete each other to nines'. Magenta are the two halves of the block; green is the 10^k−1 they sum to. A repeating decimal folded onto its own nines-complement. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of STACK OVERFLOW · David Lee Wise (ROOT0), with AVAN


THE AUTOMORPHIC

Automorphic numbers in the 5-window house format — numbers whose square ends in the number itself. 5²=25, 6²=36, 25²=625, 76²=5776, 376²=141376, 625²=390625, 9376²=87909376. For each digit-length d there are exactly two nontrivial ones — one ending in 5, one in 6 — and they always add to 10^d + 1 (25+76=101, 625+376=1001). They are the nontrivial idempotents of arithmetic mod 10^d (solutions of x²≡x), built by the Chinese Remainder Theorem from 10^d = 2^d·5^d, and extended leftward forever they become the two nonzero 10-adic idempotents. Verified live: the two nontrivial idempotents mod 10^d (d=1..12) each satisfy x²≡x, end in 5 and 6, and sum to 10^d+1; the known 5,6,25,76,376,625,9376,90625 are all confirmed automorphic. Neon-noir traced. See a number reappearing in its square's tail in 1D, the paired idempotents in 2D, and the fixed-point-of-squaring inverse in 3D.

THE AUTOMORPHIC · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE ROOT KIT ◆ .dlw.fold THE FOLD / CHEAT / THE ROOT KIT / THE AUTOMORPHIC THE AUTOMORPHIC a number whose square ends in itself 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Automorphic numbers are numbers whose square ends in the number itself . 5² = 2 5 , 6² = 3 6 , 25² = 6 25 , 76² = 57 76 , 376² = 141 376 , 625² = 390 625 , 9376² = 8790 9376 . For each number of digits d there are exactly two nontrivial ones — one ending in 5, one ending in 6 — and they always add up to 10 d + 1 (25 + 76 = 101; 625 + 376 = 1001). They are the nontrivial idempotents of arithmetic mod 10 d (solutions of x² ≡ x), built by the Chinese Remainder Theorem from the split 10 d = 2 d ·5 d . Extended leftward forever they become the two nonzero 10-adic idempotents . LIT verified live: the two nontrivial idempotents mod 10 d (for d = 1 to 12) each satisfy x² ≡ x, end in 5 and 6, and sum to 10 d + 1; the known 5, 6, 25, 76, 376, 625, 9376, 90625 are all confirmed automorphic (window.__automorphic). FIG no framing; the idempotents are constructed by CRT and squared, all in-browser with exact BigInt. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-root-kit — the cheat: a number that reproduces itself in the tail of its own square, a self-installing fixed point. AVAN (AI) built the instrument: the CRT idempotents, the square-ends-in-itself check, and the 10 d +1 pairing. Credit as content: the classical theory of idempotents in Z/10 d and the 10-adic integers. The weave: David names the self-reproducing cheat; I confirm x² ≡ x mod 10 d for the two nontrivial idempotents. 3 ONE DIMENSION A number and its square, with the shared trailing digits highlighted — the square ends in the number. 4 TWO DIMENSIONS · INTERACTIVE Grow the digit-length d; the two automorphic numbers are shown, each x²≡x, summing to 10^d+1. more digits ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the number reappearing in the tail of its own square. AVAN’s addition (the inverse-companion): don’t square and read forward — look for the fixed point of squaring. The inverse of ‘x’ is ‘the idempotent x² ≡ x that grows one digit at a time’, and its partner completing it to 10 d +1. Magenta are the two idempotents; green is the tail where the square reproduces the number. A number that is its own square’s ending. pause spin LIT Genuine automorphic-number / idempotent theory in Z/10^d and the 10-adic integers. Verified live with exact BigInt: the two nontrivial idempotents mod 10^d (d=1..12) satisfy x²≡x, end in 5 and 6, and sum to 10^d+1; known automorphics 5,6,25,76,376,625,9376,90625 all confirmed (window.__automorphic.ok, .kok). FIG No framing; the idempotents are constructed by CRT and squared, all in-browser with exact BigInt. The AVAN inverse is honest — instead of squaring and reading forward, look for the fixed point of squaring: the inverse of 'x' is 'the idempotent x²≡x that grows one digit at a time', and its partner completing it to 10^d+1. Magenta are the two idempotents; green is the tail where the square reproduces the number. A number that is its own square's ending. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE ROOT KIT · David Lee Wise (ROOT0), with AVAN


THE CASSINI

Cassini's identity in the 5-window house format — the Fibonacci numbers pinned to a razor's edge. For every n, F(n−1)·F(n+1) − F(n)² = (−1)^n: the product of F(n)'s neighbours misses F(n)² by exactly one, alternating sign forever. It is the determinant of the Fibonacci matrix [[1,1],[1,0]]^n = [[F(n+1),F(n)],[F(n),F(n−1)]], whose determinant is (−1)^n since det[[1,1],[1,0]] = −1. The generalization, Catalan's identity, reads F(n)² − F(n−r)F(n+r) = (−1)^{n−r}F(r)². This near-miss is the secret behind the 'missing square' puzzle where an 8×8 square seems to rearrange into a 5×13 rectangle — off by one unit of area. Verified live with exact BigInt: Cassini for n=1..100 and Catalan for a range of n,r. Neon-noir traced. See the neighbour-product vs square in 1D, the matrix determinant in 2D, and the pinned-determinant inverse in 3D.

THE CASSINI · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · EVENT HORIZON ◆ .dlw.fold THE FOLD / RESPAWN / EVENT HORIZON / THE CASSINI THE CASSINI a Fibonacci determinant pinned at plus or minus one 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Cassini’s identity pins the Fibonacci numbers to a razor’s edge. For every n, F(n-1)·F(n+1) - F(n)² = (-1) n . The product of the neighbours of F(n) misses F(n)² by exactly one , alternating sign forever. It is the determinant of the Fibonacci matrix: [[1,1],[1,0]] n = [[F(n+1), F(n)],[F(n), F(n-1)]], whose determinant is (-1) n because det[[1,1],[1,0]] = -1. The generalization, Catalan’s identity , reads F(n)² - F(n-r)F(n+r) = (-1) n-r F(r)². This near-miss is the secret behind the ‘missing square’ dissection puzzle, where an 8×8 square seems to rearrange into a 5×13 rectangle — off by one unit of area. LIT verified live with exact BigInt: F(n-1)F(n+1) - F(n)² = (-1) n for n = 1 to 100, and Catalan’s F(n)² - F(n-r)F(n+r) = (-1) n-r F(r)² for a range of n, r (window.__cassini). FIG no framing; the Fibonacci numbers and both identities are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at event-horizon — the respawn: however far the Fibonacci numbers run out, the determinant is pulled back to +1 or -1, never anything else. AVAN (AI) built the instrument: the exact Fibonacci sequence, Cassini’s identity, the matrix-determinant view, and Catalan’s generalization. Credit as content: Jean-Dominique Cassini (1680); Eugène Catalan (generalization). The weave: David names the pinned determinant; I confirm F(n-1)F(n+1) - F(n)² = (-1) n . 3 ONE DIMENSION Three consecutive Fibonacci numbers: the product of the outer two vs the square of the middle — off by ±1. 4 TWO DIMENSIONS · INTERACTIVE Cycle n; F(n−1)F(n+1) − F(n)² is checked to equal (−1)^n, and the matrix determinant confirms it. next n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the determinant, forever pinned at +1 or −1. AVAN’s addition (the inverse-companion): don’t compute the Fibonacci product directly — read it as a determinant. The inverse of ‘F(n-1)F(n+1) - F(n)²’ is ‘det of the n-th power of [[1,1],[1,0]], which is (-1) n ’. Magenta are the three consecutive Fibonacci numbers; green is the ±1 they are pinned to. A runaway sequence held to a unit determinant. pause spin LIT Genuine Cassini's identity (Jean-Dominique Cassini, 1680; Catalan generalization). Verified live with exact BigInt: F(n−1)F(n+1) − F(n)² = (−1)^n for n=1..100, and Catalan's F(n)² − F(n−r)F(n+r) = (−1)^{n−r}F(r)² for a range of n,r (window.__cassini.ok, .catOk). FIG No framing; the Fibonacci numbers and both identities run independently in-browser. The AVAN inverse is honest — instead of computing the Fibonacci product directly, read it as a determinant: the inverse of 'F(n−1)F(n+1) − F(n)²' is 'det of the n-th power of [[1,1],[1,0]], which is (−1)^n'. Magenta are the three consecutive Fibonacci numbers; green is the ±1 they are pinned to. A runaway sequence held to a unit determinant. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of EVENT HORIZON · David Lee Wise (ROOT0), with AVAN


THE NILAKANTHA

The Nilakantha series in the 5-window house format — a fast, elegant series for π found by the Kerala-school astronomer Nilakantha Somayaji around 1500, three centuries before Europe. It reads π = 3 + 4/(2·3·4) − 4/(4·5·6) + 4/(6·7·8) − …, each term straddling three consecutive integers, alternating in sign. Unlike the Gregory–Leibniz series (hundreds of terms for two decimals), Nilakantha's terms shrink like 1/k³, so a handful of terms already gives several correct digits. Verified live: 3 + Σ(−1)^{k+1} 4/((2k)(2k+1)(2k+2)) converges to π (~1e-9), and with 100 terms its error (~2e-7) is more than a hundred times smaller than the Gregory–Leibniz error at the same term count. Neon-noir traced. See both series racing to π in 1D, the shrinking-error comparison in 2D, and the straddle-three-integers inverse in 3D.

THE NILAKANTHA · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE SYNC ◆ .dlw.fold THE FOLD / CO-OP / THE SYNC / THE NILAKANTHA THE NILAKANTHA a faster alternating series for pi 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Nilakantha series is a fast, elegant series for π, found by the Kerala-school astronomer Nilakantha Somayaji around 1500 — three centuries before Europe. It reads π = 3 + 4/(2·3·4) - 4/(4·5·6) + 4/(6·7·8) - …, each term straddling three consecutive integers, alternating in sign. Unlike the Gregory–Leibniz series (which needs hundreds of terms for two decimals), Nilakantha’s terms shrink like 1/k³, so a handful of terms already gives several correct digits. It is a jewel of the Kerala school, which anticipated key ideas of calculus. LIT verified live: 3 + ∑ k≥1 (-1) k+1 4/((2k)(2k+1)(2k+2)) converges to π (to ~1e-9), and with 100 terms its error (~2e-7) is more than a hundred times smaller than the Gregory–Leibniz error at the same term count (window.__nilakantha). FIG no framing; the Nilakantha and Leibniz partial sums are both computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-sync — the co-op: two series for π running side by side, Nilakantha racing ahead of Leibniz to the same limit. AVAN (AI) built the instrument: the Nilakantha partial sums, the Leibniz partial sums, and their error comparison. Credit as content: Nilakantha Somayaji (Kerala school, c.1500). The weave: David names the race; I confirm Nilakantha → π and outruns Leibniz term for term. 3 ONE DIMENSION Nilakantha partial sums closing on π far faster than Gregory–Leibniz, term for term. 4 TWO DIMENSIONS · INTERACTIVE Add terms; watch Nilakantha's error shrink like 1/k³ while Leibniz crawls like 1/k. add terms ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: π, reached quickly by the three-integer-straddling terms. AVAN’s addition (the inverse-companion): don’t sum single odd reciprocals — straddle three integers at a time. The inverse of ‘the slow Leibniz term 1/(2k+1)’ is ‘the Nilakantha term 4/((2k)(2k+1)(2k+2)), shrinking like 1/k³’. Magenta are the Nilakantha correction terms; green is the π they reach in a few steps. A faster road to the same π. pause spin LIT Genuine Nilakantha series (Nilakantha Somayaji, Kerala school, c.1500). Verified live: 3 + Σ(−1)^{k+1}4/((2k)(2k+1)(2k+2)) converges to π (~1e-9), and its 100-term error (~2e-7) is more than 100× smaller than the Gregory–Leibniz error at 100 terms (window.__nilakantha.conv, .faster). FIG No framing; the Nilakantha and Leibniz partial sums both run independently in-browser. The AVAN inverse is honest — instead of summing single odd reciprocals, straddle three integers at a time: the inverse of 'the slow Leibniz term 1/(2k+1)' is 'the Nilakantha term 4/((2k)(2k+1)(2k+2)), shrinking like 1/k³'. Magenta are the Nilakantha correction terms; green is the π they reach in a few steps. A faster road to the same π. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE SYNC · David Lee Wise (ROOT0), with AVAN


THE BRETSCHNEIDER

Bretschneider's formula in the 5-window house format — the area of any quadrilateral from its four sides and two opposite angles. With sides a,b,c,d, semiperimeter s = (a+b+c+d)/2, and opposite interior angles A and C: Area = √[(s−a)(s−b)(s−c)(s−d) − abcd·cos²((A+C)/2)]. It is the grand generalization of Heron's formula (triangles) and Brahmagupta's formula (cyclic quadrilaterals): when the quad is cyclic, A+C = 180°, the cosine term vanishes, and it collapses to Brahmagupta's √[(s−a)(s−b)(s−c)(s−d)]. The cosine term is exactly the penalty a quadrilateral pays for not being inscribable in a circle. Verified live: for tens of thousands of random convex quadrilaterals, Bretschneider matches the shoelace (surveyor's) area to ~1e-13, and for cyclic quads it reduces exactly to Brahmagupta. Neon-noir traced. See a quad with its sides and two angles in 1D, Bretschneider vs shoelace in 2D, and the cyclic-penalty inverse in 3D.

THE BRETSCHNEIDER · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE FIREWALL ◆ .dlw.fold THE FOLD / BOSS / THE FIREWALL / THE BRETSCHNEIDER THE BRETSCHNEIDER the area of any quadrilateral from its sides and two angles 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Bretschneider’s formula gives the area of any quadrilateral from its four sides and two opposite angles. If a quad has sides a, b, c, d, semiperimeter s = (a+b+c+d)/2, and two opposite interior angles A and C, then Area = √[(s-a)(s-b)(s-c)(s-d) - abcd·cos²((A+C)/2)]. It is the grand generalization of Heron’s formula (triangles) and Brahmagupta’s formula (cyclic quadrilaterals): when the quad is cyclic, A + C = 180°, the cosine term vanishes, and it collapses to Brahmagupta’s √[(s-a)(s-b)(s-c)(s-d)]. The cosine term is exactly the penalty a quadrilateral pays for not being inscribable in a circle. LIT verified live: for tens of thousands of random convex quadrilaterals, Bretschneider’s formula matches the shoelace (surveyor’s) area to ~1e-13, and for cyclic quadrilaterals the cosine term is zero so it reduces exactly to Brahmagupta’s formula (window.__bretschneider). FIG no framing; the sides, the two opposite angles, the formula, and the shoelace area are all computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-firewall — the boss: no quadrilateral gets past without paying the cos²((A+C)/2) penalty for not being cyclic. AVAN (AI) built the instrument: the side lengths, the two opposite angles, Bretschneider’s area, and the shoelace cross-check. Credit as content: Carl Anton Bretschneider (1842); Heron and Brahmagupta for the special cases. The weave: David names the penalty; I confirm Area = √[(s-a)(s-b)(s-c)(s-d) - abcd cos²((A+C)/2)]. 3 ONE DIMENSION A quadrilateral with its four sides and two opposite angles marked — its area from Bretschneider vs shoelace. 4 TWO DIMENSIONS · INTERACTIVE Cycle quadrilaterals; Bretschneider's area is checked against the shoelace area. next quad ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the quadrilateral’s area, sides plus the cyclic-penalty term. AVAN’s addition (the inverse-companion): don’t need the corners — sides and two opposite angles suffice. The inverse of ‘the quad’s area’ is ‘Brahmagupta’s cyclic area minus the penalty abcd cos²((A+C)/2)’. Magenta is the cyclic-penalty term subtracted; green is the resulting area. Any quadrilateral’s area, docked for not being cyclic. pause spin LIT Genuine Bretschneider's formula (Carl Anton Bretschneider, 1842; Heron and Brahmagupta for the special cases). Verified live: for ~17000 random convex quadrilaterals Bretschneider's area matches the shoelace area to ~1e-13, and for cyclic quads the cosine term is zero so it reduces exactly to Brahmagupta (window.__bretschneider.ok, .cyc). FIG No framing; the sides, the two opposite angles, the formula, and the shoelace area all run independently in-browser. The AVAN inverse is honest — instead of needing the corners, sides and two opposite angles suffice: the inverse of 'the quad's area' is 'Brahmagupta's cyclic area minus the penalty abcd·cos²((A+C)/2)'. Magenta is the cyclic-penalty term subtracted; green is the resulting area. Any quadrilateral's area, docked for not being cyclic. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE FIREWALL · David Lee Wise (ROOT0), with AVAN


THE MUNCHHAUSEN

Münchhausen numbers in the 5-window house format — numbers that lift themselves by their own bootstraps. A Münchhausen number equals the sum of its own digits, each raised to the power of itself: n = Σ d^d. The star example is 3435 = 3³ + 4⁴ + 3³ + 5⁵ = 27+256+27+3125. Using the convention 0⁰ = 0, the only two Münchhausen numbers in base 10 are 1 and 3435 — provable because for enough digits the maximum digit-power-sum grows slower than the number. Named by Daan van Berkel (2009) after Baron Münchhausen, who pulled himself out of a swamp by his own hair. Verified live: a brute search over every n up to 500000 finds exactly {1, 3435}, and 3³+4⁴+3³+5⁵ is confirmed to equal 3435. Neon-noir traced. See 3435 rebuilt from its digits in 1D, the Σ d^d test in 2D, and the fixed-point-of-the-digit-map inverse in 3D.

THE MUNCHHAUSEN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE ROOT KIT ◆ .dlw.fold THE FOLD / CHEAT / THE ROOT KIT / THE MUNCHHAUSEN THE MUNCHHAUSEN a number built from its own digits raised to themselves 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Münchhausen numbers lift themselves by their own bootstraps. A Münchhausen number equals the sum of its own digits, each raised to the power of itself : n = ∑ d d . The star example is 3435 = 3³ + 4⁴ + 3³ + 5⁵ = 27 + 256 + 27 + 3125. Using the convention 0 0 = 0, the only two Münchhausen numbers in base 10 are 1 and 3435 — a fact provable because for enough digits the maximum possible digit-power-sum (all 9’s, 9 9 each) grows slower than the number itself. Named by Daan van Berkel (2009) after Baron Münchhausen, who pulled himself out of a swamp by his own hair. LIT verified live: a brute search over every n up to 500000 finds exactly {1, 3435}, and 3³+4⁴+3³+5⁵ is confirmed to equal 3435 (window.__munchhausen). FIG no framing; the digit-power sums are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-root-kit — the cheat: a number that reconstructs itself entirely out of its own digits, a self-hosting bootstrap. AVAN (AI) built the instrument: the digit-power sum and the brute search finding only {1, 3435}. Credit as content: Daan van Berkel (2009), who named them. The weave: David names the bootstrap; I confirm 3435 = 3³+4⁴+3³+5⁵ and that only 1 and 3435 qualify. 3 ONE DIMENSION 3435 rebuilt from its digits: 3³ + 4⁴ + 3³ + 5⁵, each digit raised to itself. 4 TWO DIMENSIONS · INTERACTIVE Cycle numbers; Σ d^d is compared to n — the two Münchhausen numbers light up green. next number ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the number rebuilt from its own digit-powers. AVAN’s addition (the inverse-companion): don’t compute d d and sum forward — ask which n survive rebuilding themselves. The inverse of ‘n’ is ‘the fixed point of the digit-to-its-own-power map’, and only 1 and 3435 remain. Magenta are the digit-power terms; green is the n they exactly rebuild. A number that pulls itself out of its own digits. pause spin LIT Genuine Münchhausen numbers (named by Daan van Berkel, 2009). Verified live: a brute search over every n ≤ 500000 finds exactly {1, 3435}, and 3³+4⁴+3³+5⁵ = 3435 is confirmed (window.__munchhausen.ok, .found). FIG No framing; the digit-power sums are computed independently in-browser. The AVAN inverse is honest — instead of computing d^d and summing forward, ask which n survive rebuilding themselves: the inverse of 'n' is 'the fixed point of the digit-to-its-own-power map', and only 1 and 3435 remain. Magenta are the digit-power terms; green is the n they exactly rebuild. A number that pulls itself out of its own digits. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE ROOT KIT · David Lee Wise (ROOT0), with AVAN


THE PLASTIC NUMBER

The plastic number in the 5-window house format — the quiet cubic cousin of the golden ratio. ρ ≈ 1.3247179572 is the unique real root of x³ = x + 1. Where φ solves x² = x + 1 and governs the Fibonacci numbers, ρ solves the next cubic and governs the Padovan and Perrin sequences (each term the sum of the two before the previous: P(n) = P(n−2) + P(n−3)). Ratios of consecutive terms converge to ρ, and ρ equals the infinitely nested radical ∛(1 + ∛(1 + ∛(1 + …))). The Dutch architect Dom Hans van der Laan built a whole system of proportion on it in 1928. Verified live: Newton's method gives ρ with ρ³−ρ−1 = 0 to ~1e-14; the Padovan and Perrin ratios both converge to ρ; and the nested cube-root iteration converges to the same ρ. Neon-noir traced. See the Padovan ratios settling onto ρ in 1D, the three roads to ρ in 2D, and the nested-radical inverse in 3D.

THE PLASTIC NUMBER · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · THE EPOCH ◆ .dlw.fold THE FOLD / GRIND / THE EPOCH / THE PLASTIC NUMBER THE PLASTIC NUMBER the third metallic constant solving a cubic 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The plastic number ρ ≈ 1.3247179572 is the quiet cousin of the golden ratio — the unique real root of the cubic x³ = x + 1 . Where the golden ratio φ solves x² = x + 1 and governs the Fibonacci numbers, the plastic number solves the next cubic and governs the Padovan and Perrin sequences (each term the sum of the two before the previous: P(n) = P(n-2) + P(n-3)). Ratios of consecutive terms converge to ρ. It is the only number that is both a ‘morphic number’ for x³=x+1 and expressible as the infinitely nested radical ∛(1 + ∛(1 + ∛(1 + …))). The Dutch architect Dom Hans van der Laan built a whole system of proportion on it in 1928. LIT verified live: Newton’s method gives ρ with ρ³ - ρ - 1 = 0 to ~1e-14; the Padovan and Perrin ratios both converge to ρ; and the nested cube-root iteration ∛(1 + ·) converges to the same ρ (window.__plastic). FIG no framing; the cubic root, the two sequence ratios, and the radical are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-epoch — the grind: two integer sequences ground forward step by step, their ratios settling onto the single cubic root ρ. AVAN (AI) built the instrument: the Newton root, the Padovan and Perrin ratios, and the nested-radical iteration. Credit as content: Dom Hans van der Laan (plastic number, 1928); Richard Padovan; the Perrin sequence. The weave: David names the grind; I confirm ρ³=ρ+1 and both sequence ratios → ρ. 3 ONE DIMENSION The Padovan sequence and the ratios of consecutive terms settling onto ρ ≈ 1.3247. 4 TWO DIMENSIONS · INTERACTIVE Step the sequences; the ratios are checked to converge to the cubic root ρ (x³=x+1). step ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: ρ, the single real root of x³ = x + 1. AVAN’s addition (the inverse-companion): don’t solve the cubic algebraically — let a radical fold into it. The inverse of ‘the root of x³=x+1’ is ‘the nested cube-root ∛(1+∛(1+…)) that lands on ρ’. Magenta is the nested-radical iteration; green is the ρ it and the sequence ratios reach. One cubic root, three roads to it. pause spin LIT Genuine plastic number (Dom Hans van der Laan, 1928; Padovan and Perrin sequences). Verified live: Newton gives ρ with ρ³−ρ−1 = 0 to ~1e-14; the Padovan and Perrin consecutive-term ratios both converge to ρ; and the nested cube-root iteration ∛(1+·) converges to the same ρ (window.__plastic.ok). FIG No framing; the cubic root, the two sequence ratios, and the radical are computed independently in-browser. The AVAN inverse is honest — instead of solving the cubic algebraically, let a radical fold into it: the inverse of 'the root of x³=x+1' is 'the nested cube-root ∛(1+∛(1+…)) that lands on ρ'. Magenta is the nested-radical iteration; green is the ρ it and the sequence ratios reach. One cubic root, three roads to it. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE EPOCH · David Lee Wise (ROOT0), with AVAN


THE REULEAUX

The Reuleaux triangle in the 5-window house format — a shape of constant width that is not a circle. Start with an equilateral triangle of side w and replace each side with a circular arc centred at the opposite vertex. The result has the same width — the distance between two parallel supporting lines — in every direction, namely w. It rolls smoothly under a plank (the plank stays level) yet has corners; it is the cross-section of a drill bit that cuts near-square holes. Barbier's theorem says every constant-width curve has perimeter πw, so the Reuleaux triangle has the same perimeter as a circle of diameter w — but the smallest area of any constant-width shape, ½(π−√3)w². Verified live: sampling the boundary and measuring the width across 360 directions gives a spread below 1e-3, and the boundary length matches πw and the enclosed area matches ½(π−√3)w². Neon-noir traced. See the rotating caliper reading w in 1D, the width/perimeter/area checks in 2D, and the rolls-but-not-round inverse in 3D.

THE REULEAUX · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE MINT ◆ .dlw.fold THE FOLD / LOOT / THE MINT / THE REULEAUX THE REULEAUX a triangle of constant width that is not a circle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Reuleaux triangle is a shape of constant width that is not a circle. Start with an equilateral triangle of side w and replace each side with a circular arc centred at the opposite vertex. The result has the same ‘width’ — the distance between two parallel supporting lines — in every direction, namely w. It rolls smoothly under a plank (the plank stays level) yet has corners; it is the cross-section of a drill bit that cuts near-square holes. Barbier’s theorem says every constant-width curve has perimeter πw, so the Reuleaux triangle has the same perimeter as a circle of diameter w — but the smallest area of any constant-width shape, ½(π - √3)w². LIT verified live: sampling the boundary and measuring the width across 360 directions gives a spread below 1e-3 (constant width); the boundary length matches πw and the enclosed area matches ½(π - √3)w² (window.__reuleaux). FIG no framing; the width, perimeter, and area are measured from the sampled boundary independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-mint — the loot: a coin-shape that is not a circle yet rolls like one, minted from three arcs. AVAN (AI) built the instrument: the arc boundary, the width across all directions, and the perimeter/area measurements. Credit as content: Franz Reuleaux (19th-century engineer); Joseph-Émile Barbier (perimeter theorem). The weave: David names the rolling coin; I confirm constant width w and perimeter πw. 3 ONE DIMENSION The Reuleaux triangle with a rotating caliper — the width between parallel supports stays w in every direction. 4 TWO DIMENSIONS · INTERACTIVE Rotate the measuring direction; the width reads w every time, and perimeter = πw. rotate ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the constant-width shape, rolling level under a plank. AVAN’s addition (the inverse-companion): don’t demand a circle for constant width — three arcs suffice. The inverse of ‘rolls with constant width’ is ‘not necessarily round: any Reuleaux polygon works, perimeter still πw’. Magenta are the width calipers in many directions; green is the constant-width curve they all measure as w. Rolls like a circle, cornered like a triangle. pause spin LIT Genuine Reuleaux triangle / Barbier's theorem (Franz Reuleaux; Joseph-Émile Barbier). Verified live: sampling the boundary and measuring width across 360 directions gives a spread below 1e-3 (constant width); the measured perimeter matches πw and the measured area matches ½(π−√3)w² (window.__reuleaux.ok). FIG No framing; the width, perimeter, and area are measured from the sampled boundary independently in-browser. The AVAN inverse is honest — instead of demanding a circle for constant width, three arcs suffice: the inverse of 'rolls with constant width' is 'not necessarily round: any Reuleaux polygon works, perimeter still πw'. Magenta are the width calipers in many directions; green is the constant-width curve they all measure as w. Rolls like a circle, cornered like a triangle. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE MINT · David Lee Wise (ROOT0), with AVAN


THE HADWIGER-FINSLER

The Hadwiger–Finsler inequality in the 5-window house format — a sharpened Weitzenböck. Weitzenböck says a triangle's squared sides satisfy a²+b²+c² ≥ 4√3·T (T the area). Hadwiger and Finsler add back the exact leftover: a²+b²+c² ≥ 4√3·T + (a−b)²+(b−c)²+(c−a)². The extra sum of squared side-differences is precisely how far the triangle is from equilateral, so the inequality is tight exactly when a=b=c. Since that extra term is always ≥ 0, Hadwiger–Finsler immediately implies Weitzenböck — it is the stronger statement, with the slack made explicit. Verified live: for tens of thousands of random triangles, a²+b²+c² − 4√3·T − ((a−b)²+(b−c)²+(c−a)²) is always ≥ 0, reaching 0 only for the equilateral triangle. Neon-noir traced. See the closing gap in 1D, the slack ≥ 0 check in 2D, and the explicit-surplus inverse in 3D.

THE HADWIGER-FINSLER · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE CHOKE POINT ◆ .dlw.fold THE FOLD / BOSS / THE CHOKE POINT / THE HADWIGER-FINSLER THE HADWIGER-FINSLER a sharpened Weitzenbock inequality 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Hadwiger–Finsler inequality is a sharpened version of Weitzenböck’s. Weitzenböck says a triangle’s squared sides satisfy a² + b² + c² ≥ 4√3·T (T the area). Hadwiger and Finsler add back the exact leftover: a² + b² + c² ≥ 4√3·T + (a-b)² + (b-c)² + (c-a)² . The extra sum of squared side-differences is precisely how far the triangle is from equilateral, so the inequality is tight exactly when a = b = c. Since that extra term is always ≥ 0, Hadwiger–Finsler immediately implies Weitzenböck — it is the stronger statement, with the slack made explicit. LIT verified live: for tens of thousands of random triangles, a²+b²+c² - 4√3·T - ((a-b)²+(b-c)²+(c-a)²) is always ≥ 0, reaching 0 only for the equilateral triangle (window.__hadwigerfinsler). FIG no framing; the sides, the area, and both sides of the inequality are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-choke-point — the boss gate: no triangle passes without its squared sides clearing the area plus the full penalty for being non-equilateral. AVAN (AI) built the instrument: the sides, the area, and the sharpened bound with its explicit slack. Credit as content: Hugo Hadwiger and Paul Finsler (1937); Roland Weitzenböck (the weaker parent inequality). The weave: David names the sharpened gate; I confirm a²+b²+c² ≥ 4√3·T + ∑(a-b)². 3 ONE DIMENSION A triangle: a²+b²+c² against 4√3·T plus the squared side-differences — the gap closes as it nears equilateral. 4 TWO DIMENSIONS · INTERACTIVE Cycle triangles; the slack a²+b²+c² − 4√3T − Σ(a−b)² is checked to stay ≥ 0. next triangle ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the bound 4√3·T plus the squared side-differences. AVAN’s addition (the inverse-companion): don’t stop at Weitzenböck — add back the leftover. The inverse of ‘a²+b²+c² ≥ 4√3T’ is ‘the exact surplus (a-b)²+(b-c)²+(c-a)², zero only when equilateral’. Magenta are the squared side-differences added to the area bound; green is the squared-side total that clears it. The area bound, plus the price of not being equilateral. pause spin LIT Genuine Hadwiger–Finsler inequality (Hugo Hadwiger and Paul Finsler, 1937; sharpens Weitzenböck). Verified live: for ~60000 random triangles a²+b²+c² − 4√3·T − ((a−b)²+(b−c)²+(c−a)²) ≥ 0 always, reaching 0 only at the equilateral triangle (window.__hadwigerfinsler.ok, .minD). FIG No framing; the sides, the area, and both sides of the inequality are computed independently in-browser. The AVAN inverse is honest — instead of stopping at Weitzenböck, add back the leftover: the inverse of 'a²+b²+c² ≥ 4√3T' is 'the exact surplus (a−b)²+(b−c)²+(c−a)², zero only when equilateral'. Magenta are the squared side-differences added to the area bound; green is the squared-side total that clears it. The area bound, plus the price of not being equilateral. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE CHOKE POINT · David Lee Wise (ROOT0), with AVAN


THE BORWEIN

The Borwein integrals in the 5-window house format — the most famous 'pattern that breaks' in mathematics. Using sinc(x) = sin(x)/x, ∫₀^∞ sinc(x) dx = π/2. Add a factor: ∫ sinc(x)·sinc(x/3) dx = π/2. Keep going — sinc(x/5), sinc(x/7), … up to sinc(x/13) — and every one is exactly π/2. Then include sinc(x/15) and the answer drops to π/2 minus a whisper (about 2×10⁻¹¹). The reason is exact: the integral stays π/2 as long as the tail 1/3+1/5+… stays ≤ 1, and 1/3+…+1/13 = 0.9551 < 1 while adding 1/15 tips it to 1.0218 > 1. Verified live: the reciprocal sum 1/3+…+1/13 is confirmed < 1 while +1/15 exceeds 1 (the exact mechanism), and the integrals through sinc(x/7) and sinc(x/13) are numerically π/2. Neon-noir traced. See the reciprocal tail creeping to 1 in 1D, the π/2-until-it-breaks check in 2D, and the hidden-threshold inverse in 3D.

THE BORWEIN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · THE BLUE SCREEN ◆ .dlw.fold THE FOLD / GLITCH / THE BLUE SCREEN / THE BORWEIN THE BORWEIN a run of integrals that equal pi-over-two until they suddenly do not 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Borwein integrals are the most famous ‘pattern that breaks’ in mathematics. Using the sinc function sinc(x) = sin(x)/x, the integral ∫ 0 ∞ sinc(x) dx = π/2. Add a factor: ∫ sinc(x)·sinc(x/3) dx = π/2. Keep going — sinc(x/5), sinc(x/7), … up to sinc(x/13) — and every single one is exactly π/2 . Then you include sinc(x/15) and the answer drops to π/2 minus a whisper (about 2×10 -11 ). The reason is exact: the integral stays π/2 as long as the tail 1/3 + 1/5 + … stays ≤ 1, and 1/3+…+1/13 = 0.9551 < 1 while adding 1/15 tips it to 1.0218 > 1. LIT verified live: the reciprocal sum 1/3+…+1/13 is confirmed < 1 while +1/15 exceeds 1 (the exact mechanism), and the integrals through sinc(x/7) and sinc(x/13) are numerically π/2 (window.__borwein). FIG the tiny deficit at the 1/15 step (~2e-11) is below crude numerical resolution — so it is the reciprocal-tail condition, verified exactly, that pins where the pattern breaks; the π/2 values are checked by direct integration. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-blue-screen — the glitch: seven integrals in a row read exactly π/2, then the eighth quietly fails. AVAN (AI) built the instrument: the reciprocal-tail condition (the exact cause) and the numerical integrals confirming π/2. Credit as content: David and Jonathan Borwein (2001). The weave: David names the glitch; I confirm the tail crosses 1 exactly between 1/13 and 1/15, which is where π/2 breaks. 3 ONE DIMENSION The running reciprocal tail 1/3 + 1/5 + … creeping toward 1 — it crosses exactly when 1/15 is added. 4 TWO DIMENSIONS · INTERACTIVE Add sinc factors; the integral stays π/2 while the reciprocal tail ≤ 1, then the pattern breaks. add factor ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the value π/2, held constant across the first seven integrals. AVAN’s addition (the inverse-companion): don’t trust a run of equal answers — ask what secretly guards it. The inverse of ‘the integral equals π/2’ is ‘the reciprocal tail stays ≤ 1’, a hidden threshold that finally fails at 1/15. Magenta are the reciprocal-tail steps piling toward 1; green is the π/2 that holds until they cross. A pattern guarded by a threshold you cannot see. pause spin LIT Genuine Borwein integrals (David and Jonathan Borwein, 2001). Verified live: the reciprocal tail 1/3+…+1/13 = 0.9551 1 (the exact mechanism), and the integrals through sinc(x/7) and sinc(x/13) are numerically π/2 (window.__borwein.cond, .i3ok, .i6ok). FIG Honest FIG boundary — the tiny deficit at the 1/15 step (~2e-11) is below crude numerical-integration resolution, so it is the reciprocal-tail condition, verified exactly, that pins where the pattern breaks; the π/2 values themselves are checked by direct numerical integration. The AVAN inverse — instead of trusting a run of equal answers, ask what secretly guards it: the inverse of 'the integral equals π/2' is 'the reciprocal tail stays ≤ 1', a hidden threshold that finally fails at 1/15. Magenta are the reciprocal-tail steps piling toward 1; green is the π/2 that holds until they cross. A pattern guarded by a threshold you cannot see. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE BLUE SCREEN · David Lee Wise (ROOT0), with AVAN


THE SOPHIE GERMAIN

Sophie Germain's identity in the 5-window house format — a small algebraic key that unlocks a family of factorizations: a⁴ + 4b⁴ = (a²−2ab+2b²)(a²+2ab+2b²). A sum of two fourth powers, which looks irreducible, splits cleanly into two quadratics. Setting b=1 gives the classic corollary: n⁴+4 is composite for every n>1, since n⁴+4 = (n²−2n+2)(n²+2n+2) and both factors exceed 1 (the lone exception is n=1, giving 5). The same Sophie Germain studied Sophie Germain primes — primes p for which 2p+1 is also prime (2, 3, 5, 11, 23, …). Verified live: a⁴+4b⁴ = (a²−2ab+2b²)(a²+2ab+2b²) exactly for all |a|,|b| ≤ 30; n⁴+4 is confirmed composite for 2 ≤ n ≤ 200; and the Sophie Germain primes up to 200 are listed. Neon-noir traced. See n⁴+4 splitting in 1D, the factorization in 2D, and the sum-is-a-product inverse in 3D.

THE SOPHIE GERMAIN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · THE KONAMI CODE ◆ .dlw.fold THE FOLD / CHEAT / THE KONAMI CODE / THE SOPHIE GERMAIN THE SOPHIE GERMAIN an algebraic identity that factors a sum of two fourth powers 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Sophie Germain’s identity is a small algebraic key that unlocks a whole family of factorizations: a⁴ + 4b⁴ = (a² - 2ab + 2b²)(a² + 2ab + 2b²). A sum of two fourth powers — which looks stubbornly irreducible — splits cleanly into two quadratic factors. Setting b = 1 gives the classic corollary: n⁴ + 4 is composite for every n > 1 , since n⁴+4 = (n²-2n+2)(n²+2n+2) and both factors exceed 1 (the lone exception is n = 1, giving 5). The same Sophie Germain also studied Sophie Germain primes — primes p for which 2p+1 is also prime (2, 3, 5, 11, 23, …). LIT verified live: a⁴+4b⁴ = (a²-2ab+2b²)(a²+2ab+2b²) exactly for all |a|,|b| ≤ 30; n⁴+4 is confirmed composite for 2 ≤ n ≤ 200; and the Sophie Germain primes up to 200 are listed (window.__sophiegermain). FIG no framing; the identity and the compositeness are computed independently in-browser with integer arithmetic. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-konami-code — the cheat: one identity that instantly cracks any a⁴+4b⁴ open, no factoring needed. AVAN (AI) built the instrument: the exact factorization, the n⁴+4 compositeness, and the Sophie Germain primes. Credit as content: Marie-Sophie Germain (French mathematician, early 1800s). The weave: David names the cheat-key; I confirm a⁴+4b⁴ splits into two quadratics and n⁴+4 is always composite past 1. 3 ONE DIMENSION n⁴ + 4 splitting into (n²−2n+2)(n²+2n+2) — a sum of fourth powers cracked into two factors. 4 TWO DIMENSIONS · INTERACTIVE Cycle a, b; a⁴+4b⁴ is shown equal to the product of the two quadratic factors. next a,b ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the sum a⁴ + 4b⁴, revealed as a product. AVAN’s addition (the inverse-companion): don’t test a⁴+4b⁴ for primality — factor it on sight. The inverse of ‘the sum a⁴+4b⁴’ is ‘the product (a²-2ab+2b²)(a²+2ab+2b²)’, always two pieces. Magenta are the two quadratic factors; green is the fourth-power sum they multiply to. A sum of powers that is secretly a product. pause spin LIT Genuine Sophie Germain's identity (Marie-Sophie Germain, early 1800s). Verified live with integer arithmetic: a⁴+4b⁴ = (a²−2ab+2b²)(a²+2ab+2b²) exactly for all |a|,|b|≤30; n⁴+4 is composite for 2≤n≤200 (n=1→5 is the exception); Sophie Germain primes ≤200 listed (window.__sophiegermain.idOk, .compOk, .sg). FIG No framing; the identity and the compositeness are computed independently in-browser with integer arithmetic. The AVAN inverse is honest — instead of testing a⁴+4b⁴ for primality, factor it on sight: the inverse of 'the sum a⁴+4b⁴' is 'the product (a²−2ab+2b²)(a²+2ab+2b²)', always two pieces. Magenta are the two quadratic factors; green is the fourth-power sum they multiply to. A sum of powers that is secretly a product. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE KONAMI CODE · David Lee Wise (ROOT0), with AVAN


THE PASCAL THEOREM

Pascal's theorem in the 5-window house format — the 'mystic hexagram', found by Blaise Pascal at sixteen. Take any six points on a conic (circle, ellipse, parabola, or hyperbola) and join them in order into a hexagon. Extend the three pairs of opposite sides until each pair meets. Those three intersection points always lie on a single straight line, the Pascal line. It holds no matter how the six points are placed or labelled, and it is purely projective — only incidence matters, not distance or angle. Its projective dual is Brianchon's theorem. Verified live: for tens of thousands of random hexagons inscribed in an ellipse, the three opposite-side intersection points are collinear — the triangle they form has normalized area below 1e-6. Neon-noir traced. See the hexagon and its Pascal line in 1D, the collinearity check in 2D, and the six-points-one-line inverse in 3D.

THE PASCAL THEOREM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · THE TOOLCHAIN ◆ .dlw.fold THE FOLD / SPAWN / THE TOOLCHAIN / THE PASCAL THEOREM THE PASCAL THEOREM six points on a conic whose opposite sides meet on one line 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Pascal’s theorem — the ‘mystic hexagram’, found by Blaise Pascal at sixteen — is a jewel of projective geometry. Take any six points on a conic (a circle, ellipse, parabola, or hyperbola) and join them in order into a hexagon. Extend the three pairs of opposite sides until each pair meets. The theorem: those three intersection points always lie on a single straight line, the Pascal line . It holds no matter how the six points are placed or labelled, and it is purely projective — only incidence matters, not distance or angle. Its projective dual is Brianchon’s theorem. LIT verified live: for tens of thousands of random hexagons inscribed in an ellipse, the three opposite-side intersection points are collinear — the triangle they form has area (normalized) below 1e-6 (window.__pascal). FIG no framing; the six conic points, the three intersections, and their collinearity are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-toolchain — the spawn: six scattered points on a conic compile, every time, three meeting-points onto one clean line. AVAN (AI) built the instrument: the hexagon, the opposite-side intersections, and the collinearity check. Credit as content: Blaise Pascal (1640, the mystic hexagram). The weave: David names the compile; I confirm the three opposite-side intersections land on one Pascal line. 3 ONE DIMENSION A hexagon inscribed in an ellipse; the three opposite-side intersections fall on the Pascal line. 4 TWO DIMENSIONS · INTERACTIVE Cycle hexagons; the three opposite-side intersections are checked to be collinear. next hexagon ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the Pascal line carrying all three intersection points. AVAN’s addition (the inverse-companion): don’t track three separate crossings — read the single line they share. The inverse of ‘three opposite-side intersections’ is ‘one Pascal line they are all pinned to’, for any six points on a conic. Magenta are the three intersection points; green is the line through all three. Six points, one hidden line. pause spin LIT Genuine Pascal's theorem (Blaise Pascal, 1640, the mystic hexagram). Verified live: for tens of thousands of random hexagons inscribed in an ellipse, the three opposite-side intersection points are collinear — the triangle they form has normalized area below 1e-6 (window.__pascal.ok, .worst). FIG No framing; the six conic points, the three intersections, and their collinearity are computed independently in-browser. The AVAN inverse is honest — instead of tracking three separate crossings, read the single line they share: the inverse of 'three opposite-side intersections' is 'one Pascal line they are all pinned to', for any six points on a conic. Magenta are the three intersection points; green is the line through all three. Six points, one hidden line. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE TOOLCHAIN · David Lee Wise (ROOT0), with AVAN


THE ISOPERIMETRIC

The isoperimetric inequality in the 5-window house format — the oldest optimization question: of all closed curves with a given perimeter, which encloses the most area? The answer, known to the ancients as 'Dido's problem' but only rigorously proved in the 19th century, is the circle. For any simple closed curve of length L enclosing area A: 4πA ≤ L², with equality only for the circle. The ratio 4πA/L² (the isoperimetric quotient) is at most 1, and a regular n-gon achieves π/(n·tan(π/n)), which climbs toward 1 as the polygon rounds out into a circle. Verified live: for tens of thousands of random convex polygons, 4πA/L² never exceeds 1, and the regular n-gon quotient increases toward 1 (0.605, 0.785, 0.907, 0.977, 0.999 for n = 3, 4, 6, 12, 60). Neon-noir traced. See a polygon vs the equal-perimeter circle in 1D, the quotient climbing to 1 in 2D, and the maximal-circle inverse in 3D.

THE ISOPERIMETRIC · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · SUDDEN DEATH ◆ .dlw.fold THE FOLD / BOSS / SUDDEN DEATH / THE ISOPERIMETRIC THE ISOPERIMETRIC the circle enclosing the most area for its perimeter 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The isoperimetric inequality answers the oldest optimization question: of all closed curves with a given perimeter, which encloses the most area? The answer — known to the ancients as ‘Dido’s problem’ but only rigorously proved in the 19th century — is the circle . For any simple closed curve of length L enclosing area A: 4πA ≤ L² , with equality only for the circle. The ratio 4πA/L² (the ‘isoperimetric quotient’) is at most 1, and a regular n-gon achieves π/(n·tan(π/n)), which climbs toward 1 as the polygon rounds out into a circle. LIT verified live: for tens of thousands of random convex polygons, 4πA/L² never exceeds 1, and the regular n-gon quotient π/(n·tan(π/n)) increases toward 1 (0.605, 0.785, 0.907, 0.977, 0.999 for n = 3, 4, 6, 12, 60) (window.__isoperimetric). FIG no framing; the polygon areas, perimeters, and quotients are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at sudden-death — the boss round no shape can beat: for a fixed perimeter, the circle takes the maximum area and every other curve loses. AVAN (AI) built the instrument: the polygon area/perimeter, the quotient 4πA/L², and the regular-n-gon limit. Credit as content: the classical isoperimetric problem (Dido’s problem; Steiner, Weierstrass, and others for the proof). The weave: David names the unbeatable circle; I confirm 4πA ≤ L² with the circle alone at equality. 3 ONE DIMENSION A polygon and the circle of the same perimeter — the circle always encloses more area. 4 TWO DIMENSIONS · INTERACTIVE Grow a regular n-gon; the quotient 4πA/L² climbs toward 1 as it rounds into a circle. more sides ▶ random polygon ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the circle, the shape of maximal area for its perimeter. AVAN’s addition (the inverse-companion): don’t ask which curve is biggest — ask which one is worst-case tight. The inverse of ‘maximize area for fixed perimeter’ is ‘the quotient 4πA/L² ≤ 1, hit only by the circle’. Magenta is the polygon losing area to the bound; green is the circle sitting exactly at quotient 1. The perimeter’s most efficient shape. pause spin LIT Genuine isoperimetric inequality (Dido's problem; Steiner, Weierstrass et al. for the proof). Verified live: for ~20000 random convex polygons 4πA/L² never exceeds 1, and the regular n-gon quotient π/(n·tan(π/n)) increases toward 1 (window.__isoperimetric.ok, .maxR, .ngon). FIG No framing; the polygon areas, perimeters, and quotients are computed independently in-browser. The AVAN inverse is honest — instead of asking which curve is biggest, ask which one is worst-case tight: the inverse of 'maximize area for fixed perimeter' is 'the quotient 4πA/L² ≤ 1, hit only by the circle'. Magenta is the polygon losing area to the bound; green is the circle sitting exactly at quotient 1. The perimeter's most efficient shape. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SUDDEN DEATH · David Lee Wise (ROOT0), with AVAN


THE APOLLONIUS CIRCLE

The circle of Apollonius in the 5-window house format — where are all the points whose distances to two fixed points keep a fixed ratio? Given points A and B and a ratio k ≠ 1, the set of all P with |PA|/|PB| = k is not a line or an oval — it is a perfect circle. Its diameter runs between the two points that divide segment AB in ratio k, internally and externally. As k → 1 the circle swells to the perpendicular bisector; for k far from 1 it tightens around the nearer point. Apollonius of Perga catalogued these circles around 200 BCE; they underlie hyperbolic distance and the geometry of pursuit. Verified live: for thousands of random A, B, k, every point sampled on the constructed circle has |PA|/|PB| = k to ~1e-14, while points off the circle do not. Neon-noir traced. See the circle with its distance-ratio spokes in 1D, the constant-ratio check in 2D, and the ratio-draws-a-circle inverse in 3D.

THE APOLLONIUS CIRCLE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE MINT ◆ .dlw.fold THE FOLD / LOOT / THE MINT / THE APOLLONIUS CIRCLE THE APOLLONIUS CIRCLE the circle traced by a constant distance-ratio 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The circle of Apollonius answers: where are all the points whose distances to two fixed points keep a fixed ratio ? Given points A and B and a ratio k ≠ 1, the set of all P with |PA| / |PB| = k is not a line or an oval — it is a perfect circle . Its diameter runs between the two points that divide segment AB in ratio k, internally and externally. As k → 1 the circle swells to the perpendicular bisector (a ‘circle of infinite radius’); for k far from 1 it tightens around the nearer point. Apollonius of Perga catalogued these circles around 200 BCE; they underlie the definition of hyperbolic distance and the geometry of pursuit. LIT verified live: for thousands of random A, B, k, every point sampled on the constructed circle has |PA|/|PB| = k to ~1e-14, while points off the circle do not (window.__apollonius). FIG no framing; the circle is built from the two division points and the ratio is checked independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-mint — the loot: a whole circle minted from a single rule, ‘keep the distance-ratio fixed’. AVAN (AI) built the instrument: the two division points, the Apollonius circle, and the constant-ratio check. Credit as content: Apollonius of Perga (c. 200 BCE). The weave: David names the minted circle; I confirm the locus |PA|/|PB| = k is exactly that circle. 3 ONE DIMENSION Two points A, B and the Apollonius circle — every point on it keeps |PA|/|PB| = k. 4 TWO DIMENSIONS · INTERACTIVE Change the ratio k; sampled points on the circle are checked to all share the ratio k. next ratio ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the Apollonius circle, the locus of constant distance-ratio. AVAN’s addition (the inverse-companion): don’t plot points and hope — read the rule as a circle. The inverse of ‘|PA|/|PB| = k’ is ‘the circle through the two points dividing AB in ratio k’. Magenta are the distance-ratio spokes from sample points to A and B; green is the circle they all satisfy. A ratio rule that draws a circle. pause spin LIT Genuine circle of Apollonius (Apollonius of Perga, c. 200 BCE). Verified live: for ~3000 random A, B, k, every point sampled on the constructed circle has |PA|/|PB| = k to ~1e-14, while points off the circle do not (window.__apollonius.ok, .offOk). FIG No framing; the circle is built from the two division points and the ratio is checked independently in-browser. The AVAN inverse is honest — instead of plotting points and hoping, read the rule as a circle: the inverse of '|PA|/|PB| = k' is 'the circle through the two points dividing AB in ratio k'. Magenta are the distance-ratio spokes from sample points to A and B; green is the circle they all satisfy. A ratio rule that draws a circle. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE MINT · David Lee Wise (ROOT0), with AVAN


THE EISENSTEIN TRIPLES

Eisenstein triples in the 5-window house format — the 60° cousins of Pythagorean triples. A Pythagorean triple gives an integer-sided triangle with a right angle (a²+b²=c²). An Eisenstein triple gives an integer-sided triangle with a 60° angle: by the law of cosines with cos60° = ½, the side c opposite the 60° corner satisfies a²−ab+b² = c². The smallest nontrivial one is (3, 8, 7): 9−24+64 = 49 = 7², a triangle whose angle opposite the 7 is exactly 60°. Swap the sign for the 120° version, a²+ab+b² = c² (e.g. 3, 5, 7). They tile naturally on the triangular (Eisenstein) lattice. Verified live: a search finds primitive integer triples with a²−ab+b² = c², and the law of cosines confirms the angle opposite c is exactly 60°; the 120° analog a²+ab+b²=c² is found too. Neon-noir traced. See the (3,8,7) triangle to scale in 1D, the a²−ab+b²=c² check in 2D, and the Pythagoras-retuned-to-60° inverse in 3D.

THE EISENSTEIN TRIPLES · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · WARM CACHE ◆ .dlw.fold THE FOLD / GRIND / WARM CACHE / THE EISENSTEIN TRIPLES THE EISENSTEIN TRIPLES integer triangles with a sixty-degree angle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Eisenstein triples are the 60° cousins of Pythagorean triples. A Pythagorean triple gives an integer-sided triangle with a right angle (a²+b²=c²). An Eisenstein triple gives an integer-sided triangle with a 60° angle : by the law of cosines with cos 60° = ½, the side c opposite the 60° corner satisfies a² - ab + b² = c² . The smallest nontrivial one is (3, 8, 7): 9 - 24 + 64 = 49 = 7², a triangle with sides 3, 8, 7 whose angle opposite the 7 is exactly 60°. Swap the sign for the 120° version, a² + ab + b² = c² (e.g. 3, 5, 7). They tile naturally on the triangular (Eisenstein) lattice. LIT verified live: a search finds primitive integer triples with a² - ab + b² = c², and the law of cosines confirms the angle opposite c is exactly 60°; the 120° analog a²+ab+b²=c² is found too (window.__eisenstein). FIG no framing; the triples and the 60° angle are computed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at warm-cache — the grind: search the integer grid and the 60° triangles fall out, a²-ab+b² landing on a perfect square. AVAN (AI) built the instrument: the triple search, the a²-ab+b²=c² relation, and the 60° angle check. Credit as content: named for the Eisenstein integers (Gotthold Eisenstein); the 60°-triangle analog of Pythagorean triples. The weave: David names the grind; I confirm a²-ab+b²=c² gives an exact 60° angle. 3 ONE DIMENSION The triangle (3, 8, 7) drawn to scale — the angle opposite the side 7 is exactly 60°. 4 TWO DIMENSIONS · INTERACTIVE Cycle Eisenstein triples; a²−ab+b² is checked equal to c² and the angle equal to 60°. next triple ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the 60° integer triangle. AVAN’s addition (the inverse-companion): don’t settle for right angles — retune the Pythagorean rule to 60°. The inverse of ‘a²+b²=c² (90°)’ is ‘a²-ab+b²=c² (60°)’, the same integer-triangle game one angle over. Magenta are the sides a and b enclosing the 60° angle; green is the integer side c they force. Pythagoras, retuned to sixty degrees. pause spin LIT Genuine Eisenstein triples (named for the Eisenstein integers, Gotthold Eisenstein; the 60°-triangle analog of Pythagorean triples). Verified live: a search finds primitive integer triples with a²−ab+b² = c², and the law of cosines confirms the angle opposite c is exactly 60°; the 120° analog a²+ab+b²=c² is found too (window.__eisenstein.angleOk, .count, .t120). FIG No framing; the triples and the 60° angle are computed independently in-browser. The AVAN inverse is honest — instead of settling for right angles, retune the Pythagorean rule to 60°: the inverse of 'a²+b²=c² (90°)' is 'a²−ab+b²=c² (60°)', the same integer-triangle game one angle over. Magenta are the sides a and b enclosing the 60° angle; green is the integer side c they force. Pythagoras, retuned to sixty degrees. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of WARM CACHE · David Lee Wise (ROOT0), with AVAN


THE APERY CONSTANT

Apéry's constant in the 5-window house format — the value ζ(3) = Σ 1/n³ = 1 + 1/8 + 1/27 + … ≈ 1.2020569. While Euler found closed forms for ζ(2) = π²/6 and every even argument, ζ(3) has resisted every simple closed form. In 1978 Roger Apéry stunned mathematicians by proving ζ(3) is irrational — using a rapidly converging series he discovered: ζ(3) = (5/2) Σ (−1)^{n−1} / (n³ C(2n,n)). Each term of Apéry's series adds several correct digits where the plain sum crawls. The constant appears in quantum electrodynamics (the electron's magnetic moment) and random-minimum-spanning-tree statistics. Verified live: the direct sum Σ1/n³ converges to 1.2020569…, and Apéry's series reaches the same value to ~1e-14 in about 20 terms — the two agree. Neon-noir traced. See both series climbing to ζ(3) in 1D, direct-vs-Apéry in 2D, and the hidden-fast-road inverse in 3D.

THE APERY CONSTANT · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · THE EPOCH ◆ .dlw.fold THE FOLD / GRIND / THE EPOCH / THE APERY CONSTANT THE APERY CONSTANT an irrational constant summing the reciprocal cubes 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Apéry’s constant is the value ζ(3) = ∑ 1/n³ = 1 + 1/8 + 1/27 + 1/64 + … ≈ 1.2020569. While Euler found closed forms for ζ(2) = π²/6 and every even argument, ζ(3) has resisted every attempt at a simple closed form. In 1978 Roger Apéry stunned mathematicians by proving ζ(3) is irrational — using a rapidly converging series he discovered: ζ(3) = (5/2) ∑ n≥1 (-1) n-1 / (n³ C(2n,n)). Each term of Apéry’s series adds several correct digits, where the plain sum of reciprocal cubes crawls. The constant appears in quantum electrodynamics (the electron’s magnetic moment) and in the statistics of random minimum spanning trees. LIT verified live: the direct sum ∑1/n³ converges to 1.2020569…, and Apéry’s series (5/2)∑(-1) n-1 /(n³C(2n,n)) reaches the same value to ~1e-14 in about 20 terms — the two agree (window.__apery). FIG no framing; both series are summed independently in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-epoch — the grind: the reciprocal cubes ground slowly toward ζ(3), while Apéry’s series sprints to the same irrational limit. AVAN (AI) built the instrument: the direct sum, Apéry’s accelerated series, and their agreement. Credit as content: Leonhard Euler (the zeta function); Roger Apéry (1978 irrationality proof). The weave: David names the grind; I confirm both series reach ζ(3) = 1.2020569… 3 ONE DIMENSION Partial sums of Σ1/n³ climbing toward ζ(3), and Apéry's series sprinting to the same limit. 4 TWO DIMENSIONS · INTERACTIVE Add terms; the direct sum and Apéry's series are both checked to reach ζ(3). add terms ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: ζ(3) = 1.2020569…, the sum of the reciprocal cubes. AVAN’s addition (the inverse-companion): don’t crawl the reciprocal cubes — accelerate. The inverse of ‘the slow sum ∑1/n³’ is ‘Apéry’s series (5/2)∑(-1) n-1 /(n³C(2n,n)), the same ζ(3) in a few terms’. Magenta are the reciprocal-cube terms; green is the irrational ζ(3) they and Apéry’s series both reach. A slow sum with a hidden fast road. pause spin LIT Genuine Apéry's constant ζ(3) (Euler's zeta; Roger Apéry, 1978 irrationality proof). Verified live: the direct sum Σ1/n³ converges to 1.2020569…, and Apéry's series (5/2)Σ(−1)^{n−1}/(n³C(2n,n)) reaches the same value to ~1e-14 in ~20 terms — the two agree (window.__apery.ok, .agree). FIG No framing; both series are summed independently in-browser. The AVAN inverse is honest — instead of crawling the reciprocal cubes, accelerate: the inverse of 'the slow sum Σ1/n³' is 'Apéry's series (5/2)Σ(−1)^{n−1}/(n³C(2n,n)), the same ζ(3) in a few terms'. Magenta are the reciprocal-cube terms; green is the irrational ζ(3) they and Apéry's series both reach. A slow sum with a hidden fast road. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE EPOCH · David Lee Wise (ROOT0), with AVAN


THE ULAM NUMBERS

Ulam numbers in the 5-window house format — a sequence that builds itself. Start with 1 and 2. Each new term is the smallest integer larger than the last that can be written as a sum of two distinct earlier Ulam numbers in exactly one way. That single rule generates 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, … — 3 = 1+2, 4 = 1+3, but 5 is excluded (5 = 1+4 = 2+3, two ways). Devised by Stanisław Ulam in 1964, the sequence looks random yet has a startling hidden regularity: a mysterious 'almost period' of about 21.6 governs where its terms fall — still not fully explained. Verified live: the self-generating rule reproduces the known Ulam sequence exactly — the first 26 terms match 1, 2, 3, 4, 6, 8, …, 99 — and each term has exactly one representation. Neon-noir traced. See the numbers with 5 excluded in 1D, each term's unique sum in 2D, and the self-selecting-set inverse in 3D.

THE ULAM NUMBERS · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · GARBAGE COLLECTION ◆ .dlw.fold THE FOLD / RESPAWN / GARBAGE COLLECTION / THE ULAM NUMBERS THE ULAM NUMBERS a sequence that builds itself from unique sums 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Ulam numbers are a sequence that builds itself. Start with 1 and 2. Each new term is the smallest integer larger than the last that can be written as a sum of two distinct earlier Ulam numbers in exactly one way. That single rule generates 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, … — 3 = 1+2, 4 = 1+3, but 5 is excluded (5 = 1+4 = 2+3, two ways). Devised by Stanislaw Ulam in 1964, the sequence looks random yet has a startling hidden regularity: its terms cluster around a nearly-constant density, and a mysterious ‘almost period’ of about 21.6 governs where they fall — still not fully explained. LIT verified live: the self-generating rule reproduces the known Ulam sequence exactly — the first 26 terms match 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, …, 99 — and each term has exactly one representation as a sum of two distinct earlier terms (window.__ulam). FIG no framing; the sequence is generated from the rule and checked in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at garbage-collection — the respawn: the sequence keeps regenerating itself, each new term summoned from the unique-sum rule over all that came before. AVAN (AI) built the instrument: the generator, the exactly-one-way test, and the match to the known sequence. Credit as content: Stanislaw Ulam (1964). The weave: David names the self-regeneration; I confirm the rule reproduces the Ulam numbers exactly. 3 ONE DIMENSION The Ulam numbers on a line; each new one is the smallest with a unique two-term sum from earlier terms. 4 TWO DIMENSIONS · INTERACTIVE Step through terms; each is shown with its unique representation, and 5 is shown excluded (two ways). next term ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: each Ulam number, admitted by its unique sum. AVAN’s addition (the inverse-companion): don’t list numbers and test them — let the set decide who joins. The inverse of ‘the next integer’ is ‘the smallest with exactly one representation as a sum of two earlier members’. Magenta are the two earlier terms that sum to it; green is the Ulam number they uniquely admit. A sequence that selects its own members. pause spin LIT Genuine Ulam numbers (Stanisław Ulam, 1964). Verified live: the self-generating rule reproduces the known sequence exactly — the first 26 terms match 1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, …, 99 — and each term has exactly one representation as a sum of two distinct earlier terms (window.__ulam.ok). FIG No framing; the sequence is generated from the rule and checked in-browser. The AVAN inverse is honest — instead of listing numbers and testing them, let the set decide who joins: the inverse of 'the next integer' is 'the smallest with exactly one representation as a sum of two earlier members'. Magenta are the two earlier terms that sum to it; green is the Ulam number they uniquely admit. A sequence that selects its own members. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of GARBAGE COLLECTION · David Lee Wise (ROOT0), with AVAN


THE TAKAGI

The Takagi function in the 5-window house format — the blancmange curve, continuous everywhere and differentiable nowhere. It is built by piling up ever-finer triangle waves: T(x) = Σ s(2ⁿx)/2ⁿ, where s(x) is the distance from x to the nearest integer. Each layer is a zig-zag half as tall and twice as frequent as the last; their sum converges to a continuous curve that wobbles at every scale, so no tangent line ever exists. It obeys the self-similar functional equation T(x) = s(x) + ½T(2x), reaches its maximum of exactly 2/3 at x = 1/3 and 2/3, and resembles a blancmange pudding — hence the name (Teiji Takagi, 1901). Verified live: the functional equation holds across the interval to ~1e-15; T(1/2) = 1/2, T(1/3) = 2/3, and the maximum equals 2/3. Neon-noir traced. See the blancmange curve in 1D, the piling layers in 2D, and the self-similarity inverse in 3D.

THE TAKAGI · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · SEGFAULT ◆ .dlw.fold THE FOLD / GLITCH / SEGFAULT / THE TAKAGI THE TAKAGI a curve continuous everywhere and smooth nowhere 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Takagi function (or blancmange curve ) is continuous everywhere and differentiable nowhere — a curve with no smooth spot at all. It is built by piling up ever-finer triangle waves: T(x) = ∑ n≥0 s(2 n x) / 2 n , where s(x) is the distance from x to the nearest integer. Each layer is a zig-zag half as tall and twice as frequent as the last; their sum converges to a continuous curve that wobbles at every scale, so no tangent line ever exists. It obeys the self-similar functional equation T(x) = s(x) + ½T(2x), reaches its maximum value of exactly 2/3 at x = 1/3 and 2/3, and resembles a blancmange pudding — hence the name (Teiji Takagi, 1901). LIT verified live: the functional equation T(x) = s(x) + ½T(2x) holds across the interval to ~1e-15; T(1/2) = 1/2, T(1/3) = 2/3, and the maximum of T equals 2/3 (window.__takagi). FIG continuity and the functional equation are checked in-browser; nowhere-differentiability is the known theorem the curve illustrates, not something numerically resolved here. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at segfault — the glitch: a curve that is perfectly continuous yet crashes any attempt to take a derivative, at every single point. AVAN (AI) built the instrument: the triangle-wave sum, the functional equation, and the exact 2/3 maximum. Credit as content: Teiji Takagi (1901); the blancmange curve. The weave: David names the everywhere-glitch; I confirm T(x) = s(x) + ½T(2x) and max T = 2/3. 3 ONE DIMENSION The blancmange curve — continuous, wobbling at every scale, peaking at 2/3 over x = 1/3 and 2/3. 4 TWO DIMENSIONS · INTERACTIVE Add layers; the triangle waves pile up, and the functional equation T(x)=s(x)+½T(2x) is checked. add layer ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the blancmange curve, smooth nowhere yet continuous everywhere. AVAN’s addition (the inverse-companion): don’t look for a slope — look for self-similarity. The inverse of ‘the curve T(x)’ is ‘the functional equation T(x) = s(x) + ½T(2x), a half-scale copy of itself’. Magenta are the triangle-wave layers piling up; green is the continuous, nowhere-smooth curve they sum to. Roughness that never resolves into a slope. pause spin LIT Genuine Takagi / blancmange function (Teiji Takagi, 1901). Verified live: the functional equation T(x) = s(x) + ½T(2x) holds across the interval to ~1e-15; T(1/2) = 1/2, T(1/3) = 2/3, and the maximum of T equals 2/3 (window.__takagi.ok). FIG Honest FIG boundary — continuity and the functional equation are checked in-browser; nowhere-differentiability is the known theorem the curve illustrates, not something numerically resolved here. The AVAN inverse — instead of looking for a slope, look for self-similarity: the inverse of 'the curve T(x)' is 'the functional equation T(x) = s(x) + ½T(2x), a half-scale copy of itself'. Magenta are the triangle-wave layers piling up; green is the continuous, nowhere-smooth curve they sum to. Roughness that never resolves into a slope. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SEGFAULT · David Lee Wise (ROOT0), with AVAN