THE RAMANUJAN SUM

Ramanujan's sum in the 5-window house format — c_q(n) adds up the primitive q-th roots of unity raised to the n-th power: c_q(n) = Σ_{gcd(a,q)=1} e^{2πi·an/q}. Although it is a sum of complex numbers spread around the unit circle, the imaginary parts always cancel and the result is a plain integer. Ramanujan showed it has a beautiful arithmetic form: c_q(n) = Σ_{d|gcd(n,q)} d·μ(q/d), a sum over the common divisors weighted by the Möbius function. It is the building block of 'Ramanujan–Fourier' expansions that turn arithmetic functions into trigonometric series. Verified live: for all q up to 60 and n up to 40, the direct primitive-root sum equals the Möbius-divisor formula to ~1e-13 and is always an integer — c_12(0)=φ(12)=4, c_9(3)=−3. Neon-noir traced. See the primitive roots summing on the circle in 1D, direct vs formula in 2D, and the arithmetic-sum inverse in 3D.

THE RAMANUJAN SUM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE STASH ◆ .dlw.fold THE FOLD / LOOT / THE STASH / THE RAMANUJAN SUM THE RAMANUJAN SUM roots of unity summing to an integer by Möbius 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Ramanujan’s sum c q (n) adds up the primitive q-th roots of unity raised to the n-th power: c q (n) = ∑ gcd(a,q)=1 e 2πi·an/q . Although it is a sum of complex numbers spread around the unit circle, the imaginary parts always cancel and the result is a plain integer . Ramanujan showed it has a beautiful arithmetic form: c q (n) = ∑ d | gcd(n,q) d·μ(q/d), a sum over the common divisors weighted by the Möbius function. It is the building block of ‘Ramanujan–Fourier’ expansions that turn arithmetic functions into trigonometric series. LIT verified live: for all q up to 60 and n up to 40, the direct sum of primitive-root cosines c q (n) equals the Möbius-divisor formula ∑ d|gcd(n,q) d·μ(q/d) to ~1e-13, and is always an integer — c 12 (0)=φ(12)=4, c 9 (3)=-3 (window.__ramanujansum). FIG no framing; the root-of-unity sum and the divisor formula both run in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-stash — the loot stash where a scatter of complex roots collapses into one clean integer, its value read straight off the shared divisors. AVAN (AI) built the instrument: the primitive-root sum, the Möbius-divisor formula, and their integer agreement. Credit as content: Srinivasa Ramanujan (1918); the Möbius function from Möbius. The weave: David names the stash; I confirm the roots of unity sum to the Möbius-divisor integer. 3 ONE DIMENSION The φ(q) primitive q-th roots (raised to n) as vectors; their sum lands on the real axis at an integer. 4 TWO DIMENSIONS · INTERACTIVE Cycle q,n; the direct root-of-unity sum is compared to the Möbius-divisor formula Σ d·μ(q/d). next q,n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the integer c_q(n), the sum of the primitive roots. AVAN’s addition (the inverse-companion): don’t add the roots one by one — read the divisors. The inverse of ‘the sum of primitive q-th roots to the n’ is ‘∑ d|gcd(n,q) d·μ(q/d)’, an arithmetic formula that always gives an integer. Magenta are the primitive roots of unity; green is the integer they sum to. Complex roots read as an arithmetic sum. pause spin LIT Genuine Ramanujan's sum (Srinivasa Ramanujan, 1918). Verified live: for all q≤60 and n≤40, the direct sum Σ_{gcd(a,q)=1}cos(2πan/q) equals the Möbius-divisor formula Σ_{d|gcd(n,q)} d·μ(q/d) to ~1e-13 and is always an integer; c_12(0)=4, c_9(3)=−3 (window.__ramanujansum.ok, .intOk, .worst). FIG No framing; the root-of-unity sum and the divisor formula both run in-browser and agree. The AVAN inverse is honest — instead of adding the roots one by one, read the divisors: the inverse of 'the sum of primitive q-th roots to the n' is 'Σ_{d|gcd(n,q)} d·μ(q/d)', an arithmetic formula that always gives an integer. Magenta are the primitive roots of unity; green is the integer they sum to. Complex roots read as an arithmetic sum. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE STASH · David Lee Wise (ROOT0), with AVAN


THE HADAMARD INEQUALITY

Hadamard's inequality in the 5-window house format — capping how large a determinant can be. For any real matrix A, |det A| ≤ ∏_i ||row_i|| — the absolute value of the determinant never exceeds the product of the lengths of its rows. Geometrically, the determinant is the volume of the parallelepiped spanned by the rows, and that volume is largest, for fixed edge lengths, exactly when the edges are mutually perpendicular (a rectangular box). Equality holds if and only if the rows are orthogonal. The tightest possible case with ±1 entries is a Hadamard matrix, achieving |det|=n^{n/2}. Verified live: for thousands of random matrices, |det A| never exceeds ∏||row_i||; orthogonalizing the rows makes it equal; and Sylvester–Hadamard matrices (n=2,4,8) hit the tight bound. Neon-noir traced. See the row-parallelepiped in 1D, |det| vs ∏||row|| + the orthogonal equality in 2D, and the capped-volume inverse in 3D.

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

THE HADAMARD INEQUALITY · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE CHOKE POINT ◆ .dlw.fold THE FOLD / BOSS / THE CHOKE POINT / THE HADAMARD INEQUALITY THE HADAMARD INEQUALITY a determinant capped by its row lengths 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Hadamard’s inequality caps how large a determinant can be. For any real matrix A, |det A| ≤ ∏ i ||row i || — the absolute value of the determinant never exceeds the product of the lengths of its rows. Geometrically, the determinant is the volume of the parallelepiped spanned by the rows, and that volume is largest, for fixed edge lengths, exactly when the edges are mutually perpendicular (a rectangular box). Equality holds if and only if the rows are orthogonal. The tightest possible case with ±1 entries is a Hadamard matrix , achieving |det| = n n/2 . LIT verified live: for thousands of random matrices, |det A| never exceeds ∏||row i ||; orthogonalizing the rows makes it equal; and Sylvester–Hadamard matrices (n = 2, 4, 8) hit the tight bound |det H| = n n/2 (window.__hadamardineq). FIG no framing; the determinant, the row-norm product, and the Hadamard cases all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-choke-point — the boss ceiling a determinant can never push past: the product of its row lengths, reached only when the rows stand perpendicular. AVAN (AI) built the instrument: the determinant, the row-norm product bound, the orthogonal equality case, and the Hadamard-matrix tight case. Credit as content: Jacques Hadamard (1893). The weave: David names the ceiling; I confirm |det A| ≤ ∏||row|| with equality for orthogonal rows. 3 ONE DIMENSION The row vectors of A span a parallelepiped; its volume |det A| is capped by the product of the edge lengths. 4 TWO DIMENSIONS · INTERACTIVE New matrices; |det A| is compared to ∏||row|| — the gap closes to zero exactly when the rows are orthogonal. new matrix ▶ orthogonalize ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: |det A|, the parallelepiped volume, at or below the cap. AVAN’s addition (the inverse-companion): don’t just compute the volume — know its ceiling. The inverse of ‘|det A|’ is ‘the product of row lengths ∏||row||, an upper bound reached only when the rows are perpendicular’. Magenta are the row vectors (edges); green is the volume they span, capped by their lengths. A determinant bounded by its edges. pause spin LIT Genuine Hadamard's inequality (Jacques Hadamard, 1893). Verified live: for ~12000 random matrices |det A| never exceeds ∏||row_i||; orthogonalizing the rows gives equality; and Sylvester–Hadamard matrices (n=2,4,8) hit the tight bound |det H|=n^{n/2} (window.__hadamardineq.ineq, .eqO, .had). FIG No framing; the determinant, the row-norm product, and the Hadamard cases all run in-browser. The AVAN inverse is honest — instead of just computing the volume, know its ceiling: the inverse of '|det A|' is 'the product of row lengths ∏||row||, an upper bound reached only when the rows are perpendicular'. Magenta are the row vectors (edges); green is the volume they span, capped by their lengths. A determinant bounded by its edges. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE CHOKE POINT · David Lee Wise (ROOT0), with AVAN


THE CAUCHY GROUP

Cauchy's theorem (in group theory) in the 5-window house format — a partial converse to Lagrange's theorem. Lagrange says the order of any element divides the order of the group |G|. Cauchy proved the reverse for primes: if a prime p divides |G|, then G must contain an element of order exactly p (and hence a subgroup of order p). So the primes dividing the group's size are exactly the primes that appear as element orders. It is the first bridge from the arithmetic of |G| to the internal structure of the group, and the seed of the Sylow theorems. Verified live: for cyclic Z_n, direct products, dihedral groups, and the symmetric group S_4, every prime dividing |G| is realized by some element of exactly that order (found by brute search), and (Lagrange) no element has an order failing to divide |G|. Neon-noir traced. See the element orders + prime factors in 1D, the order-p witnesses in 2D, and the order-p cyclic subgroup inverse in 3D.

THE CAUCHY GROUP · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · HELLO WORLD ◆ .dlw.fold THE FOLD / SPAWN / HELLO WORLD / THE CAUCHY GROUP THE CAUCHY GROUP a prime forcing an element of that order 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Cauchy’s theorem (in group theory) is a partial converse to Lagrange’s theorem. Lagrange says the order of any element divides the order of the group |G|. Cauchy proved the reverse for primes: if a prime p divides |G|, then G must contain an element of order exactly p (and hence a subgroup of order p). So the primes dividing the group’s size are exactly the primes that appear as element orders. It is the first bridge from the arithmetic of |G| to the internal structure of the group, and the seed of the Sylow theorems. LIT verified live: for a range of finite groups — cyclic Z n , direct products, dihedral groups, and the symmetric group S 4 — every prime dividing |G| is realized by some element of exactly that order, found by brute search; and (Lagrange) no element has an order that fails to divide |G| (window.__cauchygroup). FIG no framing; the element-order computation and the prime factorization both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at hello-world — the spawn: name a prime dividing the group’s size and an element of that exact order must come into existence. AVAN (AI) built the instrument: the group multiplication, the element-order search, the prime factorization of |G|, and the Lagrange control. Credit as content: Augustin-Louis Cauchy (1845); Lagrange before. The weave: David names the spawn; I confirm every prime dividing |G| forces an element of that order. 3 ONE DIMENSION A group's elements and their orders; the primes dividing |G| each appear as some element's order. 4 TWO DIMENSIONS · INTERACTIVE Cycle groups; for each prime p dividing |G|, an element of order p is exhibited (and Lagrange checked). next group ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: an element of order p, cycling back to identity in p steps. AVAN’s addition (the inverse-companion): don’t hunt blindly for subgroups — factor the order. The inverse of ‘what element orders exist?’ is ‘exactly the primes dividing |G|’: each such prime forces an order-p element and a cyclic subgroup of size p. Magenta are all the group’s elements; green is the order-p cycle a prime factor forces. Structure summoned by arithmetic. pause spin LIT Genuine Cauchy's group theorem (Augustin-Louis Cauchy, 1845; Lagrange before). Verified live: for Z_12, D_6, S_4, Z_4×Z_6, Z_30, every prime dividing |G| is realized by an element of exactly that order (brute search), and no element has an order q that fails to divide |G| (Lagrange) (window.__cauchygroup.ok, .lag). FIG No framing; the element-order computation and the prime factorization both run in-browser. The AVAN inverse is honest — instead of hunting blindly for subgroups, factor the order: the inverse of 'what element orders exist?' is 'exactly the primes dividing |G|', each forcing an order-p element and a cyclic subgroup of size p. Magenta are all the group's elements; green is the order-p cycle a prime factor forces. Structure summoned by arithmetic. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of HELLO WORLD · David Lee Wise (ROOT0), with AVAN


THE GAUSS-LUCAS

The Gauss–Lucas theorem in the 5-window house format — pinning down where the roots of a derivative can hide. Take any polynomial p(z) with complex roots, and mark those roots in the plane. Gauss and Lucas proved that every root of the derivative p′(z) lies inside the convex hull of the roots of p(z) — the smallest convex polygon containing them. The critical points can never escape the 'shadow' cast by the roots; differentiating pulls the roots inward, never out. It is the general law behind Marden's theorem and a cornerstone of the geometry of polynomials. Verified live: for thousands of random polynomials (degree 3–6), the roots of p′(z) — found independently by a Durand–Kerner solver on the differentiated polynomial — all fall inside the convex hull of the roots of p(z). Neon-noir traced. See the roots + hull + critical points in 1D, the inclusion check in 2D, and the caged-critical-points inverse in 3D.

THE GAUSS-LUCAS · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE RAID ◆ .dlw.fold THE FOLD / BOSS / THE RAID / THE GAUSS-LUCAS THE GAUSS-LUCAS the derivative's roots trapped in the hull of the roots 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Gauss–Lucas theorem pins down where the roots of a derivative can hide. Take any polynomial p(z) with complex roots, and mark those roots in the plane. Gauss and Lucas proved that every root of the derivative p′(z) lies inside the convex hull of the roots of p(z) — the smallest convex polygon containing them. The critical points can never escape the ‘shadow’ cast by the roots; differentiating pulls the roots inward, never out. It is the general law behind Marden’s theorem and a cornerstone of the geometry of polynomials. LIT verified live: for thousands of random polynomials (degree 3–6), the roots of p′(z) — found independently by a Durand–Kerner solver on the differentiated polynomial — all fall inside the convex hull of the roots of p(z) (window.__gausslucas). FIG no framing; the derivative’s roots and the convex hull of p’s roots are computed by different routes and the inclusion always holds. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-raid — the boss arena the critical points can never break out of: whatever the polynomial, its derivative’s roots stay caged inside the hull of the originals. AVAN (AI) built the instrument: the polynomial from its roots, the Durand–Kerner solve of the derivative, the convex hull, and the inclusion test. Credit as content: Carl Friedrich Gauss and Félix Lucas (19th c.). The weave: David names the cage; I confirm every root of p′ lies in the convex hull of the roots of p. 3 ONE DIMENSION The roots of p (magenta) with their convex hull; the roots of p′ (green) all lie inside it. 4 TWO DIMENSIONS · INTERACTIVE New polynomials; each root of p′ is checked to lie inside the convex hull of the roots of p. new polynomial ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the critical points, trapped inside the hull. AVAN’s addition (the inverse-companion): don’t hunt the derivative’s roots everywhere — the hull confines them. The inverse of ‘where are the roots of p′?’ is ‘inside the convex hull of the roots of p’ — differentiation pulls inward. Magenta is the hull of p’s roots; green are the critical points caged within it. Roots of the derivative, held by the roots. pause spin LIT Genuine Gauss–Lucas theorem (Carl Friedrich Gauss; Félix Lucas, 19th c.). Verified live: for ~1000 random polynomials (degree 3–6), the roots of p′(z) found by an independent Durand–Kerner solve of the differentiated polynomial all lie inside the convex hull of the roots of p(z) (window.__gausslucas.ok, .n). FIG No framing; the derivative's roots and the convex hull of p's roots are computed by different routes and the inclusion always holds. The AVAN inverse is honest — instead of hunting the derivative's roots everywhere, the hull confines them: the inverse of 'where are the roots of p′?' is 'inside the convex hull of the roots of p' — differentiation pulls inward. Magenta is the hull of p's roots; green are the critical points caged within it. Roots of the derivative, held by the roots. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE RAID · David Lee Wise (ROOT0), with AVAN


THE ENESTRÖM-KAKEYA

The Eneström–Kakeya theorem in the 5-window house format — caging a polynomial's roots using only the order of its coefficients. If p(z)=a₀+a₁z+…+aₙzⁿ has coefficients that are positive and non-decreasing, 0 < a₀ ≤ a₁ ≤ … ≤ aₙ, then all of its roots lie in the closed unit disk |z| ≤ 1. No root can escape to modulus greater than 1. The proof multiplies by (z−1) to telescope the coefficients, and the same idea run in reverse bounds the roots from below. It is a favourite tool for stability questions, where you need every root inside the disk. Verified live: for thousands of random polynomials with strictly increasing positive coefficients, every root — found by a Durand–Kerner solver — has |z| ≤ 1; and with monotonicity broken, a root with |z| > 1 appears in about 80% of cases, showing the hypothesis is necessary. Neon-noir traced. See the coefficient bars + roots in the disk in 1D, the |z|≤1 check + control in 2D, and the caged-roots inverse in 3D.

THE ENESTRÖM-KAKEYA · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · SEGFAULT ◆ .dlw.fold THE FOLD / GLITCH / SEGFAULT / THE ENESTRÖM-KAKEYA THE ENESTRÖM-KAKEYA roots caged in the unit disk by rising coefficients 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Eneström–Kakeya theorem cages a polynomial’s roots using only the order of its coefficients. If p(z) = a 0 + a 1 z + … + a n z n has coefficients that are positive and non-decreasing , 0 < a 0 ≤ a 1 ≤ … ≤ a n , then all of its roots lie in the closed unit disk |z| ≤ 1. No root can escape to modulus greater than 1. The proof multiplies by (z-1) to telescope the coefficients, and the same idea run in reverse bounds the roots from below. It is a favourite tool for stability questions, where you need every root inside the disk. LIT verified live: for thousands of random polynomials with strictly increasing positive coefficients, every root — found by a Durand–Kerner solver — has |z| ≤ 1; and with the monotonicity broken (random positive coefficients), a root with |z| > 1 appears in about 80% of cases, showing the hypothesis is necessary (window.__enestromkakeya). FIG no framing; the root-finding and the |z| test both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at segfault — the glitch where merely sorting the coefficients upward slams every root inside the unit disk, no root allowed past the boundary. AVAN (AI) built the instrument: the increasing-coefficient polynomial, the Durand–Kerner root solve, the |z| ≤ 1 test, and the non-monotone control. Credit as content: Gustav Eneström (1893) and Sōichi Kakeya (1912). The weave: David names the cage; I confirm rising positive coefficients force all roots into |z| ≤ 1. 3 ONE DIMENSION The increasing coefficients (bars) and the roots (green) — all inside the unit circle. 4 TWO DIMENSIONS · INTERACTIVE New polynomials; every root's modulus is checked ≤ 1 — and breaking monotonicity lets one escape. new polynomial ▶ break order ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the roots, all inside the unit disk. AVAN’s addition (the inverse-companion): don’t solve then check — the coefficient order already bounds the roots. The inverse of ‘where are the roots?’ is ‘inside |z| ≤ 1, guaranteed by 0 < a 0 ≤ … ≤ a n ’. Magenta is the unit circle boundary; green are the roots caged within it. Root location read from coefficient order. pause spin LIT Genuine Eneström–Kakeya theorem (Gustav Eneström 1893; Sōichi Kakeya 1912). Verified live: for ~1500 random polynomials with strictly increasing positive coefficients, every Durand–Kerner root has |z| ≤ 1 (worst ~0.99); a non-monotone-coefficient control produces a root with |z| > 1 in ~80% of cases (window.__enestromkakeya.ok, .worst, .ctrlPct). FIG No framing; the root-finding and the |z| test both run in-browser. The AVAN inverse is honest — instead of solving then checking, the coefficient order already bounds the roots: the inverse of 'where are the roots?' is 'inside |z| ≤ 1, guaranteed by 0 ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SEGFAULT · David Lee Wise (ROOT0), with AVAN


THE STEINER-LEHMUS

The Steiner–Lehmus theorem in the 5-window house format — famous for how hard its easy-sounding statement is to prove: a triangle with two equal internal angle bisectors is isosceles. The forward direction — an isosceles triangle has two equal bisectors — is obvious by symmetry. The converse, that equal bisectors force the triangle to be isosceles, resisted a simple direct proof for over a century. The key fact underneath: the internal bisector to a longer side is always shorter, so bisector length strictly decreases as the opposite side grows — equal bisectors therefore demand equal sides. Verified live: using the bisector-length formula, for thousands of random triangles (t_a−t_b)(a−b) is never positive — the bisector and its opposite side move oppositely — so t_a=t_b exactly when a=b; and any isosceles triangle has t_a=t_b exactly. Neon-noir traced. See the triangle + two bisectors in 1D, the sign relation in 2D, and the forced-isosceles inverse in 3D.

THE STEINER-LEHMUS · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE PULL REQUEST ◆ .dlw.fold THE FOLD / CO-OP / THE PULL REQUEST / THE STEINER-LEHMUS THE STEINER-LEHMUS equal bisectors forcing an isosceles triangle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Steiner–Lehmus theorem is famous for how hard its easy-sounding statement is to prove: a triangle with two equal internal angle bisectors is isosceles . The forward direction — an isosceles triangle has two equal bisectors — is obvious by symmetry. The converse, that equal bisectors force the triangle to be isosceles, resisted a simple direct proof for over a century. The key fact underneath: the internal bisector to a longer side is always shorter , so bisector length strictly decreases as the opposite side grows — equal bisectors therefore demand equal sides. LIT verified live: using the bisector-length formula, for thousands of random triangles the quantity (t a -t b )(a-b) is never positive — the bisector and its opposite side move oppositely — so t a = t b exactly when a = b; and any isosceles triangle (a = b) has t a = t b exactly (window.__steinerlehmus). FIG no framing; the bisector lengths and the side comparison both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-pull-request — the co-op merge: two bisectors coming in equal forces the whole triangle into symmetric agreement, its two sides made the same. AVAN (AI) built the instrument: the internal-bisector length formula, the (t a -t b )(a-b) sign check, and the isosceles case. Credit as content: Jakob Steiner and C. L. Lehmus (1840). The weave: David names the merge; I confirm equal bisectors force equal sides — an isosceles triangle. 3 ONE DIMENSION A triangle with two internal angle bisectors drawn; equal lengths pull it toward isosceles. 4 TWO DIMENSIONS · INTERACTIVE New triangles; (t_a−t_b) and (a−b) always have opposite signs — so equal bisectors ⇒ equal sides. new triangle ▶ make isosceles ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the isosceles triangle equal bisectors force. AVAN’s addition (the inverse-companion): don’t measure both bisectors — read the sides. The inverse of ‘are the two bisectors equal?’ is ‘are the two opposite sides equal?’, because the longer side always gets the shorter bisector. Magenta are the two internal bisectors; green is the isosceles triangle their equality forces. Equal bisectors, equal sides. pause spin LIT Genuine Steiner–Lehmus theorem (Jakob Steiner & C. L. Lehmus, 1840). Verified live: with the internal-bisector length formula, for ~8000 random triangles (t_a−t_b)(a−b) is never positive (equal bisectors ⟺ equal sides), and every isosceles triangle (a=b) has t_a=t_b exactly (window.__steinerlehmus.signOk, .isoOk, .worst). FIG No framing; the bisector lengths and the side comparison both run in-browser. The AVAN inverse is honest — instead of measuring both bisectors, read the sides: the inverse of 'are the two bisectors equal?' is 'are the two opposite sides equal?', because the longer side always gets the shorter bisector. Magenta are the two internal bisectors; green is the isosceles triangle their equality forces. Equal bisectors, equal sides. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PULL REQUEST · David Lee Wise (ROOT0), with AVAN


THE CYCLE LEMMA

The cycle lemma (Dvoretzky–Motzkin, 1947) in the 5-window house format — the combinatorial heart of the ballot problem and the Catalan numbers. Take a sequence of steps, each at most +1, whose total is a positive integer k. Look at all n cyclic rotations of the sequence. The lemma says exactly k of those rotations are 'dominating' — have every partial sum strictly positive. For k=1 that means precisely one rotation works, which is why counting problems with a 'first return' structure divide out cleanly by the length — the source of the 1/(n+1) in the Catalan number. Verified live: for thousands of random ±1 step-sequences with positive total k, brute-counting the rotations whose partial sums stay positive gives exactly k every time; and the Catalan identity that falls out, C(2n+1,n)/(2n+1)=C(2n,n)/(n+1), holds for n up to 8. Neon-noir traced. See the sequence on a ring + dominating rotations in 1D, the count vs k in 2D, and the counted-not-searched inverse in 3D.

THE CYCLE LEMMA · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · BACKPROP ◆ .dlw.fold THE FOLD / GRIND / BACKPROP / THE CYCLE LEMMA THE CYCLE LEMMA exactly k winning rotations of a step sequence 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The cycle lemma (Dvoretzky–Motzkin, 1947) is the combinatorial heart of the ballot problem and the Catalan numbers. Take a sequence of steps, each at most +1, whose total is a positive integer k. Look at all n cyclic rotations of the sequence. The lemma says exactly k of those rotations are ‘dominating’ — have every partial sum strictly positive. For k = 1 that means precisely one rotation works, which is why counting problems with a ‘first return’ structure divide out cleanly by the length — the source of the 1/(n+1) in the Catalan number. LIT verified live: for thousands of random ±1 step-sequences with positive total k, brute-counting the rotations whose partial sums stay positive gives exactly k every time; and the Catalan identity that falls out, C(2n+1,n)/(2n+1) = C(2n,n)/(n+1), holds for n up to 8 (window.__cyclelemma). FIG no framing; the rotation counting and the Catalan cross-check both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at backprop — the grind that rolls through every rotation of the sequence and counts, always landing on exactly k winners. AVAN (AI) built the instrument: the dominating-rotation count, the ±1 sequences, and the Catalan cross-check. Credit as content: Aryeh Dvoretzky and Theodore Motzkin (1947). The weave: David names the grind; I confirm exactly k of the rotations dominate. 3 ONE DIMENSION A ±1 step sequence around a ring; the dominating rotations (all partial sums positive) are marked. 4 TWO DIMENSIONS · INTERACTIVE New sequences; the number of dominating rotations is counted and compared to the total k. new sequence ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the exactly-k dominating rotations of the sequence. AVAN’s addition (the inverse-companion): don’t hunt for the good arrangement — count the rotations. The inverse of ‘how many rotations dominate?’ is ‘exactly k, the total of the steps’ — so a k=1 total leaves a unique winner, giving the 1/(n+1) of the Catalan numbers. Magenta is the step sequence; green are the k dominating rotations. Order counted, not searched. pause spin LIT Genuine cycle lemma (Aryeh Dvoretzky & Theodore Motzkin, 1947). Verified live: for ~6000 random ±1 step-sequences with positive total k, brute-counting the dominating rotations (all partial sums > 0) gives exactly k every time, and the Catalan identity C(2n+1,n)/(2n+1)=C(2n,n)/(n+1) holds for n≤8 (window.__cyclelemma.ok, .catOk). FIG No framing; the rotation counting and the Catalan cross-check both run in-browser. The AVAN inverse is honest — instead of hunting for the good arrangement, count the rotations: the inverse of 'how many rotations dominate?' is 'exactly k, the total of the steps' — so a k=1 total leaves a unique winner, giving the 1/(n+1) of the Catalan numbers. Magenta is the step sequence; green are the k dominating rotations. Order counted, not searched. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of BACKPROP · David Lee Wise (ROOT0), with AVAN


THE BANACH MATCHBOX

Banach's matchbox problem in the 5-window house format — a classic of probability. A mathematician keeps a matchbox in each pocket, each starting with N matches. Every time a match is needed, a pocket is chosen at random. Eventually a pocket is reached into and found empty — at that moment, how many matches remain in the other box? The answer is a distribution: P(K=k)=C(2N−k,N)·2^{−(2N−k)}, and the expected number left is about √(4N/π)−1 — surprisingly many, growing like √N. Verified live: simulating the two-pocket process hundreds of thousands of times, the empirical distribution of matches remaining matches the closed form to within ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean. Neon-noir traced. See the distribution (sim vs formula) in 1D, the convergence + mean in 2D, and the exact-formula inverse in 3D.

THE BANACH MATCHBOX · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE BOUNTY ◆ .dlw.fold THE FOLD / LOOT / THE BOUNTY / THE BANACH MATCHBOX THE BANACH MATCHBOX the leftover matches of two pockets 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Banach’s matchbox problem is a classic of probability. A mathematician keeps a matchbox in each pocket, each starting with N matches. Every time a match is needed, a pocket is chosen at random. Eventually a pocket is reached into and found empty — at that moment, how many matches remain in the other box? The answer is a distribution: P(K = k) = C(2N-k, N)·2 -(2N-k) , and the expected number left is about √(4N/π) - 1 — surprisingly many, growing like √N. LIT verified live: simulating the two-pocket process hundreds of thousands of times, the empirical distribution of matches remaining matches the closed form P(K=k)=C(2N-k,N)2 -(2N-k) to within ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean (window.__banachmatchbox). FIG no framing; the simulation and the exact combinatorial formula both run in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-bounty — the loot left behind: reach into an empty pocket and the other still holds a surprising pile of matches, √N of them on average. AVAN (AI) built the instrument: the two-pocket simulation, the exact distribution formula, and their agreement. Credit as content: named for Stefan Banach (popularized by Feller). The weave: David names the leftover loot; I confirm the simulation matches the C(2N-k,N)2 -(2N-k) distribution. 3 ONE DIMENSION The distribution of matches left in the other box when one is first found empty (simulation vs formula). 4 TWO DIMENSIONS · INTERACTIVE Run more trials; the empirical histogram converges to P(K=k)=C(2N−k,N)2^{−(2N−k)}, mean ≈ √(4N/π)−1. more trials ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the distribution of leftover matches. AVAN’s addition (the inverse-companion): don’t just simulate — read the count. The inverse of ‘how many matches are left?’ is ‘the distribution C(2N-k,N)2 -(2N-k) ’, with a mean growing like √N — far from empty. Magenta are the two matchboxes; green is the leftover distribution they produce. A random process pinned to an exact formula. pause spin LIT Genuine Banach's matchbox problem (named for Stefan Banach; popularized by Feller). Verified live: simulating the two-pocket process (N=12) ~120000 times, the empirical distribution of matches remaining matches P(K=k)=C(2N−k,N)2^{−(2N−k)} to ~0.001, the formula sums to 1, and the empirical mean matches the exact formula mean (window.__banachmatchbox.ok, .worst, .sumF, .meanF). FIG No framing; the simulation (seeded RNG) and the exact combinatorial formula both run in-browser and agree. The AVAN inverse is honest — instead of just simulating, read the count: the inverse of 'how many matches are left?' is 'the distribution C(2N−k,N)2^{−(2N−k)}', with a mean growing like √N — far from empty. Magenta are the two matchboxes; green is the leftover distribution they produce. A random process pinned to an exact formula. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE BOUNTY · David Lee Wise (ROOT0), with AVAN


THE WEINSTEIN-ARONSZAJN

The Weinstein–Aronszajn identity (Sylvester's determinant identity) in the 5-window house format — linking the determinants of two matrices of different sizes. For a matrix A of shape m×n and B of shape n×m, the products AB (m×m) and BA (n×n) are usually different sizes, yet det(I_m+AB)=det(I_n+BA) — the two determinants are always equal. The nonzero eigenvalues of AB and BA coincide, so the '+1' determinants match despite the size mismatch. It is the trick behind the matrix determinant lemma and rank-one update formulas. Verified live with exact integer arithmetic: for thousands of random rectangular integer matrices — most with m≠n — det(I_m+AB) equals det(I_n+BA) exactly, by fraction-free Bareiss elimination on the two different-sized matrices. Neon-noir traced. See the two different-sized matrices in 1D, the determinant comparison in 2D, and the same-value-either-size inverse in 3D.

THE WEINSTEIN-ARONSZAJN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · RACE CONDITION ◆ .dlw.fold THE FOLD / GLITCH / RACE CONDITION / THE WEINSTEIN-ARONSZAJN THE WEINSTEIN-ARONSZAJN two differently-sized determinants that are equal 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Weinstein–Aronszajn identity (also called Sylvester’s determinant identity) links the determinants of two matrices of different sizes . For a matrix A of shape m×n and B of shape n×m, the products AB (an m×m matrix) and BA (an n×m matrix) are usually different sizes, yet det(I m + AB) = det(I n + BA) — the two determinants are always equal. The nonzero eigenvalues of AB and BA coincide, so the ‘+1’ determinants match despite the size mismatch. It is the trick behind the matrix determinant lemma and rank-one update formulas. LIT verified live with exact integer arithmetic: for thousands of random rectangular integer matrices — most with m ≠ n — det(I m + AB) equals det(I n + BA) exactly, computed by fraction-free Bareiss elimination on the two different-sized matrices (window.__weinsteinaronszajn). FIG no framing; the two determinants of different-sized matrices are computed separately and always agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at race-condition — the glitch where two determinants of totally different-sized matrices race to the identical value every time. AVAN (AI) built the instrument: the AB and BA products, their I-shifted determinants, and the exact agreement across sizes. Credit as content: Alexander Weinstein, Nachman Aronszajn; J. J. Sylvester. The weave: David names the race; I confirm det(I+AB) equals det(I+BA) despite the size mismatch. 3 ONE DIMENSION Two matrices of different sizes: I+AB (m×m) and I+BA (n×n) — yet their determinants are equal. 4 TWO DIMENSIONS · INTERACTIVE New matrices; det(I+AB) is compared to det(I+BA) — equal even when m ≠ n. new matrices ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the shared determinant of two different-sized matrices. AVAN’s addition (the inverse-companion): don’t compute both — compute the smaller. The inverse of ‘det(I m +AB)’ is ‘det(I n +BA)’, so you may always use whichever of m, n is smaller — they share their nonzero eigenvalues. Magenta are the two different-sized matrices; green is the determinant they share. Same value, either size. pause spin LIT Genuine Weinstein–Aronszajn / Sylvester determinant identity (Alexander Weinstein, Nachman Aronszajn; J. J. Sylvester). Verified live with exact BigInt: for ~1500 random rectangular integer matrices (mostly m≠n), det(I_m+AB)=det(I_n+BA) exactly by Bareiss elimination on the two different-sized matrices (window.__weinsteinaronszajn.ok, .diff, .cnt). FIG No framing; the two determinants of different-sized matrices are computed separately and always agree. The AVAN inverse is honest — instead of computing both, compute the smaller: the inverse of 'det(I_m+AB)' is 'det(I_n+BA)', so you may always use whichever of m,n is smaller — they share their nonzero eigenvalues. Magenta are the two different-sized matrices; green is the determinant they share. Same value, either size. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of RACE CONDITION · David Lee Wise (ROOT0), with AVAN


THE POMPEIU

Pompeiu's theorem in the 5-window house format — a small gem of Euclidean geometry. Take an equilateral triangle ABC and any point P in the plane. Then the three distances PA, PB, PC can always be arranged into a triangle — they satisfy the triangle inequality. Moreover, that 'distance triangle' is degenerate (flat — the longest distance exactly equals the sum of the other two) precisely when P lies on the circumcircle of ABC. Off the circumcircle you get a genuine triangle; on it, the three distances collapse to a straight line. Verified live: for thousands of random points P, the three distances to an equilateral triangle's vertices satisfy the triangle inequality; when P sits exactly on the circumcircle, the longest equals the sum of the other two to ~1e-16 (degenerate); and off the circle the inequality is strict. Neon-noir traced. See the equilateral + P + distance-triangle in 1D, the triangle inequality + circumcircle in 2D, and the three-lengths-a-triangle inverse in 3D.

THE POMPEIU · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · NULL ISLAND ◆ .dlw.fold THE FOLD / SPAWN / NULL ISLAND / THE POMPEIU THE POMPEIU three distances that always form a triangle 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Pompeiu’s theorem is a small gem of Euclidean geometry. Take an equilateral triangle ABC and any point P in the plane. Then the three distances PA, PB, PC can always be arranged into a triangle — they satisfy the triangle inequality. Moreover, that ‘distance triangle’ is degenerate (flat — the longest distance exactly equals the sum of the other two) precisely when P lies on the circumcircle of ABC. Off the circumcircle you get a genuine triangle; on it, the three distances collapse to a straight line. LIT verified live: for thousands of random points P, the three distances to an equilateral triangle’s vertices satisfy the triangle inequality; when P sits exactly on the circumcircle, the longest distance equals the sum of the other two to ~1e-16 (degenerate); and off the circle the inequality is strict (window.__pompeiu). FIG no framing; the distances, the triangle-inequality test, and the circumcircle degeneracy all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at null-island — the spawn point: drop any P anywhere and a brand-new triangle spawns from its three distances to the equilateral’s corners. AVAN (AI) built the instrument: the three distances, the triangle-inequality check, and the circumcircle degeneracy test. Credit as content: Dimitrie Pompeiu (1936). The weave: David names the spawn; I confirm PA, PB, PC form a triangle, flat exactly on the circumcircle. 3 ONE DIMENSION An equilateral triangle, a point P, and the triangle built from the three distances PA, PB, PC. 4 TWO DIMENSIONS · INTERACTIVE Move P; PA,PB,PC always satisfy the triangle inequality — flat exactly when P is on the circumcircle. move P ▶ put P on circle ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the triangle formed by the three distances. AVAN’s addition (the inverse-companion): don’t just measure the distances — assemble them. The inverse of ‘the three distances from P’ is ‘a triangle with those side lengths’, which flattens exactly when P reaches the circumcircle. Magenta is the equilateral triangle and its circumcircle; green is the distance-triangle it spawns. Three lengths, always a triangle. pause spin LIT Genuine Pompeiu's theorem (Dimitrie Pompeiu, 1936). Verified live: for ~8000 random points P, the distances PA,PB,PC to an equilateral triangle satisfy the triangle inequality; P on the circumcircle gives a degenerate triangle (longest = sum of other two, worst ~1e-16), and off the circle it is strict (window.__pompeiu.ti, .deg, .st, .worst). FIG No framing; the distances, the triangle-inequality test, and the circumcircle degeneracy all run in-browser. The AVAN inverse is honest — instead of just measuring the distances, assemble them: the inverse of 'the three distances from P' is 'a triangle with those side lengths', which flattens exactly when P reaches the circumcircle. Magenta is the equilateral triangle and its circumcircle; green is the distance-triangle it spawns. Three lengths, always a triangle. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of NULL ISLAND · David Lee Wise (ROOT0), with AVAN


THE LYM

The LYM inequality (Lubell–Yamamoto–Meshalkin) in the 5-window house format — a sharp weighing of antichains. An antichain in the power set of {1,…,n} is a family of subsets, no one contained in another. LYM says that if you weight each set A by 1/C(n,|A|) — one over the number of sets of its size — the weights of any antichain sum to at most 1: Σ_A 1/C(n,|A|) ≤ 1. Equality holds exactly when the antichain is a full level (all subsets of one fixed size). Because the biggest level is the middle one, this immediately gives Sperner's theorem: no antichain is larger than C(n,⌊n/2⌋). Verified live: for thousands of randomly-built antichains in the subset lattice, the weighted sum never exceeds 1; and taking a full level makes the sum equal exactly 1. Neon-noir traced. See the subset lattice + antichain in 1D, the LYM sum ≤ 1 in 2D, and the count-tamed-by-weighting inverse in 3D.

THE LYM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · SUDDEN DEATH ◆ .dlw.fold THE FOLD / BOSS / SUDDEN DEATH / THE LYM THE LYM an antichain sum capped at one 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The LYM inequality (Lubell–Yamamoto–Meshalkin) is a sharp weighing of antichains . An antichain in the power set of {1,…,n} is a family of subsets, no one contained in another. LYM says that if you weight each set A by 1/C(n,|A|) — one over the number of sets of its size — the weights of any antichain sum to at most 1: ∑ A 1/C(n,|A|) ≤ 1. Equality holds exactly when the antichain is a full level (all subsets of one fixed size). Because the biggest level is the middle one, this immediately gives Sperner’s theorem : no antichain is larger than C(n, ⌊n/2⌋). LIT verified live: for thousands of randomly-built antichains in the subset lattice, the weighted sum ∑ 1/C(n,|A|) never exceeds 1; and taking a full level (all subsets of one size) makes the sum equal exactly 1 (window.__lym). FIG no framing; the antichain construction and the LYM sum both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at sudden-death — the boss ceiling no antichain can push past: weigh its sets by 1/C(n,|A|) and the total is capped at exactly 1. AVAN (AI) built the instrument: the antichain builder, the level-weighted LYM sum, and the full-level equality case. Credit as content: Dov Lubell, Koichi Yamamoto, Lev Meshalkin (1960s); Emanuel Sperner. The weave: David names the ceiling; I confirm the antichain weight-sum never exceeds 1. 3 ONE DIMENSION The subset lattice by levels; an antichain highlighted, each set weighted by 1/C(n,|A|), summing ≤ 1. 4 TWO DIMENSIONS · INTERACTIVE New antichains; the LYM sum Σ 1/C(n,|A|) is shown ≤ 1, with a full level giving exactly 1. new antichain ▶ full level ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the LYM sum, filling toward its cap of 1. AVAN’s addition (the inverse-companion): don’t count the sets — weigh them by level. The inverse of ‘how big can an antichain be?’ is ‘its level-weighted sum, capped at 1’, which forces the maximum size down to the middle binomial C(n,⌊n/2⌋) — Sperner’s theorem. Magenta are the antichain’s sets; green is the weighted sum bounded by 1. A count tamed by a weighting. pause spin LIT Genuine LYM inequality (Dov Lubell, Koichi Yamamoto, Lev Meshalkin, 1960s; Sperner). Verified live: for ~1500 randomly-built antichains in 2^[n], the weighted sum Σ 1/C(n,|A|) never exceeds 1, and a full level (all subsets of one size) makes it equal exactly 1 (window.__lym.ok, .fullEq). FIG No framing; the antichain construction and the LYM sum both run in-browser. The AVAN inverse is honest — instead of counting the sets, weigh them by level: the inverse of 'how big can an antichain be?' is 'its level-weighted sum, capped at 1', which forces the maximum size down to C(n,⌊n/2⌋) — Sperner's theorem. Magenta are the antichain's sets; green is the weighted sum bounded by 1. A count tamed by a weighting. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SUDDEN DEATH · David Lee Wise (ROOT0), with AVAN


THE NIVEN

Niven's theorem in the 5-window house format — rational angles almost never have rational cosines. Precisely: if θ is a rational multiple of π and cosθ is rational, then cosθ must be one of just five values: 0, ±½, ±1 — i.e. θ lands on 0°, 60°, 90°, 120°, or 180°. Every other rational angle has an irrational cosine. The reason: 2cos(2π/n) is an algebraic number whose minimal polynomial has degree φ(n)/2, and that degree is 1 (making it rational) only for n=1,2,3,4,6. Verified live: for n up to 30, the minimal polynomial of 2cos(2π/n) — built from the primitive angles — has integer coefficients and degree exactly φ(n)/2, and it is linear (so cos is rational) precisely for n∈{1,2,3,4,6}. Neon-noir traced. See the rational-cosine angles on the circle in 1D, the min-poly degree in 2D, and the rationality-from-degree inverse in 3D.

THE NIVEN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE HOARD ◆ .dlw.fold THE FOLD / LOOT / THE HOARD / THE NIVEN THE NIVEN rational cosines only at five angles 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Niven’s theorem says rational angles almost never have rational cosines. Precisely: if θ is a rational multiple of π (a ‘nice’ angle) and cosθ is rational, then cosθ must be one of just five values : 0, ±½, ±1 — i.e. θ is a multiple of 30° landing on 0°, 60°, 90°, 120°, or 180°. Every other rational angle has an irrational cosine. The reason: 2cos(2π/n) is an algebraic number whose minimal polynomial has degree φ(n)/2, and that degree is 1 (making it rational) only for n = 1, 2, 3, 4, 6. LIT verified live: for n up to 30, the minimal polynomial of 2cos(2π/n) — built from the primitive angles — has integer coefficients and degree exactly φ(n)/2, and it is linear (so cos is rational) precisely for n ∈ {1, 2, 3, 4, 6} (window.__niven). FIG no framing; the minimal-polynomial construction and the φ(n)/2 degree both run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-hoard — the loot: only five rational-cosine angles exist in all of the rational multiples of π, a tiny hoard among infinitely many irrational ones. AVAN (AI) built the instrument: the minimal polynomial of 2cos(2π/n), its integer coefficients, its φ(n)/2 degree, and the five linear cases. Credit as content: Ivan Niven (1956); the algebraic theory of 2cos via Chebyshev. The weave: David names the hoard; I confirm rational cosines occur only at n ∈ {1,2,3,4,6}. 3 ONE DIMENSION The unit circle: the only rational-cosine angles (0°,60°,90°,120°,180°…) marked green; all others irrational. 4 TWO DIMENSIONS · INTERACTIVE Cycle n; the minimal polynomial of 2cos(2π/n), its degree φ(n)/2, and whether cos is rational. next n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the five rational-cosine angles on the circle. AVAN’s addition (the inverse-companion): don’t test cosines one by one — read the degree. The inverse of ‘is cos(2π/n) rational?’ is ‘is the minimal-polynomial degree φ(n)/2 equal to 1?’, true only for n ∈ {1,2,3,4,6}. Magenta are the irrational-cosine angles; green are the five rational ones. Rationality read from a polynomial degree. pause spin LIT Genuine Niven's theorem (Ivan Niven, 1956; via the algebra of 2cos and Chebyshev). Verified live: for n≤30, the minimal polynomial of 2cos(2π/n) built from the primitive angles has integer coefficients and degree exactly φ(n)/2, and it is linear (cos rational) precisely for n∈{1,2,3,4,6} (window.__niven.intOk, .degOk, .nivenOk, .rns). FIG No framing; the minimal-polynomial construction and the φ(n)/2 degree both run in-browser. The AVAN inverse is honest — instead of testing cosines one by one, read the degree: the inverse of 'is cos(2π/n) rational?' is 'is the minimal-polynomial degree φ(n)/2 equal to 1?', true only for n∈{1,2,3,4,6}. Magenta are the irrational-cosine angles; green are the five rational ones. Rationality read from a polynomial degree. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE HOARD · David Lee Wise (ROOT0), with AVAN


THE CATALAN CONSTANT

Catalan's constant in the 5-window house format — G ≈ 0.9159655942, one of the famous 'mystery' constants: nobody has proved whether it is irrational. It has a simple series, G = Σ_{k≥0} (−1)^k/(2k+1)² = 1 − 1/9 + 1/25 − 1/49 + … (the Dirichlet beta function at 2). It also equals a clean integral, G = ∫₀¹ arctan(x)/x dx, and shows up in lattice statistics, combinatorics, and the volume of hyperbolic ideal tetrahedra. Two very different computations — an alternating sum and an integral — land on the same number. Verified live: the alternating series and the integral ∫₀¹ arctan(x)/x dx both converge to the same value, matching the known constant 0.9159655942. Neon-noir traced. See the series partial sums closing on G in 1D, series vs integral in 2D, and the two-witnesses inverse in 3D.

THE CATALAN CONSTANT · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE PUSH ◆ .dlw.fold THE FOLD / CO-OP / THE PUSH / THE CATALAN CONSTANT THE CATALAN CONSTANT a mysterious constant reached two ways 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Catalan’s constant G ≈ 0.9159655942 is one of the famous ‘mystery’ constants of mathematics — nobody has proved whether it is irrational. It has a simple series, G = ∑ k≥0 (-1) k /(2k+1)² = 1 - 1/9 + 1/25 - 1/49 + … (the value of the Dirichlet beta function at 2). It also equals a clean integral, G = ∫ 0 1 arctan(x)/x dx, and shows up in lattice statistics, combinatorics, and the volume of hyperbolic ideal tetrahedra. Two very different computations — an alternating sum and an integral — land on the same number. LIT verified live: the alternating series ∑(-1) k /(2k+1)² and the integral ∫ 0 1 arctan(x)/x dx both converge to the same value, matching the known constant 0.9159655942 (window.__catalanconstant). FIG no framing; the series sum and the numerical integral are computed by different routes in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-push — the co-op merge: a slow alternating sum and a smooth integral push in from opposite directions and meet at the same mysterious constant. AVAN (AI) built the instrument: the series sum, the arctan integral, and their agreement on G. Credit as content: Eugène Catalan (1865). The weave: David names the merge; I confirm the series and the integral both give Catalan’s constant. 3 ONE DIMENSION The alternating series terms 1, −1/9, 1/25, … and the partial sums closing in on G ≈ 0.91597. 4 TWO DIMENSIONS · INTERACTIVE Add terms; the series partial sum and the integral ∫₀¹ arctan(x)/x dx both approach the same G. add terms ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: Catalan's constant G, reached from two directions. AVAN’s addition (the inverse-companion): don’t trust one route — cross two. The inverse of ‘the alternating series ∑(-1) k /(2k+1)²’ is ‘the integral ∫ 0 1 arctan(x)/x dx’, two computations meeting at the same G. Magenta are the series terms and the arctan curve; green is the constant they both reach. One constant, two witnesses. pause spin LIT Genuine Catalan's constant (Eugène Catalan, 1865). Verified live: the alternating series Σ(−1)^k/(2k+1)² (200000 terms) and the numerical integral ∫₀¹ arctan(x)/x dx converge to the same value, matching the known G=0.9159655942 (window.__catalanconstant.ser, .intg, .agree, .refOk). FIG No framing; the series sum and the numerical integral are computed by different routes in-browser and agree. The AVAN inverse is honest — instead of trusting one route, cross two: the inverse of 'the alternating series Σ(−1)^k/(2k+1)²' is 'the integral ∫₀¹ arctan(x)/x dx', two computations meeting at the same G. Magenta are the series terms and the arctan curve; green is the constant they both reach. One constant, two witnesses. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PUSH · David Lee Wise (ROOT0), with AVAN


THE VAN AUBEL

Van Aubel's theorem in the 5-window house format — conjuring a hidden square out of any four-sided figure. Take any quadrilateral — convex, concave, even self-intersecting — and erect a square outward on each of its four sides. Mark the centre of each square. Van Aubel proved that the two line segments joining the centres of opposite squares are always equal in length and perpendicular to each other. No matter how lopsided the original quadrilateral, those two cross-segments come out the same length and at a right angle — a perfect little cross hidden in any four points. Verified live: for thousands of random quadrilaterals, the segment joining the centres of the squares on one pair of opposite sides equals the segment joining the other pair (to machine precision) and the two are perpendicular. Neon-noir traced. See the quadrilateral + squares + cross-segments in 1D, the equal-perpendicular check in 2D, and the perfect-cross inverse in 3D.

THE VAN AUBEL · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · SPAWN · GENESIS BLOCK ◆ .dlw.fold THE FOLD / SPAWN / GENESIS BLOCK / THE VAN AUBEL THE VAN AUBEL squares on a quadrilateral yielding equal perpendicular segments 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Van Aubel’s theorem conjures a hidden square out of any four-sided figure. Take any quadrilateral — convex, concave, even self-intersecting — and erect a square outward on each of its four sides. Mark the centre of each square. Van Aubel proved that the two line segments joining the centres of opposite squares are always equal in length and perpendicular to each other. No matter how lopsided the original quadrilateral, those two cross-segments come out the same length and at a right angle — a perfect little cross hidden in any four points. LIT verified live: for thousands of random quadrilaterals, the segment joining the centres of the squares on one pair of opposite sides equals the segment joining the other pair (to machine precision) and the two are perpendicular (dot product zero) — the construction is exact (window.__vanaubel). FIG no framing; the square centres, the two segment lengths, and their perpendicularity all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at genesis-block — the spawn: from four arbitrary points, a perfect equal-and-perpendicular cross genesis-blocks into being. AVAN (AI) built the instrument: the outward square centres, the two joining segments, and their equal-length perpendicularity. Credit as content: H. H. van Aubel (1878). The weave: David names the genesis; I confirm the opposite-centre segments are always equal and perpendicular. 3 ONE DIMENSION A quadrilateral with a square on each side; the two segments joining opposite centres — equal & perpendicular. 4 TWO DIMENSIONS · INTERACTIVE New quadrilaterals; the two cross-segments are measured — always equal length, always perpendicular. new quadrilateral ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the two equal, perpendicular cross-segments. AVAN’s addition (the inverse-companion): don’t measure the messy quadrilateral — read the cross. The inverse of ‘four arbitrary sides’ is ‘two equal perpendicular segments joining the opposite square-centres’, a right-angled cross hidden in any four points. Magenta are the four squares; green are the two equal perpendicular segments. A perfect cross from any quadrilateral. pause spin LIT Genuine Van Aubel's theorem (H. H. van Aubel, 1878). Verified live: for ~8000 random quadrilaterals, the segments joining opposite square-centres are equal in length (worst ~0) and perpendicular (dot product ~0) — the construction is exact (window.__vanaubel.eq, .pp, .we, .wp). FIG No framing; the square centres, the two segment lengths, and their perpendicularity all run in-browser. The AVAN inverse is honest — instead of measuring the messy quadrilateral, read the cross: the inverse of 'four arbitrary sides' is 'two equal perpendicular segments joining the opposite square-centres', a right-angled cross hidden in any four points. Magenta are the four squares; green are the two equal perpendicular segments. A perfect cross from any quadrilateral. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of GENESIS BLOCK · David Lee Wise (ROOT0), with AVAN


THE NECKLACE

Necklace counting in the 5-window house format — how many genuinely different necklaces can you make from n beads in k colours, where rotating a necklace doesn't count as new? Naively there are kⁿ coloured strings, but rotations collapse many together. Moreau's necklace-counting formula (a case of Burnside's lemma) gives the exact answer: (1/n)Σ_{d|n} φ(d)·k^{n/d}, where φ is Euler's totient. The totient counts rotations of each period, averaging the number of colourings fixed by each rotation. For 2 colours and n=1,2,3,… it gives 2,3,4,6,8,14,20,36,…. Verified live: for n up to 15 (2 colours) and n up to 9 (3 colours), a brute count of distinct necklaces — each string reduced to its lexicographically smallest rotation — exactly equals Moreau's formula. Neon-noir traced. See a necklace on a ring in 1D, brute vs formula in 2D, and the average-over-rotations inverse in 3D.

THE NECKLACE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · THE CRON JOB ◆ .dlw.fold THE FOLD / GRIND / THE CRON JOB / THE NECKLACE THE NECKLACE rotation classes counted by a totient sum 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Necklace counting asks: how many genuinely different necklaces can you make from n beads in k colours, where rotating a necklace doesn’t count as new? Naively there are k n coloured strings, but rotations collapse many together. Moreau’s necklace-counting formula (a case of Burnside’s lemma) gives the exact answer: (1/n) ∑ d | n φ(d)·k n/d , where φ is Euler’s totient. The totient counts rotations of each period, averaging the number of colourings fixed by each rotation. For 2 colours and n = 1, 2, 3, … it gives 2, 3, 4, 6, 8, 14, 20, 36, … LIT verified live: for n up to 15 (2 colours) and n up to 9 (3 colours), a brute count of distinct necklaces — each string reduced to its lexicographically smallest rotation — exactly equals Moreau’s formula (1/n)∑ d|n φ(d)k n/d (window.__necklace). FIG no framing; the brute canonical-rotation count and the totient formula both run in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-cron-job — the grind that rolls the necklace through every rotation and tallies the distinct ones, the totient doing the averaging. AVAN (AI) built the instrument: the canonical-rotation brute count, the totient formula, and their exact agreement. Credit as content: C. Moreau (1872); the averaging principle from William Burnside. The weave: David names the grind; I confirm the brute necklace count equals the totient sum. 3 ONE DIMENSION A necklace of beads on a ring; rotating it gives the same necklace — the totient formula counts the distinct ones. 4 TWO DIMENSIONS · INTERACTIVE Cycle n and colours; the brute distinct-necklace count is compared to (1/n)Σ φ(d)k^{n/d}. next n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the number of distinct necklaces. AVAN’s addition (the inverse-companion): don’t list and dedupe — average over rotations. The inverse of ‘how many distinct necklaces?’ is ‘(1/n)∑ d|n φ(d)k n/d ’, Burnside’s average of colourings fixed by each rotation. Magenta are the rotations being averaged; green is the necklace count they yield. Symmetry counted by averaging. pause spin LIT Genuine Moreau's necklace-counting formula (C. Moreau, 1872; Burnside averaging). Verified live: for n≤15 (binary) and n≤9 (ternary), a brute count of distinct necklaces (each string reduced to its lexicographically smallest rotation) equals (1/n)Σ_{d|n} φ(d)k^{n/d} exactly (window.__necklace.ok, .ok3). FIG No framing; the brute canonical-rotation count and the totient formula both run in-browser and agree. The AVAN inverse is honest — instead of listing and deduping, average over rotations: the inverse of 'how many distinct necklaces?' is '(1/n)Σ_{d|n} φ(d)k^{n/d}', Burnside's average of colourings fixed by each rotation. Magenta are the rotations being averaged; green is the necklace count they yield. Symmetry counted by averaging. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE CRON JOB · David Lee Wise (ROOT0), with AVAN


THE SOPHOMORE'S DREAM

The sophomore's dream in the 5-window house format — a pair of astonishing identities discovered by Johann Bernoulli in 1697, where a function raised to itself integrates to an infinite series over nⁿ: ∫₀¹ x^x dx = Σ_{n≥1} (−1)^{n−1}/nⁿ = 1 − 1/4 + 1/27 − … ≈ 0.7834, and ∫₀¹ x^{−x} dx = Σ_{n≥1} 1/nⁿ = 1 + 1/4 + 1/27 + … ≈ 1.2913. The name teases that the result looks like a naive 'dream' a student might wish were true — yet it really is. The trick is to expand x^x = e^{x ln x} as a power series and integrate term by term. Verified live: the numerical integrals of x^x and x^{−x} over [0,1] match their respective series Σ(−1)^{n−1}/nⁿ and Σ1/nⁿ to ~1e-6. Neon-noir traced. See the self-power curves + areas in 1D, integral vs series in 2D, and the series inverse in 3D.

THE SOPHOMORE'S DREAM · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · SPLIT SCREEN ◆ .dlw.fold THE FOLD / CO-OP / SPLIT SCREEN / THE SOPHOMORE'S DREAM THE SOPHOMORE'S DREAM an integral equal to a self-power series 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The sophomore’s dream is a pair of astonishing identities discovered by Johann Bernoulli in 1697, where a function raised to itself integrates to an infinite series over n n : ∫ 0 1 x x dx = ∑ n≥1 (-1) n-1 /n n = 1 - 1/4 + 1/27 - … ≈ 0.7834, and ∫ 0 1 x -x dx = ∑ n≥1 1/n n = 1 + 1/4 + 1/27 + … ≈ 1.2913. The name teases that the result looks like a naive ‘dream’ a student might wish were true — yet it really is. The trick is to expand x x = e x ln x as a power series and integrate term by term. LIT verified live: the numerical integrals of x x and x -x over [0, 1] match their respective series ∑(-1) n-1 /n n and ∑1/n n to ~1e-6 (window.__sophomore). FIG no framing; the numerical integrals and the self-power series are computed by different routes in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at split-screen — two panels: a smooth self-power integral on one, an n n series on the other, landing on the same dreamlike value. AVAN (AI) built the instrument: the numerical integrals of x x and x -x , the n n series, and their agreement. Credit as content: Johann Bernoulli (1697). The weave: David names the split screen; I confirm ∫x ±x equals the self-power series. 3 ONE DIMENSION The curves x^x (dipping to a minimum) and x^{−x} on [0,1]; their areas are the two n^n series. 4 TWO DIMENSIONS · INTERACTIVE Add series terms; the n^n partial sums converge to the integrals ∫₀¹ x^x dx and ∫₀¹ x^{−x} dx. add terms ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the two integral values, equal to their n^n series. AVAN’s addition (the inverse-companion): don’t integrate the self-power — expand it. The inverse of ‘∫ 0 1 x ±x dx’ is ‘the series ∑ (±1) n-1 /n n ’, obtained by expanding e ±x ln x and integrating term by term. Magenta are the self-power curves; green are the n n series they equal. A self-power integral read as a clean series. pause spin LIT Genuine sophomore's dream (Johann Bernoulli, 1697). Verified live: the numerical integrals ∫₀¹ x^x dx and ∫₀¹ x^{−x} dx match the series Σ(−1)^{n−1}/nⁿ and Σ1/nⁿ respectively to ~1e-6 (window.__sophomore.okA, .okB, .Ip, .In). FIG No framing; the numerical integrals and the self-power series are computed by different routes in-browser and agree. The AVAN inverse is honest — instead of integrating the self-power, expand it: the inverse of '∫₀¹ x^{±x} dx' is 'the series Σ(±1)^{n−1}/nⁿ', obtained by expanding e^{±x ln x} and integrating term by term. Magenta are the self-power curves; green are the nⁿ series they equal. A self-power integral read as a clean series. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of SPLIT SCREEN · David Lee Wise (ROOT0), with AVAN


THE LANDER-PARKIN

The Lander–Parkin counterexample in the 5-window house format — it demolished a 200-year-old conjecture of Euler. Extending Fermat's Last Theorem, Euler conjectured in 1769 that summing fewer than k perfect k-th powers can never equal a k-th power — e.g. you'd need at least five fifth-powers to make a fifth-power. In 1966, using an early computer, Lander and Parkin found: 27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵ — just four fifth-powers. Euler was wrong. Later Noam Elkies and Roger Frye found a fourth-power version with only three terms: 95800⁴ + 217519⁴ + 414560⁴ = 422481⁴. Verified live with exact big-integer arithmetic: 27⁵+84⁵+110⁵+133⁵ equals 144⁵ exactly (four terms), and 95800⁴+217519⁴+414560⁴ equals 422481⁴ exactly (three terms); a nearby altered sum is not a perfect fifth power. Neon-noir traced. See the four fifth-powers stacking to one in 1D, the exact identity + control in 2D, and the one-witness-refutes inverse in 3D.

THE LANDER-PARKIN · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CHEAT · NOCLIP ◆ .dlw.fold THE FOLD / CHEAT / NOCLIP / THE LANDER-PARKIN THE LANDER-PARKIN a counterexample refuting Euler's conjecture 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Lander–Parkin counterexample demolished a 200-year-old conjecture of Euler. Extending Fermat’s Last Theorem, Euler conjectured in 1769 that summing fewer than k perfect k-th powers can never equal a k-th power — e.g. you’d need at least five fifth-powers to make a fifth-power. In 1966, using an early computer, Lander and Parkin found: 27 5 + 84 5 + 110 5 + 133 5 = 144 5 — just four fifth-powers. Euler was wrong. Later Noam Elkies and Roger Frye found a fourth-power version with only three terms: 95800 4 + 217519 4 + 414560 4 = 422481 4 . LIT verified live with exact big-integer arithmetic: 27 5 +84 5 +110 5 +133 5 equals 144 5 exactly (four terms, refuting Euler), and 95800 4 +217519 4 +414560 4 equals 422481 4 exactly (three terms); a nearby altered sum is not a perfect fifth power (window.__landerparkin). FIG no framing; the exact arbitrary-precision arithmetic runs in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at noclip — the cheat that clips straight through a 200-year-old conjecture: a single explicit sum walks past the wall Euler thought was there. AVAN (AI) built the instrument: the exact big-integer fifth- and fourth-power sums, and a control near-miss. Credit as content: L. J. Lander & T. R. Parkin (1966); Noam Elkies and Roger Frye (fourth powers). The weave: David names the noclip; I confirm the exact equalities that refute Euler’s conjecture. 3 ONE DIMENSION Four fifth-powers 27⁵, 84⁵, 110⁵, 133⁵ stacking up to exactly 144⁵ — Euler said you'd need five. 4 TWO DIMENSIONS · INTERACTIVE The exact big-integer identity, and a control that alters one base and breaks the equality. toggle example ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the exact equality of four fifth-powers with one. AVAN’s addition (the inverse-companion): don’t trust the conjecture — search for a witness. The inverse of ‘can fewer than k k-th powers sum to a k-th power?’ is ‘yes — here is an explicit counterexample’, and one witness is enough to refute a universal claim. Magenta are the four summand powers; green is the single power they equal. A conjecture broken by one example. pause spin LIT Genuine Lander–Parkin counterexample to Euler's sum-of-powers conjecture (L. J. Lander & T. R. Parkin, 1966; Elkies/Frye for 4th powers). Verified live with exact BigInt: 27⁵+84⁵+110⁵+133⁵ = 144⁵ (four terms) and 95800⁴+217519⁴+414560⁴ = 422481⁴ (three terms), while a control near-miss (133→134) is not a perfect fifth power (window.__landerparkin.ok5, .ok4, .ctrl). FIG No framing; the exact arbitrary-precision arithmetic runs in-browser. The AVAN inverse is honest — instead of trusting the conjecture, search for a witness: the inverse of 'can fewer than k k-th powers sum to a k-th power?' is 'yes — here is an explicit counterexample', and one witness is enough to refute a universal claim. Magenta are the four summand powers; green is the single power they equal. A conjecture broken by one example. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of NOCLIP · David Lee Wise (ROOT0), with AVAN


THE SYLVESTER INERTIA

Sylvester's law of inertia in the 5-window house format — a symmetric matrix has an unchangeable 'signature'. Any real symmetric matrix M can be transformed by congruence — M → PᵀMP for an invertible P — into many different-looking matrices. But the counts of positive, negative, and zero eigenvalues (the signature n₊, n₋, n₀) never change. You can rescale and mix the coordinates however you like; the number of 'plus' and 'minus' directions of the quadratic form is a fixed invariant. It is what lets us classify quadratic forms and read the character (definite, indefinite) of a form from any convenient basis. Verified live: for thousands of random symmetric matrices, the signature from eigenvalue signs is unchanged after a random congruence PᵀMP; and the number of negative eigenvalues equals the sign changes in the leading principal minors (Jacobi's criterion) — two independent computations of the same signature. Neon-noir traced. See the matrix + eigenvalue signs in 1D, invariance + Jacobi minors in 2D, and the surviving-invariant inverse in 3D.

THE SYLVESTER INERTIA · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · RESPAWN · THE RESURRECT ◆ .dlw.fold THE FOLD / RESPAWN / THE RESURRECT / THE SYLVESTER INERTIA THE SYLVESTER INERTIA a signature invariant under congruence 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Sylvester’s law of inertia says a symmetric matrix has an unchangeable ‘signature’. Any real symmetric matrix M can be transformed by congruence — M → P T MP for an invertible P — into many different-looking matrices. But the counts of positive, negative, and zero eigenvalues (the signature n + , n - , n 0 ) never change. You can rescale and mix the coordinates however you like; the number of ‘plus’ and ‘minus’ directions of the quadratic form is a fixed invariant. It is what lets us classify quadratic forms and read the character (definite, indefinite) of a form from any convenient basis. LIT verified live: for thousands of random symmetric matrices, the signature computed from the eigenvalue signs is unchanged after a random congruence P T MP; and the number of negative eigenvalues equals the number of sign changes in the sequence of leading principal minors (Jacobi’s criterion) — two independent computations of the same signature (window.__sylvesterinertia). FIG no framing; the eigenvalue signature, the congruence, and the minor-sign-change count all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-resurrect — the invariant that resurrects unchanged after any congruence: mangle the matrix, and its signature comes back exactly as it was. AVAN (AI) built the instrument: the eigenvalue signature, the random congruence, and the leading-minor sign-change cross-check. Credit as content: James Joseph Sylvester (1852); Carl Gustav Jacob Jacobi (minor criterion). The weave: David names the invariant; I confirm the signature survives congruence and matches the minor sign-changes. 3 ONE DIMENSION A symmetric matrix and its eigenvalue signs — the signature (n₊, n₋, n₀) that congruence can never change. 4 TWO DIMENSIONS · INTERACTIVE New matrices; the signature is shown unchanged after a random congruence PᵀMP, and matched to the minor sign-changes. new matrix ▶ apply congruence ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the signature, invariant under every congruence. AVAN’s addition (the inverse-companion): don’t read the matrix entries — count the signs. The inverse of ‘which symmetric matrix?’ is ‘its signature (n + , n - , n 0 )’, the one thing congruence cannot touch — also read off the sign changes in the leading minors. Magenta are the congruence-transformed matrices; green is the signature they all share. The invariant that survives every basis change. pause spin LIT Genuine Sylvester's law of inertia (James Joseph Sylvester, 1852; Jacobi minor criterion). Verified live: for ~1200 random symmetric matrices, the signature (n₊,n₋,n₀) from eigenvalue signs is unchanged after a random congruence PᵀMP, and n₋ equals the number of sign changes in the leading principal minors (Jacobi) — two independent computations (window.__sylvesterinertia.inv, .jac). FIG No framing; the eigenvalue signature, the congruence, and the minor-sign-change count all run in-browser. The AVAN inverse is honest — instead of reading the matrix entries, count the signs: the inverse of 'which symmetric matrix?' is 'its signature (n₊,n₋,n₀)', the one thing congruence cannot touch — also read off the sign changes in the leading minors. Magenta are the congruence-transformed matrices; green is the signature they all share. The invariant that survives every basis change. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE RESURRECT · David Lee Wise (ROOT0), with AVAN


THE CAUCHY INTERLACING

Cauchy's interlacing theorem in the 5-window house format — pinning the eigenvalues of a submatrix between those of the whole. Take a symmetric n×n matrix M with eigenvalues λ₁≥λ₂≥…≥λₙ, and delete one row and the matching column to get an (n−1)×(n−1) principal submatrix B with eigenvalues μ₁≥…≥μ_{n−1}. Cauchy proved they interlace: λ_i ≥ μ_i ≥ λ_{i+1} for every i. Each submatrix eigenvalue is trapped in the gap between two consecutive eigenvalues of the full matrix. It is the backbone of eigenvalue algorithms, Sturm sequences, and Sylvester's law of inertia. Verified live: for thousands of random symmetric matrices, the eigenvalues of a principal submatrix (computed independently by the Jacobi method) always satisfy λ_i ≥ μ_i ≥ λ_{i+1} — the interlacing never fails. Neon-noir traced. See the two eigenvalue sets on a line in 1D, the interlacing check in 2D, and the nested-eigenvalues inverse in 3D.

THE CAUCHY INTERLACING · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · BOSS · THE FINAL BOSS ◆ .dlw.fold THE FOLD / BOSS / THE FINAL BOSS / THE CAUCHY INTERLACING THE CAUCHY INTERLACING submatrix eigenvalues interlacing the whole 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Cauchy’s interlacing theorem pins the eigenvalues of a submatrix between those of the whole. Take a symmetric n×n matrix M with eigenvalues λ 1 ≥ λ 2 ≥ … ≥ λ n , and delete one row and the matching column to get an (n-1)×(n-1) principal submatrix B with eigenvalues μ 1 ≥ … ≥ μ n-1 . Cauchy proved they interlace : λ i ≥ μ i ≥ λ i+1 for every i. Each submatrix eigenvalue is trapped in the gap between two consecutive eigenvalues of the full matrix. It is the backbone of eigenvalue algorithms, Sturm sequences, and Sylvester’s law of inertia. LIT verified live: for thousands of random symmetric matrices, the eigenvalues of a principal submatrix (computed independently by the Jacobi method) always satisfy λ i ≥ μ i ≥ λ i+1 — the interlacing never fails (window.__cauchyinterlacing). FIG no framing; the two eigenvalue sets are computed separately and the interlacing inequalities always hold. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-final-boss — the arena where every submatrix eigenvalue is pinned between two walls it can never cross: λ i above, λ i+1 below. AVAN (AI) built the instrument: the Jacobi eigenvalues of the matrix and its submatrix, and the interlacing test. Credit as content: Augustin-Louis Cauchy. The weave: David names the walls; I confirm each submatrix eigenvalue is trapped between consecutive eigenvalues of the whole. 3 ONE DIMENSION The eigenvalues of M (green) and of its submatrix (magenta) on a line — the magenta ones interlace the green. 4 TWO DIMENSIONS · INTERACTIVE New matrices; each submatrix eigenvalue μ_i is checked to lie in [λ_{i+1}, λ_i]. new matrix ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the interlaced eigenvalues, each submatrix value in its gap. AVAN’s addition (the inverse-companion): don’t recompute from scratch — bound with the whole. The inverse of ‘the submatrix’s eigenvalues’ is ‘the gaps between the full matrix’s eigenvalues that trap them’. Magenta are the submatrix eigenvalues; green are the full matrix’s eigenvalues that sandwich them. Eigenvalues nested inside eigenvalues. pause spin LIT Genuine Cauchy interlacing theorem (Augustin-Louis Cauchy). Verified live: for ~4000 random symmetric matrices, the eigenvalues μ_i of a principal submatrix (independent Jacobi computation) always satisfy λ_i ≥ μ_i ≥ λ_{i+1} where λ are M's eigenvalues — worst violation 0 (window.__cauchyinterlacing.ok, .worst). FIG No framing; the two eigenvalue sets are computed separately and the interlacing inequalities always hold. The AVAN inverse is honest — instead of recomputing from scratch, bound with the whole: the inverse of 'the submatrix's eigenvalues' is 'the gaps between the full matrix's eigenvalues that trap them'. Magenta are the submatrix eigenvalues; green are the full matrix's eigenvalues that sandwich them. Eigenvalues nested inside eigenvalues. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE FINAL BOSS · David Lee Wise (ROOT0), with AVAN


THE BRAHMAGUPTA

Brahmagupta's formula in the 5-window house format — the area of a cyclic quadrilateral (vertices on a circle) from its side lengths alone: Area = √((s−a)(s−b)(s−c)(s−d)), where s=(a+b+c+d)/2 is the semiperimeter. It is the four-sided generalization of Heron's triangle formula — and remarkably, among all quadrilaterals with those four side lengths, the cyclic one has the largest possible area. So Brahmagupta's value is not just the cyclic area but the maximum area achievable with those sides. Verified live: for thousands of quadrilaterals with vertices on a circle, the shoelace (coordinate) area equals √((s−a)(s−b)(s−c)(s−d)) to ~1e-14; and any non-cyclic quadrilateral with the same side lengths has a strictly smaller area. Neon-noir traced. See the inscribed quadrilateral in 1D, shoelace vs formula + maximality in 2D, and the biggest-area-from-sides inverse in 3D.

THE BRAHMAGUPTA · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · LOOT · THE INVENTORY ◆ .dlw.fold THE FOLD / LOOT / THE INVENTORY / THE BRAHMAGUPTA THE BRAHMAGUPTA a cyclic quadrilateral's maximal area from its sides 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Brahmagupta’s formula gives the area of a cyclic quadrilateral (one whose four vertices lie on a circle) from its side lengths alone: Area = √((s-a)(s-b)(s-c)(s-d)), where s = (a+b+c+d)/2 is the semiperimeter. It is the four-sided generalization of Heron’s triangle formula — and remarkably, among all quadrilaterals with those four side lengths, the cyclic one has the largest possible area . So Brahmagupta’s value is not just the cyclic area but the maximum area achievable with those sides. LIT verified live: for thousands of quadrilaterals with vertices placed on a circle, the shoelace (coordinate) area equals √((s-a)(s-b)(s-c)(s-d)) to ~1e-14; and any non-cyclic quadrilateral with the same side lengths has a strictly smaller area — the cyclic case is the maximum (window.__brahmagupta). FIG no framing; the coordinate area, the sides-only formula, and the maximality control all run in-browser. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-inventory — the loot: for a fixed set of four sides, the cyclic arrangement yields the biggest area you can bag. AVAN (AI) built the instrument: the on-circle shoelace area, the Brahmagupta sides-only formula, and the non-cyclic maximality control. Credit as content: Brahmagupta (628 CE); Heron for the triangle case. The weave: David names the biggest haul; I confirm the cyclic area equals the formula and is the maximum for those sides. 3 ONE DIMENSION A quadrilateral inscribed in a circle; its area is √((s−a)(s−b)(s−c)(s−d)) from the four sides alone. 4 TWO DIMENSIONS · INTERACTIVE New cyclic quadrilaterals; the shoelace area is compared to Brahmagupta's formula (and its maximality). new quadrilateral ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the cyclic quadrilateral's area, the maximum for its sides. AVAN’s addition (the inverse-companion): don’t place the corners — read the sides. The inverse of ‘the area of a cyclic quadrilateral’ is ‘√((s-a)(s-b)(s-c)(s-d)) from the sides alone’, which is also the greatest area those four sides can enclose. Magenta is the circle the vertices lie on; green is the maximal area they bound. Biggest area, from the sides. pause spin LIT Genuine Brahmagupta's formula (Brahmagupta, 628 CE; Heron for triangles). Verified live: for ~3000 quadrilaterals with vertices on a circle, the shoelace area equals √((s−a)(s−b)(s−c)(s−d)) to ~1e-14, and non-cyclic quadrilaterals with the same sides have strictly smaller area (window.__brahmagupta.eq, .mx, .worst). FIG No framing; the coordinate area, the sides-only formula, and the maximality control all run in-browser. The AVAN inverse is honest — instead of placing the corners, read the sides: the inverse of 'the area of a cyclic quadrilateral' is '√((s−a)(s−b)(s−c)(s−d)) from the sides alone', which is also the greatest area those four sides can enclose. Magenta is the circle the vertices lie on; green is the maximal area they bound. Biggest area, from the sides. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE INVENTORY · David Lee Wise (ROOT0), with AVAN


THE TAXICAB

1729, the taxicab number, in the 5-window house format — the smallest positive integer expressible as a sum of two positive cubes in two different ways: 1729 = 1³+12³ = 9³+10³. Its fame comes from a 1919 anecdote: when G. H. Hardy visited the ailing Srinivasa Ramanujan and remarked that his taxi's number, 1729, seemed rather dull, Ramanujan instantly replied that it was very interesting — the smallest number expressible as a sum of two cubes two ways. It is the second 'taxicab number' Ta(2); the next such number is 4104 = 2³+16³ = 9³+15³. Verified live: a brute search over all sums of two positive cubes finds that 1729 is the smallest integer with two distinct such representations, and the next one is 4104. Neon-noir traced. See 1729 built two ways in 1D, the brute search flagging it in 2D, and the number-hiding-two-cubes inverse in 3D.

THE TAXICAB · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GLITCH · THE BLUE SCREEN ◆ .dlw.fold THE FOLD / GLITCH / THE BLUE SCREEN / THE TAXICAB THE TAXICAB the smallest two-way sum of two cubes 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION 1729, the taxicab number , is the smallest positive integer expressible as a sum of two positive cubes in two different ways : 1729 = 1³ + 12³ = 9³ + 10³. Its fame comes from a 1919 anecdote: when G. H. Hardy visited the ailing Srinivasa Ramanujan and remarked that his taxi’s number, 1729, seemed rather dull, Ramanujan instantly replied that it was very interesting — the smallest number expressible as a sum of two cubes two ways. It is the second ‘taxicab number’ Ta(2); the next such number is 4104 = 2³ + 16³ = 9³ + 15³. LIT verified live: a brute search over all sums of two positive cubes finds that 1729 is the smallest integer with two distinct such representations (1³+12³ and 9³+10³), and the next one is 4104 (window.__taxicab). FIG no framing; the exhaustive cube-sum search runs in-browser and confirms 1729 as the smallest. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-blue-screen — the glitch made famous: a ‘dull’ taxi number that turns out to hide two cube-sums, crashing the assumption that it was boring. AVAN (AI) built the instrument: the exhaustive two-cube-sum search and the confirmation that 1729 is the smallest two-way case. Credit as content: G. H. Hardy & Srinivasa Ramanujan (1919); the taxicab-number concept. The weave: David names the glitch; I confirm 1729 is the smallest sum of two cubes two ways. 3 ONE DIMENSION 1729 built two ways: 1³+12³ and 9³+10³ — the two cube-pairs that reach the same total. 4 TWO DIMENSIONS · INTERACTIVE Scan upward; the brute cube-sum search flags 1729 as the first number with two representations. next two-way number ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: 1729, met by two different cube-pairs. AVAN’s addition (the inverse-companion): don’t judge a number dull — factor it into cubes. The inverse of ‘the number 1729’ is ‘the two cube-pairs 1³+12³ and 9³+10³ that both reach it’, the smallest such coincidence. Magenta are the two cube-pairs; green is the number they share. A dull number hiding two cubes. pause spin LIT Genuine Hardy–Ramanujan taxicab number 1729 (anecdote 1919; taxicab-number concept). Verified live: an exhaustive search over sums of two positive cubes confirms 1729 is the smallest integer with two distinct representations (1³+12³ and 9³+10³), and the next is 4104 (window.__taxicab.smallest, .next, .ok). FIG No framing; the exhaustive cube-sum search runs in-browser and confirms 1729 as the smallest. The AVAN inverse is honest — instead of judging a number dull, factor it into cubes: the inverse of 'the number 1729' is 'the two cube-pairs 1³+12³ and 9³+10³ that both reach it', the smallest such coincidence. Magenta are the two cube-pairs; green is the number they share. A dull number hiding two cubes. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE BLUE SCREEN · David Lee Wise (ROOT0), with AVAN


THE WALD

Wald's identity in the 5-window house format — a clean law for random sums that stop at a random time. Suppose you add up independent, identically distributed steps X₁, X₂, …, and you keep a rule that decides when to stop — a stopping time N (it may depend on the steps seen so far, but not the future). Wald proved that the expected total equals the expected number of steps times the expected step: E[S_N] = E[N]·E[X], where S_N = X₁+…+X_N. Even though N is random and correlated with the walk, the average total factors perfectly. Verified live: simulating a walk with steps uniform on {1,2,3} (E[X]=2), stopping the first time the total reaches 50, the empirical average final total E[S_N] matches E[N]·E[X] to within a fraction of a percent over hundreds of thousands of runs. Neon-noir traced. See the stopping walk in 1D, E[S_N] vs E[N]·E[X] in 2D, and the factors-in-the-mean inverse in 3D.

THE WALD · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · CO-OP · THE HANDOFF ◆ .dlw.fold THE FOLD / CO-OP / THE HANDOFF / THE WALD THE WALD an expected sum equal to expected count times expected step 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION Wald’s identity is a clean law for random sums that stop at a random time. Suppose you add up independent, identically distributed steps X 1 , X 2 , …, and you keep a rule that decides when to stop — a stopping time N (it may depend on the steps seen so far, but not the future). Wald proved that the expected total equals the expected number of steps times the expected step: E[S N ] = E[N]·E[X] , where S N = X 1 + … + X N . Even though N is random and correlated with the walk, the average total factors perfectly. LIT verified live: simulating a walk with steps uniform on {1, 2, 3} (so E[X] = 2), stopping the first time the running total reaches 50, the empirical average final total E[S N ] matches E[N]·E[X] to within a fraction of a percent over hundreds of thousands of runs (window.__wald). FIG no framing; the stopping-time simulation and the E[N]·E[X] product both run in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at the-handoff — the co-op pass: the random total and the product E[N]·E[X] hand off to the same expected value, whatever the stopping rule. AVAN (AI) built the instrument: the stopping-time walk simulation, the expected total, and the E[N]·E[X] product. Credit as content: Abraham Wald (1944). The weave: David names the handoff; I confirm E[S N ] equals E[N]·E[X]. 3 ONE DIMENSION A random walk of steps {1,2,3} climbing until it reaches the threshold; N steps, total S_N. 4 TWO DIMENSIONS · INTERACTIVE Run more trials; the empirical E[S_N] converges to E[N]·E[X]. more trials ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the expected total, equal to E[N]·E[X]. AVAN’s addition (the inverse-companion): don’t track the whole random sum — factor it. The inverse of ‘the expected stopped total E[S N ]’ is ‘E[N]·E[X]’, the average count times the average step — the randomness of N and the walk decouple in the mean. Magenta are the random walk paths; green is the expected total they share with E[N]·E[X]. A random sum that factors in the mean. pause spin LIT Genuine Wald's identity (Abraham Wald, 1944). Verified live: simulating a walk with steps uniform on {1,2,3} (E[X]=2), stopped the first time the total reaches 50, the empirical E[S_N] matches E[N]·E[X] to within FIG No framing; the stopping-time simulation and the E[N]·E[X] product both run in-browser and agree. The AVAN inverse is honest — instead of tracking the whole random sum, factor it: the inverse of 'the expected stopped total E[S_N]' is 'E[N]·E[X]', the average count times the average step — the randomness of N and the walk decouple in the mean. Magenta are the random walk paths; green is the expected total they share with E[N]·E[X]. A random sum that factors in the mean. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE HANDOFF · David Lee Wise (ROOT0), with AVAN


THE DURFEE SQUARE

The Durfee square in the 5-window house format — the largest square that fits in the top-left corner of a partition's Young diagram. For a partition of n drawn as rows of boxes, its Durfee square has side d = the largest number such that the partition has at least d parts each of size ≥ d. This single number splits every partition into three pieces: the d×d square, a partition to its right (parts ≤ d), and a partition below (at most d parts). That decomposition gives a beautiful generating-function identity for the partition numbers: Σ_n p(n)qⁿ = Σ_{d≥0} q^{d²}/∏_{i=1}^d(1−qⁱ)² — sorting all partitions by their Durfee-square size. Verified live: expanding Σ_{d≥0} q^{d²}/∏_{i=1}^d(1−qⁱ)² as a power series, the coefficient of qⁿ equals the partition number p(n) for every n up to 45 — p(40)=37338, p(45)=89134. Neon-noir traced. See a Young diagram with its Durfee square in 1D, the generating function vs p(n) in 2D, and the split-by-square inverse in 3D.

THE DURFEE SQUARE · WORLD II — THE FOLD ◀ THE FOLD 0ROOT.AI // WORLD II · GRIND · WARM CACHE ◆ .dlw.fold THE FOLD / GRIND / WARM CACHE / THE DURFEE SQUARE THE DURFEE SQUARE a square hidden in every partition 1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION The Durfee square is the largest square that fits in the top-left corner of a partition’s Young diagram. For a partition of n drawn as rows of boxes, its Durfee square has side d = the largest number such that the partition has at least d parts each of size ≥ d. This single number splits every partition into three pieces: the d×d square, a partition to its right (parts ≤ d), and a partition below (at most d parts). That decomposition gives a beautiful generating-function identity for the partition numbers: ∑ n p(n)q n = ∑ d≥0 q d² / ∏ i=1 d (1-q i )² — sorting all partitions by their Durfee-square size. LIT verified live: expanding ∑ d≥0 q d² /∏ i=1 d (1-q i )² as a power series, the coefficient of q n equals the partition number p(n) for every n up to 45 — p(40)=37338, p(45)=89134 (window.__durfee). FIG no framing; the Durfee-square generating function and a brute partition count both run in-browser and agree. 2 HOW IT WAS WEAVED · AI + HUMAN David (human) seated this at warm-cache — the grind that rolls through every partition and reads off its Durfee square, sorting the whole pile by that one number. AVAN (AI) built the instrument: the Durfee-square generating function, the brute partition count, and their coefficient-by-coefficient agreement. Credit as content: William Durfee (a student of J. J. Sylvester, 1880s). The weave: David names the grind; I confirm the Durfee generating function reproduces the partition numbers. 3 ONE DIMENSION A partition's Young diagram with its Durfee square shaded — the biggest square in the top-left corner. 4 TWO DIMENSIONS · INTERACTIVE Cycle n; the Durfee-square generating function's q^n coefficient is compared to the partition count p(n). next n ▶ verify ▶ 5 THREE DIMENSIONS + AVAN’S INVERSE The green forward object: the partition number p(n), summed over Durfee-square sizes. AVAN’s addition (the inverse-companion): don’t count partitions blindly — sort them by their square. The inverse of ‘p(n)’ is ‘∑ d q d² /∏(1-q i )²’, grouping partitions by the size of their Durfee square. Magenta are the Durfee squares of each size; green is the partition count they assemble. Every partition split by its square. pause spin LIT Genuine Durfee square identity (William Durfee, a student of J. J. Sylvester, 1880s). Verified live: expanding Σ_{d≥0} q^{d²}/∏_{i=1}^d(1−qⁱ)² as a power series, the coefficient of qⁿ equals the brute partition count p(n) for every n=0..45; p(40)=37338, p(45)=89134 (window.__durfee.ok, .p40, .p45). FIG No framing; the Durfee-square generating function and a brute partition count both run in-browser and agree. The AVAN inverse is honest — instead of counting partitions blindly, sort them by their square: the inverse of 'p(n)' is 'Σ_d q^{d²}/∏(1−qⁱ)²', grouping partitions by the size of their Durfee square. Magenta are the Durfee squares of each size; green is the partition count they assemble. Every partition split by its square. ◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of WARM CACHE · David Lee Wise (ROOT0), with AVAN