Values of the Double Sine Function

Shin-ya Koyama and Nobushige Kurokawa

Abstract.We calculate basic values of the double sine function. The algebraicity for some special values is proved. Its behavior in the fundamental domain is also studied. Especially we show that it has just two extremes: one relative maximum and one relative minimum. Asymptotic formulas for these extremal values are proved.

Key words: Double sine function; double Hurwitz zeta function; extremal value; algebraicity AMS Classification 11M06

1 Introduction

The double sine function was discovered by Hölder [H] in 1886 as a generalization of the usual sine function. It was defined by the infinite product

F (x) = e^{x} ∏_{n=1}^{∞} ((\frac{1 − \frac{x}{n}}{1 + \frac{x}{n}} )^{n} e^{2x}) .

Hölder proved basic properties of F (x) containing the periodicity and the multiplication formula. About ninety years later, Shintani [S] constructed the normalized double sine function F (x, (ω_{1}, ω_{2})) to investigate Kronecker’s Jugendtraum for real quadratic fields. Hölder’s double sine function F (x) is expressed by Shintani’s double sine function as F (x) = F (1 − x, (1, 1)). We studied these double sine functions in previous papers [K1]-[K4] [KK1]-[KK5] [KW1] [KW2] with various applications and generalizations. There also exists a physical application such as Jimbo-Miwa [JM]. Thus, the double sine function is an important mathematical object, but its study is still in a primitive stage. For example we do not know the precise behaviors concerning extermal values. In this paper we investigate this basic problem. The difficulty of this problem is coming from the fact that we do not have sufficient knowledge about the derivatives of the double sine function in general. We refer to [KK1] [KW1] [K3] for related results.

For later use, we recall the general construction of normalized multiple sine functions following [K1] [K2] [KK1]. The normalized multiple sine function of period ω = (ω_{1}, ..., ω_{r}) for ω_{1}, ..., ω_{r} > 0 is defined as

S_{r}(x, ω) = ( ∏ ∐_{n_{1},...,n_{r}=0}^{∞}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + x))

× ( ∏ ∐_{m_{1},...,m_{r}=1}^{∞}(m_{1}ω_{1} + ⋯ + m_{r}ω_{r} − x))^{(−1)^{r−1}} ,

where the zeta reguralized product ∏ ∐ of Deninger [D] is used. Alternatively, S_{r}(x, ω) is written as

S_{r}(x, ω) = Γ_{r}(x, ω)^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − x, ω)^{(−1)^{r}}

in terms of the regularized multiple gamma function

Γ_{r}(x, ω) = ( ∏ ∐_{n_{1},...,n_{r}=0}^{∞}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + x))^{−1}

= exp ( \frac{∂}{∂s}ζ_{r}(s, x, ω)|_{s=0}) .

Here

ζ_{r}(s, x, ω) = ∑_{n_{1},...,n_{r}=0}^{∞}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + x)^{−s}

is the multiple Hurwitz zeta function defined by Barnes [B]. The case r = 1 is reduced to the usual gamma function and sine function:

Γ_{1}(x, ω) = \frac{Γ(x/ω)}{\sqrt{2π}}ω^{\frac{x}{ω} − \frac{1}{2}}

and

S_{1}(x, ω) = \frac{2π}{Γ(\frac{x}{ω})Γ(1 − \frac{x}{ω})} = 2 sin (\frac{πx}{ω}) .

For simplicity we write

Γ_{r}(x) = Γ_{r}(x, (1, ..., 1))

and

S_{r}(x) = S_{r}(x, (1, ..., 1)).

As proved in [KK1] the double sine function F (x) of Hölder [H] is identified as

F (x) = S_{2}(1 − x, (1, 1))
= S_{2}(1 − x)
= S_{2}(x)^{−1}S_{1}(x),

and the double sine function F (x, (ω_{1}, ω_{2})) of Shintani [S] is nothing but

F (x, (ω_{1}, ω_{2})) = S_{2}(x, (ω_{1}, ω_{2})).

On the other hand another type of multiple sine function 𝒮_{r}(x), which we call the primitive multiple sine function in [KK1], is defined as

𝒮_{r}(x) = exp (\frac{x^{r−1}}{r − 1}) ∏_{n=1}^{∞} (P_{r} (\frac{x}{n}) P_{r} (−\frac{x}{n})^{(−1)^{r−1}})^{n^{r−1}}

for r ≥ 2 with

P_{r}(u) = (1 − u) exp (u + \frac{u^{2}}{2} + ⋯ + \frac{u^{r}}{r} ) .

Hölder’s double sine function is F (x) = 𝒮_{2}(x) and 𝒮_{r}(x) is also written via S_{r}(x) explicitly (see [KK1]).

Now we describe our results. We look at the behavior of S_{2}(x, (ω_{1}, ω_{2})) on the real line. From the basic periodicity

S_{2}(x + ω_{1} + ω_{2}, (ω_{1}, ω_{2})) = −\frac{S_{2}(x, (ω_{1}, ω_{2}))}{4 sin(\frac{πx}{ω_{1}}) sin(\frac{πx}{ω_{2}})}

proved in [KK1] in a more generalized form, we may restrict ourselves to the fundamental domain 0 ≤ x < ω_{1} + ω_{2}. Since S_{2}(x, (ω_{2}, ω_{1})) = S_{2}(x, (ω_{1}, ω_{2})) we may assume 0 < ω_{1} ≤ ω_{2} without the loss of generality.

First the basic values are given in the following theorem.

Theorem 1

(1)

S_{2} (\frac{ω_{1} + ω_{2}}{2} , (ω_{1}, ω_{2})) = 1.

S_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = S_{2} (\frac{ω_{2}}{2} , (ω_{1}, ω_{2})) = \sqrt{2}.

(3)

S_{2} (ω_{1} + \frac{ω_{2}}{2} , (ω_{1}, ω_{2})) = S_{2} (ω_{2} + \frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = \frac{\sqrt{2}}{2}.

(4)

S_{2}(ω_{1}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{2}}{ω_{1}}} .

(5)

S_{2}(ω_{2}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{1}}{ω_{2}}} .

We find extremal values as follows: Theorem 2 Let 0 < ω_{1} ≤ ω_{2}. Then we have:

(1) The double sine function S_{2}(x, (ω_{1}, ω_{2})) has just two extremes S_{2}(α_{max}(ω_{1}, ω_{2}), (ω_{1}, ω_{2}))

and S_{2}(α_{min}(ω_{1}, ω_{2}), (ω_{1}, ω_{2})) in 0 ≤ x < ω_{1} + ω_{2}, where

0 < α_{max}(ω_{1}, ω_{2}) < α_{min}(ω_{1}, ω_{2}) < ω_{1} + ω_{2}.

(2) S_{2}(α_{max}(ω_{1}, ω_{2}), (ω_{1}, ω_{2})) is a relative maximum and

\frac{ω_{1}}{2} ≤ α_{max}(ω_{1}, ω_{2}) ≤ \frac{ω_{2}}{2} .

(3) S_{2}(α_{min}(ω_{1}, ω_{2}), (ω_{1}, ω_{2})) is a relative minimum and

ω_{1} + \frac{ω_{2}}{2} ≤ α_{min}(ω_{1}, ω_{2}) ≤ ω_{2} + \frac{ω_{1}}{2} .

(4)

α_{max}(ω_{1}, ω_{2}) + α_{min}(ω_{1}, ω_{2}) = ω_{1} + ω_{2}.

(5)

S_{2}(α_{max}(ω_{1}, ω_{2}), (ω_{1}, ω_{2}))S_{2}(α_{min}(ω_{1}, ω_{2}), (ω_{1}, ω_{2})) = 1.

The asymptotic results for the extremal values are shown as follows:

Theorem 3 (1)

lim_{t↓0} α_{max}(1, t) = \frac{1}{6}.

(2)

lim_{t→∞} \frac{α_{max}(1, t)}{t} = \frac{1}{6}.

(3)

lim_{t↓0} α_{min}(1, t) = \frac{5}{6}.

(4)

lim_{t→∞} \frac{α_{min}(1, t)}{t} = \frac{5}{6}.

(5)

lim_{\frac{ω_{1}}{ω_{2}} →0,∞} \frac{α_{max}(ω_{1}, ω_{2})}{ω_{1} + ω_{2}} = \frac{1}{6}.

(6)

lim_{\frac{ω_{1}}{ω_{2}} →0,∞} \frac{α_{min}(ω_{1}, ω_{2})}{ω_{1} + ω_{2}} = \frac{5}{6}.

Theorem 4 (1)

lim sup_{t↓0} \frac{log S_{2}(α_{max}(1, t), (1, t))}{1/t} ≤ \frac{1}{6}.

(2)

lim inf_{t↓0} \frac{log S_{2}(α_{max}(1, t), (1, t))}{1/t} ≥ \frac{1}{6} log (\frac{3e}{π}) .

(3)

lim sup_{t→∞} \frac{log S_{2}(α_{max}(1, t), (1, t))}{t} ≤ \frac{1}{6}.

(4)

lim inf_{t→∞} \frac{log S_{2}(α_{max}(1, t), (1, t))}{t} ≥ \frac{1}{6} log (\frac{3e}{π}) .

(5)

lim inf_{t↓0} \frac{log S_{2}(α_{min}(1, t), (1, t))}{1/t} ≥ −\frac{1}{6}.

lim sup_{t↓0} \frac{log S_{2}(α_{min}(1, t), (1, t))}{1/t} ≤ −\frac{1}{6} log (\frac{3e}{π}) .

(7)

lim inf_{t→∞} \frac{log S_{2}(α_{min}(1, t), (1, t))}{t} ≥ −\frac{1}{6}.

(8)

lim sup_{t→∞} \frac{log S_{2}(α_{min}(1, t), (1, t))}{t} ≤ −\frac{1}{6} log (\frac{3e}{π}) .

We show the following algebraicity result: Theorem 5 Let N_{1} and N_{2} be positive integers with (N_{1}, N_{2}) = 1 or 2. Then the value S_{2}(m, (N_{1}, N_{2})) is an algebraic number in Q (sin \frac{2π}{N_{i}} , cos \frac{2π}{N_{i}} , \sqrt{N_{i}} | i = 1, 2) for m = 1, ..., N_{1} + N_{2} − 1. Examples

S_{2}(1, (4, 6)) = 1,  S_{2}(2, (4, 6)) = \sqrt{2},  S_{2}(3, (4, 6)) = \sqrt{2},
S_{2}(4, (4, 6)) = \sqrt{\frac{3}{2}},  S_{2}(5, (4, 6)) = 1,  S_{2}(6, (4, 6)) = \sqrt{\frac{2}{3}},
S_{2}(7, (4, 6)) = \frac{1}{\sqrt{2}},  S_{2}(8, (4, 6)) = \frac{1}{\sqrt{2}},  S_{2}(9, (4, 6)) = 1.

We obtain the following result in the case of a period of a special type. Theorem 6

(1) For an integer n ≥ 1,

S_{2}(n, (1, 6n)) = S_{2}(n + 1, (1, 6n)) = max{S_{2}(m, (1, 6n)) | m = 1, ..., 6n} and

n < α_{max}(1, 6n) < n + 1.

lim sup_{n→∞} \frac{log S_{2}(n, (1, 6n))}{n} ≤ 1.

(3)

lim inf_{n→∞} \frac{log S_{2}(n, (1, 6n))}{n} ≥ log (\frac{3e}{π}) .

We make a few remarks. Algebraic natures of special values of the double sine function are very interesting from arithmetic view points. Values in Theorems 5 and 6 are examples of N-division values of the double sine function for N = 1 and 2. General N-division values would be important to realize Kronecker’s Jugendtraum as indicated first by Shintani [S]. Moreover, extremal values appearing in Theorems 2 and 4 are quite mysterious. Our results can be generalized to the case of the multiple sine function to some extent. We will treat generalizations on another opportunity.

2 Special values: Proof of Theorem 1

Proof of Theorem 1

(1) This central value comes from the definition

S_{2} (\frac{ω_{1} + ω_{2}}{2} , (ω_{1}, ω_{2})) = \frac{Γ_{2} (\frac{ω_{1}+ω_{2}}{2} , (ω_{1}, ω_{2}))}{Γ_{2} (\frac{ω_{1}+ω_{2}}{2} , (ω_{1}, ω_{2}))} = 1.

(2) We have

S_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = \frac{Γ_{2} (\frac{ω_{1}}{2} + ω_{2}, (ω_{1}, ω_{2}))}{Γ_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2}))} = Γ_{1} (\frac{ω_{1}}{2} , ω_{1})^{−1} ,

where we used the (quasi-)periodicity

Γ_{2}(x + ω_{2}, (ω_{1}, ω_{2})) = Γ_{2}(x, (ω_{1}, ω_{2}))Γ_{1}(x, ω_{1})^{−1}.

Notice that

Γ_{1}(x, ω) = \frac{Γ(\frac{x}{ω})}{\sqrt{2π}}ω^{\frac{x}{ω} − \frac{1}{2}} .

Hence

Γ_{1} (\frac{ω}{2} , ω) = \frac{Γ(\frac{1}{2})}{\sqrt{2π}}ω^{\frac{1}{2} − \frac{1}{2}} = \frac{1}{\sqrt{2}}.

Thus

S_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = \sqrt{2}.

By symmetry we also have

S_{2} (\frac{ω_{2}}{2} , (ω_{1}, ω_{2})) = \sqrt{2}.

(3) By the reflection S_{2}(ω_{1} + ω_{2} − x, (ω_{1}, ω_{2})) = S_{2}(x, (ω_{1}, ω_{2}))^{−1} we have

S_{2} (ω_{1} + \frac{ω_{2}}{2} , (ω_{1}, ω_{2})) = S_{2} (\frac{ω_{2}}{2} , (ω_{1}, ω_{2}))^{−1} = \frac{\sqrt{2}}{2}

and

S_{2} (ω_{2} + \frac{ω_{1}}{2} , (ω_{1}, ω_{2})) = S_{2} (\frac{ω_{1}}{2} , (ω_{1}, ω_{2}))^{−1} = \frac{\sqrt{2}}{2}.

(4) First, we remark that

S_{2} (ω_{1}, (ω_{1}, ω_{2})) = \frac{Γ_{2}(ω_{2}, (ω_{1}, ω_{2}))}{Γ_{2}(ω_{1}, (ω_{1}, ω_{2}))}
= lim_{x→0} \frac{Γ_{2}(ω_{2} + x, (ω_{1}, ω_{2}))}{Γ_{2}(ω_{1} + x, (ω_{1}, ω_{2}))}.

Here we use

\frac{Γ_{2}(ω_{2} + x, (ω_{1}, ω_{2}))}{Γ_{2}(ω_{1} + x, (ω_{1}, ω_{2}))} = \frac{Γ_{2}(x, (ω_{1}, ω_{2}))Γ_{1}(x, ω_{1})^{−1}}{Γ_{2}(x, (ω_{1}, ω_{2}))Γ_{1}(x, ω_{2})^{−1}}
= \frac{Γ_{1}(x, ω_{2})}{Γ_{1}(x, ω_{1})}

= \frac{Γ(\frac{x}{ω_{2}} )ω_{2}^{\frac{x}{ω_{2}} − \frac{1}{2}}}{Γ(\frac{x}{ω_{1}} )ω_{1}^{\frac{x}{ω_{1}} − \frac{1}{2}}} .

Hence we obtain

S_{2} (ω_{1}, (ω_{1}, ω_{2})) = lim_{x→0} \frac{Γ(\frac{x}{ω_{2}} )ω_{2}^{\frac{x}{ω_{2}} − \frac{1}{2}}}{Γ(\frac{x}{ω_{1}} )ω_{1}^{\frac{x}{ω_{1}} − \frac{1}{2}}}
= lim_{x→0} \frac{\frac{ω_{2}}{x} ω_{2}^{− \frac{1}{2}}}{\frac{ω_{1}}{x} ω_{1}^{− \frac{1}{2}}}
= \sqrt{\frac{ω_{2}}{ω_{1}}} .

By symmetry (or from S_{2} (ω_{2}, (ω_{1}, ω_{2})) = S_{2} (ω_{1}, (ω_{1}, ω_{2}))^{−1}) we have

S_{2} (ω_{2}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{1}}{ω_{2}}} .

Remark From the relation

S_{2} (ω_{1}, (ω_{1}, ω_{2})) = lim_{x→0} S_{2} (ω_{1} + x, (ω_{1}, ω_{2}))
= lim_{x→0} \frac{S_{2} (x, (ω_{1}, ω_{2}))}{S_{1}(x, ω_{2})}
= \frac{S′_{2}(0, (ω_{1}, ω_{2}))}{S′_{1}(0, ω_{2})}
= \frac{ω_{2}}{2π} S′_{2}(0, (ω_{1}, ω_{2}))

we see that

S_{2} (ω_{1}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{2}}{ω_{1}}} ⟺ S′_{2}(0, (ω_{1}, ω_{2})) = \frac{2π}{\sqrt{ω_{1}ω_{2}}}
⟺_{symmetry} S_{2} (ω_{2}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{1}}{ω_{2}}} .

We refer to [K3] for a direct proof of S′_{2}(0, (ω_{1}, ω_{2})) = \frac{2π}{\sqrt{ω_{1}ω_{2}}} .

3 Extermal points: Proof of Theorems 2 and 3

Proof of Theorem 2 To simplify the notation we put S(x) = S_{2}(x, (ω_{1}, ω_{2})) here. Then

S(x) = Γ_{2}(x, (ω_{1}, ω_{2}))^{−1}Γ_{2}(ω_{1} + ω_{2} − x, (ω_{1}, ω_{2})) with

Γ_{2}(x, (ω_{1}, ω_{2})) = exp (\frac{∂}{∂s}ζ_{2}(0, x, (ω_{1}, ω_{2})))

for

ζ_{2}(s, x, (ω_{1}, ω_{2})) = ∑_{n_{1},n_{2}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + x)^{−s}.

We notice that ζ_{2}(s, x, (ω_{1}, ω_{2})) converges absolutely in Re(s) > 2 and it has a meromorphic continuation to all s ∈ C. Moreover ζ_{2}(s, x, (ω_{1}, ω_{2})) is holomorphic at s = 0.

Since

log S(x) = − log Γ_{2}(x, (ω_{1}, ω_{2})) + log Γ_{2}(ω_{1} + ω_{2} − x, (ω_{1}, ω_{2}))
= −\frac{∂}{∂s}ζ_{2}(0, x, (ω_{1}, ω_{2})) + \frac{∂}{∂s}ζ_{2}(0, ω_{1} + ω_{2} − x, (ω_{1}, ω_{2}))

we have

\frac{S′}{S} (x) = −\frac{∂^{2}}{∂s∂x}ζ_{2}(0, x, (ω_{1}, ω_{2})) − \frac{∂^{2}}{∂s∂x}ζ_{2}(0, ω_{1} + ω_{2} − x, (ω_{1}, ω_{2})).

Hence

(\frac{S′}{S} )′ (x) = −\frac{∂^{3}}{∂s∂x^{2}} ζ_{2}(0, x, (ω_{1}, ω_{2})) + \frac{∂^{3}}{∂s∂x^{2}} ζ_{2}(0, ω_{1} + ω_{2} − x, (ω_{1}, ω_{2}))

and

(\frac{S′}{S} )′′ (x) = −\frac{∂^{4}}{∂s∂x^{3}} ζ_{2}(0, x, (ω_{1}, ω_{2})) − \frac{∂^{4}}{∂s∂x^{3}} ζ_{2}(0, ω_{1} + ω_{2} − x, (ω_{1}, ω_{2}))

= 2 ( ∑_{n_{1},n_{2}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + x)^{−3} + ∑_{m_{1},m_{2}≥1}(m_{1}ω_{1} + m_{2}ω_{2} − x)^{−3}) ,

where we use the relation

\frac{∂^{3}}{∂x^{3}} ζ_{2}(s, x, (ω_{1}, ω_{2})) = (−s)(−s − 1)(−s − 2)ζ_{2}(s + 3, x, (ω_{1}, ω_{2}))

coming from

\frac{∂}{∂x}ζ_{2}(s, x, (ω_{1}, ω_{2})) = \frac{∂}{∂x} ( ∑_{n_{1},n_{2}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + x)^{−s})
= −s ∑_{n_{1},n_{2}≥0}(n_{1}ω_{1} + n_{2}ω_{2} + x)^{−s−1}
= −sζ_{2}(s + 1, x, (ω_{1}, ω_{2})).

Thus

(\frac{S′}{S} )′′ (x) > 0

in 0 < x < ω_{1} + ω_{2} and

(\frac{S′}{S} )′ (\frac{ω_{1} + ω_{2}}{2} ) = 0.

This shows Figure 1. We show that S′ (\frac{ω_{1}+ω_{2}}{2} ) < 0. Suppose that

S′ (\frac{ω_{1} + ω_{2}}{2} ) ≥ 0.

Then

\frac{S′}{S} (\frac{ω_{1} + ω_{2}}{2} ) ≥ 0

and

\frac{S′}{S} (x) ≥ 0 in 0 < x < ω_{1} + ω_{2}

since (\frac{S′}{S} )′ (x) > 0 in \frac{ω_{1}+ω_{2}}{2} < x < ω_{1} + ω_{2} and (\frac{S′}{S} )′ (x) < 0 in 0 < x < \frac{ω_{1}+ω_{2}}{2} . Hence

S′(x) ≥ 0 in 0 < x < ω_{1} + ω_{2}.

Thus, especially

S (\frac{ω_{1}}{2} ) ≤ S (\frac{ω_{1} + ω_{2}}{2} ) .

But, this contradicts Theorem 1(1)(2): \sqrt{2} > 1. So we conclude that

S′ (\frac{ω_{1} + ω_{2}}{2} ) < 0.

Thus we get Figure 2. Hence there exist uniquely α and β in 0 < α < β < ω_{1} + ω_{2} such that S′(α) = S′(β) = 0; notice that S(0) = 0 and lim_{x↑ω_{1}+ω_{2}} S(x) = +∞. Now, inequalities

\frac{ω_{1}}{2} ≤ α ≤ \frac{ω_{2}}{2} and ω_{1} + \frac{ω_{2}}{2} ≤ β ≤ ω_{2} + \frac{ω_{1}}{2}

are seen with Theorem 1(1)(3) considered. Moreover

S′(ω_{1} + ω_{2} − x) = −\frac{S′(x)}{S(x)^{2}}

shows that β = ω_{1} + ω_{2} − α. Thus we get Theorem 2. Consequently we have Figure 3. Proof of Theorem 3 Properties (5) and (6) follow from (1) (2) and (3) (4), respectively. From the symmetry

S_{2}(x, (ω_{1}, ω_{2})) = S_{2}(x, (ω_{2}, ω_{1})) and the homogeneity

S_{2}(x, (ω_{1}, ω_{2})) = S_{2}(cx, (cω_{1}, cω_{2})) for c > 0 (see [KK1]) we see that

α_{max}(ω_{1}, ω_{2}) = α_{max}(ω_{2}, ω_{1}) and

α_{max}(ω_{1}, ω_{2}) = \frac{α_{max}(cω_{1}, cω_{2})}{c}.

In particular

α_{max} (1, \frac{1}{t}) = α_{max} (\frac{1}{t}, 1) = \frac{α_{max}(1, t)}{t}.

Hence we find that (1) implies (2). As we also have

α_{min}(1, t) = 1 + t − α_{max}(1, t) we see (1) implies (3) and that (2) implies (4). Therefore it suffices to show (1). \underline{Proof of (1)} Let 0 < t < \frac{1}{6} and put N(t) = [\frac{1}{6t}] + 1. We show that

(N(t) − 1)t ≤ α_{max}(1, t) ≤ (N(t) + 1)t. If this is proved, we have

\frac{1}{6} − t < α_{max}(1, t) ≤ \frac{1}{6} + 2t

from

\frac{1}{6t} < N(t) ≤ \frac{1}{6t} + 1.

Hence we get

lim_{t→0} α_{max}(1, t) = \frac{1}{6}.

Proof of α_{max}(1, t) ≥ (N(t) − 1)t From 0 < (N(t) − 1)t ≤ \frac{1}{6}, we see that

\frac{S_{2}(N(t)t, (1, t))}{S_{2}((N(t) − 1)t, (1, t))} = \frac{1}{2 sin(π(N(t) − 1)t)} ≥ 1.

Hence

S_{2}(N(t)t, (1, t)) ≥ S_{2}((N(t) − 1)t, (1, t)). Suppose that (N(t) − 1)t > α_{max}(1, t). Then

α_{max}(1, t) < (N(t) − 1)t
< N(t)t
≤ \frac{1}{6} + t
< \frac{1}{2}.

So the decreasing nature of S_{2}(x, (1, t)) on (α_{max}(1, t), \frac{1}{2}) implies

S_{2}(N(t)t, (1, t)) < S_{2}((N(t) − 1)t, (1, t)). This gives a contradiction. Hence

(N(t) − 1)t ≤ α_{max}(1, t).

Proof of α_{max}(1, t) ≤ (N(t) + 1)t From \frac{1}{2} > N(t)t > \frac{1}{6}, we get

\frac{S_{2}((N(t) + 1)t, (1, t))}{S_{2}(N(t)t, (1, t))} = \frac{1}{2 sin(πN(t)t)} < 1.

S_{2}((N(t) + 1)t, (1, t)) < S_{2}(N(t)t, (1, t)).

Suppose that (N(t)+1)t < α_{max}(1, t). Then the increasing nature of S_{2}(x, (1, t)) on (0, α_{max}(1, t)) implies

S_{2}(N(t)t, (1, t)) < S_{2}((N(t) + 1)t, (1, t)).

This gives the contradiction. Hence

α_{max}(1, t) ≤ (N(t) + 1)t.

4 Proof of Theorem 4

It suffices to show (1) and (2), since (3) and (4) follow from them by the change of variable t to \frac{1}{t}; (5)-(8) come from (1)-(4) by

S_{2}(α_{min}(1, t), (1, t)) = S_{2}(α_{max}(1, t), (1, t))^{−1}.

To prove (1) and (2) we show the following facts: Let 0 < t < \frac{1}{6} and put N(t) = [\frac{1}{6t}] + 1 as in the proof of Theorem 3. Then

(a)

S_{2}(α_{max}(1, t), (1, t)) ≤ \frac{1}{\sqrt{t}4 sin(πt) sin(2πt)} (6^{N(t)−2}t^{N(t)−2}(N(t) − 2)!)^{−1} ,

(b)

S_{2}(α_{max}(1, t), (1, t)) ≥ \frac{1}{\sqrt{t}} (\frac{1}{2}(2π)^{N(t)−2}t^{N(t)−2}N(t)!)^{−1} .

Moreover

(c)

lim_{t↓0} \frac{(N(t) − 2) log \frac{1}{t} − (N(t) − 2) log 6 − log((N(t) − 2)!)}{1/t} = \frac{1}{6},

(d)

lim_{t↓0} \frac{(N(t) − 2) log \frac{1}{t} − (N(t) − 2) log(2π) − log(N(t)!)}{1/t} = \frac{1}{6} log (\frac{3e}{π}) .

First we check that (1) and (2) follow from (a)-(d). This is easy since

lim sup_{t↓0} \frac{log S_{2}(α_{max}(1, t), (1, t))}{1/t}
≤_{(a)} lim sup_{t↓0} \frac{(N(t) − 2) log \frac{1}{t} − (N(t) − 2) log 6 − log((N(t) − 2)!) − log(\sqrt{t}4 sin(πt) sin(2πt))}{1/t}
=_{(c)} \frac{1}{6}

and

lim inf_{t↓0} \frac{log S_{2}(α_{max}(1, t), (1, t))}{1/t}
≥_{(b)} lim inf_{t↓0} \frac{(N(t) − 2) log \frac{1}{t} − (N(t) − 2) log(2π) − log(N(t)!) − log(\sqrt{t}\frac{1}{2})}{1/t}
=_{(d)} \frac{1}{6} log (\frac{3e}{π}) .

Now we show (a)-(d). Proof of (a)(b) Using

S_{2}(α_{max}(1, t) − (l − 1)t, (1, t)) = S_{2}(α_{max}(1, t) − lt, (1, t))S_{1}(α_{max}(1, t) − lt)^{−1} for l = 1, ..., N(t) − 2, we have

S_{2}(α_{max}(1, t), (1, t)) = S_{2}(α_{max}(1, t) − (N(t) − 2)t, (1, t)) ∏_{l=1}^{N(t)−2} S_{1}(α_{max}(1, t) − lt)^{−1}.

From the inequalites

(N(t) − 1)t ≤ α_{max}(1, t) ≤ (N(t) + 1)t shown in the proof of Theorem 3, we see that

t ≤ α_{max}(1, t) − (N(t) − 2)t ≤ 3t. Hence

S_{2}(α_{max}(1, t), (1, t)) ≤ S_{2}(3t, (1, t)) ∏_{l=1}^{N(t)−2} S_{1}((N(t) − 1 − l)t)^{−1}

and

S_{2}(α_{max}(1, t), (1, t)) ≥ S_{2}(t, (1, t)) ∏_{l=1}^{N(t)−2} S_{1}((N(t) + 1 − l)t)^{−1}.

Here we notice that

S_{2}(t, (1, t)) = \frac{1}{\sqrt{t}}

and

S_{2}(3t, (1, t)) = \frac{1}{\sqrt{t}4 sin(πt) sin(2πt)}.

The former identity is Theorem 1(5) and the latter is obtained from

S_{2}(3t, (1, t)) = S_{2}(2t, (1, t))S_{1}(2t)^{−1}
= S_{2}(t, (1, t))S_{1}(t)^{−1}S_{1}(2t)^{−1}.

Remarking

S_{1}(x) = 2 sin(πx) ≤ 2πx for 0 < x < \frac{1}{2}

and

S_{1}(x) = 2 sin(πx) ≥ 6x for 0 < x < \frac{1}{6}

we see that

∏_{l=1}^{N(t)−2} S_{1}((N(t) − 1 − l)t) ≥ ∏_{l=1}^{N(t)−2}(6(N(t) − 1 − l)t)
= 6^{N(t)−2}t^{N(t)−2}(N(t) − 2)!,

where we used

0 < (N(t) − 2)t < \frac{1}{6},

and that

∏_{l=1}^{N(t)−2} S_{1}((N(t) + 1 − l)t) ≤ ∏_{l=1}^{N(t)−2}(2π(N(t) + 1 − l)t)
= \frac{1}{2}(2π)^{N(t)−2}t^{N(t)−2}N(t)!.

Combining these inequalities we get (a) and (b). Proof of (c)(d) First, reformulate numerators of (c) and (d) as

(N(t) − 2) log \frac{1}{t} − (N(t) − 2) log 6 − log((N(t) − 2)!)

= (N(t) − 2) log \frac{1}{6tN(t)} + (N(t) − 2) log N(t) − log((N(t))!) + log(N(t)(N(t) − 1))
= N(t) + (N(t) − 2) log \frac{1}{6tN(t)} − log \frac{N(t)!}{N(t)^{N(t)+ \frac{1}{2}} e^{−N(t)}} − \frac{5}{2} log N(t) + log(N(t)(N(t) − 1))

and (N(t) − 2) log \frac{1}{t} − (N(t) − 2) log(2π) − log(N(t)!)

= (N(t) − 2) log \frac{6}{2π} + (N(t) − 2) log \frac{1}{t} − (N(t) − 2) log 6 − log((N(t))!)
= (N(t) − 2) log \frac{3}{π} + N(t) + (N(t) − 2) log \frac{1}{6tN(t)} − log \frac{N(t)!}{N(t)^{N(t)+ \frac{1}{2}} e^{−N(t)}} − \frac{5}{2} log N(t).

Now the estimation

1 < 6tN(t) < 1 + 6t shows that

lim_{t↓0} \frac{N(t)}{1/t} = \frac{1}{6},

lim_{t↓0} log \frac{1}{6tN(t)} = 0,

and

lim_{t↓0} \frac{log N(t)}{1/t} = 0.

Moreover, Stirling’s formula

lim_{n→∞} \frac{n!}{n^{n+ \frac{1}{2}} e^{−n}} = \sqrt{2π}

implies that

lim_{t↓0} \frac{log \frac{N(t)!}{N(t)^{N(t)+ \frac{1}{2}} e^{−N(t)}}}{1/t} = 0.

Hence we get (c) and (d).

5 Proof of Theorem 5

\underline{(1) (N_{1}, N_{2}) = 1 case:} Take integers M_{1}, M_{2} ≥ 1 such that M_{1}N_{1} − M_{2}N_{2} = 1. Then

mM_{1}N_{1} = m + mM_{2}N_{2} and

S_{2}(mM_{1}N_{1}, (N_{1}, N_{2})) = S_{2}(m + mM_{2}N_{2}, (N_{1}, N_{2}))
= S_{2}(m + (mM_{2} − 1)N_{2}, (N_{1}, N_{2}))S_{1}(m + (mM_{2} − 1)N_{2}, N_{1})^{−1}
= S_{2}(m + (mM_{2} − 1)N_{2}, (N_{1}, N_{2}))S_{1} (\frac{m + (mM_{2} − 1)}{N_{1}})^{−1}

with

S_{1}(x) = 2 sin πx.

Hence, inductively we have

S_{2}(mM_{1}N_{1}, (N_{1}, N_{2})) = S_{2}(m, (N_{1}, N_{2})) ∏_{l=0}^{mM_{2}−1} S_{1} (\frac{m + lN_{2}}{N_{1}} )^{−1} .

Hence we get

S_{2}(m, (N_{1}, N_{2})) = S_{2}(mM_{1}N_{1}, (N_{1}, N_{2})) ∏_{l=0}^{mM_{2}−1} S_{1} (\frac{m + lN_{2}}{N_{1}} ) .

On the other hand

S_{2}(mM_{1}N_{1}, (N_{1}, N_{2})) = S_{2}((mM_{1}^{−1})N_{1}, (N_{1}, N_{2}))S_{1}((mM_{1} − 1)N_{1}, N_{2})^{−1}
= S_{2}(N_{1}, (N_{1}, N_{2})) ∏_{k=1}^{mM_{1}−1} S_{1} (\frac{kN_{1}}{N_{2}} )^{−1}
= \sqrt{\frac{N_{2}}{N_{1}}} ∏_{k=1}^{mM_{1}−1} S_{1} (\frac{kN_{1}}{N_{2}} )^{−1} .

Thus

S_{2}(m, (N_{1}, N_{2})) = \sqrt{\frac{N_{2}}{N_{1}}} ∏_{l=0}^{mM_{2}−1} S_{1} (\frac{m + lN_{2}}{N_{1}} ) ∏_{k=1}^{mM_{1}−1} S_{1} (\frac{kN_{1}}{N_{2}} )^{−1} ,

which is an algebraic number in Q (sin \frac{2π}{N_{i}} , cos \frac{2π}{N_{i}} , \sqrt{N_{i}} | i = 1, 2).

\underline{(2) (N_{1}, N_{2}) = 2 case:}

Since (\frac{N_{1}}{2} , \frac{N_{2}}{2} ) = 1, the case of even m is reduced to (1) via

S_{2}(m, (N_{1}, N_{2})) = S_{2} (\frac{m}{2}, (\frac{N_{1}}{2} , \frac{N_{2}}{2} )) .

Now let m be an odd integer. Since either \frac{N_{1}}{2} or \frac{N_{2}}{2} is odd, we may assume that \frac{N_{1}}{2} is odd without the loss of generality. Then, m − \frac{N_{1}}{2} is even, so we may express

m − \frac{N_{1}}{2} = k_{1}N_{1} + k_{2}N_{2}

for integers k_{1} and k_{2}. Hence the algebraicity of

S_{2}(m, (N_{1}, N_{2})) = S_{2} (\frac{N_{1}}{2} + k_{1}N_{1} + k_{2}N_{2}, (N_{1}, N_{2}))

is reduced to

S_{2} (\frac{N_{1}}{2} , (N_{1}, N_{2})) = \sqrt{2}

via the (quasi-)periodicity exactly similar to the case (1).

Proof of Examples.: Calculation of S_{2}(m, (4, 6)) for m = 1, 2, ..., 9.

From the reflection relation

S_{2}(ω_{1} + ω_{2} − x, (ω_{1}, ω_{2})) = S_{2}(x, (ω_{1}, ω_{2}))^{−1}

it is sufficient to deal with the cases with m = 1, ..., 5. First, cases m = 2, 3 are shown from Theorem 1(2):

S_{2} (\frac{ω_{i}}{2} , (ω_{1}, ω_{2})) = \sqrt{2}.

Secondly, cases m = 4 and m = 5 are deduced from Theorem 1(4) and (1), respectively:

S_{2} (ω_{1}, (ω_{1}, ω_{2})) = \sqrt{\frac{ω_{2}}{ω_{1}}} .

and

S_{2} (\frac{ω_{1} + ω_{2}}{2} , (ω_{1}, ω_{2})) = 1.

Lastly, it remains to show the case m = 1. Let x = 1, ω_{1} = 4, ω_{2} = 6 in the periodicity

S_{2}(x + ω_{1}, (ω_{1}, ω_{2})) = S_{2}(x, (ω_{1}, ω_{2}))S_{1}(x, ω_{2})^{−1}
= S_{2}(x, (ω_{1}, ω_{2})) (2 sin \frac{πx}{ω_{2}})^{−1} .

Then we have

S_{2}(5, (4, 6)) = S_{2}(1, (4, 6)) (2 sin \frac{π}{6} )^{−1}
= S_{2}(1, (4, 6)).

Hence

S_{2}(1, (4, 6)) = S_{2}(5, (4, 6)) = 1.

6 Proof of Theorem 6

Notice that (1) is seen from

S_{2}(n, (1, 6n)) = S_{2}(n + 1, (1, 6n))
= \sqrt{6n} (∏_{k=1}^{n} 2 sin \frac{kπ}{6n} )^{−1} .

Hence, using

\frac{k}{n} ≤ 2 sin \frac{kπ}{6n} ≤ \frac{kπ}{3n}

for k = 1, ..., n, we have

\frac{n!}{n^{n}} ≤ ∏_{k=1}^{n} 2 sin \frac{kπ}{6n} ≤ (\frac{π}{3} )^{n} \frac{n!}{n^{n}} .

Thus

\sqrt{6}\frac{n^{n+ \frac{1}{2}}}{n!} (\frac{3}{π} )^{n} ≤ S_{2}(n, (1, 6n)) ≤ \sqrt{6}\frac{n^{n+ \frac{1}{2}}}{n!} .

So, the asymptotic formula of Stirling

\sqrt{6}\frac{n^{n+ \frac{1}{2}}}{n!} ∼ \sqrt{\frac{3}{π}} e^{n}

and

\sqrt{6}\frac{n^{n+ \frac{1}{2}}}{n!} (\frac{3}{π} )^{n} ∼ \sqrt{\frac{3}{π}} (\frac{3e}{π})^{n}

give (2) and (3).

O y x ω_{1} + ω_{2}_{\frac{ω_{1}+ω_{2}}{2}} y = (\frac{S′}{S} )′ (x) Figure 1.

22

O y  x α  β  ω_{1} + ω_{2}_{\frac{ω_{1}+ω_{2}}{2}} y = \frac{S′}{S} (x) Figure 2.

23 O y

x

1

\sqrt{2}

ω_{1} + ω_{2} \frac{ω_{1}+ω_{2}}{2} β α \frac{ω_{1}}{2} \frac{ω_{2}}{2}

ω_{1} + \frac{ω_{2}}{2}

ω_{2} + \frac{ω_{1}}{2}

\frac{\sqrt{2}}{2}

y = S_{2}(x, (ω_{1}, ω_{2})) (0 < ω_{1} ≤ ω_{2})

Figure 3.

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Zeta Institute 2-5-27, Hayabuchi, Tsuzuki-ku, Yokohama 224-0025, Japan koyama@tmtv.ne.jp

Department of Mathematics, Tokyo Institute of Technology Oo-okayama, Meguro-ku, Tokyo 152-8551, Japan kurokawa@math.titech.ac.jp