proofs
Provedphase-diagrampolynomial-methodattentiondiagram

e_poly(x, c) = 2c·log(1 + x/2c) + x·log(1 + 2c/x) is increasing in x > 0

For all c,x,y∈R\htmlData{mx=1}{c, x, y \in \htmlData{mx=0}{\mathbb{R}}}, if 0<c0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{c}, 0<x0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{x} and x≤y\htmlData{mx=0}{x} \mathrel{\htmlData{mx=2}{\leq}} \htmlData{mx=1}{y}, then

2c⋅log⁡(1+x2c)+x⋅log⁡(1+2cx)≤2c⋅log⁡(1+y2c)+y⋅log⁡(1+2cy)\begin{aligned} & \htmlData{mx=18}{\htmlData{mx=9}{\htmlData{mx=2}{2\htmlData{mx=1}{c}} \cdot \htmlData{mx=8}{\log \left(\htmlData{mx=7}{1 + \htmlData{mx=6}{\frac{\htmlData{mx=3}{x}}{\htmlData{mx=5}{2\htmlData{mx=4}{c}}}}}\right)}} + \htmlData{mx=17}{\htmlData{mx=10}{x} \cdot \htmlData{mx=16}{\log \left(\htmlData{mx=15}{1 + \htmlData{mx=14}{\frac{\htmlData{mx=12}{2\htmlData{mx=11}{c}}}{\htmlData{mx=13}{x}}}}\right)}}} \\ & \quad \htmlData{mx=0}{\leq} \htmlData{mx=36}{\htmlData{mx=27}{\htmlData{mx=20}{2\htmlData{mx=19}{c}} \cdot \htmlData{mx=26}{\log \left(\htmlData{mx=25}{1 + \htmlData{mx=24}{\frac{\htmlData{mx=21}{y}}{\htmlData{mx=23}{2\htmlData{mx=22}{c}}}}}\right)}} + \htmlData{mx=35}{\htmlData{mx=28}{y} \cdot \htmlData{mx=34}{\log \left(\htmlData{mx=33}{1 + \htmlData{mx=32}{\frac{\htmlData{mx=30}{2\htmlData{mx=29}{c}}}{\htmlData{mx=31}{y}}}}\right)}}} \end{aligned}
What the notation means · or point at any symbol
RReal numbers
The real numbers: every point on the number line. (ℝ)
x ∈ TVariables
c, x, y range over real numbers: the statement must hold for every choice. ((c x y : ℝ))
<Less than
0 is strictly smaller than c. (0 < c)
≤At most
x is less than or equal to y. (x ≤ y)
·Multiplication
2 times c. (2 * c)
logNatural logarithm
The logarithm of 1 + x/(2 · c) to base e: the power e must be raised to to give 1 + x/(2 · c). (Real.log (1 + x / (2 * c)))
+Addition
1 plus x/(2 · c). (1 + x / (2 * c))
a/bDivision
x divided by 2 · c. In Lean, dividing by 0 gives 0. (x / (2 * c))

Makes the Phase-Diagram formula a proven lower bound on the polynomial-method exponent: the true Taylor degree is κ(α)·α·ln n with κ ≥ e (the self-correction of the conjecture), and since e_poly is monotone, evaluating it at κα only raises the exponent. Proof via log u ≤ u − 1 applied twice (chord bounds), no calculus. Part of 'Phase-Diagram conjecture: sampling versus the polynomial method in dimension c ln n'.

How it was proved

Proved
4 October 2026
By
Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by
Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms
Classical.choice, Quot.sound, propext
Proof
57 lines of Lean
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