Tao Section 5: summing the Type I envelope, given the Vinogradov lemma

Summing the Type I envelope, given the Vinogradov lemma. Let $q\ge4$ with $(a,q)=1$, let $4\alpha=\frac aq+\beta$ with $|\beta|\le q^{-2}$, let $x>0$ and $U,V\ge40$ with $UV\le\frac x4$. Assume the Vinogradov-type lemma with $B=4(\log2)\log2x$, in the form of the previous display. Let $W$ be any nonnegative function satisfying, for every positive odd $d\le UV$,

(the second alternative dropped when the sine vanishes). Then

This is the combinatorial half of the source's Type I estimate, with the constants that its own chain of estimates yields. Writing $C=4(\log2)\log2x$: for $d\le\frac q2$ one has $\|4d\alpha\|_{\mathbb R/\mathbb Z}\ge\frac1q-\frac{q/2}{q^2}=\frac1{2q}$, hence $\frac1{|\sin(2\pi d\alpha)|}\le2q$, and the odd-restricted Vinogradov lemma bounds that range by $2qC+\frac1\pi Cq\log4q$. Each block $2jq+\frac q2<d\le2(j+1)q+\frac q2$ has length exactly $2q$, so that lemma applies with prefactor $\lfloor\frac{2q}{2q}\rfloor+1=2$ and the first alternative frozen at the block's left endpoint, giving $2\frac{x}{2jq+q/2}\log x+4C+\frac4\pi Cq\log4q$. Summing the harmonic part by the integral test — including its $j=0$ term, which the source's display omits — produces the first term, and the rest is at most $\frac1\pi C(2UV+5q)(\pi+\log4q)$, which is within $1.78(UV+\frac52q)(8+\log q)\log2x$ because $\frac{8\log2}\pi\le1.78$ and $\pi+\log4q\le8+\log q$.

Formalization Note $W$ is an arbitrary nonnegative function on $\mathbb N$, constrained only on the positive odd $d\le UV$, so the statement says nothing about exponential sums. The index set is the platform's theorem51Divisors U V and the sine is written $\sin(\pi\cdot2\alpha\cdot d)$. The degenerate range $UV<\frac q2$, where every divisor is small, is covered separately.