Explicit Asymptotics for Signed Binomial Sums and Applications to the Carnevale-Voll

Conjecture

Laurent Habsieger Université de Lyon, CNRS UMR 5208 Université Claude Bernard Lyon 1 Institut Camille Jordan 43 boulevard du 11 novembre 1918 69622 Villeurbanne Cedex France

                 habsieger@math.univ-lyon1.fr

Abstract Carnevale and Voll conjectured that ∑_{j}(−1)^{j}\binom{λ_{1}}{j}\binom{λ_{2}}{j} ≠ 0 when λ_{1} and λ_{2} are two distinct integers. We check the conjecture when either λ_{2} or λ_{1} − λ_{2} is small. We investigate the asymptotic behaviour of the sum when the ratio r := λ_{1}/λ_{2} is fixed and λ_{2} goes to infinity. We find an explicit range r ≥ 5.8362 on which the conjecture is true. We show that the conjecture is almost surely true for any fixed r. For r close to 1, we give several explicit intervals on which the conjecture is also true.

1 Introduction

Carnevale and Voll [1] studied Dirichlet series enumerating orbits of Cartesian products of maps whose orbits distributions are modelled on the distributions of finite index subgroups of free abelian groups of finite ranks. For Cartesian products of more than three maps they establish a natural boundary for meromorphic continuation. For products of two maps, they formulate two combinatorial conjectures that prove the existence of such a natural boundary. These conjectures state that some explicit polynomials have no unitary factors, i.e., polynomial factors that, for a suitable choice of variables, are univariate and have all their zeroes on the unit circle. The polynomials related to their Conjecture A [1] are given by:

C_{λ_{1},λ_{2}}(x, 1) = ∑_{j=0}^{λ_{2}} \binom{λ_{1}}{j}\binom{λ_{2}}{j}x^{j}

for positive integers λ_{1} ≥ λ_{2}, and the conjectured property is the following.

Conjecture 1. Let λ_{1} > λ_{2} be two positive integers. Then C_{λ_{1},λ_{2}}(−1, 1) ≠ 0.

Note that

C_{λ,λ}(−1, 1) = ∑_{j=0}^{λ}(−1)^{j}\binom{λ}{j}\binom{λ}{λ − j} = ∮ (1 − z)^{λ}(1 + z)^{λ} \frac{dz}{2iπz^{λ+1}} = ∮ (1 − z^{2})^{λ} \frac{dz}{2iπz^{λ+1}}

= \begin{cases}(−1)^{λ/2}\binom{λ}{λ/2}, & if λ is even; \\ 0, & if λ is odd. \end{cases}

This explains why the case λ_{1} = λ_{2} is excluded.

Carnevale and Voll [1] reported that Stanton pointed out the following property: the alternating summands have increasing absolute values for λ_{1} > λ_{2}(λ_{2} + 1) − 1, which shows the next result.

Proposition 2. For all λ_{2} and λ_{1} > λ_{2}(λ_{2} + 1) − 1, we have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0.

For a fixed λ_{2}, the sum C_{λ_{1},λ_{2}}(−1, 1) is a polynomial in λ_{1} of degree λ_{2}. The first values are C_{λ_{1},0}(−1, 1) = 1, C_{λ_{1},1}(−1, 1) = 1 − λ_{1}. Moreover, for 2 ≤ λ_{2} ≤ 240, we checked with Maple that it is an irreducible polynomial over ℚ when λ_{2} is even, and that it is the product of λ_{1} − λ_{2} by an irreducible polynomial over ℚ when λ_{2} is odd. The even case required much less time (238 seconds) than the odd case (63908 seconds). Since an irreducible polynomial of degree at least 2 cannot have an integer zero, we deduce that the conjecture is true for the first values of λ_{2}.

Proposition 3. For all 1 ≤ λ_{2} ≤ 240 and λ_{1} > λ_{2}, we have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0.

The aim of this paper is to study the size of C_{λ_{1},λ_{2}}(−1, 1) when λ_{1} and λ_{2} are large enough. We shall give explicit estimates in order to extend the range of validity of Conjecture 1.

We start by relating C_{λ_{1},λ_{2}}(x, 1) to a complex integral formula:

C_{λ_{1},λ_{2}}(x, 1) = ∮ (1 + z)^{λ_{1}}(z + x)^{λ_{2}} \frac{dz}{2iπz^{λ_{2}+1}} ,

where the path is a simple one around 0. This will always be the case from now on. We thus have

(−1)^{λ_{2}}C_{λ_{1},λ_{2}}(−1, 1) = ∮ (1 + z)^{λ_{1}}(1 − z)^{λ_{2}} \frac{dz}{2iπz^{λ_{2}+1}} .

Put λ = λ_{2} and r = λ_{1}/λ_{2} > 1. We get

(−1)^{λ_{2}}C_{λ_{1},λ_{2}}(−1, 1) = ∮ exp(λf(z))\frac{dz}{2iπz} ,  (1)

with

f(z) = f(r, z) := r log(1 + z) + log(1 − z) − log z .  (2)

We need to find the right path to be able to find the asymptotic behaviour of this kind of integral when λ goes to infinity. Let us take z = ρe^{iθ} with −π ≤ θ ≤ π. The parameter ρ will be optimal when f′ vanishes on the path. Since

f′(z) = \frac{r}{1 + z} − \frac{1}{1 − z} − \frac{1}{z} = \frac{−rz^{2} + (r − 1)z − 1}{z(1 − z^{2})} ,  (3)

and (r − 1)^{2} − 4r = r^{2} − 6r + 1 = (r − 3 − 2\sqrt{2})(r − 3 + 2\sqrt{2}), we need to distinguish several cases: 1 < r < 3 + 2\sqrt{2}, r = 3 + 2\sqrt{2} and r > 3 + 2\sqrt{2}. We thus define

ρ = ρ(r) := \begin{cases}\frac{r−1−\sqrt{r^{2}−6r+1}}{2r}, & if r ≥ 3 + 2\sqrt{2}; \\ \frac{1}{\sqrt{r}} = |\frac{r−1±i\sqrt{−r^{2}+6r−1}}{2r} | & if r ≤ 3 + 2\sqrt{2}. \end{cases}  (4)

By (1) we want to study the integral

I(λ) = I(r, λ) := ∫_{−π}^{π} exp(λf(ρe^{iθ}))\frac{dθ}{2π} .  (5)

In the case r > 3 + 2\sqrt{2}, we find f(ρe^{iθ}) = f(ρ) − M \frac{θ^{2}}{2} + o(θ^{3}) for some positive real

number M, when θ goes to 0. We therefore find I(λ) ∼ ∫_{−∞}^{∞} exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{2π}. We prove an effective version of this equivalence.

Theorem 4. For r > 3 + 2\sqrt{2}, put M = M(r) := ρ^{2}f′′(ρ). Then M > 0 and we have

|\frac{\sqrt{2πλM}}{exp(λf(ρ))}I(λ) − 1| ≤ \frac{3(3 + 2\sqrt{2})π^{5}}{256λM^{2}} + \frac{5(3 + 2\sqrt{2})}{24λM^{3}} + \frac{\sqrt{2} exp (−λM \frac{π^{2}}{2} )}{π^{3/2}\sqrt{λM}} .

This theorem shows that, for all r, we have I(λ) ≠ 0 for λ large enough. We use Proposition 3 and further tools to deduce a large range of validity for Conjecture 1.

Theorem 5. For λ_{1} ≥ 5.8362λ_{2} > 0, we have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0.

Note that the value 5.8362 is close to 3 + 2\sqrt{2} = 5.82842 …, the limit of the method.

Since r = λ_{1}/λ_{2} is a rational number, the case r = 3 + 2\sqrt{2} cannot occur, so we shall not give it in detail. The same method would provide an effective version of the equality

\frac{2^{2/3}3^{1/6}π(\sqrt{2} + 1)^{1/3}}{Γ(1/3)} × \frac{λ^{1/3}}{2^{(2+\sqrt{2})λ}} I(3 + 2\sqrt{2}, λ) = 1 + O (\frac{1}{λ^{1/6}} ) .

In the case 1 < r < 3 + 2\sqrt{2}, the situation is quite different. There are two conjugate points on the integrating circle where f′ vanishes. Their contributions partially cancel each other, so we cannot get an exact equivalent term: for some choices of (λ_{1}, λ_{2}), the implied constant may be really small. The analog of Theorem 4 has indeed the following form.

Theorem 6. For 1 < r < 3 + 2\sqrt{2}, define γ_{1} = arccos \frac{3r−1}{2\sqrt{2}r}, γ_{2} = − arccos \frac{r−3}{2\sqrt{2}} and γ_{3} =

\frac{1}{2} arcsin \frac{(r−1)^{2}}{4r}. For λ ≥ \frac{512r^{3/2}}{(r+1)(−r^{2}+6r−1)^{3/2}} , we have

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{2^{1+ \frac{(r+1)λ}{2}}}I(λ) − cos ((rγ_{1} + γ_{2})λ + γ_{3})| ≤ \frac{16336}{\sqrt{λ}(−r^{2} + 6r − 1)^{11/4}} .

It seems quite difficult to find a lower bound for | cos (λ_{1}γ_{1} + λ_{2}γ_{2} + γ_{3}) | for every λ_{1}, λ_{2}, and thus to show that I(λ) ≠ 0. However, we can upper bound the number of possible exceptions.

Theorem 7. For r > 1, we have

#{(λ_{1}, λ_{2}) : λ_{1} = rλ_{2} , C_{λ_{1},λ_{2}}(−1, 1) ≠ 0 , λ_{2} ≤ x}

≤ \begin{cases}O_{r}(1), & if r > 3 + 2\sqrt{2}; \\ _{\frac{102644}{(−r^{2}+6r−1)^{11/4} log( \frac{1+\sqrt{5}}{2} )}} \sqrt{x} log x + O_{r}(\sqrt{x}), & if 1 < r < 3 + 2\sqrt{2}. \end{cases}

Note that this theorem implies that, for a fixed r > 1, we have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0 almost surely. We can get more explicit estimates when r is close to 1, following a suggestion of Dennis Stanton. This case is also of special interest since C_{λ_{1},λ_{2}}(−1, 1) can be reduced to a sum with at most ⌊\frac{λ_{1}}{2} ⌋ − ⌈\frac{λ_{2}}{2} ⌉ terms, using a hypergeometric transformation. This enables us to prove the following analogue of Proposition 3.

Proposition 8. For all 1 ≤ λ_{1} − λ_{2} ≤ 701, we have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0.

We then prove the following theorems.

Theorem 9. For 1 < λ_{1}/λ_{2} < 3 + 2\sqrt{2}, define γ_{1} = arccos \frac{3λ_{1}−λ_{2}}{2\sqrt{2}λ_{1}} and γ_{2} = − arccos \frac{λ_{1}−3λ_{2}}{2\sqrt{2}λ_{2}} .

Assume λ_{1} − λ_{2} ≥ 702. We then have |\frac{\sqrt{πλ_{2}}}{2^{\frac{λ_{1}+λ_{2}+1}{2}}} I(λ_{2}) − cos (λ_{1}γ_{1} + λ_{2}γ_{2})| ≤ 0.0165 for

λ_{1} − λ_{2} ≤ \sqrt{8πλ_{2}}, and

\sqrt{λ}_{2} |\frac{\sqrt{πλ_{2}}}{2^{\frac{λ_{1}+λ_{2}+1}{2}}} I(λ_{2}) − cos (λ_{1}γ_{1} + λ_{2}γ_{2})| ≤ \begin{cases}1.05882, & if log λ_{2} ≤ λ_{1} − λ_{2} ≤ \sqrt{πλ_{2}}; \\ 1.30775, & if \sqrt{πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{2πλ_{2}}; \\ 1.50929, & if \sqrt{2πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{3πλ_{2}}; \\ 1.68876, & if \sqrt{3πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{4πλ_{2}}; \\ 1.85482, & if \sqrt{4πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{5πλ_{2}}; \\ 2.01189, & if \sqrt{5πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{6πλ_{2}}; \\ 2.1626, & if \sqrt{6πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{7πλ_{2}}; \\ 2.30865, & if \sqrt{7πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{8πλ_{2}}. \end{cases}

Theorem 10. We have C_{λ_{1},λ_{2}}(−1, 1) ≠ 0 in the following cases:

• λ_{1} + λ_{2} ≡ 0 (mod 4): λ_{2} < λ_{1} ≤ λ_{2} + \sqrt{2πλ_{2}} − 1.0443 or λ_{2} + \sqrt{2πλ_{2}} + 3.1407 ≤ λ_{1} ≤ λ_{2} + \sqrt{6πλ_{2}} − 0.9275;

• λ_{1} + λ_{2} ≡ 1 (mod 4): λ_{2} < λ_{1} ≤ λ_{2} + \sqrt{3πλ_{2}} − 0.984 or λ_{2} + \sqrt{3πλ_{2}} + 3.8433 ≤ λ_{1} ≤ \sqrt{7πλ_{2}} − 0.9231;

• λ_{1} + λ_{2} ≡ 2 (mod 4), λ_{2} + max (702, 2.0582λ_{2}^{1/4}) ≤ λ_{1} ≤ λ_{2} + \sqrt{2πλ_{2}} − 0.9535 or

2\sqrt{πλ_{2}} + 4.5938 ≤ λ_{1} − λ_{2} ≤ 2\sqrt{2πλ_{2}} − 0.9218;

• λ_{1} + λ_{2} ≡ 3 (mod 4): λ_{2} < λ_{1} ≤ λ_{2} + \sqrt{πλ_{2}} − 1.1958 or λ_{2} + \sqrt{πλ_{2}} + 2.5913 ≤ λ_{1} ≤ λ_{2} + \sqrt{5πλ_{2}} − 0.9367.

In Theorem 10 we study what happens before and after the first gap. The results obtained show that we miss at most six values in the first gap, which is quite small.

In the next section we study the case r > 3+2\sqrt{2}. We investigate the case 1 < r < 3+2\sqrt{2} in Section 3, and focus on the case r close to 1 in Section 4. We end this paper with some remarks and conjectures.

Before starting our studies, let us note that

\begin{aligned}f′′(z) & = −\frac{r}{(1 + z)^{2}} − \frac{1}{(1 − z)^{2}} + \frac{1}{z^{2}} , \\ f′′′(z) & = \frac{2r}{(1 + z)^{3}} − \frac{2}{(1 − z)^{3}} − \frac{2}{z^{3}} , \end{aligned}  (6)

and let us define the even function g(θ) = g(r, θ) := ℜ(f(ρe^{iθ})) and the odd function h(θ) = h(r, θ) := ℑ(f(ρe^{iθ})).

Since \overline{f(ρe^{iθ})} = f(ρe^{−iθ}), we have the useful expressions

∫_{−π}^{π} exp(λf(ρe^{iθ}))\frac{dθ}{2π} = ∫_{−π}^{π} exp(λg(θ)) cos (λh(θ)) \frac{dθ}{2π} = ∫_{0}^{π} exp(λg(θ)) cos (λh(θ)) \frac{dθ}{π} .

(7)

2 The case r > 3 + 2\sqrt{2}

2.1 General properties

A straightforward calculation shows that

ρ′(r) = \frac{\sqrt{1 − 6r + r^{2}} + 1 − 3r}{2r^{2}\sqrt{1 − 6r + r^{2}}} < 0 ,  (8)

and we get in this case

0 < ρ < ρ(3 + 2\sqrt{2}) = \sqrt{2} − 1 .  (9)

From (3) we have

r = \frac{1 + ρ}{ρ(1 − ρ)}  (10)

and from (6) and (9) we get that

M = M(r) := ρ^{2}f′′(ρ) = \frac{rρ}{(1 + ρ)^{2}} − \frac{ρ}{(1 − ρ)^{2}} = \frac{1 − 2ρ − ρ^{2}}{(1 + ρ)(1 − ρ)^{2}} > 0 .  (11)

Another straightforward calculation shows that

M′(r) = −ρ′(r)\frac{1 + ρ + 5ρ^{2} + ρ^{3}}{(1 + ρ)^{2}(1 − ρ)^{3}} > 0 ,  (12)

and we get

0 < M < 1 .  (13)

Let us now state a key lemma, which shows what kind of estimates are needed and how to use them to get results for I(λ). Lemma 11. Let δ ∈ [0, π]. Assume that

1. g(θ) − g(0) ≤ −KM \frac{θ^{2}}{2} , for some constant 0 < K ≤ 1,

2. |g(θ) − g(0) + M \frac{θ^{2}}{2} | ≤ C_{g}θ^{4}, for some positive constant C_{g},

3. |h(θ)| ≤ C_{h}|θ|^{3}, for some positive constant C_{h}, for 0 ≤ |θ| ≤ δ. We then have

|\begin{matrix}√ & \overline{2πλM} \\ exp & (λf(ρ)) \end{matrix}I(λ) − 1|

< \frac{3C_{g}}{λK^{5/2}M^{2}} + \frac{15C_{h}^{2}}{2λM^{3}} + \frac{\sqrt{2}}{δ\sqrt{πλM}} exp (−λM \frac{δ^{2}}{2} ) + (1 − \frac{δ}{π} ) \sqrt{2πλM} exp (−KM \frac{δ^{2}}{2} ) .

Proof. Note that

|e^{x} cos y − 1| ≤ |(e^{x} − 1) cos y| + 1 − cos y ≤ |x|e^{max(x,0)} + \frac{y^{2}}{2}  (14)

for all real numbers x and y.

Let us use this property with x = λ (g(θ) − g(0) + M \frac{θ^{2}}{2} ) and y = λh(θ). Condition 1 provides the upper bound max(x, 0) ≤ λ(1 − K)M \frac{θ^{2}}{2} . Conditions 2 and 3 imply |x| ≤ λC_{g}θ^{4} and |y| ≤ λC_{h}|θ|^{3}, respectively. From (14) and these bounds we deduce that

exp (−λf(ρ)) |∫_{−δ}^{δ} exp(λf(ρe^{iθ}))\frac{dθ}{2π} − ∫_{−δ}^{δ} exp (λ (f(ρ) − M \frac{θ^{2}}{2} )) \frac{dθ}{2π} |

= |∫_{−δ}^{δ} (e^{x} cos y − 1) exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{2π} |

≤ ∫_{−δ}^{δ} λC_{g}θ^{4} exp (−λKM \frac{θ^{2}}{2} ) \frac{dθ}{2π} + ∫_{−δ}^{δ} \frac{λ^{2}C_{h}^{2}θ^{6}}{2} exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{2π}

< \frac{λC_{g}Γ(5/2)}{2π(λKM/2)^{5/2}} + \frac{λ^{2}C_{h}^{2}Γ(7/2)}{4π(λM/2)^{7/2}} = \frac{3C_{g}}{\sqrt{2π}λ^{3/2}(KM)^{5/2}} + \frac{15C_{h}^{2}}{2\sqrt{2π}λ^{3/2}M^{7/2}} .

Since

∫_{δ≤|θ|≤+∞} exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{2π} ≤ ∫_{δ}^{+∞} \frac{θ}{δ} exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{π} = \frac{exp (−λM \frac{δ^{2}}{2} )}{δλMπ}

and

∫_{−∞}^{+∞} exp (−λM \frac{θ^{2}}{2} ) \frac{dθ}{2π} = \frac{1}{\sqrt{2πλM}} ,

we obtain

|\begin{matrix}√ & \overline{2πλM} \\ exp & (λf(ρ)) \end{matrix} ∫_{−δ}^{δ} exp(λf(ρe^{iθ}))\frac{dθ}{2π} − 1| < \frac{3C_{g}}{λK^{5/2}M^{2}} + \frac{15C_{h}^{2}}{2λM^{3}} + \frac{\sqrt{2}}{δ\sqrt{πλM}} exp (−λM \frac{δ^{2}}{2} ) .

To deal with the remaining integral ∫_{δ≤|θ|≤π} exp(λf(ρe^{iθ}))\frac{dθ}{2π}, we first show that g is increasing on [−π, 0] and decreasing on [0, π]. From the definition

g(θ) = ℜ (f (ρe^{iθ})) = r log |1 + ρe^{iθ}| + log |1 − ρe^{iθ}| − log ρ

and (10) we deduce that

g(θ) = \frac{1 + ρ}{2ρ(1 − ρ)} log (1 + 2 cos θρ + ρ^{2}) + \frac{1}{2} log (1 − 2 cos θρ + ρ^{2}) − log ρ

and

g′(θ) = −\frac{1 + ρ}{2ρ(1 − ρ)} × \frac{2ρ sin θ}{1 + 2ρ cos θ + ρ^{2}} + \frac{1}{2} × \frac{2ρ sin θ}{1 − 2ρ cos θ + ρ^{2}}

= −\frac{sin θ(2ρ(1 + 2ρ − ρ^{2})(1 − cos θ) + (1 − ρ^{2})(1 − 2ρ − ρ^{2}))}{(1 − ρ)((1 + ρ^{2})^{2} − 4ρ^{2} cos^{2} θ)} ≤ 0 ,

which proves this intermediate result. We therefore have

\frac{1}{exp (λf(ρ))} |∫_{δ≤|θ|≤π} exp(λf(ρe^{iθ}))\frac{dθ}{2π} | ≤ ∫_{δ≤|θ|≤π} exp (λ(g(θ) − g(0))) \frac{dθ}{2π}

≤ \frac{π − δ}{2π} (exp (λ(g(δ) − g(0))) + exp (λ(g(−δ) − g(0)))) ≤ \frac{π − δ}{π} exp (−KM \frac{δ^{2}}{2} ) ,

and the lemma follows.

2.2 Estimates for g and h

In this subsection, we obtain explicit versions of Conditions 1-3 in Lemma 11. We shall present two kinds of inequalities: a general inequality valid for r > 3 + 2\sqrt{2}, and a more precise one only valid when r is close to 3 + 2\sqrt{2}. The first ones will provide applications when r is large enough, and the second ones when r is close to 3 + 2\sqrt{2}.

Lemma 12. For r > 3 + 2\sqrt{2} and |θ| ≤ π, we have

g(θ) − g(0) ≤ −\frac{2}{π^{2}} Mθ^{2} .

For 3 + 2\sqrt{2} < r ≤ 7.686899 and |θ| ≤ π/3, we have

g(θ) − g(0) < −M \frac{θ^{2}}{2} .

Proof. From the definition of g, we get

g(θ) − g(0) = \frac{r}{2} log (\frac{1 + 2ρ cos θ + ρ^{2}}{(1 + ρ)^{2}} ) + \frac{1}{2} log (\frac{1 − 2ρ cos θ + ρ^{2}}{(1 − ρ)^{2}} )

= \frac{r}{2} log (1 − \frac{2ρ(1 − cos θ)}{(1 + ρ)^{2}}) + \frac{1}{2} log (1 + \frac{2ρ(1 − cos θ)}{(1 − ρ)^{2}}) .

(15)

Without loss of generality we may assume θ > 0. Since log(1 + x) ≤ x for x > −1, from (15) we obtain the following upper bound:

g(θ) − g(0) ≤ (\frac{ρ}{(1 − ρ)^{2}} − \frac{rρ}{(1 + ρ)^{2}} ) (1 − cos θ) = −2M sin^{2} (\frac{θ}{2}) ≤ −2M (\frac{θ}{π} )^{2} , (16)

which proves the first part of the lemma.

From log(1 + x) ≤ x − \frac{x^{2}}{2} + \frac{x^{3}}{3} for x > −1, from (11) and (15) we deduce that

g(θ) − g(0) ≤ −M(1 − cos θ) − (\frac{ρ^{2}}{(1 − ρ)^{4}} + \frac{rρ^{2}}{(1 + ρ)^{4}} ) (1 − cos θ)^{2}

+ (\frac{ρ^{3}}{(1 − ρ)^{6}} − \frac{rρ^{3}}{(1 + ρ)^{6}} ) \frac{4(1 − cos θ)^{3}}{3} .

(17)

Let us put

φ_{1}(θ) := cos θ − 1 + \frac{θ^{2}}{2} − \frac{(1 − cos θ)^{2}}{6} − \frac{2(1 − cos θ)^{3}}{45} ,

so that φ_{1}(0) = φ′_{1}(0) = 0 and φ′′_{1}(θ) = \frac{2(1−cos θ)^{3}}{5} . By Taylor’s formula and using the parity of φ_{1}, there exists t_{θ} ∈ [0, |θ|] such that

0 ≤ φ_{1}(θ) = \frac{2(1 − cos t_{θ})^{3}}{5} × \frac{θ^{2}}{2} ≤ \frac{θ^{2}(1 − cos θ)^{3}}{5} .  (18)

From (17), we get

g(θ) − g(0) ≤ M (−\frac{θ^{2}}{2} + \frac{(1 − cos θ)^{2}}{6} + \frac{2(1 − cos θ)^{3}}{45} + \frac{θ^{2}(1 − cos θ)^{3}}{5} )

− (\frac{ρ^{2}}{(1 − ρ)^{4}} + \frac{rρ^{2}}{(1 + ρ)^{4}} ) (1 − cos θ)^{2} + (\frac{ρ^{3}}{(1 − ρ)^{6}} − \frac{rρ^{3}}{(1 + ρ)^{6}} ) \frac{4(1 − cos θ)^{3}}{3}

= −M \frac{θ^{2}}{2} + c_{1}(ρ)(1 − cos θ)^{2} + c_{2}(ρ, θ)(1 − cos θ)^{3} .

with

\begin{aligned}c_{1}(ρ) : & = \frac{M}{6} − (\frac{ρ^{2}}{(1 − ρ)^{4}} + \frac{rρ^{2}}{(1 + ρ)^{4}} ) = \frac{1 − 8ρ + 9ρ^{2} − 32ρ^{3} − 9ρ^{4} − 8ρ^{5} − ρ^{6}}{6(1 + ρ)^{3}(1 − ρ)^{4}} \\ c_{2}(ρ, θ) : & = M (\frac{2}{45} + \frac{θ^{2}}{5} ) + \frac{4}{3} (\frac{ρ^{3}}{(1 − ρ)^{6}} − \frac{rρ^{3}}{(1 + ρ)^{6}} ) . \end{aligned}

For 3 + 2\sqrt{2} < r ≤ 7.686899, we have 0.19186222 ≤ ρ(r) < \sqrt{2} − 1, and we check that

c_{1}(ρ) + (1 − cos θ)c_{2}(ρ, θ) ≤ c_{1}(ρ) + \frac{c_{2}(ρ, π/3)}{2} < 0

for |θ| ≤ π/3. The lemma follows. Lemma 13. Let c ∈ [−1, 1]. For r > 3 + 2\sqrt{2} and cos θ ≥ c, we have

|g(θ) − g(0) + M \frac{θ^{2}}{2} | ≤ C_{g}θ^{4} ,

with

C_{g} := max (\frac{1 − 2ρ − ρ^{2}}{24(1 + ρ)(1 − ρ)^{2}} , \frac{ρ^{4} + 6ρ^{3} + 2ρ^{2} + 4ρ − 1}{24(1 + ρ)(1 − ρ)^{4}} + \frac{ρ}{4(1 − ρ^{2})(1 + ρ^{2} + 2ρc)}) .

Moreover, for r ≤ 7.494, we have

C_{g} = \frac{ρ^{4} + 6ρ^{3} + 2ρ^{2} + 4ρ − 1}{24(1 + ρ)(1 − ρ)^{4}} + \frac{ρ}{4(1 − ρ^{2})(1 + ρ^{2} + 2ρc)} .

Proof. It follows from (16) that g(θ) − g(0) ≤ M(cos θ − 1). The upper bound

g(θ) − g(0) + M \frac{θ^{2}}{2} ≤ M \frac{θ^{4}}{24} = \frac{1 − 2ρ − ρ^{2}}{24(1 + ρ)(1 − ρ)^{2}} θ^{4}  (19)

follows from (11) and from cos θ ≤ 1 − \frac{θ^{2}}{2} + \frac{θ^{4}}{24}.

Since log(1−x) ≥ −x− \frac{x^{2}}{2(1−x)} and log(1+x) ≥ x− \frac{x^{2}}{2} for x ∈ (0, 1), from (15) we deduce the lower bound

g(θ) − g(0) ≥ −M(1 − cos θ) − (\frac{rρ^{2}}{(1 + ρ)^{4}(1 − \frac{2ρ(1−cos θ)}{(1+ρ)^{2}})} + \frac{ρ^{2}}{(1 − ρ)^{4}} ) (1 − cos θ)^{2} .

From (18) we get

cos θ − 1 ≥ −\frac{θ^{2}}{2} + \frac{(1 − cos θ)^{2}}{6} + \frac{2(1 − cos θ)^{3}}{45} ≥ −\frac{θ^{2}}{2} + \frac{(1 − cos θ)^{2}}{6} ,

and hence g(θ) − g(0) + M \frac{θ^{2}}{2} ≥ −φ_{2}(ρ, c)(1 − cos θ)^{2}, with

\begin{aligned}φ_{2}(ρ, c) : & = −\frac{M}{6} + \frac{rρ^{2}}{(1 + ρ)^{4} (1 − \frac{2ρ(1−c)}{(1+ρ)^{2}})} + \frac{ρ^{2}}{(1 − ρ)^{4}} \\ & = \frac{ρ^{4} + 6ρ^{3} + 2ρ^{2} + 4ρ − 1}{6(1 + ρ)(1 − ρ)^{4}} + \frac{ρ}{(1 − ρ^{2})(1 + ρ^{2} + 2ρc)} . \end{aligned}

From (1 − cos θ)^{2} ≤ θ^{4}/4, we obtain g(θ) − g(0) + M \frac{θ^{2}}{2} ≥ − max (φ_{2}(ρ, c), 0) \frac{θ^{4}}{4} and therefore

|g(θ) − g(0) + M \frac{θ^{2}}{2} | ≤ max (\frac{φ_{2}(ρ, c)}{4}, 0, \frac{1 − 2ρ − ρ^{2}}{24(1 + ρ)(1 − ρ)^{2}} ) θ^{4} = C_{g}θ^{4} .

We note that

φ_{2}(ρ, c) − \frac{M}{6} ≥ φ_{2}(ρ, 1) − \frac{M}{6} = \frac{ρ^{6} + 5ρ^{5} + 3ρ^{4} + 14ρ^{3} − 3ρ^{2} + 5ρ − 1}{2(1 − ρ)^{4}(1 + ρ)^{3}} ≥ 0

for ρ ≥ 0.2002734. The second part of the lemma follows.

Lemma 14. For r > 3 + 2\sqrt{2} and cos θ ≥ c, we have |h(θ)| ≤ C_{h}|θ|^{3} with

C_{h} = C_{h}(r, c) := max (\frac{(1 + ρ^{2})(ρ^{2} + 4ρ − 1)}{6(1 − ρ)^{3}(1 + ρ)^{2}}, \frac{(1 + ρ^{2})(1 − 2ρ(1 + c) − ρ^{2})}{6(1 − ρ)((1 + ρ^{2})^{2} − 4ρ^{2}c^{2})}) .

Moreover, for r ≤ 6.537 and c ≥ 1/2, we have

C_{h} = \frac{(1 + ρ^{2})(ρ^{2} + 4ρ − 1)}{6(1 − ρ)^{3}(1 + ρ)^{2}} .

Proof. We have

h(θ) = \frac{f (ρe^{iθ}) − f (ρe^{−iθ})}{2i}

so that h(0) = 0 and

h′(θ) = \frac{ρe^{iθ}f′ (ρe^{iθ}) + ρe^{−iθ}f′ (ρe^{−iθ})}{2} = ℜ (ρe^{iθ}f′ (ρe^{iθ}))

= ℜ (\frac{rρe^{iθ}}{1 + ρe^{iθ}} − \frac{ρe^{iθ}}{1 − ρe^{iθ}} − 1) = \frac{rρ(cos θ + ρ)}{1 + ρ^{2} + 2ρ cos θ} − \frac{ρ(cos θ − ρ)}{1 + ρ^{2} − 2ρ cos θ} − 1 .

Since h′(0) = 0, we find

h′(θ) = \frac{rρ(cos θ + ρ)}{1 + ρ^{2} + 2ρ cos θ} − \frac{rρ}{1 + ρ} − \frac{ρ(cos θ − ρ)}{1 + ρ^{2} − 2ρ cos θ} + \frac{ρ}{1 − ρ}

= −\frac{rρ(1 − ρ)(1 − cos θ)}{(1 + ρ)(1 + ρ^{2} + 2ρ cos θ)} + \frac{ρ(1 + ρ)(1 − cos θ)}{(1 − ρ)(1 + ρ^{2} − 2ρ cos θ)}

= (\frac{ρ(1 + ρ)}{(1 − ρ)(1 + ρ^{2} − 2ρ cos θ)} − \frac{1}{1 + ρ^{2} + 2ρ cos θ}) × (1 − cos θ) .

Since

\frac{ρ(1 + ρ)}{(1 − ρ)(1 + ρ^{2} − 2ρ cos θ)} − \frac{1}{1 + ρ^{2} + 2ρ cos θ}

≤ \frac{ρ(1 + ρ)}{(1 − ρ)^{3}} − \frac{1}{(1 + ρ)^{2}} = \frac{(1 + ρ^{2})(ρ^{2} + 4ρ − 1)}{(1 − ρ)^{3}(1 + ρ)^{2}} =: φ_{3}(ρ) ,

we deduce that h′(θ) ≤ max(φ_{3}(ρ), 0)\frac{θ^{2}}{2} .

Similarly we get h′(θ) ≥ −φ_{4}(ρ, c)(1 − cos θ) ≥ − max(φ_{4}(ρ, c), 0)\frac{θ^{2}}{2} with

φ_{4}(ρ, c) := \frac{1}{1 + ρ^{2} + 2ρc} − \frac{ρ(1 + ρ)}{(1 − ρ)(1 + ρ^{2} − 2ρc)} = \frac{(1 + ρ^{2})(1 − 2ρ(1 + c) − ρ^{2})}{(1 − ρ)((1 + ρ^{2})^{2} − 4ρ^{2}c^{2})} .

We thus obtain

|h′(θ)| ≤ max (φ_{3}(ρ), 0, φ_{4}(ρ, c)) \frac{|θ|^{2}}{2} .

Since −φ_{4} ≤ φ_{3}, the first part of the lemma follows by integrating.

For c ≥ 1/2, note that

φ_{3}(ρ) − φ_{4}(ρ, c) ≥ φ_{3}(ρ) − φ_{4}(ρ, 1/2) = \frac{(1 + ρ^{2})(2ρ^{6} + 7ρ^{5} − 3ρ^{4} − 2ρ^{3} + 3ρ^{2} + 7ρ − 2)}{(1 − ρ)^{3}(1 + ρ)^{2}(ρ^{2} + ρ + 1)(ρ^{2} − ρ + 1)} ≥ 0

for ρ ≥ 0.26101. The second part of the lemma follows.

2.3 Proof of Theorem 4

In view of Lemma 11, we just need to estimate K, C_{g}, and C_{h}, when δ = π. Because of Lemma 12, we may choose K = 4/π^{2}. By Lemma 13, and using the notation in its proof, we may choose C_{g} = max (\frac{M}{24}, \frac{φ_{2}(ρ,−1)}{4}). By (13) we already know M/24 < 1/24. Since

\frac{∂φ_{2}}{∂ρ} (ρ, −1) = \frac{ρ^{5} + 15ρ^{4} + 10ρ^{3} + 46ρ^{2} + 17ρ + 7}{(1 + ρ)^{2}(1 − ρ)^{5}} > 0

and φ_{2}(\sqrt{2} − 1, −1) = (3 + 2\sqrt{2})/2, we find C_{g} ≤ (3 + 2\sqrt{2})/8.

With the notation in the proof of Lemma 14, we may put C_{h} = max(φ_{3}(ρ), φ_{4}(ρ, −1))/6. Since

φ_{3}(ρ) − φ_{4}(ρ, −1) = \frac{2(1 + ρ^{2})(ρ^{2} + 2ρ − 1)}{(1 − ρ)^{3}(1 + ρ)^{2}} ≤ 0

and

\frac{∂φ_{4}}{∂ρ} (ρ, −1) = \frac{1 + 5ρ + ρ^{2} + ρ^{3}}{6(1 + ρ)^{2}(1 − ρ)^{3}} ≥ 0 ,

we find C_{h} ≤ φ_{4}(\sqrt{2} − 1, −1)/6 = (\sqrt{2} + 1)/6.

These estimates prove the theorem.

2.4 Proof of Theorem 5

Let Φ_{1}(r, λ) denote the upper bound given in Theorem 4. It follows from (12) that Φ_{1} is a decreasing function of r. The function Φ_{1} is also obviously a decreasing function of λ. We thus obtain

Φ_{1}(r, λ) ≤ Φ_{1}(5.941893, 241) < 0.9999978502 ,

for r ≥ 5.941893 and λ ≥ 241. Theorem 4 therefore implies that I(r, λ) ≠ 0 for r ≥ 5.941893 and λ ≥ 241. Proposition 3 ensures us that I(r, λ) ≠ 0 for r and λ ≤ 240. We thus get I(r, λ) ≠ 0 for r ≥ 5.941893 and positive integers λ.

We can now assume r ≤ 5.941893. Let us get a better version of Theorem 4 in this case. By Lemma 12, we may choose K = 1 when δ ≤ \frac{π}{3}. We also put c = cos δ in the definitions

of C_{g} and C_{h}. We then get |\frac{\sqrt{2πλM}}{exp(λf(ρ))}I(λ) − 1| ≤ Φ_{2}(r, λ, δ), where

Φ_{2}(r, λ, δ) := \frac{3C_{g}(r, c)}{λM^{2}} + \frac{15C_{h}(r, c)^{2}}{2λM^{3}} + (\frac{\sqrt{2}}{δ\sqrt{πλM}} + (1 − \frac{δ}{π} ) \sqrt{2πλM}) exp (−λM \frac{δ^{2}}{2} ) .

Let us show that Φ_{2} is a decreasing function of r, to get results on an interval rather than at a point. Let us study each term defining Φ_{2}.

We find

\frac{∂C_{g}}{∂ρ} (ρ, c) = \frac{P (ρ, c) + 7(1 − ρ)}{6(1 + ρ)^{2}(1 − ρ)^{5}(1 + ρ^{2} + 2ρc)^{2}} ,

where P is a polynomial in c and ρ with nonnegative coefficients. We obtain \frac{∂C_{g}}{∂ρ} (ρ, c) ≥ 0, and the first term in Φ_{2} is therefore a decreasing function of r by (8) and (12).

We also find

\frac{∂C_{h}}{∂ρ} (ρ) = \frac{ρ^{5} + 9ρ^{4} + 8ρ^{3} + 28ρ^{2} − ρ + 3}{6(1 − ρ)^{4}(1 + ρ)^{3}} ≥ 0 ,

and the second term in Φ_{2} is also a decreasing function of r by (8) and (12).

The third term is easily seen to be a decreasing function of r by (12) when λMδ^{2} ≥ 1. Therefore, for λMδ^{2} ≥ 1, the function Φ_{2}(r, λ, δ) is a decreasing function of r and λ, the monotonicity in λ being easy under the condition λMδ^{2} ≥ 1. Computations then shows Φ_{2}(5.8478, 241, 0.75) < 0.9936 and Φ_{2}(5.8362, 980, 0.5) < 0.9999, while λMδ^{2} > 20 in these two cases. We thus proved I(λ) ≠ 0 for r ≥ 5.8362, except when 241 ≤ λ ≤ 979 and 5.8362 ≤ r < 5.8478. There are a finite number of possible exceptions, corresponding to the cases 241 ≤ λ_{2} ≤ 979 and 5.8362λ_{2} ≤ λ_{1} < 5.8478λ_{2}. We checked these cases with Maple in 5470 seconds. Here the limitation comes from the size of λ_{2} that must be handled by Maple in the computations. For our program, the limitation is λ_{2} ≤ 998.

3 The case r < 3 + 2\sqrt{2}

3.1 General properties

In this case, the derivative f′ has exactly two zeroes. These zeroes are the conjugate complex numbers

z_{r} = \frac{r − 1 + i\sqrt{−r^{2} + 6r − 1}}{2r} = ρe^{iα} and z̄_{r} = \frac{r − 1 − i\sqrt{−r^{2} + 6r − 1}}{2r} = ρe^{−iα} ,

with ρ = \frac{1}{\sqrt{r}} and α ∈ (0, π/2). Note that

cos α = \frac{r − 1}{2\sqrt{r}} and sin α = \frac{\sqrt{−r^{2} + 6r − 1}}{2\sqrt{r}} .  (20)

Many properties from the previous section can be rewritten. We still have

r = \frac{1 + z_{r}}{z_{r}(1 − z_{r})} and f′′(z_{r}) = \frac{1 − 2z_{r} − z_{r}^{2}}{(1 + z_{r})z_{r}^{2}(1 − z_{r})^{2}} ,

from which we find

f′′(z_{r})z_{r}^{2} = \frac{\sqrt{−r^{2} + 6r − 1}}{2} (\frac{(r + 1)\sqrt{−r^{2} + 6r − 1} − i(r − 1)^{2}}{4r} ) = \frac{\sqrt{−r^{2} + 6r − 1}}{2}e^{−iβ}

(21)

with β ∈ (0, π/2). Note that

cos β = \frac{(r + 1)\sqrt{−r^{2} + 6r − 1}}{4r} and sin β = \frac{(r − 1)^{2}}{4r} .  (22)

We also use (3) to compute

\frac{∂}{∂θ} (f (ρe^{iθ})) = i\frac{−rρ^{2}e^{2iθ} + (r − 1)ρe^{iθ} − 1}{1 − ρ^{2}e^{2iθ}} = ir\frac{−e^{2iθ} + 2e^{iθ} cos α − 1}{r − e^{2iθ}} .

We then obtain the useful expressions

g′(θ) + ih′(θ) = ir\frac{(e^{iθ} − e^{iα})(e^{−iα} − e^{iθ})}{r − e^{2iθ}} = 2ir\frac{cos α − cos θ}{(r − 1) cos θ − i(r + 1) sin θ} . (23)

As in the previous section, we now state our key lemma. Lemma 15. Let δ ∈ [0, α]. Assume that

1. |f (ρe^{iθ}) − f (ρe^{iα}) + \frac{\sqrt{−r^{2}+6r−1}}{4}e^{−iβ}(θ − α)^{2}| ≤ C_{f}|θ − α|^{3}, for some positive constant C_{f},

2. g(θ) − g(α) ≤ −\frac{(r+1)(−r^{2}+6r−1)}{16r}(θ − α)^{2} + C_{g}|θ − α|^{3}, for some constant C_{g} > 0, for α − δ ≤ θ ≤ α + δ.

We then have

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{2^{1+ \frac{r+1}{2} λ}}I(λ) − cos ((rγ_{1} + γ_{2})λ + \frac{β}{2} )| ≤ \frac{128r^{2}C_{f}e^{C_{g}λδ^{3}}}{\sqrt{πλ}(r + 1)^{2}(−r^{2} + 6r − 1)^{7/4}}

+\frac{8re^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}}}{δ\sqrt{πλ}(r + 1)(−r^{2} + 6r − 1)^{3/4}} + \frac{(π − 2δ)(−r^{2} + 6r − 1)^{1/4}\sqrt{λ}}{2\sqrt{π}}e^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}+C_{g}λδ^{3}} ,

with γ_{1} = arccos \frac{3r−1}{2\sqrt{2}r}, and γ_{2} = − arccos \frac{r−3}{2\sqrt{2}}. Proof. Instead of (14), we use the bound

|e^{z} − 1| ≤ |z|e^{max(ℜz,0)} ,  (24)

with z = f (ρe^{iθ}) − f (ρe^{iα}) + \frac{\sqrt{−r^{2}+6r−1}}{4}e^{−iβ}(θ − α)^{2}. Conditions 1 and 2 then imply |z| ≤ C_{f}λ|θ − α|^{3} and ℜz ≤ C_{g}λ|θ − α|^{3}, using (22). We thus find

|∫_{α−δ}^{α+δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} − ∫_{α−δ}^{α+δ} e^{−^{\frac{\sqrt{−r^{2}+6r−1}}{4}} e^{−iβ}λ(θ−α)^{2}} \frac{dθ}{2π} |

= |∫_{α−δ}^{α+δ}(e^{z} − 1)e^{−^{\frac{\sqrt{−r^{2}+6r−1}}{4}} e^{−iβ}λ(θ−α)^{2}} \frac{dθ}{2π} |

≤ ∫_{α−δ}^{α+δ} C_{f}λ|θ − α|^{3} exp (λ max (g(θ) − g(α), −\frac{\sqrt{−r^{2} + 6r − 1}}{4} cos β(θ − α)^{2})) \frac{dθ}{2π}

≤ C_{f}λ ∫_{−δ}^{δ} |u|^{3} exp (−\frac{(r + 1)(−r^{2} + 6r − 1)}{16r}λu^{2} + C_{g}λ|u|^{3}) \frac{du}{2π}

≤ C_{f}λe^{C_{g}λδ^{3}} ∫_{0}^{π} u^{3} exp (−\frac{(r + 1)(−r^{2} + 6r − 1)}{16r}λu^{2}) \frac{du}{π}

= C_{f}e^{C_{g}λδ^{3}} \frac{128r^{2}}{π(r + 1)^{2}(−r^{2} + 6r − 1)^{2}λ} .

Since

\begin{aligned}| ∫_{|θ−α|>δ} e^{−^{\frac{\sqrt{−r^{2}+6r−1}}{4}} e^{−iβ}λ(θ−α)^{2}} \frac{dθ}{2π} | & ≤ ∫_{δ}^{+∞} \frac{u}{δ}e^{−^{\frac{\sqrt{−r^{2}+6r−1}}{4}} cos βλu^{2}} \frac{du}{π} \\ & = \frac{8re^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}}}{πδλ(r + 1)(−r^{2} + 6r − 1)} . \end{aligned}

and

∫_{−∞}^{+∞} e^{−^{\frac{\sqrt{−r^{2}+6r−1}}{4}} e^{−iβ}λ(θ−α)^{2}} \frac{dθ}{2π} = \frac{e^{iβ/2}}{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}

we get

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{e^{λg(α)}} ∫_{α−δ}^{α+δ} e^{λf(ρe^{iθ})} \frac{dθ}{2π} − e^{iλh(α)+iβ/2}|

≤ \frac{128r^{2}C_{f}e^{C_{g}λδ^{3}}}{\sqrt{πλ}(r + 1)^{2}(−r^{2} + 6r − 1)^{7/4}} + \frac{8re^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}}}{δ\sqrt{πλ}(r + 1)(−r^{2} + 6r − 1)^{3/4}} .

By (23) we have

g′(θ) = −\frac{2r(r + 1) sin θ}{|(r − 1) cos θ + i(r + 1) sin θ|^{2}} (cos α − cos θ) ,

which shows that g is increasing on [0, α] and decreasing on [α, π]. From Condition 2 we deduce that

\begin{aligned}|∫_{0}^{α−δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | & ≤ ∫_{0}^{α−δ} e^{λ(g(θ)−g(α))} \frac{dθ}{2π} ≤ \frac{α − δ}{2π} e^{λ(g(α−δ)−g(α))} \\ & ≤ \frac{α − δ}{2π} e^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}+C_{g}λδ^{3}} \end{aligned}

and

|∫_{α+δ}^{π} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | ≤ \frac{π − α − δ}{2π} e^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}+C_{g}λδ^{3}} .

We obtain

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{e^{λg(α)}} ∫_{0}^{π} e^{λf(ρe^{iθ})} \frac{dθ}{2π} − e^{iλh(α)+iβ/2}| ≤ \frac{128r^{2}C_{f}e^{C_{g}λδ^{3}}}{\sqrt{πλ}(r + 1)^{2}(−r^{2} + 6r − 1)^{7/4}}

+\frac{8re^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}}}{δ\sqrt{πλ}(r + 1)(−r^{2} + 6r − 1)^{3/4}} + \frac{(π − 2δ)(−r^{2} + 6r − 1)^{1/4}\sqrt{λ}}{2\sqrt{π}}e^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}+C_{g}λδ^{3}} .

By (7) we have I(λ) = 2ℜ (∫_{0}^{π} e^{λf(ρe^{iθ})} \frac{dθ}{2π}). and the lemma then follows from (22) and

\begin{aligned}e^{iγ_{1}} & = \frac{1 + z_{r}}{|1 + z_{r}|} = \frac{3r − 1 + i\sqrt{−r^{2} + 6r − 1}}{2\sqrt{2}r} , \\ e^{iγ_{2}} & = \frac{z_{r}^{−1} − 1}{|z_{r}^{−1} − 1|} = \frac{r − 3 − i\sqrt{−r^{2} + 6r − 1}}{2\sqrt{2}} , \\ e^{λg(α)} & = |1 + z_{r}|^{rλ}|z_{r}^{−1} − 1|^{λ} = \sqrt{2}^{(r+1)λ} . \end{aligned}

3.2 Estimates for f and g

Lemma 16. For α/2 ≤ θ ≤ π − α/2, we have

|f (ρe^{iθ}) − f (ρe^{iα}) + \frac{\sqrt{−r^{2} + 6r − 1}}{4}e^{−iβ}(θ − α)^{2}| ≤ 0.33846\frac{(r + 1)^{2}}{r^{2}} |θ − α|^{3} .

Proof. From (23) we know that

g′(θ) + ih′(θ) = ir\frac{(e^{iθ} − e^{iα})(e^{−iα} − e^{iθ})}{r − e^{2iθ}} .

We also have

\frac{e^{−iα} − e^{iθ}}{r − e^{2iθ}} = \frac{e^{−iα} − e^{iα}}{r − e^{2iα}} + \frac{(e^{iθ} − e^{iα})(e^{i(θ−α)} + 1 − r − e^{i(α+θ)})}{(r − e^{2iθ})(r − e^{2iα})} ,

which gives

|g′(θ) + ih′(θ) − ir\frac{e^{−iα} − e^{iα}}{r − e^{2iα}} (e^{iθ} − e^{iα})| ≤ \frac{\sqrt{−r^{2} + 6r − 1} + \sqrt{r}(r − 1)}{2\sqrt{r^{2} + 1 + \sqrt{r}(r − 1)}}(θ − α)^{2}

since

|e^{i(θ−α)} + 1 − r − e^{i(α+θ)}| ≤ r − 1 + 2 sin α = r − 1 + \begin{matrix}√ & \overline{−r^{2} + 6r − 1} \\ √ & \overline{r} \end{matrix} ,
|r − e^{2iα}|^{2} = r^{2} + 1 − 2r cos(2α) = r^{2} + 1 − 2r (2\frac{(r − 1)^{2}}{4r} − 1) = 4r ,
|r − e^{2iθ}|^{2} = r^{2} + 1 − 2r cos(2θ) ≥ r^{2} + 1 − 2r cos(α) = r^{2} + 1 − \sqrt{r}(r − 1) .

Moreover we find

|\frac{e^{−iα} − e^{iα}}{r − e^{2iα}} (e^{iθ} − e^{iα}) − \frac{i(1 − e^{2iα})}{r − e^{2iα}}(θ − α)| ≤ \frac{sin α}{|r − e^{2iα}|}(θ − α)^{2} = \frac{\sqrt{−r^{2} + 6r − 1}}{4r}(θ − α)^{2}

and therefore

|g′(θ) + ih′(θ) + r\frac{1 − e^{2iα}}{r − e^{2iα}} (θ − α)|

≤ (\frac{\sqrt{−r^{2} + 6r − 1} + \sqrt{r}(r − 1)}{2\sqrt{r^{2} + 1 + \sqrt{r}(r − 1)}} + \frac{\sqrt{−r^{2} + 6r − 1}}{4}) (θ − α)^{2} .

We obtain the lemma by using (20), (21), (22), and checking the inequality

\frac{\sqrt{−r^{2} + 6r − 1} + \sqrt{r}(r − 1)}{2\sqrt{r^{2} + 1 + \sqrt{r}(r − 1)}} + \frac{\sqrt{−r^{2} + 6r − 1}}{4} ≤ 1.01537\frac{(r + 1)^{2}}{r^{2}} .

Lemma 17. For α/2 ≤ θ ≤ π − α/2, we have

g(θ) − g(α) ≤ −\frac{(r + 1)(−r^{2} + 6r − 1)}{16r}(θ − α)^{2} + \frac{r + 1}{4}|θ − α|^{3} .

Proof. We find

g(θ) − g(α) = \frac{r}{2} log (\frac{1 + ρ^{2} + 2ρ cos θ}{1 + ρ^{2} + 2ρ cos α}) + \frac{1}{2} log (\frac{1 + ρ^{2} − 2ρ cos θ}{1 + ρ^{2} − 2ρ cos α})

= \frac{r}{2} log (1 + \frac{2ρ(cos θ − cos α)}{1 + ρ^{2} + 2ρ cos α}) + \frac{1}{2} log (1 − \frac{2ρ(cos θ − cos α)}{1 + ρ^{2} − 2ρ cos α})

= \frac{r}{2} log (1 + ρ(cos θ − cos α)) + \frac{1}{2} log (1 − ρ^{−1}(cos θ − cos α))

since 2ρ cos α = 1 − ρ^{2}. We deduce the upper bound

g(θ) − g(α) ≤ −\frac{r + 1}{4}(cos θ − cos α)^{2} + \frac{1 − r^{2}}{6\sqrt{r}} (cos θ − cos α)^{3}

= \frac{r + 1}{4} (−(cos θ − cos α)^{2} − \frac{4 cos α}{3}(cos θ − cos α)^{3}) .

Let us define φ_{5}(θ) = sin^{2} α(θ − α)^{2} − (cos θ − cos α)^{2} − \frac{4 cos α}{3}(cos θ − cos α)^{3}, so that

φ′′_{5}(θ) = 2(cos θ − cos α) (6 cos α cos^{2} θ + 2(1 − cos^{2} α) cos θ − 3 cos α)
=: 2(cos θ − cos α)p(cos θ, cos α) .

We check that

p(cos θ, cos α) ≤ p(cos(α/2), cos α) = 3 + (1 − cos^{2} α) (\sqrt{2(1 + cos α)} − 3) ≤ 3 ,
p(cos θ, cos α) ≥ min_{{−1≤x≤1 \atop 0≤y≤1}} p(x, y) = p(0, 1) = −3 .

This gives |φ′′_{5}(θ)| ≤ 6|θ − α| and φ_{5}(θ) ≤ |θ − α|^{3}, and the lemma follows using (20).

3.3 Proof of Theorem 6

We use Lemmas 15, 16, 17 to get

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{2^{1+ \frac{r+1}{2} λ}}I(λ) − cos ((rγ_{1} + γ_{2})λ_{2} + \frac{β}{2} )| ≤ \frac{24.45e^{(r+1)λ \frac{δ^{3}}{4}}}{\sqrt{λ}(−r^{2} + 6r − 1)^{7/4}}

+\frac{8re^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}}}{δ\sqrt{πλ}(r + 1)(−r^{2} + 6r − 1)^{3/4}} + \frac{(π − 2δ)(−r^{2} + 6r − 1)^{1/4}\sqrt{λ}}{2\sqrt{π}}e^{− \frac{(r+1)(−r^{2}+6r−1)}{16r} λδ^{2}+(r+1)λ \frac{δ^{3}}{4}} ,

for δ ≤ α/2. We apply the inequalities xe^{−x} ≤ 1 and x^{3}e^{−x} ≤ 27e^{−3} to the second and third term of the right hand side respectively to get

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{2^{1+ \frac{r+1}{2} λ}}I(λ) − cos ((rγ_{1} + γ_{2})λ_{2} + \frac{β}{2} )| ≤ \frac{24.45e^{(r+1)λ \frac{δ^{3}}{4}}}{\sqrt{λ}(−r^{2} + 6r − 1)^{7/4}}

+\frac{128r^{2}}{δ^{3}λ^{3/2}\sqrt{π}(r + 1)^{2}(−r^{2} + 6r − 1)^{7/4}} + \frac{55296\sqrt{π}r^{3}e^{(r+1)λ \frac{δ^{3}}{4}}}{e^{3}δ^{6}λ^{5/2}(r + 1)^{3}(−r^{2} + 6r − 1)^{11/4}} .

We now choose δ = (\frac{(r+1)λ}{8})^{−1/3}. By hypothesis, we have δ ≤ \frac{sin α}{2} ≤ \frac{α}{2}. We obtain

|\frac{(−r^{2} + 6r − 1)^{1/4}\sqrt{πλ}}{2^{1+ \frac{r+1}{2} λ}}I(λ) − cos ((rγ_{1} + γ_{2})λ_{2} + \frac{β}{2} )|

\begin{matrix}≤ \frac{1}{\sqrt{λ}(−r^{2} + 6r − 1)^{11/4}} (24.45e^{2}(−r^{2} + 6r − 1) + \frac{16r^{2}(−r^{2} + 6r − 1)}{\sqrt{π}(r + 1)} + \frac{864\sqrt{π}r^{3}}{e(r + 1)} ) \\ ≤ \frac{16336}{\sqrt{λ}(−r^{2} + 6r − 1)^{11/4}} , \end{matrix}

which proves the theorem.

3.4 Proof of Theorem 7

For r > 3 + 2\sqrt{2}, Theorem 4 shows that I(λ) ≠ 0 for λ large enough, the implied bound only depending on M(r). The first part of the theorem follows.

For 1 < r < 3+2\sqrt{2}, define S_{r} as the set of non negative integers λ such that I(r, λ) = 0. As usual, let ∥x∥ denote the distance of x to the nearest integer. For λ ∈ S_{r}, we have

∥(rγ_{1} + γ_{2})λ + \frac{π + β}{2} ∥ ≤ \frac{π}{2} |sin ((rγ_{1} + γ_{2})λ + \frac{π + β}{2} )| ≤ \frac{25661}{\sqrt{λ}(−r^{2} + 6r − 1)^{11/4}} ,

by Theorem 6. For λ, λ′ ∈ S_{r}, λ < λ′ < λ + ^{\frac{(−r^{2}+6r−1)^{11/4}\sqrt{λ}}{102644}}, we find

∥(rγ_{1} + γ_{2})(λ′ − λ)∥ ≤ \frac{51322}{\sqrt{λ}(−r^{2} + 6r − 1)^{11/4}} ≤ \frac{1}{2(λ′ − λ)} .

By Legendre’s theorem [3, pp. 27–29], this implies that λ′ −λ is a denominator q_{n} in the continued fraction expansion of rγ_{1} + γ_{2}. Since q_{n} ≥ F_{n+1} = \frac{1}{\sqrt{5}} ((\frac{1+\sqrt{5}}{2})^{n+1} − (\frac{1−\sqrt{5}}{2})^{n+1}) ≥ \frac{1}{\sqrt{5}} (\frac{1+\sqrt{5}}{2})^{n}, the number of such denominators less than q is upper bounded by \frac{log(q\sqrt{5})}{log( \frac{1+\sqrt{5}}{2} )}, and we get for x ≥ 1:

#S_{r} ∩ [x, x + \frac{(−r^{2} + 6r − 1)^{11/4}\sqrt{x}}{102644}]

≤ \begin{matrix}log & \frac{(−r^{2}+6r−1)^{11/4}\sqrt{5x}}{102644} \\ log & (\frac{1+\sqrt{5}}{2}) \end{matrix} = φ_{6} (x + \frac{(−r^{2} + 6r − 1)^{11/4}\sqrt{x}}{102644}) − φ_{6}(x) ,

with

φ_{6}(x) = \frac{102644}{(−r^{2} + 6r − 1)^{11/4} log (\frac{1+\sqrt{5}}{2})}\sqrt{x} log x + O_{r}(\sqrt{x}) .

The second part of the theorem follows.

4 The case r close to 1

When r goes to 1, the angles α and β go to π/2 and 0 respectively. So we shall prove specific estimates in this case. Before that, we establish Proposition 8.

4.1 Small values of λ_{1} − λ_{2}

We have

C_{λ_{1},λ_{2}}(−1, 1) = ∮ (z + 1)^{λ_{1}}(1 − z)^{λ_{2}} \frac{dz}{2iπz^{λ_{1}+1}} = ∮ (1 − z^{2})^{λ_{2}}(z + 1)^{λ_{1}−λ_{2}} \frac{dz}{2iπz^{λ_{1}+1}}

= ∑_{λ_{2}≤2j≤λ_{1}}(−1)^{j}\binom{λ_{2}}{j}\binom{λ_{1} − λ_{2}}{λ_{1} − 2j}

= \frac{λ_{2}!}{⌊\frac{λ_{1}}{2} ⌋! ⌊\frac{λ_{2}}{2} ⌋!} ∑_{⌈\frac{λ_{2}}{2} ⌉≤j≤⌊\frac{λ_{1}}{2} ⌋}(−1)^{j}\binom{λ_{1} − λ_{2}}{λ_{1} − 2j}\frac{⌊\frac{λ_{1}}{2} ⌋!}{j!} \frac{⌊\frac{λ_{2}}{2} ⌋!}{(λ_{2} − j)!}

= \frac{λ_{2}!(−1)^{⌈^{\frac{λ_{2}}{2}} ⌉}}{⌊\frac{λ_{1}}{2} ⌋! ⌊\frac{λ_{2}}{2} ⌋!} ∑_{0≤j≤⌊\frac{λ_{1}}{2} ⌋−⌈\frac{λ_{2}}{2} ⌉}(−1)^{j}\binom{λ_{1} − λ_{2}}{2j + ⌈\frac{λ_{2}}{2} ⌉ − ⌊\frac{λ_{2}}{2} ⌋}\frac{⌊\frac{λ_{1}}{2} ⌋!}{(⌈\frac{λ_{2}}{2} ⌉ + j)!} \frac{⌊\frac{λ_{2}}{2} ⌋!}{(⌊\frac{λ_{2}}{2} ⌋ − j)!} .

When λ_{1} − λ_{2} is fixed, the last sum is a polynomial C̃ in ⌊\frac{λ_{2}}{2} ⌋ of degree at most ⌊\frac{λ_{1}}{2} ⌋ − ⌈\frac{λ_{2}}{2} ⌉, also depending on the parity of λ_{1} and λ_{2}. More precisely we consider the four families of polynomials

C̃_{l,0,0}(k) = ∑_{0≤j≤l}(−1)^{j}\binom{2l}{2j}\frac{(k + l)!}{(k + j)!} \frac{k!}{(k − j)!}  for (λ_{2}, λ_{1}) = (2k, 2k + 2l),
C̃_{l,0,1}(k) = ∑_{0≤j≤l}(−1)^{j}\binom{2l + 1}{2j}\frac{(k + l)!}{(k + j)!} \frac{k!}{(k − j)!}  for (λ_{2}, λ_{1}) = (2k, 2k + 2l + 1),
C̃_{l,1,0}(k) = ∑_{0≤j≤l−1}(−1)^{j}\binom{2l}{2j + 1}\frac{(k + l)!}{(k + 1 + j)!} \frac{k!}{(k − j)!} for (λ_{2}, λ_{1}) = (2k + 1, 2k + 2l + 1),
C̃_{l,1,1}(k) = ∑_{0≤j≤l}(−1)^{j}\binom{2l + 1}{2j + 1}\frac{(k + l + 1)!}{(k + 1 + j)!} \frac{k!}{(k − j)!} for (λ_{2}, λ_{1}) = (2k + 1, 2k + 2l + 2).

The first values are given by C̃_{0,0,0}(k) = C̃_{1,0,0}(k) = C̃_{0,0,1}(k) = C̃_{0,1,1}(k) = 1, C̃_{0,1,0}(k) = 0, C̃_{1,1,0}(k) = 2, C̃_{2,1,0}(k) = 8, C̃_{1,0,1}(k) = −2k + 1, C̃_{1,1,1}(k) = 2k + 6. We can compute the leading terms of each polynomial to show that the degree of C̃_{l,0,0}(k), C̃_{l,0,1}(k), C̃_{l,1,1}(k) is at least 2 for l ≥ 2, and that the degree of C̃_{l,1,0}(k) is at least 2 for l ≥ 3:

C̃_{l,0,0}(k) = ∑_{0≤j≤l}(−1)^{i}\binom{2l}{2j} (k^{l} + (\binom{l + 1}{2} − j^{2}) k^{l−1} + ⋯)

= ℜ ( ∑_{0≤m≤2l} i^{m}\binom{2l}{m} (k^{l} + (\binom{l + 1}{2} − \frac{m^{2}}{4} ) k^{l−1} + ⋯))

= ℜ ((1 + i)^{2l}k^{l} + (\binom{l + 1}{2} − \frac{l(1 + 2li)}{4}) (1 + i)^{2l}k^{l−1} + ⋯)

C̃_{l,0,1}(k) = ∑_{0≤j≤l}(−1)^{j}\binom{2l + 1}{2j} (k^{l} + ⋯) = ℜ ( ∑_{0≤m≤2l+1} i^{m}\binom{2l + 1}{m} (k^{l} + ⋯))

= ℜ ((1 + i)^{2l+1}k^{l} + ⋯) = ±2^{l}k^{l} + ⋯

C̃_{l,1,0}(k) = ∑_{0≤j≤l−1}(−1)^{j}\binom{2l}{2j + 1} (k^{l−1} + (\binom{l + 1}{2} − j^{2} − j − 1) k^{l−2} + ⋯)

= ℑ ( ∑_{0≤m≤2l} i^{m}\binom{2l}{m} (k^{l−1} + (\binom{l + 1}{2} − \frac{m^{2} + 3}{4}) k^{l−2} + ⋯))

= ℑ ((1 + i)^{2l}k^{l−1} + (\binom{l + 1}{2} − \frac{3 + l(1 + 2li)}{4}) (1 + i)^{2l}k^{l−2} + ⋯)

C̃_{l,1,1}(k) = ∑_{0≤j≤l}(−1)^{j}\binom{2l + 1}{2j + 1} (k^{l} + ⋯) = ℑ ( ∑_{0≤m≤2l+1} i^{m}\binom{2l + 1}{m} (k^{l} + ⋯))

= ℑ ((1 + i)^{2l+1}k^{l} + ⋯) = ±2^{l}k^{l} + · · · .

For each of the four cases, we checked with Maple that C̃_{l} is an irreducible polynomial for 2 ≤ l ≤ 350, which proves Proposition 8 since C̃_{l} then has no integer zero. Each case required between 95000 and 98000 seconds.

4.2 The approach

We give a more specific version of Lemma 15.

Lemma 18. Let δ, δ_{1} be two positive real numbers with δ < min (2/3, 3(3 − r)/4) and δ ≤ δ_{1} ≤ α. Assume that

1. |f (ρe^{iθ}) − f (ρe^{iα}) + \frac{(θ−α)^{2}}{2} | ≤ \frac{|θ−α|^{3}}{3} + \frac{r−1}{4}(θ − α)^{2}, for α − δ ≤ θ ≤ α + δ,

2. g(θ) − g(α) ≤ −\frac{(θ−α)^{2}}{2} + \frac{|θ−α|^{3}}{2} , for α − δ_{1} ≤ θ ≤ α + δ_{1}.

Define γ_{1} = arccos \frac{3r−1}{2\sqrt{2}r} and γ_{2} = − arccos \frac{r−3}{2\sqrt{2}}. We then have

|\frac{\sqrt{πλ}}{2^{\frac{1+(r+1)λ}{2}}} I(λ) − cos ((rγ_{1} + γ_{2})λ)| ≤ \frac{1}{3\sqrt{2πλ} (\frac{3−r}{4} − \frac{δ}{3})^{2}} + \frac{r − 1}{2^{7/2} (\frac{3−r}{4} − \frac{δ}{3})^{3/2}} + \frac{\sqrt{2}e^{−λ \frac{δ^{2}}{2}}}{\sqrt{πλ}δ}

+ (\frac{π − 2δ_{1}}{\sqrt{2π}} + \frac{2^{3/2}}{\sqrt{π}δ(2 − 3δ)}) \sqrt{λ}e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} .

Proof. We follow the proof of Lemma 15, and use Conditions 1 and 2:

|∫_{α−δ}^{α+δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} − ∫_{α−δ}^{α+δ} e^{−λ \frac{(θ−α)^{2}}{2}} \frac{dθ}{2π} |

\begin{matrix}≤ ∫_{α−δ}^{α+δ} λ (\frac{|θ − α|^{3}}{3} + \frac{r − 1}{4}(θ − α)^{2}) e^{− \frac{3−r}{4} λ(θ−α)^{2}+λ \frac{|θ−α|^{3}}{3}} \frac{dθ}{2π} \\ ≤ λ ∫_{0}^{δ} (\frac{u^{3}}{3} + \frac{r − 1}{4}u^{2}) e^{−λu^{2}(\frac{3−r}{4} − \frac{δ}{3})} \frac{du}{π} < λ ∫_{0}^{∞} (\frac{u^{3}}{3} + \frac{r − 1}{4}u^{2}) e^{−λu^{2}(\frac{3−r}{4} − \frac{δ}{3})} \frac{du}{π} , \end{matrix}

that is

|∫_{α−δ}^{α+δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} − ∫_{α−δ}^{α+δ} e^{−λ \frac{(θ−α)^{2}}{2}} \frac{dθ}{2π} | ≤ \frac{1}{6πλ (\frac{3−r}{4} − \frac{δ}{3})^{2}} + \frac{r − 1}{16\sqrt{πλ} (\frac{3−r}{4} − \frac{δ}{3})^{3/2}} .

(25)

We still have

|∫_{|θ−α|>δ} e^{−λ \frac{(θ−α)^{2}}{2}} \frac{dθ}{2π} | ≤ \frac{e^{−λ \frac{δ^{2}}{2}}}{πλδ}  (26)

and

∫_{−∞}^{∞} e^{−λ \frac{(θ−α)^{2}}{2}} \frac{dθ}{2π} = \frac{1}{\sqrt{2πλ}} .  (27)

Let us now assume 0 ≤ θ ≤ α − δ. As in the proof of Lemma 15, we deduce the following from (23) and Condition 2:

|∫_{0}^{α−δ_{1}} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | ≤ \frac{α − δ_{1}}{2π} e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} .  (28)

Put u(θ) = −\frac{(θ−α)^{2}}{2} − \frac{(θ−α)^{3}}{2} , so that g(θ) ≤ u(θ) for θ ≤ α. We check that u′(θ) = \frac{3}{2}(α − θ)(\frac{2}{3}−α+θ) ≥ 0 for α−2/3 ≤ θ ≤ α, and u′(θ)−u′(α−δ) = −(θ+α+δ)(3(θ−α−δ)/2+1) ≥ 0 for 0 ≤ θ ≤ α − δ. We thus get \frac{u′(θ)}{u′(α−δ)} ≥ 1 for 0 ≤ θ ≤ α − δ. We get

|∫_{α−δ_{1}}^{α−δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | ≤ ∫_{α−δ_{1}}^{α−δ} \frac{u′(θ)}{u′(α − δ)}e^{λu(θ)} \frac{dθ}{2π} = \frac{e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} − e^{−λ \frac{δ^{2}}{2} +λ \frac{δ^{3}}{2}}}{3πδ(2/3 − δ)} . (29)

The estimates (28) and (29) give

|∫_{0}^{α−δ} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | ≤ \frac{α − δ_{1}}{2π} e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} + \frac{e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}}}{πδ(2 − 3δ)} .  (30)

Similarly we have

|∫_{α+δ}^{π} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} | ≤ \frac{π − α − δ_{1}}{2π} e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} + \frac{e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}}}{πδ(2 − 3δ)} .  (31)

From (25), (26), (27), (30), (31) we obtain

|∫_{0}^{π} e^{λ(f(ρe^{iθ})−f(ρe^{iα}))} \frac{dθ}{2π} − \frac{1}{\sqrt{2πλ}}| ≤ \frac{1}{6πλ (\frac{3−r}{4} − \frac{δ}{3})^{2}} + \frac{r − 1}{16\sqrt{πλ} (\frac{3−r}{4} − \frac{δ}{3})^{3/2}} + \frac{e^{−λ \frac{δ^{2}}{2}}}{πλδ}

+\frac{π − 2δ_{1}}{2π} e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}} + \frac{2e^{−λ \frac{δ^{2}_{1}}{2} +λ \frac{δ^{3}_{1}}{2}}}{πδ(2 − 3δ)} ,

and the lemma follows from

ℜ (∫_{0}^{π} e^{λ(f(ρe^{iθ})−g(α))} \frac{dθ}{2π} − \frac{e^{iλh(α)}}{\sqrt{2πλ}} ) = \frac{I(λ)}{2} − \frac{cos(λh(α))}{\sqrt{2πλ}} .

4.3 Estimates for f and g

Lemma 19. Assume 1 ≤ r ≤ 2.282. For α − \sqrt{r − 1} ≤ θ ≤ α + \sqrt{r − 1}, we have

|f (ρe^{iθ}) − f (ρe^{iα}) + \frac{(θ − α)^{2}}{2} | ≤ \frac{|θ − α|^{3}}{3} + \frac{r − 1}{4}(θ − α)^{2} .

Proof. We first notice that \sqrt{r − 1} ≤ α = arccos (\frac{r−1}{2\sqrt{r}}), for 1 ≤ r ≤ 2.282459. This implies 0 ≤ θ ≤ 2α ≤ π.

By (23) we have

\begin{aligned}g′(θ) + ih′(θ) & = ir\frac{(e^{iθ} − e^{iα})(e^{−iα} − e^{iθ})}{r − e^{2iθ}} = ir\frac{(e^{i(θ−α)} − 1) (1 − e^{i(θ+α)})}{r − e^{2iθ}} \\ & = i (e^{i(θ−α)} − 1) (1 + \frac{e^{i(θ+α)} (e^{i(θ−α)} − r)}{r − e^{2iθ}}) , \end{aligned}

which gives

|g′(θ) + ih′(θ) − i (e^{i(θ−α)} − 1)| ≤ |θ − α|\frac{|e^{i(θ−α)} − 1| + r − 1}{|r − e^{2iθ}|} ≤ \frac{(θ − α)^{2} + (r − 1)|θ − α|}{|r − e^{2iθ}|} .

Since 2α − 2\sqrt{r − 1} ≤ 2π − 2α − 2\sqrt{r − 1}, we also get

|r − e^{2iθ}|^{2} = r^{2} + 1 − 2r cos(2θ) ≥ r^{2} + 1 − 2r cos (2α − 2\sqrt{r − 1})
≥ r^{2} + 1 − 2r cos(2α) − 4r\sqrt{r − 1} sin(2α) + 4r(r − 1) cos(2α)
= 4r − 2(r − 1)^{3/2}\sqrt{−r^{2} + 6r − 1} + 2(r − 1)(r^{2} − 4r + 1)
= 4 + 2(r − 1)^{3/2} (\sqrt{−r^{2} + 6r − 1} + (r − 3)\sqrt{r − 1}) ≥ 4 ,

which leads to the upper bound

|g′(θ) + ih′(θ) − i (e^{i(θ−α)} − 1)| ≤ \frac{(θ − α)^{2}}{2} + \frac{r − 1}{2}|θ − α| .

We deduce

|g′(θ) + ih′(θ) + (θ − α)| ≤ \frac{(θ − α)^{2}}{2} + \frac{r − 1}{2}|θ − α| + \frac{(θ − α)^{2}}{2} ,

and the lemma follows by integrating. Lemma 20. For α/2 ≤ θ ≤ 3α/2 and 1 < r ≤ 2.11952, we have

g(θ) − g(α) ≤ −\frac{(θ − α)^{2}}{2} + \frac{|θ − α|^{3}}{2} .

Proof. From Lemma 17 we get

g(θ) − g(α) ≤ −\frac{(θ − α)^{2}}{2} + \frac{|θ − α|^{3}}{2} + (r − 1)(θ − α)^{2} (\frac{r^{2} − 4r − 1}{8r} + \frac{|θ − α|}{4}) .

We check that

\frac{r^{2} − 4r − 1}{16r} + \frac{α}{8} ≤ 0

for 1 < r ≤ 2.119518 …, and the lemma follows.

4.4 Proof of Theorem 9

For (r − 1)λ ≥ 702 and r − 1 ≤ \sqrt{\frac{8π}{λ}}, we have λ ≥ ⌈\frac{702^{2}}{8π} ⌉ = 19609 and therefore r ≤

1 + \sqrt{\frac{8π}{19609}} < 1.03581. We thus can apply Lemmas 19 and 20, and we set δ = \sqrt{r − 1} and δ_{1} = α/2. We then have δ ≤ 0.19 < 2/3 < 3(3 − r)/4. We therefore can use Lemma 18, and we obtain

|\frac{\sqrt{πλ}}{2^{\frac{1+(r+1)λ}{2}}} I(λ) − cos ((rγ_{1} + γ_{2})λ)| ≤ \frac{1}{3\sqrt{2πλ} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{2}} + \frac{r − 1}{2^{7/2} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{3/2}}

+\frac{\sqrt{2}e^{− \frac{λ(r−1)}{2}}}{\sqrt{πλ(r − 1)}} + (\frac{π − α}{\sqrt{2π}} + \frac{2^{3/2}}{\sqrt{π}δ(2 − 3δ)}) \sqrt{λ}e^{−λ \frac{α^{2}}{8} +λ \frac{α^{3}}{16}} .

Since δ(2 − 3δ) ≥ \sqrt{\frac{702}{λ}}(2 − 3\sqrt{0.03581}) ≥ \frac{37.949}{\sqrt{λ}} and α ≤ π/2, we find

|\frac{\sqrt{πλ}}{2^{\frac{1+(r+1)λ}{2}}} I(λ) − cos ((rγ_{1} + γ_{2})λ)| ≤ \frac{1}{3\sqrt{2πλ} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{2}} + \frac{r − 1}{2^{7/2} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{3/2}}

+\frac{\sqrt{2}e^{− \frac{λ(r−1)}{2}}}{\sqrt{πλ(r − 1)}} + (0.6267 + 0.04206\sqrt{λ}) \sqrt{λ}e^{−λ(1− \frac{π}{4})\frac{α^{2}}{8}} .

Let Φ_{4}(r, λ) denote this last upper bound. The first two terms and the fourth term are increasing functions of r ≤ 1.03581 and decreasing functions of λ ≥ 19609, the third term is a decreasing function of λ(r − 1) ≥ 702: we already get this way Φ_{4}(r, λ) ≤ 0.0165, which proves the first inequality in the theorem.

Further assume log λ ≤ (r − 1)λ ≤ 1.67007\sqrt{λ}. We find λ ≥ λ_{0} := ⌈\frac{702^{2}}{1.67007^{2}} ⌉ = 176688,

r ≤ 1 + \frac{1.67007}{λ_{0}}, α ≥ α_{0} := arccos (\frac{\sqrt{1.67007^{2}/λ_{0}}}{2(1+\sqrt{1.67007^{2}/λ_{0}})^{1/2}} ) and

\sqrt{λ}Φ_{4}(r, λ) ≤ \frac{1}{3\sqrt{2π} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{2}} + \frac{1.67007}{2^{7/2} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{3/2}}

+\frac{1}{\sqrt{351π}} + (0.6267 + 0.04206\sqrt{λ}_{0}) λ_{0}e^{−λ_{0}(1− \frac{π}{4})\frac{α^{2}_{0}}{8}} < 1.05882 ,

which proves the second inequality in the theorem for log λ ≤ (r − 1)λ ≤ 1.67007\sqrt{λ}. Assume now c_{1}\sqrt{λ} ≤ (r − 1)λ ≤ c_{2}\sqrt{λ}, so that λ ≥ λ_{0} := ⌈\frac{702^{2}}{c_{2}} ⌉. We find here

\sqrt{λ}Φ_{4}(r, λ) ≤ \frac{1}{3\sqrt{2π} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{2}} + \frac{c_{2}}{2^{7/2} (\frac{3−r}{4} − \frac{\sqrt{r−1}}{3})^{3/2}}

+\frac{\sqrt{2}e^{−^{\frac{c_{1}\sqrt{λ}_{0}}{2}}}}{\sqrt{πc_{1}\sqrt{λ}_{0}}} + (0.6267 + 0.04206\sqrt{λ}_{0}) λ_{0}e^{−λ_{0}(1− \frac{π}{4})\frac{α^{2}_{0}}{8}} ,

which gives

\sqrt{λ}Φ_{4}(r, λ) ≤ \begin{cases}1.05882, & if 1.67007\sqrt{λ} ≤ (r − 1)λ ≤ \sqrt{πλ}; \\ 1.30775, & if \sqrt{πλ} ≤ (r − 1)λ ≤ \sqrt{2πλ}; \\ 1.50929, & if \sqrt{2πλ} ≤ (r − 1)λ ≤ \sqrt{3πλ}; \\ 1.68876, & if \sqrt{3πλ} ≤ (r − 1)λ ≤ \sqrt{4πλ}; \\ 1.85482, & if \sqrt{4πλ} ≤ (r − 1)λ ≤ \sqrt{5πλ}; \\ 2.01189, & if \sqrt{5πλ} ≤ (r − 1)λ ≤ \sqrt{6πλ}; \\ 2.1626, & if \sqrt{6πλ} ≤ (r − 1)λ ≤ \sqrt{7πλ}; \\ 2.30865, & if \sqrt{7πλ} ≤ (r − 1)λ ≤ \sqrt{8πλ}. \end{cases}

This completes the proof of the theorem.

We now need lower bounds for | cos(λ_{1}γ_{1} + λ_{2}γ_{2})|, and this is the aim of the next subsection.

4.5 Additional lemmas

Lemma 21. For 1 ≤ r ≤ \frac{9+\sqrt{73}}{4} = 4.386 …, we have

0 ≤ rγ_{1} + γ_{2} − (r − 3)\frac{π}{4} + \frac{(r − 1)^{2}}{4} ≤ \frac{(r − 1)^{3}}{8} .

Proof. Define γ(r) = rγ_{1} + γ_{2}. We have γ_{1}(1) = \frac{π}{4}, γ_{2}(1) = −\frac{3π}{4}, γ′_{1}(r) = −\frac{1}{r\sqrt{−r^{2}+6r−1}}, γ′_{2}(r) = \frac{1}{\sqrt{−r^{2}+6r−1}}, and therefore γ(1) = −\frac{π}{2}, γ′ = γ_{1}, γ′′(r) = −\frac{1}{r\sqrt{−r^{2}+6r−1}}, γ′′(1) = −\frac{1}{2},

γ^{(3)}(r) = \frac{−2r^{2} + 9r − 1}{r^{2}(−r^{2} + 6r − 1)^{3/2}} and γ^{(4)}(r) = −2\frac{3r^{4} − 27r^{3} + 70r^{2} − 15r + 1}{r^{3}(−r^{2} + 6r − 1)^{5/2}} .

Since 3r^{4} − 27r^{3} + 70r^{2} − 15r + 1 > 0, the function γ^{(3)} is decreasing on [1, 3 + 2\sqrt{2}). From γ^{(3)}(1) = 3/4 and γ^{(3)}(\frac{9+\sqrt{73}}{4}) = 0, we get 0 ≤ γ^{(3)}(r) ≤ 3/4 for 1 ≤ r ≤ \frac{9+\sqrt{73}}{4}, and the required inequality follows.

Lemma 22. For 1 ≤ r ≤ 3, we have

• For λ_{1} + λ_{2} ≡ 0 (mod 4):

| cos(λ(rγ_{1} + γ_{2}))|

≥ \begin{cases}cos (\frac{(r−1)^{2}}{4} λ) , & if λ\frac{(r−1)^{2}}{4} ≤ \frac{π}{2}; \\ min (cos (π − \frac{(r−1)^{2}}{4} λ) , cos (π − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ)) , & if \frac{π}{3−r} ≤ λ\frac{(r−1)^{2}}{4} ≤ \frac{3π}{2}. \end{cases}

• For λ_{1} + λ_{2} ≡ 1 (mod 4):

| cos(λ(rγ_{1} + γ_{2}))|

≥ \begin{cases}min (\frac{1}{\sqrt{2}}, cos (\frac{π}{4} − λ\frac{(r−1)^{2}}{4} )) , & if λ\frac{(r−1)^{2}}{4} ≤ \frac{3π}{4}; \\ min (cos (\frac{5π}{4} − \frac{(r−1)^{2}}{4} λ) , cos (\frac{5π}{4} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ)) , & if \frac{3π}{2(3−r)} ≤ λ\frac{(r−1)^{2}}{4} ≤ \frac{7π}{4}. \end{cases}

• For λ_{1} + λ_{2} ≡ 2 (mod 4):

| cos(λ(rγ_{1} + γ_{2}))|

≥ \begin{cases}min (cos (\frac{π}{2} − λ\frac{(r−1)^{2}}{4} ) , cos (\frac{π}{2} − \frac{3−r}{2}λ\frac{(r−1)^{2}}{4} )) , & if λ\frac{(r−1)^{2}}{4} ≤ π; \\ min (cos (\frac{3π}{2} − λ\frac{(r−1)^{2}}{4} ) , cos (\frac{3π}{2} − \frac{3−r}{2}λ\frac{(r−1)^{2}}{4} )) , & if \frac{2π}{3−r} ≤ λ\frac{(r−1)^{2}}{4} ≤ 2π. \end{cases}

• For λ_{1} + λ_{2} ≡ 3 (mod 4):

| cos(λ(rγ_{1} + γ_{2}))|

≥ \begin{cases}cos (\frac{π}{4} + λ\frac{(r−1)^{2}}{4} ) , & if λ\frac{(r−1)^{2}}{4} ≤ \frac{π}{4}; \\ min (cos (\frac{3π}{4} − λ\frac{(r−1)^{2}}{4} ) , cos (\frac{3π}{4} − \frac{3−r}{2}λ\frac{(r−1)^{2}}{4} )) , & if \frac{π}{2(3−r)} ≤ λ\frac{(r−1)^{2}}{4} ≤ \frac{5π}{4}. \end{cases}

Proof. From Lemma 21 we deduce the inequalities

0 ≤ λ_{1}γ_{1} + λ_{2}γ_{2} − (λ_{1} − 3λ_{2})\frac{π}{4} + λ_{2} \frac{(r − 1)^{2}}{4} ≤ λ_{2} \frac{(r − 1)^{3}}{8} .

Let us define η = λ_{1}γ_{1} + λ_{2}γ_{2} − (λ_{1} − 3λ_{2})\frac{π}{4} so that we have

−\frac{(r − 1)^{2}}{4} λ_{2} ≤ η ≤ \frac{r − 3}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ 0 ,

for 1 ≤ r ≤ 3. We then find

| cos(λ_{1}γ_{1} + λ_{2}γ_{2})| = \begin{cases}cos η, & if λ_{1} + λ_{2} ≡ 0 (mod 4) and −\frac{π}{2} ≤ η ≤ 0; \\ cos (π + η) , & if λ_{1} + λ_{2} ≡ 0 (mod 4) and −\frac{3π}{2} ≤ η ≤ −\frac{π}{2}; \\ cos (\frac{π}{4} + η) , & if λ_{1} + λ_{2} ≡ 1 (mod 4) and −\frac{3π}{4} ≤ η ≤ 0; \\ cos (\frac{5π}{4} + η) , & if λ_{1} + λ_{2} ≡ 1 (mod 4) and −\frac{7π}{4} ≤ η ≤ −\frac{3π}{4}; \\ cos (\frac{π}{2} + η) , & if λ_{1} + λ_{2} ≡ 2 (mod 4) and −π ≤ η ≤ 0; \\ cos (\frac{3π}{2} + η) , & if λ_{1} + λ_{2} ≡ 2 (mod 4) and −2π ≤ η ≤ −π; \\ cos (−\frac{π}{4} + η) , & if λ_{1} + λ_{2} ≡ 3 (mod 4) and −\frac{π}{4} ≤ η ≤ 0; \\ cos (\frac{3π}{4} + η) , & if λ_{1} + λ_{2} ≡ 3 (mod 4) and −\frac{5π}{4} ≤ η ≤ −\frac{π}{4}. \end{cases}

The lemma follows.

4.6 Proof of Theorem 10

In order to get a contradiction, assume I(λ_{2}) = 0 with λ_{1} − λ_{2} ≥ 702, by Proposition 8. Theorem 9 then gives an upper bound for | cos(λ_{1}γ_{1} + λ_{2}γ_{2})| that may be smaller than the lower bound given in Lemma 22. Because of the form of Theorem 9 and Lemma 22, we need to distinguish several cases, according to the residue class of λ_{1} + λ_{2} modulo 4 and to the size of \frac{(λ_{1}−λ_{2})^{2}}{4λ_{2}} . When needed, we shall use the upper bound r ≤ 1.03581 and the estimate arccos x ≥ \frac{π}{2} − \frac{x}{\sqrt{1−x^{2}}} for 0 < x < 1.

4.6.1 The case λ_{1} + λ_{2} ≡ 0 (mod 4)

• For 702 ≤ λ_{1} − λ_{2} ≤ log λ_{2}, we obtain the bounds λ_{2} ≥ e^{702} and \frac{(λ_{1}−λ_{2})^{2}}{4λ_{2}} ≤ \frac{702^{2}}{4e^{702}} < \frac{π}{2}. From Theorem 9 and Lemma 22 we get the following contradiction:

0.9999 ≤ cos (\frac{702^{2}}{4e^{702}} ) ≤ cos (\frac{(λ_{1} − λ_{2})^{2}}{4λ_{2}} ) ≤ | cos(λ(rγ_{1} + γ_{2}))| ≤ 0.0165 .

• For max(702, log λ_{2}) ≤ λ_{1} − λ_{2} ≤ \sqrt{2πλ_{2}} − 1.0443, we obtain the lower bound λ_{2} ≥ ⌈\frac{703.0443^{2}}{2π} ⌉ = 78666. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{(λ_{1}−λ_{2})^{2}}{4λ_{2}} ) ≤ \frac{1.30775}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{(λ_{1} − λ_{2})^{2}}{4λ_{2}} ≥ arccos \frac{1.30775}{\sqrt{λ}_{2}} ≥ \frac{π}{2} − \frac{1.30775}{\sqrt{λ_{2} − 1.30775^{2}}} > (\sqrt{\frac{π}{2}} − \frac{0.52215}{\sqrt{λ}_{2}})^{2} .

• For max(702, \sqrt{2πλ_{2}} + 3.1407) ≤ λ_{1} − λ_{2} ≤ \sqrt{3πλ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{3π} ⌉ = 52289. From Theorem 9 and Lemma 22 we deduce the inequality cos (π − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.50929}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{1.50929}{\sqrt{λ_{2} − 1.50929^{2}}} ≤ π − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ π − \frac{π}{2} (1 − \frac{\sqrt{3π}}{2\sqrt{λ_{2}}} ) (1 + \frac{3.1407}{\sqrt{2πλ_{2}}} )^{2} .

• For max(702, \sqrt{3πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{\frac{16}{5−r}πλ_{2}} < \sqrt{12.68λ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{12.68}⌉ = 38865. From Theorem 9 and Lemma 22 we deduce the inequalities cos (π − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.85482}{\sqrt{λ}_{2}} < 0.00941, and we get the contradiction

1.5613 < arccos(0.00941) < π − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ π − \frac{3 − r}{8}3π ≤ 0.8276 .

• For max (702, \sqrt{\frac{16}{5−r}πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{6πλ_{2}} − 0.9275, we obtain the lower bound λ_{2} ≥ ⌈\frac{702.9275^{2}}{6π} ⌉ = 26214. From Theorem 9 and Lemma 22 we deduce the inequality cos (π − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{2.01189}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{2.01189}{\sqrt{λ_{2} − 2.01189^{2}}} ≤ λ_{2} \frac{(r − 1)^{2}}{4} − π ≤ \frac{3π}{2} (1 − \frac{0.9275}{\sqrt{6πλ_{2}}} )^{2} − π .

4.6.2 The case λ_{1} + λ_{2} ≡ 1 (mod 4)

• For 703 ≤ λ_{1}−λ_{2} ≤ \sqrt{2πλ_{2}}, we obtain the bound λ_{2} ≥ ⌈\frac{703^{2}}{2π} ⌉ = 78656. From Theorem 9 and Lemma 22 we obtain the following contradiction:

\frac{1}{\sqrt{2}} ≤ | cos(λ(rγ_{1} + γ_{2}))| ≤ min (0.0165, \frac{1.30775}{\sqrt{λ_{2}}}) .

• For max(703, \sqrt{2πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{3πλ_{2}} − 0.984, we obtain the bound λ_{2} ≥ ⌈\frac{703.984^{2}}{3π} ⌉ = 52585. The inequality cos (\frac{π}{4} − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.50929}{\sqrt{λ}_{2}} follows from Theorem 9 and Lemma 22, and we get the contradiction

\frac{(λ_{1} − λ_{2})^{2}}{4λ_{2}} ≥ \frac{π}{4} + arccos (\frac{1.50929}{\sqrt{λ}_{2}}) ≥ \frac{3π}{4} − \frac{1.50929}{\sqrt{λ_{2} − 1.50929^{2}}} > (\sqrt{\frac{3π}{4}} − \frac{0.492}{\sqrt{λ}_{2}})^{2} .

• For max(703, \sqrt{3πλ_{2}} + 3.8433) ≤ λ_{1} − λ_{2} ≤ \sqrt{4πλ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{4π} ⌉ = 39217. The inequality cos (\frac{5π}{4} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.68876}{\sqrt{λ}_{2}} follows from Theorem 9 and Lemma 22 , and we get the contradiction

\frac{π}{2} − \frac{1.68876}{\sqrt{λ_{2} − 1.68876^{2}}} ≤ \frac{5π}{4} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{5π}{4} − \frac{3π}{4} (1 − \begin{matrix}√ & \overline{π} \\ √ & \overline{λ_{2}} \end{matrix} ) (1 + \frac{3.8433}{\sqrt{3πλ_{2}}} )^{2} .

• For max(703, \sqrt{4πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{\frac{20}{5−r}πλ_{2}} < \sqrt{15.85λ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{703^{2}}{15.85}⌉ = 31181. From Theorem 9 and Lemma 22 we deduce the inequalities cos (\frac{5π}{4} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{2.01189}{\sqrt{λ}_{2}} < 0.0114, and we get the contradiction

1.5593 < arccos(0.0114) < \frac{5π}{4} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{5π}{4} − \frac{3 − r}{8}4π < 0.8417 .

• For max (703, \sqrt{\frac{20}{5−r}πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{7πλ_{2}} − 0.9231, we obtain the lower bound λ_{2} ≥ ⌈\frac{703.9231^{2}}{7π} ⌉ = 22533. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{5π}{4} − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{2.1626}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{2.1626}{\sqrt{λ_{2} − 2.1626^{2}}} ≤ \frac{(r − 1)^{2}}{4} λ_{2} − \frac{5π}{4} ≤ \frac{7π}{4} (1 − \frac{0.9231}{\sqrt{7πλ_{2}}} )^{2} − \frac{5π}{4} .

4.6.3 The case λ_{1} + λ_{2} ≡ 2 (mod 4)

• For max (702, 2.0582λ_{2}^{1/4}) ≤ λ_{1}−λ_{2} ≤ \sqrt{πλ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{π} ⌉ = 156865. The inequality cos (\frac{π}{2} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.05582}{\sqrt{λ}_{2}} follows from Theorem 9 and Lemma 22 , and we get the contradiction

\frac{π}{2} − \frac{1.05882}{\sqrt{λ_{2} − 1.05882^{2}}} ≤ \frac{π}{2} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{π}{2} − (1 − \frac{2.0582}{2λ_{2}^{3/4}}) \frac{2.0582^{2}}{4\sqrt{λ}_{2}} .

• For max (702, \sqrt{πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{\frac{8}{5−r}πλ_{2}} < \sqrt{6.34λ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{6.34} ⌉ = 77730. From Theorem 9 and Lemma 22 we deduce the inequalities cos (\frac{π}{2} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.50929}{\sqrt{λ}_{2}} < 0.00542, and we get the contradiction

1.5653 < arccos(0.00542) < \frac{π}{2} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{π}{2} − \frac{3 − r}{8}π < 0.7995 .

• For max (702, \sqrt{\frac{8}{5−r}πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ 2\sqrt{πλ_{2}} − 0.9535, we obtain the lower bound λ_{2} ≥ ⌈\frac{702.9535^{2}}{4π} ⌉ = 39323. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{π}{2} − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.68876}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{1.68876}{\sqrt{λ_{2} − 1.68876^{2}}} ≤ \frac{(r − 1)^{2}}{4} λ_{2} − \frac{π}{2} ≤ π (1 − \frac{0.9535}{2\sqrt{πλ_{2}}} )^{2} − \frac{π}{2} .

• For max (702, 2\sqrt{πλ_{2}} + 4.5938) ≤ λ_{1} − λ_{2} ≤ \sqrt{5πλ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{5π} ⌉ = 31373. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{3π}{2} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.85482}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{1.85482}{\sqrt{λ_{2} − 1.85482^{2}}} ≤ \frac{3π}{2} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{3π}{2} −π (1 − \frac{\sqrt{5π}}{2\sqrt{λ_{2}}} ) (1 + \frac{4.5938}{2\sqrt{πλ_{2}}} )^{2} .

• For \sqrt{5πλ_{2}} ≤ λ_{1} − λ_{2} ≤ \sqrt{\frac{24}{5−r}πλ_{2}} < \sqrt{19.02λ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{702^{2}}{19.02}⌉ = 25910. The inequalities cos (\frac{3π}{2} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{2.1626}{\sqrt{λ}_{2}} < 0.01344 follow from Theorem 9 and Lemma 22, and we get the contradiction

1.5573 < arccos(0.01344) < \frac{3π}{2} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{3π}{2} − \frac{3 − r}{8}5π < 0.8558 .

• For max (702, \sqrt{\frac{24}{5−r}πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ 2\sqrt{2πλ_{2}} − 0.9218, we obtain the lower bound λ_{2} ≥ ⌈\frac{702.9218^{2}}{8π} ⌉ = 19660. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{3π}{2} − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{2.30865}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{2.30865}{\sqrt{λ_{2} − 2.30865^{2}}} ≤ \frac{(r − 1)^{2}}{4} λ_{2} − \frac{3π}{2} ≤ 2π (1 − \frac{0.9218}{2\sqrt{2πλ_{2}}} )^{2} − \frac{3π}{2} .

4.6.4 The case λ_{1} + λ_{2} ≡ 3 (mod 4)

• For 703 ≤ λ_{1} − λ_{2} ≤ log λ_{2}, we still find \frac{(λ_{1}−λ_{2})^{2}}{4λ_{2}} ≤ \frac{703^{2}}{4e^{703}} < \frac{π}{2} and we get the contradiction cos (\frac{π}{4} + \frac{703^{2}}{4e^{703}} ) ≤ 0.0165.

• For max(703, log λ_{2}) ≤ λ_{1}−λ_{2} ≤ \sqrt{πλ_{2}}−1.1958, we obtain the bound λ_{2} ≥ ⌈\frac{704.1958^{2}}{π} ⌉ = 157848. The inequality cos (\frac{π}{4} + \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.05882}{\sqrt{λ}_{2}} follows from Theorem 9 and Lemma 22, and we get the contradiction

\frac{(λ_{1} − λ_{2})^{2}}{4λ_{2}} ≥ arccos (\frac{1.05882}{\sqrt{λ}_{2}}) − \frac{π}{4} ≥ \frac{π}{4} − \frac{1.05882}{\sqrt{λ_{2} − 1.05882^{2}}} > (\sqrt{\frac{π}{4}} − \frac{0.5979}{\sqrt{λ}_{2}})^{2} .

• For max(703, \sqrt{πλ_{2}} + 2.5913) ≤ λ_{1} − λ_{2} ≤ \sqrt{2πλ_{2}}, we obtain the bound λ_{2} ≥ ⌈\frac{703^{2}}{2π} ⌉ = 78656. The inequality cos (\frac{3π}{4} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.30775}{\sqrt{λ}_{2}} follows from Theorem 9 and Lemma 22 , and we get the contradiction

\frac{π}{2} − \frac{1.30775}{\sqrt{λ_{2} − 1.30775^{2}}} ≤ \frac{3π}{4} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{3π}{4} − \frac{π}{4} (1 − \begin{matrix}√ & \overline{π} \\ √ & \overline{2λ_{2}} \end{matrix} ) (1 + \frac{2.5913}{\sqrt{πλ_{2}}})^{2} .

• For max(703, \sqrt{2πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{\frac{12}{5−r}πλ_{2}} < \sqrt{9.51λ_{2}}, we obtain the lower bound λ_{2} ≥ ⌈\frac{703^{2}}{9.51} ⌉ = 51968. From Theorem 9 and Lemma 22 we deduce the inequalities cos (\frac{3π}{4} − \frac{3−r}{2} \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.68876}{\sqrt{λ}_{2}} < 0.00741, and we get the contradiction

1.5633 < arccos(0.00741) < \frac{3π}{4} − \frac{3 − r}{2} \frac{(r − 1)^{2}}{4} λ_{2} ≤ \frac{3π}{4} − \frac{3 − r}{8}2π < 0.8136 .

• For max (703, \sqrt{\frac{12}{5−r}πλ_{2}}) ≤ λ_{1} − λ_{2} ≤ \sqrt{5πλ_{2}} − 0.9367, we obtain the bound λ_{2} ≥ ⌈\frac{703.9367^{2}}{5π} ⌉ = 31547. From Theorem 9 and Lemma 22 we deduce the inequality cos (\frac{3π}{4} − \frac{(r−1)^{2}}{4} λ_{2}) ≤ \frac{1.85482}{\sqrt{λ}_{2}}, and we get the contradiction

\frac{π}{2} − \frac{1.85482}{\sqrt{λ_{2} − 1.85482^{2}}} ≤ \frac{(r − 1)^{2}}{4} λ_{2} − \frac{3π}{4} ≤ \frac{5π}{4} (1 − \frac{0.9367}{\sqrt{5πλ_{2}}} )^{2} − \frac{3π}{4} .

5 Concluding remarks

In the introduction, we discussed the irreducibility of C_{X,λ_{2}}, and noticed how different are the cases λ_{2} even and λ_{2} odd. We checked both cases for λ_{2} ≤ 240, but we can go further in the even case. Using Maple during 35101 seconds, we showed that C_{X,λ_{2}} is irreducible over ℚ when λ_{2} ≤ 600 is even. This motivates the following conjecture.

Conjecture 23. For λ_{2} ≥ 2 even, the polynomial C_{X,λ_{2}} is irreducible over ℚ. For λ_{2} ≥ 3 odd, the polynomial C_{X,λ_{2}} is the product of X − λ_{2} by an irreducible polynomial over ℚ.

We used the same technics, together with hypergeometric transformations, to study the case λ_{1}−λ_{2} small and to prove Proposition 8. We introduced the four families of polynomials and checked their irreducibility over ℚ for l ≤ 350. It is quite likely that this property holds for all larger values of l.

Conjecture 24. For l ≥ 3, the polynomials C̃_{l,0,0}(X), C̃_{l,0,1}(X), C̃_{l,1,0}(X) and C̃_{l,1,1}(X) are irreducible over ℚ.

Let us now discuss the results obtained in Theorem 10. The first thing we noticed is that the case λ_{1} + λ_{2} ≡ 2 (mod 4) differs from the other cases. It would be nice to be nice to deal with the interval 702 ≤ λ_{1} − λ_{2} ≤ 2.0582λ_{2}^{1/4} for any λ_{2}, to fill the initial gap. Secondly we chose to reach the second explicit intervals with no solutions. How far could we go with this method? It would be nice to get improvements that enable to break the \sqrt{λ}_{2} barrier and to go up to λ_{2}^{1/2+ϵ} for some ϵ > 0.

6 Acknowledgments

The author wishes to thank the anonymous referee for substantially helping to improve the presentation of this paper in many ways.

References

[1] A. Carnevale and C. Voll, Orbit Dirichlet series and multiset permutations, Monatsh. Math. 186 (2018), 215–233.

[2] J. Dieudonné, Calcul Infinitésimal, Hermann, 1968.

[3] A. M. Legendre, Essai Sur la Théorie des Nombres, Duprat, Paris, 1798.

2010 Mathematics Subject Classification: Primary 11B65, Secondary 05A10 11B83. Keywords: binomial sum, asymptotics Received July 3 2019; revised versions received February 5 2020; February 9 2020; May 27 2020. Published in Journal of Integer Sequences, June 10 2020.

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