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Tao Theorem 5.1, Type I half

Proved
TaoFivePrimes.theorem51_typeI_envelope

by Hartmann_Psi · Sep 14, 2026 · Mathlib 0df444a (Lean v4.33.1)

analytic-number-theoryexponential-sumsgoldbachminor-arcsnumber-theory

The Type I half of Tao's Theorem 5.1. Let q≥4q\ge4q≥4 with (a,q)=1(a,q)=1(a,q)=1, let 4α=aq+β4\alpha=\frac aq+\beta4α=qa​+β with ∣β∣≤q−2|\beta|\le q^{-2}∣β∣≤q−2, and let U,V≥40U,V\ge40U,V≥40 with U,V<xU,V<xU,V<x, UV≤x4UV\le\frac x4UV≤4x​ and UV2≥xUV^2\ge xUV2≥x. Let (cd)(c_d)(cd​) be any complex coefficients with ∣cd∣≤1|c_d|\le1∣cd​∣≤1 for every positive odd d≤UVd\le UVd≤UV, and let

TI(x,α,U,V;c)=∑d≤UVd odd∣∑m odd(log⁡m+cdlog⁡d)η0 ⁣(dmx)e(αdm)∣T_I(x,\alpha,U,V;c)=\sum_{\substack{d\le UV\\ d\text{ odd}}}\Bigl|\sum_{m\text{ odd}}\bigl(\log m+c_d\log d\bigr)\eta_0\!\Bigl(\frac{dm}{x}\Bigr)e(\alpha dm)\Bigr|TI​(x,α,U,V;c)=d≤UVd odd​∑​​m odd∑​(logm+cd​logd)η0​(xdm​)e(αdm)​

be the Type I sum produced by the variant of Vaughan's identity. Then

TI(x,α,U,V;c) ≤ 0.5 xq(log⁡x)log⁡(2UVq+4)+0.89(UV+52q)(8+log⁡q)log⁡(2x).T_I(x,\alpha,U,V;c)\ \le\ 0.5\,\frac xq(\log x)\log\Bigl(\frac{2UV}{q}+4\Bigr)+0.89\Bigl(UV+\frac52q\Bigr)(8+\log q)\log(2x).TI​(x,α,U,V;c) ≤ 0.5qx​(logx)log(q2UV​+4)+0.89(UV+25​q)(8+logq)log(2x).

This is the first half of the source's Section 5: the two terms on the right are exactly the first two terms of Theorem 5.1. The argument is summation by parts in mmm over blocks of length 2q2q2q, the odd-restricted Vinogradov-type lemma on each block, and an integral test on the resulting harmonic sum over blocks.

Note for anyone attacking this Two of the source's intermediate displays in this passage do not come out as written. The integral test for the block sum drops an additive 444: its j=0j=0j=0 term alone contributes x2q⋅4\frac{x}{2q}\cdot42qx​⋅4, so the display fails once UV/q<e4−42≈25.3UV/q<\frac{e^4-4}2\approx25.3UV/q<2e4−4​≈25.3 (at q=1q=1q=1, UV=10UV=10UV=10 the two sides are 2.8937x2.8937x2.8937x and 1.5890x1.5890x1.5890x). And the per-block application of the odd-restricted Vinogradov lemma uses the factor 111 where that lemma gives ⌊2q2q⌋+1=2\lfloor\frac{2q}{2q}\rfloor+1=2⌊2q2q​⌋+1=2, since the blocks have length exactly 2q2q2q. The statement above is the source's, unmodified; a proof must recover that slack from the two terms rather than transcribe the chain.

Formalization Note The Type I sum, the divisor set and the smoothed cutoff are the platform definitions imported from Def_TaoFivePrimes_Theorem51Sums; the inner sum is over all odd integers m=2n+1m=2n+1m=2n+1, n∈Zn\in\mathbb Zn∈Z, made finite by the compact support of η0\eta_0η0​.

Preamble
import Mathlib
import Definitions.Def_TaoFivePrimes_Theorem51Sums

open Finset
Formal statement
theorem TaoFivePrimes.theorem51_typeI_envelope
    (x alpha beta : ℝ) (a : ℤ) (q : ℕ) (hq : 4 ≤ q)
    (haq : Nat.Coprime a.natAbs q)
    (halpha : 4 * alpha = (a : ℝ) / q + beta)
    (hbeta : |beta| ≤ 1 / (q : ℝ) ^ 2)
    (U V : ℝ) (hU40 : 40 ≤ U) (hV40 : 40 ≤ V) (hUx : U < x) (hVx : V < x)
    (hUV : U * V ≤ x / 4) (hUV2 : x ≤ U * V ^ 2)
    (c : ℕ → ℂ) (hc : ∀ d ∈ TaoFivePrimes.theorem51Divisors U V, ‖c d‖ ≤ 1) :
    TaoFivePrimes.theorem51TypeI x alpha U V c ≤
      0.5 * (x / q) * Real.log x * Real.log (2 * U * V / q + 4)
        + 0.89 * (U * V + (5 / 2) * q) * (8 + Real.log q) * Real.log (2 * x) := by sorry
Source
Terence Tao, "Every odd number greater than 1 is the sum of at most five primes", Mathematics of Computation 83 (2014), 997-1038; arXiv:1201.6656, https://arxiv.org/abs/1201.6656, Section 5, proof of Theorem 5.1, the Type I estimate (the first two terms)

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