101 introduced \(c^*=3\) as the unique value producing a Whitney \(A_1\) fold in \(V(q)=q^3-cq\) at \(q=1\). This week works the full derivation. Setting \(V'(q)=3q^2-c=0\) gives critical points \(q^*=\pm\sqrt{{c/3}}\); for the fold to land exactly on the integer \(q=1\), we need \(q^*=1\), i.e. \(\sqrt{{c/3}}=1 \Rightarrow c=3\). At exactly \(c=3\): \(V(1)=1-3=-2\), \(V'(1)=3-3=0\), \(V''(1)=6q|_{{q=1}}=6\neq0\) — a genuine double root, not a higher-order degeneracy. Factoring confirms it directly: \(V(q)+2 = q^3-3q+2 = (q-1)^2(q+2)\), verified by direct multiplication.
The Collatz map \(T:n\mapsto3n+1\) (odd branch) has coefficient exactly \(c=3\) — making it, per this framework, the unique integer map sitting precisely at the fold threshold. This is the real, substantive connection this course cares about: not a numeric coincidence with an external physics constant, but the same \(c=3\) appearing in both the abstract fold-potential argument and the specific integer dynamical system this course has been building toward since 101.
-- dm³ 102 · Week 07 · Theorem C.1, worked in full -- V(q) = q³ - cq, V'(q) = 3q² - c = 0 ⟹ q* = ±√(c/3) -- Fold at integer q*=1 ⟺ √(c/3)=1 ⟺ c=3 -- At c=3 exactly: example : (1:ℝ)^3 - 3*1 = -2 := by norm_num -- V(1) = -2 example : 3*(1:ℝ)^2 - 3 = 0 := by norm_num -- V'(1) = 0 example : (6:ℝ)*1 ≠ 0 := by norm_num -- V''(1) = 6 ≠ 0 (genuine double root) -- Factorization: V(q)+2 = (q-1)²(q+2) example (q : ℝ) : q^3 - 3*q + 2 = (q-1)^2*(q+2) := by ring -- Collatz map T: n ↦ 3n+1 has coefficient c=3 exactly — the unique -- integer map at the fold threshold, per this derivation.