4 Exponential sum growth exponents

4.1 Phase functions

Definition 4.1 Phase function

A phase function is a (variable) smooth function $F : [1, 2] \to R$ . A phase function $F$ will be called a model phase function if there exists a fixed exponent $σ > 0$ with the property that

F^{(p + 1)} (u) - \frac{d^{p}}{d u^{p}} u^{- σ} = o (1)

for all (variable) $u \in [1, 2]$ and all fixed $p \geq 0$ , where $F^{(p + 1)}$ denotes the $(p + 1)^{st}$ derivative of $F$ .

For instance, $u \mapsto \log u$ is a model phase function (with $σ = 1$ ), and for any fixed $σ \neq 1$ , $u \mapsto u^{1 - σ} / (1 - σ)$ is also a model phase function. Informally, a model phase function is a function which asymptotically behaves like $u \mapsto \log u$ (for $σ = 1$ ) or $u \mapsto u^{1 - σ} / (1 - σ)$ (for $σ \neq 1$ ), up to constants. This turns out to be a good class for exponential sum estimates, as it is stable under Weyl differencing and Legendre transforms, which show up in the van der Corput A-process and B-process respectively.

Note from Proposition 2.1 that the $o (1)$ decay rate in 1 can be made uniform, after passing to a subsequence if necessary.

4.2 Exponential sum exponent

The main purpose of this chapter is to introduce and establish the basic properties of the following exponent function.

Definition 4.2 Exponent sum growth exponent

For any fixed $α \geq 0$ , let $β (α) \in R$ denote the least possible (fixed) exponent for which the following claim holds: whenever $N, T \geq 1$ are (variable) quantities with $T$ unbounded and $N = T^{α + o (1)}$ , $F$ is a model phase function, and $I \subset [N, 2 N]$ is an interval, then

\sum_{n \in I} e (T F (n / N)) ≪ T^{β (α) + o (1)} .

Implemented at bound_beta.py as:
Bound_beta

It is easy to see that the set of possible candidates for $β (α)$ is closed (thanks to underspill), non-empty, and bounded from below, so $β$ is well-defined as a (fixed) function from $[0, + \infty)$ to $R$ . Specializing to the logarithmic phase $F (u) = \log u$ , and performing a complex conjugation, we see in particular that

\sum_{n \in I} n^{- i T} ≪ T^{β (α) + o (1)}

whenever $T$ is unbounded, $N = T^{α + o (1)}$ , and $I$ is an interval in $[N, 2 N]$ . Thus it is clear that knowledge of $β$ is of relevance to understanding the Riemann zeta function.

The quantity $β (α)$ can also be formulated without asymptotic notation, but at the cost of introducing some “epsilon and delta” parameters:

Lemma 4.3 Non-asymptotic definition of

β

Let $α \geq 0$ and $\overset{―}{β} \in R$ be fixed. Then the following are equivalent:

$β (α) \leq \overset{―}{β}$ .
For every (fixed) $ε > 0$ and $σ > 0$ there exists (fixed) $δ > 0$ , $P \geq 1$ , $C \geq 1$ with the following property: if $T \geq C$ , $T^{α - δ} \leq N \leq T^{α + δ}$ are (fixed) real numbers, $I \subset [N, 2 N]$ is a (fixed) interval, and $F$ is a (fixed) phase function such that

$| F^{(p + 1)} (u) - \frac{d^{p}}{d u^{p}} u^{- σ} | \leq δ$
3

for all (fixed) $0 \leq p \leq P$ and $u \in [1, 2]$ , then

$| \sum_{n \in I} e (T F (n / N)) | \leq C T^{\overset{―}{β} + ε} .$

Proof ▶

It is easy to see that (ii) implies (i) by expanding out all the definitions (and using Proposition 2.1 to resolve any uniformity issues). Conversely, suppose that (ii) fails. Carefully negating all the quantifiers, we conclude that there exists a fixed $ε, σ > 0$ such that for any (fixed) natural number $i$ , one can find real numbers $T = T_{i} \geq i$ , $T^{α - 1 / i} \leq N = N_{i} \leq T^{α + 1 / i}$ , an interval $I = I_{i} \subset [N_{i}, 2 N_{i}]$ , and a phase function $F = F_{i}$ such that

| F_{i}^{(p + 1)} (u) - \frac{d^{p}}{d u^{p}} u^{- σ} | \leq 1 / i

for all (fixed) $0 \leq p \leq i$ and $u \in [1, 2]$ , but that

| \sum_{n \in I} e (T F (n / N)) | \geq i T^{\overset{―}{β} + ε} .

But then $F = F_{i}$ is a model phase function which gives a counterexample to the claim $β (α) \leq \overset{―}{β}$ .

We will however work with the asymptotic formulation of $β$ throughout this database, as it makes the proofs somewhat cleaner.

We record the trivial bounds on $β$ :

Lemma 4.4 Trivial bounds on

β

For any fixed $α > 1$ , we have

β (α) = α - 1.

For fixed $0 \leq α \leq 1$ , we have

\frac{α}{2} \leq β (α) \leq α .

In particular

β (0) = 0.

Implemented at bound_beta.py as:
trivial_beta_bound_1
trivial_beta_bound_2

Proof ▶

Let $T > 1$ be unbounded, $N = T^{α + o (1)}$ , $I \subset [N, 2 N]$ an interval, and $F$ a model phase function.

For $α > 1$ , the Euler–Maclaurin formula (see e.g. [ 217 , (2.1.2) ] ) gives

| \sum_{N \leq n \leq 2 N} n^{i T} | = | \frac{2^{1 + i T} - 1}{1 + i T} N^{1 + i T} + O (1) | ≍ \frac{N}{T}

which gives the lower bound $β (α) \geq α - 1$ ; applying the Euler–Maclaurin formula for model phase functions $F$ then gives the matching upper bound.

The triangle inequality bound

\sum_{n \in I} e (T F (n / N)) ≪ N

gives the upper bound $β (α) \leq α$ . Next, if $0 \leq α \leq 1$ , then from Lemma 3.1 (and the $≫ 1 / N$ -separated nature of the $F (n / N)$ for model phase functions $F$ , after passing to a subsequence if necessary) that

\int_{T^{2 T}} {| \sum_{n \in [N, 2 N]} e (t F (n / N)) |}^{2} d t ≍ T N

for $N = c T^{α}$ for $c$ a fixed small enough constant, which by the pigeonhole principle implies that

{| \sum_{n \in [N, 2 N]} e (t F (n / N)) |}^{2} d t ≫ N^{1 / 2} = T^{α / 2}

for at least one $t ≍ T$ , giving the claim.

As we shall see, the exponent pair conjecture is equivalent to the lower bound here being sharp, thus it is conjectured that

β (α) = {\begin{cases} α / 2, & 0 \leq α \leq 1 \\ α - 1, & α > 1 \end{cases} .

Note the discontinuity at $1$ . Despite this, we have:

Lemma 4.5 Upper semicontinuity

$β$ is an upper semicontinuous function.

Proof ▶

We record the classical bounds on $β$ :

Lemma 4.6 Van der Corput

A

process for

β

If $0 \leq α \leq 2 / 3$ and $h \geq 0$ then

2 β (α) \leq max (2 α - h, 2 h, α - h + sup_{2 α - 1 \leq h^{'} \leq h} ((h^{'} + 1 - α) β (\frac{α}{h^{'} + 1 - α}) + h^{'})) .

Implemented at bound_beta.py as:
apply_van_der_corput_process_for_beta(bounds)

Proof ▶

By definition, there exists an unbounded $T$ , $N = T^{α + o (1)}$ , $F$ a model phase function, and $I \subset [N, 2 N]$ such that

\sum_{n \in I} e (T F (n / N)) = T^{β (α) + o (1)} .

Applying [ 110 , (2.54) ] with $H := T^{h}$ , as well as the ensuing computations to dispose of the $j ≪ T / N^{2}$ terms, one then has

T^{2 β (α) + o (1)} ≪ N^{2} H^{- 1} + H^{2} + N H^{- 1} \sum_{T / N^{2} ≪ j ≪ H} | \sum_{n \in I \cap I - j} e (T (F ((n + j) / N) - F (n / N))) |

and hence by the pigeonhole principle

T^{2 β (α) + o (1)} ≪ N^{2} H^{- 1} + H^{2} + T^{o (1)} N H^{- 1} \sum_{j = T^{h^{'} + o (1)}} | \sum_{n \in I \cap I - h} e (T (F ((n + j) / N) - F (n / N))) |

for some $2 α - 1 \leq h^{'} \leq h$ (one can delete this term if $h < 2 α - 1$ ). One can verify that $- \frac{1}{σ} \frac{N}{j} (F (u + j / N) - F (u))$ is a model phase function. Thus, by Definition 4.2, one has

\sum_{n \in I \cap I - h} e (T (F ((n + j) / N) - F (n / N))) ≪ (T^{1 + h^{'} + o (1)} / N)^{β (α / (h^{'} + 1 - α)) + o (1)},

and the claim follows after evaluating all terms as powers of $T$ .

Proposition 4.7 Van der Corput inequality

For any natural number $k \geq 2$ and any $α > 0$ , one has

β (α) \leq max (α + \frac{1 - k α}{2^{k} - 2}, (1 - 2^{2 - k}) α - \frac{1 - α}{2^{k} - 2}) .

Thus for instance when $k = 2$ we have

β (α) \leq max (\frac{1}{2}, \frac{2 α - 1}{2}),

so in particular

β (1) = \frac{1}{2},

by Lemma 4.4, when $k = 3$ one has

β (α) \leq max (\frac{1 + 3 α}{6}, \frac{6 α - 1}{3}),

and when $k = 4$ one has

β (α) \leq max (\frac{10 α + 1}{14}, \frac{29 α - 2}{28}) .

This form of upper bound of $β (α)$ - as the maximum of a finite number of linear functions of $α$ - is extremely common in the literature.

Proof ▶

Follows from [ 113 , Theorem 8.20 ] . It is also possible to prove this by induction on $k$ using Lemma 4.6.

Corollary 4.8 Optimizing the van der Corput inequality

For any $α > 0$ one has

β (α) \leq inf_{k \in N : k \geq 2} α + \frac{1 - k α}{2^{k} - 2} .

Thus for instance

β (α) \leq min (\frac{1}{2}, \frac{1 + 3 α}{6}, \frac{10 α + 1}{14}) .

Proof ▶

Let $β_{k} (α) = α + (1 - k α) / (2^{k} - 2)$ and

α_{k} = \frac{2^{k}}{(k - 1) 2^{k} + 2} .

Via a routine computation, $β_{k + 1} (α) \leq β_{k} (α)$ for $α \geq α_{k}$ and any $k \geq 2$ . Thus, to verify that $β (α) \leq β_{k} (α)$ for $0 \leq α \leq 1 / 2$ , it suffices to just show that the same result holds for $0 \leq α \leq α_{k}$ . However, for $0 \leq α \leq α_{k}$ and $k \geq 2$ , we have

0 \leq α \leq α_{k} \leq \frac{2^{k + 1}}{(k - 3) 2^{k} + 8}

which rearranges to give

α + \frac{1 - k α}{2^{k} - 2} \geq (1 - 2^{2 - k}) α - \frac{1 - α}{2^{k} - 2}, (0 \leq α \leq α_{k}, k \geq 2)

which completes the proof in view of Proposition 4.7. See Figure 4.1.

We can remove the role of $I$ in the definition of $β$ :

Lemma 4.9

In Definition 4.2, one can take the interval $I$ to be $[N, 2 N]$ .

Proof ▶

Suppose that $α, \overset{―}{β}$ are fixed quantities such that the bounds in Definition 4.2 hold just for $I = [N, 2 N]$ , thus whenever $T > 1$ is unbounded, $N = T^{α + o (1)}$ , and $F$ is a model phase function one has

\sum_{N \leq n \leq 2 N} e (T F (n / N)) ≪ T^{\overset{―}{β} + o (1)} .

Our task is then to show that

\sum_{n \in I} e (T f F (n / N)) ≪ T^{\overset{―}{β} + o (1)}

under the same hypotheses. Similarly with $α = 1$ we can use the proof of Lemma 4.7 to obtain $\overset{―}{β} \geq 1 / 2$ , and we are again done. Thus we may assume that $α < 1$ .

For $n \in [N, 2 N]$ , the constraint $n \in I$ is equivalent to restricting $F (n / N)$ to an interval $J$ of length $O (1)$ , which we can also smooth out by $O (1 / N)$ without affecting the sum. Applying a Fourier expansion and the triangle inequality, we can thus bound the left-hand side by

≪ T^{o (1)} + \int_{- N^{1 + o (1)}}^{N^{1 + o (1)}} | \sum_{n \in [N, 2 N]} e (T F (n / N) - t F (n / N)) | \frac{d t}{1 + | t |} .

Since $α > 1$ , we have $| t - T | \leq T / 2$ for all $t$ in the integral if $T$ is large enough. From hypothesis 9 (with $T$ replaced by $T - t$ )we have

| \sum_{n \in [N, 2 N]} e (T F (n / N) - t F (n / N)) | ≪ T^{\overset{―}{β} + o (1)}

for all such $t$ , and the claim follows. See also Sargos [ 202 , p 310 ] .

Lemma 4.10 Reflection

For any $0 < α < 1$ , we have $β (α) - \frac{α}{2} = β (1 - α) - \frac{1 - α}{2}$ , or equivalently $β (1 - α) = \frac{1}{2} - α + β (α)$ .

TODO: implement this in python

Proof ▶

This is the van der Corput $B$ -process. See e.g., [ 92 , p 370 ] .

4.3 Known bounds on $β$

\includegraphics[width=0.5\linewidth ]{chapter/van_der_corput_beta.png} — Figure 4.1 The bounds in Proposition 4.7 for $k = 2, 3, 4, 5$ , compared against the optimized bound in Corollary 4.8.

\includegraphics[width=0.5\linewidth ]{chapter/van_der_corput_beta_vs_conjectured.png} — Figure 4.2 The bound in Corollary 4.8, compared against the trivial upper and lower bounds in Lemma 4.4.

We remark that this corollary also follows from Proposition 5.10.

Theorem 4.11 1989 Watt bound

For any $3 / 7 \leq α \leq 1 / 2$ , one has

β (α) \leq \frac{89}{560} + \frac{1}{2} α .

Recorded in literature.py as:
add_beta_bound_watt_1989()

Proof ▶

Theorem 4.12 1991 Huxley–Kolesnik bound

For any $2 / 5 \leq α \leq 1 / 2$ one has

β (α) \leq max (\frac{1 + 8 α}{22}, \frac{11 + 112 α}{158}, \frac{1 + 17 α}{22}) .

Recorded in literature.py as:
add_beta_bound_huxley_kolesnik_1991()

Proof ▶

Theorem 4.13 1993 Huxley bound

For any $0 \leq α \leq 49 / 114$ , one has

β (α) \leq max (\frac{13}{60} + \frac{7}{20} α, \frac{11}{120} + \frac{13}{20} α) .

Furthermore, for any $49 / 114 \leq α \leq 1 / 2$ , one has

β (α) \leq \frac{89}{570} + \frac{1}{2} α .

Recorded in literature.py as:
add_beta_bound_huxley_1993()

Proof ▶

Theorem 4.14 Second 1993 Huxley bound

If $0 \leq α \leq 1$ , then $β (α)$ is bounded by

\begin{aligned} \frac{1}{146} (13 + 94 α) & for α \leq \frac{87}{275} \\ \frac{1}{244} (11 + 191 α) & for \frac{87}{275} \leq α \leq \frac{423}{1295} \\ \frac{1}{1282} (89 + 908 α) & for \frac{423}{1295} \leq α \leq \frac{227}{601} \\ \frac{1}{280} (29 + 173 α) & for \frac{227}{601} \leq α \leq \frac{12}{31} \\ \frac{1}{128} (4 + 103 α) & for \frac{12}{31} \leq α \leq 1. \end{aligned}

Recorded in literature.py as:
add_beta_bound_huxley_1993_3()

Proof ▶

Theorem 4.15 1995 Sargos bound

[ 202 , Th é or è me 2.4, Lemme 2.6 ] For any $0 \leq α \leq 1$ , one has

β (α) \leq max (α + \frac{3 (1 - 4 α)}{40}, \frac{7}{8} α, \frac{1}{3} α - \frac{1 - 4 α}{6}, 0)

and

β (α) \leq max (α + \frac{1 - 4 α}{14}, \frac{5}{6} α, \frac{1}{3} α - \frac{1 - 4 α}{6}, 0) .

Recorded in literature.py as:
add_beta_bound_sargos_1995()

Table 4.1 Huxley table 17.1.

$β_{0} (α)$	$X$	$Y$
$\frac{4 + 39 α}{60}$	$\frac{7}{12}$	$\frac{517}{873} = 0.5922 \dots$
$\frac{29 + 42 α}{120}$	$\frac{65}{114}$	$\frac{7}{12} = 0.5833 \dots$
$\frac{89 + 285 α}{570}$	$\frac{49}{114}$	$\frac{65}{114} = 0.5701 \dots$
$\frac{11 + 78 α}{120}$	$\frac{5}{12}$	$\frac{49}{114} = 0.4298 \dots$
$\frac{13 + 21 α}{60}$	$\frac{356}{873}$	$\frac{5}{12} = 0.4166 \dots$
$\frac{4 + 103 α}{128}$	$\frac{12}{31}$	$\frac{356}{873} = 0.4546 \dots$
$\frac{29 + 173 α}{280}$	$\frac{227}{601}$	$\frac{12}{31} = 0.3870 \dots$
$\frac{89 + 908 α}{1282}$	$\frac{423}{1295}$	$\frac{227}{601} = 0.3777 \dots$
$\frac{11 + 191 α}{244}$	$\frac{87}{275}$	$\frac{423}{1295} = 0.3266 \dots$
$\frac{13 + 94 α}{146}$	$\frac{1424}{4747}$	$\frac{87}{275} = 0.3163 \dots$
$\frac{4 + 235 α}{264}$	$\frac{120}{419}$	$\frac{1424}{4747} = 0.2999 \dots$
$\frac{49 + 1351 α}{1614}$	$\frac{967}{3428}$	$\frac{120}{419} = 0.2863 \dots$
$\frac{29 + 464 α}{600}$	$\frac{199}{716}$	$\frac{967}{3428} = 0.2820 \dots$
$\frac{89 + 2243 α}{2706}$	$\frac{19}{74}$	$\frac{199}{716} = 0.2779 \dots$
$\frac{11 + 428 α}{492}$	$\frac{161}{646}$	$\frac{19}{74} = 0.2567 \dots$
$\frac{13 + 253 α}{318}$	$\frac{2848}{12173} = 0.2339 \dots$	$\frac{161}{646} = 0.2492 \dots$

Table 4.2 Huxley table 19.2.

$β_{0} (α)$	$X$	$Y$
$\frac{89 + 285 α}{570}$	$\frac{106822}{246639}$	$\frac{139817}{246639} = 0.5668 \dots$
$\frac{2387 + 17972 α}{27290}$	$\frac{675}{1574}$	$\frac{106822}{246639} = 0.4331 \dots$
$\frac{2819 + 19177 α}{29855}$	$\frac{699371}{1647930}$	$\frac{675}{1574} = 0.4288 \dots$
$\frac{11897 + 88442 α}{134680}$	$\frac{156527}{370694}$	$\frac{699371}{1647930} = 0.4243 \dots$
$\frac{113 + 897 α}{1345}$	$\frac{263}{638}$	$\frac{156527}{370694} = 0.4222 \dots$
$\frac{491 + 3624 α}{5530}$	$\frac{143}{349}$	$\frac{263}{638} = 0.4122 \dots$
$\frac{569 + 1053 α}{2800}$	$\frac{307}{761}$	$\frac{143}{349} = 0.4097 \dots$
$\frac{1273 + 2484 α}{6410}$	$\frac{68682}{171139}$	$\frac{307}{761} = 0.4034 \dots$
$\frac{4 + 103 α}{128}$	$\frac{12}{31}$	$\frac{68682}{171139} = 0.4013 \dots$
$\frac{29 + 173 α}{280}$	$\frac{227}{601} = 0.3777 \dots$	$\frac{12}{31} = 0.3870 \dots$

Theorem 4.16 1996 Huxley table

One can bound $β (α)$ by $β_{0} (α)$ for $X \leq α \leq Y$ for $β_{0}, X, Y$ given by Tables 4.1, 4.2.

Recorded in literature.py as:
add_beta_bound_huxley_1996()

add_beta_bound_huxley_1996_2()

Proof ▶

Theorem 4.17 2001 Huxley–Kolesnik bound

For any $2 / 5 \leq α \leq 1 / 2$ one has

β (α) \leq max (\frac{7}{80} + \frac{79}{120} α, \frac{3}{32} + \frac{103}{160} α, \frac{9}{40} + \frac{13}{40} α) .

Recorded in literature.py as:
add_beta_bound_huxley_kolesnik_2001()

Proof ▶

Theorem 4.18 2002 Robert–Sargos bound

For any $α > 0$ one has

β (α) \leq max (α + \frac{1 - 4 α}{13}, - \frac{7 (1 - 4 α)}{13}) .

Recorded in literature.py as:
add_beta_bound_robert_sargos_2002()

Proof ▶

Theorem 4.19 Sargos 2003 bound

For any $α > 0$ one has

β (α) \leq max (α + \frac{1 - 8 α}{204}, - \frac{95 (1 - 8 α)}{204})

and

β (α) \leq max (α + \frac{7 (1 - 9 α)}{2640}, - \frac{1001 (1 - 9 α)}{2640}) .

Recorded in literature.py as:
add_beta_bound_sargos_2003()

Proof ▶

Theorem 4.20 Huxley bound

For any $1 / 3 \leq α \leq 1 / 2$ , one has

β (α) \leq max (\frac{37 + 59 α}{170}, \frac{63 + 449 α}{690}) .

Recorded in literature.py as:
add_beta_bound_huxley_2005()

Proof ▶

Theorem 4.21 2016 Robert bound

For any $0 < α \leq 3 / 7$ one has

β (α) \leq max (α + \frac{1 - 4 α}{12}, \frac{11}{12} α) .

Recorded in literature.py as:
add_beta_bound_robert_2016()

Proof ▶

Theorem 4.22 Second 2016 Robert bound

If $k \geq 4$ and $α \geq - (1 - k α) \frac{k - 1}{2 k - 3}$ then

β (α) \leq α + max (\frac{1 - k α}{2 (k - 1) (k - 2)}, - \frac{1}{2 (k - 1) (k - 2)}) .

Recorded in literature.py as:
add_beta_bound_robert_2016_2(Constants.BETA_TRUNCATION)

Proof ▶

Theorem 4.23 2017 Heath-Brown bound

For any $α > 0$ and any natural number $k \geq 3$ one has

β (α) \leq α + max (\frac{1 - k α}{k (k - 1)}, - \frac{α}{k (k - 1)}, - \frac{2 α}{k (k - 1)} - \frac{2 (1 - k α)}{k^{2} (k - 1)}) .

Recorded in literature.py as:
add_beta_bound_heath_brown_2017(Constants.BETA_TRUNCATION)

Proof ▶

Theorem 4.24 2017 Bourgain bound

One has

β (α) \leq {\begin{cases} \frac{2}{9} + \frac{1}{3} α, & \frac{1}{3} < α \leq \frac{5}{12}, \\ \frac{1}{12} + \frac{2}{3} α, & \frac{5}{12} < α \leq \frac{3}{7}, \\ \frac{13}{84} + \frac{1}{2} α, & \frac{3}{7} < α \leq \frac{1}{2} . \end{cases}

Recorded in literature.py as:
add_beta_bound_bourgain_2017()

Proof ▶

Theorem 4.25 2020 Heath-Brown bound

[ 46 , Theorem 11.2 ] If $α$ is fixed with $1 \leq 4 α - 1 \leq 2$ (i.e., $1 / 2 \leq α \leq 3 / 4$ ), then

β (α) \leq max (α (1 - \frac{4 α - 1}{4 (4 α - 1) + 8}), \frac{8}{9} α) .

TODO: implement this in python

Theorem 4.26 Combined bound

For $X \leq α \leq Y$ , one has $β (α) \leq β_{0} (α)$ , where $β_{0}, X, Y$ are given by Table 4.3.

Proof ▶

Table 4.3 Bounds on

β (α)

of the form

β (α) \leq β_{0} (α), (X \leq α \leq Y)

$β_{0} (α)$	$X$	$Y$	Reference
$\frac{13}{414} + \frac{346}{414} α$	$0$	$\frac{2848}{12173} = 0.2339 \dots$	Exponent pair $A^{2} (\frac{13}{84}, \frac{55}{84})$
$\frac{13}{318} + \frac{253}{318} α$	$\frac{2848}{12173}$	$\frac{161}{646} = 0.2492 \dots$	Theorem 4.16
$\frac{11}{492} + \frac{107}{123} α$	$\frac{161}{646}$	$\frac{19}{74} = 0.2567 \dots$	Theorem 4.16
$\frac{89}{2706} + \frac{2243}{2706} α$	$\frac{19}{74}$	$\frac{199}{716} = 0.2779 \dots$	Theorem 4.16
$\frac{29}{600} + \frac{58}{75} α$	$\frac{199}{716}$	$\frac{967}{3428} = 0.2820 \dots$	Theorem 4.16
$\frac{49}{1614} + \frac{1351}{1614} α$	$\frac{967}{3428}$	$\frac{120}{419} = 0.2863 \dots$	Theorem 4.16
$\frac{1}{66} + \frac{235}{264} α$	$\frac{120}{419}$	$\frac{1328}{4447} = 0.2986 \dots$	Theorem 4.16
$\frac{13}{194} + \frac{139}{194} α$	$\frac{1328}{4447}$	$\frac{104}{343} = 0.3032 \dots$	Exponent pair $A (\frac{13}{84}, \frac{55}{84})$
$\frac{13}{146} + \frac{47}{73} α$	$\frac{104}{343}$	$\frac{87}{275} = 0.3163 \dots$	Theorem 4.16
$\frac{11}{244} + \frac{191}{244} α$	$\frac{87}{275}$	$\frac{423}{1295} = 0.3266 \dots$	Theorem 4.16
$\frac{89}{1282} + \frac{454}{641} α$	$\frac{423}{1295}$	$\frac{227}{601} = 0.3777 \dots$	Theorem 4.16
$\frac{29}{280} + \frac{173}{280} α$	$\frac{227}{601}$	$\frac{12}{31} = 0.3870 \dots$	Theorem 4.16
$\frac{1}{32} + \frac{103}{128} α$	$\frac{12}{31}$	$\frac{1508}{3825} = 0.3942 \dots$	Theorem 4.16
$\frac{18}{199} + \frac{521}{796} α$	$\frac{1508}{3825}$	$\frac{62831}{155153} = 0.4049 \dots$	Exponent pair $D (\frac{13}{84}, \frac{55}{84})$
$\frac{569}{2800} + \frac{1053}{2800} α$	$\frac{62831}{155153}$	$\frac{143}{349} = 0.4097 \dots$	Theorem 4.16
$\frac{491}{5530} + \frac{1812}{2765} α$	$\frac{143}{349}$	$\frac{263}{638} = 0.4122 \dots$	Theorem 4.16
$\frac{113}{1345} + \frac{897}{1345} α$	$\frac{263}{638}$	$\frac{1673}{4038} = 0.4143 \dots$	Theorem 4.16
$\frac{2}{9} + \frac{1}{3} α$	$\frac{1673}{4038}$	$\frac{5}{12} = 0.4166 \dots$	Theorem 4.24
$\frac{1}{12} + \frac{2}{3} α$	$\frac{5}{12}$	$\frac{3}{7} = 0.4285 \dots$	Theorem 4.24
$\frac{13}{84} + \frac{1}{2} α$	$\frac{3}{7}$	$\frac{1}{2}$	Theorem 4.24

Recorded in literature.py as:
add_beta_bound_trudgian_yang_2024()

\includegraphics[width=0.5\linewidth ]{chapter/van_der_corput_vs_best_beta.png} — Figure 4.3 The bounds in Proposition 4.7, compared against the best-known bound on $β (α)$ .

4 Exponential sum growth exponents

4.1 Phase functions

4.2 Exponential sum exponent

4.3 Known bounds on β

4.3 Known bounds on $β$