Introduction
The study of zeros of quasimodular forms has largely developed in two directions: the distribution of zeros and the transcendence of zeros. In this paper, we study the distribution and transcendence of zeros of quasimodular forms defined by iterated sums.
The zeros of Eisenstein series and RSD method
Rankin and Swinnerton-Dyer proved in [25 ] that, for even k ≥ 4 k \ge4 k ≥ 4
G k ( τ ) : = ∑ m , n ∈ Z ( m , n ) ≠ ( 0 , 0 ) 1 ( m τ + n ) k G_k(\tau) := \sum_{\substack{m,n\in\mathbb{Z}\\(m,n)\ne(0,0)}} \frac{1}{(m\tau+n)^k} G k ( τ ) := m , n ∈ Z ( m , n ) = ( 0 , 0 ) ∑ ( m τ + n ) k 1 in the standard fundamental domain
F : = { τ ∈ H ∣ − 1 / 2 ≤ Re τ ≤ 0 , ∣ τ ∣ ≥ 1 } ∪ { τ ∈ H ∣ 0 < Re τ < 1 / 2 , ∣ τ ∣ > 1 } \mathcal{F} := \{\tau\in\mathbb{H}\mid-1/2 \le\operatorname{Re}\tau\le0,\ |\tau| \ge1\} \cup\{\tau\in\mathbb{H}\mid0 < \operatorname{Re}\tau< 1/2,\ |\tau| > 1\} F := { τ ∈ H ∣ − 1/2 ≤ Re τ ≤ 0 , ∣ τ ∣ ≥ 1 } ∪ { τ ∈ H ∣ 0 < Re τ < 1/2 , ∣ τ ∣ > 1 } lie on the arc of the unit circle given by { e i t ∣ π / 2 ≤ t ≤ 2 π / 3 } \{e^{it}\mid\pi/2 \le t \le2\pi/3\} { e i t ∣ π /2 ≤ t ≤ 2 π /3 } H \mathbb{H} H F k : R → R F_k:\mathbb{R}\to\mathbb{R} F k : R → R F k ( t ) = e i k t / 2 G k ( e i t ) F_k(t)=e^{ikt/2}G_k(e^{it}) F k ( t ) = e ik t /2 G k ( e i t ) m ∈ [ k / 4 , k / 3 ] m\in[k/4,k/3] m ∈ [ k /4 , k /3 ] ( − 1 ) m F k ( 2 m π / k ) > 0 (-1)^mF_k(2m\pi/k)>0 ( − 1 ) m F k ( 2 mπ / k ) > 0 G k G_k G k k / 12 k/12 k /12
Miezaki, Nozaki, and Shigezumi in [22 ] applied RSD method to the study of zeros of Eisenstein series for the Fricke groups Γ 0 ∗ ( 2 ) \Gamma^*_0(2) Γ 0 ∗ ( 2 ) Γ 0 ∗ ( 3 ) \Gamma^*_0(3) Γ 0 ∗ ( 3 ) [12 ] adapted the RSD method to prove that, for even k ≥ 4 k \ge4 k ≥ 4 G k ′ G'_k G k ′ F \mathcal{F} F Re ( τ ) = − 1 / 2 \operatorname{Re}(\tau)=-1/2 Re ( τ ) = − 1/2 Γ 0 ∗ ( 2 ) \Gamma^*_0(2) Γ 0 ∗ ( 2 )
As another interesting development, Wittersum and Ringeling in [viR25] obtained a result analogous to that of [12 ] for Eisenstein series of odd weight, which are not even quasimodular forms.
Our Theorem 1.1 is regarded as an analogue of their result. However, a novel feature of our result is that it is obtained without using RSD method employed in their work.
The zeros of depth one quasimodular forms
Although relatively little is known about the zeros of quasimodular forms of arbitrary depth, several results have been obtained in the depth-one case.
El Basraoui and Sebbar [8 ] studied the zeros of G 2 G_2 G 2 { τ ∈ H ∣ − 1 / 2 ≤ Re ( τ ) < 1 / 2 } \{\tau\in\mathbb{H}\mid-1/2\le\operatorname{Re}(\tau)<1/2\} { τ ∈ H ∣ − 1/2 ≤ Re ( τ ) < 1/2 } S L 2 ( Z ) SL_2(\mathbb{Z}) S L 2 ( Z ) [15 ] and, independently, Wood and Young [33 ] refined these results by showing that each Ford circle contains exactly one simple zero of G 2 G_2 G 2 G 2 G_2 G 2 [29 ] , [5 ] , and [21 ] .
Valence formulas for derivatives of Eisenstein series, which provide typical examples of depth-one quasimodular forms, were subsequently established in [12 Sug26] . Van Ittersum and Ringeling [31 ] established a valence formula for arbitrary quasimodular forms of depth one and also considered quasimodular forms of arbitrary depth. More recently, Im and Lee [16 ] studied real zeros of depth-one quasimodular forms and obtained several results extending those of [12 ] .
quasimodular forms and multiple Eisenstein series
quasimodular forms were introduced by Kaneko and Zagier [20 ] . The ring of quasimodular forms for S L 2 ( Z ) SL_{2}(\mathbb{Z}) S L 2 ( Z ) M ~ = C [ G 2 , G 4 , G 6 ] \widetilde{\mathcal{M}}=\mathbb{C}[G_{2},G_{4},G_{6}] M = C [ G 2 , G 4 , G 6 ] G 2 , G 4 , G 6 G_{2},G_{4},G_{6} G 2 , G 4 , G 6 C \mathbb{C} C
C [ X , Y , Z ] ⟶ M ~ , P ( X , Y , Z ) ⟼ P ( G 2 , G 4 , G 6 ) , \mathbb{C}[X,Y,Z]\longrightarrow\widetilde{\mathcal{M}},\qquad P(X,Y,Z)\longmapsto P(G_{2},G_{4},G_{6}), C [ X , Y , Z ] ⟶ M , P ( X , Y , Z ) ⟼ P ( G 2 , G 4 , G 6 ) , is an isomorphism. In particular, every quasimodular form admits a unique expression as a polynomial in G 2 , G 4 , G 6 G_{2},G_{4},G_{6} G 2 , G 4 , G 6 G 2 G_{2} G 2 X X X C [ X , Y , Z ] \mathbb{C}[X,Y,Z] C [ X , Y , Z ] f f f k k k f = P ( G 2 , G 4 , G 6 ) f=P(G_{2},G_{4},G_{6}) f = P ( G 2 , G 4 , G 6 ) P ( X , Y , Z ) ∈ C [ X , Y , Z ] P(X,Y,Z)\in\mathbb{C}[X,Y,Z] P ( X , Y , Z ) ∈ C [ X , Y , Z ] k k k P ( t 2 X , t 4 Y , t 6 Z ) = t k P ( X , Y , Z ) P(t^{2}X,t^{4}Y,t^{6}Z)=t^{k}P(X,Y,Z) P ( t 2 X , t 4 Y , t 6 Z ) = t k P ( X , Y , Z ) f f f P P P X X X
Multiple Eisenstein series were introduced in [11 ] as a generalization of the ordinary Eisenstein series G k G_{k} G k r ∈ Z > 0 r\in\mathbb{Z}_{>0} r ∈ Z > 0 k 1 , … , k r ∈ Z ≥ 2 k_{1},\ldots,k_{r}\in\mathbb{Z}_{\geq2} k 1 , … , k r ∈ Z ≥ 2 τ ∈ H \tau\in\mathbb{H} τ ∈ H Z N : = { n ∈ Z ∣ ∣ n ∣ < N } \mathbb{Z}_{N}:=\{n\in\mathbb{Z}\mid|n|<N\} Z N := { n ∈ Z ∣ ∣ n ∣ < N } ≺ \prec ≺ Z τ + Z \mathbb{Z}\tau+\mathbb{Z} Z τ + Z
0 ≺ m τ + n ⟺ d e f { m > 0 , m = 0 and n > 0. 0\prec m\tau+n\overset{\mathrm{def}}{\Longleftrightarrow}
\begin{cases}
m>0,\\
m=0\ \text{and}\ n>0.
\end{cases} 0 ≺ m τ + n ⟺ def { m > 0 , m = 0 and n > 0. and
m 1 τ + n 1 ≺ m 2 τ + n 2 ⟺ d e f 0 ≺ ( m 2 − m 1 ) τ + ( n 2 − n 1 ) . m_{1}\tau+n_{1}\prec m_{2}\tau+n_{2}\overset{\mathrm{def}}{\Longleftrightarrow}0\prec(m_{2}-m_{1})\tau+(n_{2}-n_{1}). m 1 τ + n 1 ≺ m 2 τ + n 2 ⟺ def 0 ≺ ( m 2 − m 1 ) τ + ( n 2 − n 1 ) . The multiple Eisenstein series of depth r r r ( k 1 , … , k r ) (k_{1},\ldots,k_{r}) ( k 1 , … , k r )
G k 1 , … , k r ( τ ) : = lim M → ∞ lim N → ∞ ∑ w s ∈ M Z τ + Z N 0 ≺ w 1 ≺ ⋯ ≺ w r 1 w 1 k 1 ⋯ w r k r . G_{k_{1},\ldots,k_{r}}(\tau):=\lim_{M\to\infty}\lim_{N\to\infty}\sum_{\substack{w_{s}\in M\mathbb{Z}\tau+\mathbb{Z}_{N}\\0\prec w_{1}\prec\cdots\prec w_{r}}}\frac{1}{w_{1}^{k_{1}}\cdots w_{r}^{k_{r}}}. G k 1 , … , k r ( τ ) := M → ∞ lim N → ∞ lim w s ∈ M Z τ + Z N 0 ≺ w 1 ≺ ⋯ ≺ w r ∑ w 1 k 1 ⋯ w r k r 1 . The space spanned by these functions has been studied from various perspectives; see, for example, [4 ] and the references therein. In this paper, we consider only multiple Eisenstein series G { 2 } n G_{\{2\}^{n}} G { 2 } n [14 ] , as the definition of G { k } n G_{\{k\}^{n}} G { k } n
∑ n = 0 ∞ G { k } n ( τ ) X n = exp ( ∑ m = 1 ∞ ( − 1 ) m − 1 G k m ( τ ) m X m ) . (1) \sum_{n=0}^{\infty}G_{\{k\}^{n}}(\tau)X^{n}=\exp\left(\sum_{m=1}^{\infty}(-1)^{m-1}\frac{G_{km}(\tau)}{m}X^{m}\right). \tag*{(1)} n = 0 ∑ ∞ G { k } n ( τ ) X n = exp ( m = 1 ∑ ∞ ( − 1 ) m − 1 m G k m ( τ ) X m ) . (1) Here, for an integer k ≥ 2 k\geq2 k ≥ 2
G k ( τ ) : = ζ ( k ) + ( 2 π i ) k ( k − 1 ) ! ∑ n = 1 ∞ σ k − 1 ( n ) q n , where σ k − 1 ( n ) = ∑ d ∣ n d k − 1 G_{k}(\tau):=\zeta(k)+\frac{(2\pi i)^{k}}{(k-1)!}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^{n},
\qquad\text{where}\qquad
\sigma_{k-1}(n)=\sum_{\substack{d\mid n}}d^{k-1} G k ( τ ) := ζ ( k ) + ( k − 1 )! ( 2 π i ) k n = 1 ∑ ∞ σ k − 1 ( n ) q n , where σ k − 1 ( n ) = d ∣ n ∑ d k − 1 and q = e 2 π i τ q=e^{2\pi i\tau} q = e 2 π i τ 2 G k = G ~ k 2G_{k}=\widetilde{G}_{k} 2 G k = G k k k k
G { 2 } 0 = 1 , G 2 , 2 = 1 2 ( G 2 2 − G 4 ) , G 2 , 2 , 2 = 1 6 G 2 3 − 1 2 G 2 G 4 + 1 3 G 6 . G_{\{2\}^{0}}=1,\qquad G_{2,2}=\frac{1}{2}(G_{2}^{2}-G_{4}),\qquad G_{2,2,2}=\frac{1}{6}G_{2}^{3}-\frac{1}{2}G_{2}G_{4}+\frac{1}{3}G_{6}. G { 2 } 0 = 1 , G 2 , 2 = 2 1 ( G 2 2 − G 4 ) , G 2 , 2 , 2 = 6 1 G 2 3 − 2 1 G 2 G 4 + 3 1 G 6 . It follows from (1) that
d d G 2 G { 2 } n = G { 2 } n − 1 (2) \frac{d}{dG_{2}}G_{\{2\}^{n}}=G_{\{2\}^{n-1}} \tag*{(2)} d G 2 d G { 2 } n = G { 2 } n − 1 (2) and hence G { 2 } n G_{\{2\}^{n}} G { 2 } n 2 n 2n 2 n n n n
Q [ G 2 , G 4 , G 6 ] = Q [ G 2 , G 2 , 2 , G 2 , 2 , 2 ] . \mathbb{Q}[G_{2},G_{4},G_{6}]=\mathbb{Q}[G_{2},G_{2,2},G_{2,2,2}]. Q [ G 2 , G 4 , G 6 ] = Q [ G 2 , G 2 , 2 , G 2 , 2 , 2 ] . Transcendence of quasimodular forms
Transcendence questions for quasimodular forms have been studied from various perspectives. A classical Schneider theorem [27 ] states that if τ ∈ H \tau\in\mathbb{H} τ ∈ H j ( τ ) j(\tau) j ( τ ) [23 ] established a remarkable algebraic independence result for Eisenstein series. These results have become fundamental tools in the transcendence theory of quasimodular forms. These ideas have been further developed to study special values and zeros of quasimodular forms.
Summary of our results
In this article, we are interested in the zeros of the quasimodular form G { 2 } n G_{\{2\}^n} G { 2 } n 2 n 2n 2 n n n n G { 2 } n G_{\{2\}^n} G { 2 } n n n n η 3 \eta^3 η 3
G { 2 } n ( τ ) = G { 2 } n ( τ + 1 ) and G { 2 } n ( − τ ‾ ) = G { 2 } n ( τ ) ‾ G_{\{2\}^n}(\tau) = G_{\{2\}^n}(\tau+1) \quad\text{and}\quad G_{\{2\}^n}(-\overline{\tau}) = \overline{G_{\{2\}^n}(\tau)} G { 2 } n ( τ ) = G { 2 } n ( τ + 1 ) and G { 2 } n ( − τ ) = G { 2 } n ( τ ) that the zeros of G { 2 } n G_{\{2\}^n} G { 2 } n 1 1 1
Theorem 1.1. For each integer n > 0 n>0 n > 0 G { 2 } n G_{\{2\}^n} G { 2 } n n n n { i y ∣ y ∈ R > 0 } \{iy \mid y \in\mathbb{R}_{>0}\} { i y ∣ y ∈ R > 0 } { 1 / 2 + i y ∣ y ∈ R > 0 } \{1/2+iy \mid y \in\mathbb{R}_{>0}\} { 1/2 + i y ∣ y ∈ R > 0 }
The proof of the result does not use RSD method, which has been highly effective in previous studies. This is a notable difference from related works. A key ingredient in our proof is a result of Kwaśnicki [19 ] concerning bell-shaped functions.
Interlacing of zeros arises naturally in the study of Eisenstein series. Nozaki [24 ] proved that the zeros of E k E_k E k E k + 12 E_{k+12} E k + 12 [10 ] to pairs E k E_k E k E k + a E_{k+a} E k + a A , B ⊂ R A,B\subset\mathbb{R} A , B ⊂ R ∣ ∣ A ∣ − ∣ B ∣ ∣ = 1 \bigl||A|-|B|\bigr|=1 ∣ A ∣ − ∣ B ∣ = 1 A A A B B B
Corollary 1.2. Set S 0 ( n ) : = { y ∈ R > 0 : G { 2 } n ( i y ) = 0 } S_0^{(n)}:=\{y\in\mathbb{R}_{>0}:G_{\{2\}^n}(iy)=0\} S 0 ( n ) := { y ∈ R > 0 : G { 2 } n ( i y ) = 0 } S 1 / 2 ( n ) : = { y ∈ R > 0 : G { 2 } n ( 1 / 2 + i y ) = 0 } S_{1/2}^{(n)}:=\{y\in\mathbb{R}_{>0}:G_{\{2\}^n}(1/2+iy)=0\} S 1/2 ( n ) := { y ∈ R > 0 : G { 2 } n ( 1/2 + i y ) = 0 } n ≥ 1 n\ge1 n ≥ 1 ∗ ∈ { 0 , 1 / 2 } \ast\in\{0,1/2\} ∗ ∈ { 0 , 1/2 } S ∗ ( n ) S_\ast^{(n)} S ∗ ( n ) S ∗ ( n + 1 ) S_\ast^{(n+1)} S ∗ ( n + 1 )
Let i y 1 ( 0 , n ) , … , i y n ( 0 , n ) iy_1^{(0,n)},\ldots,iy_n^{(0,n)} i y 1 ( 0 , n ) , … , i y n ( 0 , n ) 1 2 + i y 1 ( 1 / 2 , n ) , … , 1 2 + i y n ( 1 / 2 , n ) \frac{1}{2}+iy_1^{(1/2,n)},\ldots,\frac{1}{2}+iy_n^{(1/2,n)} 2 1 + i y 1 ( 1/2 , n ) , … , 2 1 + i y n ( 1/2 , n ) y 1 ( ∗ , n ) > ⋯ > y n ( ∗ , n ) y_1^{(\ast,n)}>\cdots>y_n^{(\ast,n)} y 1 ( ∗ , n ) > ⋯ > y n ( ∗ , n ) ∗ ∈ { 0 , 1 / 2 } \ast\in\{0,1/2\} ∗ ∈ { 0 , 1/2 } § 4 , for each fixed r r r [17 ] implies that the sequence { y r ( ∗ , n ) / n } n ≥ r \{y_r^{(\ast,n)}/n\}_{n\ge r} { y r ( ∗ , n ) / n } n ≥ r
Theorem 1.3. Set κ = 1 \kappa=1 κ = 1 ∗ = 0 \ast=0 ∗ = 0 κ = 2 \kappa=2 κ = 2 ∗ = 1 / 2 \ast=1/2 ∗ = 1/2 r > 0 r>0 r > 0
lim n → ∞ y r ( ∗ , n ) n = 1 π κ r log ( 2 κ r + 1 2 κ r − 1 ) . \lim_{n\to\infty}\frac{y_r^{(\ast,n)}}{n}=\frac{1}{\pi\kappa r}\log\left(\frac{2\kappa r+1}{2\kappa r-1}\right). n → ∞ lim n y r ( ∗ , n ) = π κ r 1 log ( 2 κ r − 1 2 κ r + 1 ) . More precisely, we have
lim n → ∞ ( y r ( ∗ , n ) − ( n + 1 2 ) 1 π κ r log ( 2 κ r + 1 2 κ r − 1 ) ) = 0. \lim_{n\to\infty}\left(y_r^{(\ast,n)}-\left(n+\frac{1}{2}\right)\frac{1}{\pi\kappa r}\log\left(\frac{2\kappa r+1}{2\kappa r-1}\right)\right)=0. n → ∞ lim ( y r ( ∗ , n ) − ( n + 2 1 ) π κ r 1 log ( 2 κ r − 1 2 κ r + 1 ) ) = 0. More generally, we have the following.
Theorem 1.4. For any integers r > 0 r>0 r > 0 j j j { z r , j ( n ) } n ≥ r \{z_{r,j}^{(n)}\}_{n\ge r} { z r , j ( n ) } n ≥ r G { 2 } n G_{\{2\}^n} G { 2 } n
lim n → ∞ ( z r , j ( n ) − ( j r + i ( n + 1 2 ) 1 π r log ( 2 r + 1 2 r − 1 ) ) ) = 0. \lim_{n\to\infty}\left(z_{r,j}^{(n)}-\left(\frac{j}{r}+i\left(n+\frac{1}{2}\right)\frac{1}{\pi r}\log\left(\frac{2r+1}{2r-1}\right)\right)\right)=0. n → ∞ lim ( z r , j ( n ) − ( r j + i ( n + 2 1 ) π r 1 log ( 2 r − 1 2 r + 1 ) ) ) = 0. The following two theorems generalize the results of [8 ] .
Theorem 1.5. For each integer n > 0 n>0 n > 0 G { 2 } n G_{\{2\}^n} G { 2 } n
Theorem 1.6. Let f f f k k k k / 2 k/2 k /2 f f f f f f S L 2 ( Z ) \mathrm{SL}_2(\mathbb{Z}) SL 2 ( Z ) G { 2 } n G_{\{2\}^n} G { 2 } n n > 0 n>0 n > 0
The following conjecture follows immediately from the transformation law of G 2 G_2 G 2 n = 1 n=1 n = 1 [8 Proposition 3.3] ). However, for general n n n n = 2 n=2 n = 2
Conjecture 1.7. For each integer n > 0 n>0 n > 0 G { 2 } n G_{\{2\}n} G { 2 } n S L 2 ( Z ) \mathrm{SL}_2(\mathbb{Z}) SL 2 ( Z )
As a consequence of [13 Theorem 1] , every zero of G k G_k G k k ≥ 4 k \ge4 k ≥ 4 [18 ] conjectured that all zeros of G 2 G_2 G 2 [7 ] or [23 ] , we obtain the following result. However, it remains unknown whether G { 2 } n G_{\{2\}n} G { 2 } n § 6.3 .
Theorem 1.8. Let f ∈ Q ‾ [ G 2 , G 4 , G 6 ] f \in\overline{\mathbb{Q}}[G_2,G_4,G_6] f ∈ Q [ G 2 , G 4 , G 6 ] k k k k / 2 k/2 k /2 f f f H \mathbb{H} H n > 0 n>0 n > 0 G { 2 } n G_{\{2\}n} G { 2 } n
Furthermore, by the Gelfond–Schneider theorem [9 , 26 ] , log a / π \log a/\pi log a / π a > 0 a>0 a > 0 a ≠ 1 a\ne1 a = 1
1 π κ r log ( 2 κ r + 1 2 κ r − 1 ) , κ ∈ { 1 , 2 } , \frac{1}{\pi\kappa r}\log\left(\frac{2\kappa r+1}{2\kappa r-1}\right), \qquad\kappa\in\{1,2\}, π κ r 1 log ( 2 κ r − 1 2 κ r + 1 ) , κ ∈ { 1 , 2 } , are transcendental (see Corollary 4.4).
Finally, we discuss the location of the zeros of G { 2 } n G_{\{2\}^n} G { 2 } n a c ∈ Q \frac{a}{c} \in\mathbb{Q} c a ∈ Q c > 0 c > 0 c > 0 a c \frac{a}{c} c a a c \frac{a}{c} c a 1 2 c 2 \frac{1}{2c^2} 2 c 2 1 [15 ] and [33 ] should also hold for G { 2 } n G_{\{2\}^n} G { 2 } n G { 2 } n G_{\{2\}^n} G { 2 } n n ≥ 3 n \ge3 n ≥ 3 G { 2 } n G_{\{2\}^n} G { 2 } n n n n G { 2 } n G_{\{2\}^n} G { 2 } n G { 2 } n G_{\{2\}^n} G { 2 } n G 2 , 2 G_{2,2} G 2 , 2 Figure 1 illustrates the zeros of G { 2 } 5 G_{\{2\}^5} G { 2 } 5 n = 5 n = 5 n = 5 ( n + 1 ) / 2 (n+1)/2 ( n + 1 ) /2
Figure 1. Zeros of G { 2 } 5 G_{\{2\}5} G { 2 } 5
Theorem 1.9. Each Ford circle contains exactly two distinct (and simple) zeros of G 2 , 2 G_{2,2} G 2 , 2
The set { G 2 , G 4 , G 6 } \{G_2,G_4,G_6\} { G 2 , G 4 , G 6 } G 4 G_4 G 4 G 6 G_6 G 6 { G 2 , G 2 , 2 , G 2 , 2 , 2 } \{G_2,G_{2,2},G_{2,2,2}\} { G 2 , G 2 , 2 , G 2 , 2 , 2 }
Organization of the paper
In § 4 , we give a proof of Theorem 1.1. That is achieved by showing that η ( i y ) 3 \eta(iy)^3 η ( i y ) 3 § 5 , we give proofs of Theorems 1.3 and 1.4. This is obtained by explicitly examining the Fourier expansion of η ( τ ) 3 \eta(\tau)^3 η ( τ ) 3 § 6 , we prove Theorems 1.5 , 1.6 and 1.8, and in § 7 , we prove Theorem 1.9. These results are established as applications of previous work.
Use of AI
The authors used ChatGPT 5.6 Sol as a supplementary tool in developing some of the mathematical statements and proofs presented in this work. While several ideas, computations, and intermediate arguments were explored through discussions with the model, the mathematical results, their formulation, and the final proofs are due to the authors. All statements and proofs were independently checked and verified by the authors, who take full responsibility for the content of this paper.
Acknowledgments
The authors would like to thank Professor Abdellah Sebbar for pointing out an issue in an earlier version of this paper. The first named author expresses his gratitude to his academic advisor Professor Masanobu Kaneko and Professor Hiroyuki Ochiai. The first named author was supported by JST SPRING, Japan Grant Number JPMJSP2136. The second named author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (RS-2025-25415913).
Bell-shaped functions and number of zeros
In this section, we prove Theorem 1.1. Before giving the proof, we first recall the notion of bell-shaped functions and some of their properties, following [19 , 17 ] .
Recall the bell-shaped functions
We say that a function f : R → R f:\mathbb{R}\to\mathbb{R} f : R → R n n n n + 1 n+1 n + 1 m m m x 1 < x 2 < ⋯ < x m x_1 < x_2 < \cdots< x_m x 1 < x 2 < ⋯ < x m R \mathbb{R} R f ( x j ) f ( x j + 1 ) < 0 f(x_j)f(x_{j+1}) < 0 f ( x j ) f ( x j + 1 ) < 0 j = 1 , … , m − 1 j = 1,\ldots,m-1 j = 1 , … , m − 1 f f f f ′ ( x ) ≠ 0 f'(x) \ne0 f ′ ( x ) = 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 f f f n n n n n n
Definition. A smooth function f : R → R f:\mathbb{R}\to\mathbb{R} f : R → R strictly bell-shaped if, for every n = 0 , 1 , 2 , … n = 0,1,2,\ldots n = 0 , 1 , 2 , … n n n f ( n ) f^{(n)} f ( n ) ± ∞ \pm\infty ± ∞ f ( n ) f^{(n)} f ( n ) n n n f : R → R f:\mathbb{R}\to\mathbb{R} f : R → R f f f R \mathbb{R} R weakly bell-shaped if, for every t > 0 t > 0 t > 0
( f ∗ G t ) ( x ) = ∫ − ∞ ∞ f ( y ) G t ( x − y ) d y (f * \mathcal{G}_t)(x) = \int_{-\infty}^{\infty} f(y)\mathcal{G}_t(x-y)\,dy ( f ∗ G t ) ( x ) = ∫ − ∞ ∞ f ( y ) G t ( x − y ) d y is well-defined and strictly bell-shaped. Here, G t \mathfrak{G}_t G t
G t ( x ) = 1 4 π t e − x 2 / ( 4 t ) . \mathfrak{G}_t(x) = \frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}. G t ( x ) = 4 π t 1 e − x 2 / ( 4 t ) . Theorem 2.1 (Corollary 4.4 in [19 ] ). Every strictly bell-shaped function is weakly bell-shaped.
Theorem 2.2 (Theorem 1.1 in [19 ] ). Suppose that f f f ± ∞ \pm\infty ± ∞ ∞ \infty ∞ − ∞ -\infty − ∞ ξ ∈ R ∖ { 0 } \xi\in\mathbb{R} \setminus\{0\} ξ ∈ R ∖ { 0 } f f f
L f ( i ξ ) : = ∫ − ∞ ∞ e − i ξ x f ( x ) d x = exp ( − a ξ 2 − i b ξ + c + ∫ − ∞ ∞ K ( i ξ , s ) φ ( s ) d s ) . (3) \mathcal{L}f(i\xi) := \int_{-\infty}^{\infty} e^{-i\xi x}f(x)\,\mathrm{d}x = \exp\left(-a\xi^2 - ib\xi+ c + \int_{-\infty}^{\infty} K(i\xi,s)\varphi(s)\,\mathrm{d}s\right). \tag*{(3)} L f ( i ξ ) := ∫ − ∞ ∞ e − i ξ x f ( x ) d x = exp ( − a ξ 2 − ib ξ + c + ∫ − ∞ ∞ K ( i ξ , s ) φ ( s ) d s ) . (3) where
K ( i ξ , s ) : = 1 i ξ + s − ( 1 s − i ξ s 2 ) 1 R ∖ ( − 1 , 1 ) ( s ) K(i\xi,s) := \frac{1}{i\xi+s} - \left(\frac{1}{s} - \frac{i\xi}{s^2}\right)\mathbf{1}_{\mathbb{R}\setminus(-1,1)}(s) K ( i ξ , s ) := i ξ + s 1 − ( s 1 − s 2 i ξ ) 1 R ∖ ( − 1 , 1 ) ( s ) (with the former integral understood as an improper integral); here a ≥ 0 a \ge0 a ≥ 0 b , c ∈ R b,c \in\mathbb{R} b , c ∈ R φ : R → R \varphi:\mathbb{R}\to\mathbb{R} φ : R → R
(a) For every k ∈ Z k \in\mathbb{Z} k ∈ Z φ ( s ) − k \varphi(s)-k φ ( s ) − k k = 0 k=0 k = 0 s = 0 s=0 s = 0
φ ( s ) ≥ 0 for s > 0 , φ ( s ) ≤ 0 for s < 0. \varphi(s) \ge0 \quad\text{for } s>0,\qquad\varphi(s) \le0 \quad\text{for } s<0. φ ( s ) ≥ 0 for s > 0 , φ ( s ) ≤ 0 for s < 0. (b) We have
( ∫ − ∞ − 1 + ∫ 1 ∞ ) ∣ φ ( s ) ∣ ∣ s ∣ 3 d s < ∞ . \left(\int_{-\infty}^{-1}+\int_1^\infty\right)\frac{|\varphi(s)|}{|s|^3}\,\mathrm{d}s < \infty. ( ∫ − ∞ − 1 + ∫ 1 ∞ ) ∣ s ∣ 3 ∣ φ ( s ) ∣ d s < ∞. (c) We have
∫ − 1 1 Re L f ( i ξ ) d ξ < ∞ and lim ξ → 0 Im L f ( i ξ ) = 0. \int_{-1}^{1}\operatorname{Re}\mathcal{L}f(i\xi)\,\mathrm{d}\xi< \infty\qquad\text{and} \qquad\lim_{\xi\to0}\operatorname{Im}\mathcal{L}f(i\xi)=0. ∫ − 1 1 Re L f ( i ξ ) d ξ < ∞ and ξ → 0 lim Im L f ( i ξ ) = 0. Then f f f f f f f f f a , b , c a,b,c a , b , c φ \varphi φ L f ( i ξ ) \mathcal{L}f(i\xi) L f ( i ξ ) (3) ) correspond to some weakly bell-shaped function f f f b b b
Theorem 2.3 (Corollary 1.6 in [17 ] ). If f f f p p p q q q n = 1 , 2 , … n=1,2,\ldots n = 1 , 2 , … f ( n ) f^{(n)} f ( n ) [ p n , q n ] [pn,qn] [ p n , q n ]
Proof of thm:num-zero
First, we give the expression of G { 2 } n G_{\{2\}^n} G { 2 } n η \eta η
η ( τ ) : = q 1 / 24 ∏ m = 1 ∞ ( 1 − q m ) , q : = e 2 π i τ . \eta(\tau):=q^{1/24}\prod_{m=1}^{\infty}(1-q^m),\qquad q:=e^{2\pi i\tau}. η ( τ ) := q 1/24 m = 1 ∏ ∞ ( 1 − q m ) , q := e 2 π i τ . Let D D D q d d q = 1 2 π i d d τ q\frac{\mathrm{d}}{\mathrm{d}q}=\frac{1}{2\pi i}\frac{\mathrm{d}}{\mathrm{d}\tau} q d q d = 2 π i 1 d τ d
Lemma 2.4. For each integer n ≥ 0 n\ge0 n ≥ 0
G { 2 } n ( τ ) = 8 n π 2 n ( 2 n + 1 ) ! D n ( η ( τ ) 3 ) η ( τ ) 3 . (4) G_{\{2\}^n}(\tau)=\frac{8^n\pi^{2n}}{(2n+1)!}\frac{D^n(\eta(\tau)^3)}{\eta(\tau)^3}. \tag*{(4)} G { 2 } n ( τ ) = ( 2 n + 1 )! 8 n π 2 n η ( τ ) 3 D n ( η ( τ ) 3 ) . (4) Proof. The equation is essentially the same as Equation (3.3) in [2 ] (see also [3 ] ):
2 n ( 2 n + 1 ) ! D n ( η ( τ ) 3 ) η ( τ ) 3 = 1 ( 2 π i ) 2 n ∑ ( 1 m 1 , … , n m n ) ⊢ n ∏ j = 1 n 1 m j ! ( − 1 j G 2 j ( τ ) ) m j . \frac{2^n}{(2n+1)!}\frac{D^n(\eta(\tau)^3)}{\eta(\tau)^3}
=
\frac{1}{(2\pi i)^{2n}}
\sum_{(1^{m_1},\ldots,n^{m_n})\vdash n}
\prod_{j=1}^{n}\frac{1}{m_j!}\left(-\frac{1}{j}G_{2j}(\tau)\right)^{m_j}. ( 2 n + 1 )! 2 n η ( τ ) 3 D n ( η ( τ ) 3 ) = ( 2 π i ) 2 n 1 ( 1 m 1 , … , n m n ) ⊢ n ∑ j = 1 ∏ n m j ! 1 ( − j 1 G 2 j ( τ ) ) m j . Here, ⊢ \vdash ⊢ ( 1 m 1 , … , n m n ) (1^{m_1},\ldots,n^{m_n}) ( 1 m 1 , … , n m n ) n n n (1) , we have
∑ n = 0 ∞ G { 2 } n ( τ ) X n = exp ( ∑ m = 1 ∞ ( − 1 ) m − 1 G 2 m ( τ ) m X m ) = ∑ k = 0 ∞ ( ∑ ( 1 m 1 , … , k m k ) ⊢ k ∏ j = 1 k 1 m j ! ( − 1 j G 2 j ( τ ) ) m j ( − X ) k ) , \begin{aligned}
\sum_{n=0}^{\infty}G_{\{2\}n}(\tau)X^n=\exp\left(\sum_{m=1}^{\infty}(-1)^{m-1}\frac{G_{2m}(\tau)}{m}X^m\right) \\
&=\sum_{k=0}^{\infty}\left(\sum_{(1^{m_1},\ldots,k^{m_k})\vdash k}\prod_{j=1}^{k}\frac{1}{m_j!}\left(-\frac{1}{j}G_{2j}(\tau)\right)^{m_j}(-X)^k\right),
\end{aligned} n = 0 ∑ ∞ G { 2 } n ( τ ) X n = exp ( m = 1 ∑ ∞ ( − 1 ) m − 1 m G 2 m ( τ ) X m ) = k = 0 ∑ ∞ ( 1 m 1 , … , k m k ) ⊢ k ∑ j = 1 ∏ k m j ! 1 ( − j 1 G 2 j ( τ ) ) m j ( − X ) k , where the second equality follows from the generating function for Pólya’s cycle index polynomials; see, for example, Lemma 2.1 in [1 ] . Thus, we have
G { 2 } n ( τ ) = ( − 1 ) n ∑ ( 1 m 1 , … , n m n ) ⊢ n ∏ j = 1 n 1 m j ! ( − 1 j G 2 j ( τ ) ) m j = 8 n π 2 n ( 2 n + 1 ) ! D n ( η ( τ ) 3 ) η ( τ ) 3 . G_{\{2\}n}(\tau)=(-1)^n\sum_{(1^{m_1},\ldots,n^{m_n})\vdash n}\prod_{j=1}^{n}\frac{1}{m_j!}\left(-\frac{1}{j}G_{2j}(\tau)\right)^{m_j}=\frac{8^n\pi^{2n}}{(2n+1)!}\frac{D^n(\eta(\tau)^3)}{\eta(\tau)^3}. G { 2 } n ( τ ) = ( − 1 ) n ( 1 m 1 , … , n m n ) ⊢ n ∑ j = 1 ∏ n m j ! 1 ( − j 1 G 2 j ( τ ) ) m j = ( 2 n + 1 )! 8 n π 2 n η ( τ ) 3 D n ( η ( τ ) 3 ) . This completes the proof. □
By Lemma 2.4 and the fact that the η \eta η H \mathbb{H} H G { 2 } n G_{\{2\}n} G { 2 } n D n ( η 3 ) D^n(\eta^3) D n ( η 3 )
η ( τ ) 3 = ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) q ( 2 m + 1 ) 2 / 8 . (5) \eta(\tau)^3=\sum_{m=0}^{\infty}(-1)^m(2m+1)q^{(2m+1)^2/8}. \tag*{(5)} η ( τ ) 3 = m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) q ( 2 m + 1 ) 2 /8 . (5) Define the functions F 0 , F 1 / 2 : R → R F_0,F_{1/2}:\mathbb{R}\to\mathbb{R} F 0 , F 1/2 : R → R
F 0 ( y ) : = { 4 π η ( 4 i y π ) 3 = 4 π ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) e − y ( 2 m + 1 ) 2 ( y > 0 ) , 0 ( y ≤ 0 ) . F_0(y):=
\begin{cases}
\frac{4}{\pi}\eta\left(\frac{4iy}{\pi}\right)^3=\frac{4}{\pi}\sum_{m=0}^{\infty}(-1)^m(2m+1)e^{-y(2m+1)^2} & (y>0),\\
0 & (y\le0).
\end{cases} F 0 ( y ) := { π 4 η ( π 4 i y ) 3 = π 4 ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) e − y ( 2 m + 1 ) 2 0 ( y > 0 ) , ( y ≤ 0 ) . and
F 1 / 2 ( y ) : = { 2 2 π e π i / 8 η ( 1 2 + 4 i y π ) 3 = 2 2 π ∑ m = 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ ( 2 m + 1 ) e − y ( 2 m + 1 ) 2 ( y > 0 ) , 0 ( y ≤ 0 ) . F_{1/2}(y):=
\begin{cases}
\frac{2\sqrt{2}}{\pi e^{\pi i/8}}\eta\left(\frac{1}{2}+\frac{4iy}{\pi}\right)^3=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^{\lfloor m/2\rfloor}(2m+1)e^{-y(2m+1)^2} & (y>0),\\
0 & (y\le0).
\end{cases} F 1/2 ( y ) := { π e π i /8 2 2 η ( 2 1 + π 4 i y ) 3 = π 2 2 ∑ m = 0 ∞ ( − 1 ) ⌊ m /2 ⌋ ( 2 m + 1 ) e − y ( 2 m + 1 ) 2 0 ( y > 0 ) , ( y ≤ 0 ) . Since η 3 \eta^3 η 3 3 / 2 3/2 3/2 α \alpha α
lim τ → α D n ( η 3 ) ( τ ) = 0. \lim_{\tau\to\alpha}D^n(\eta^3)(\tau)=0. τ → α lim D n ( η 3 ) ( τ ) = 0. Hence, the functions F 0 F_0 F 0 F 1 / 2 F_{1/2} F 1/2 R \mathbb{R} R F 0 ( y ) F_0(y) F 0 ( y ) F 1 / 2 ( y ) F_{1/2}(y) F 1/2 ( y ) G { 2 } n G_{\{2\}n} G { 2 } n n n n Re ( τ ) = 1 / 2 \operatorname{Re}(\tau)=1/2 Re ( τ ) = 1/2
Lemma 2.5. Define
φ 0 ( s ) = ∑ m = 0 ∞ 1 [ ( 2 m + 1 ) 2 , ∞ ) ( s ) . \varphi_0(s)=\sum_{m=0}^{\infty}\mathbf{1}_{[(2m+1)^2,\infty)}(s). φ 0 ( s ) = m = 0 ∑ ∞ 1 [( 2 m + 1 ) 2 , ∞ ) ( s ) . Then, F 0 F_0 F 0 φ 0 \varphi_0 φ 0 Theorem 2.2 , and we have
L F 0 ( i ξ ) = exp ( − π 2 8 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s ) . \mathcal{L}F_0(i\xi)=\exp\left(-\frac{\pi^2}{8}\xi+\int_{-\infty}^{\infty}K(i\xi,s)\varphi_0(s)\,ds\right). L F 0 ( i ξ ) = exp ( − 8 π 2 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s ) . Consequently, F 0 F_0 F 0 F 0 F_0 F 0
Proof. We only verify that F 0 F_0 F 0 φ 0 \varphi_0 φ 0 (3) and condition (c) in Theorem 2.2 . The remaining conditions are straightforward to check. For ξ ∈ R \xi\in\mathbb{R} ξ ∈ R
L F 0 ( i ξ ) = ∫ − ∞ ∞ e − i ξ y F 0 ( y ) d y = 4 π lim ε → + 0 ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) ∫ ε ∞ e − ( i ξ + ( 2 m + 1 ) 2 ) y d y = 4 π lim ε → + 0 e − i ξ ε ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 i ξ + ( 2 m + 1 ) 2 e − ( 2 m + 1 ) 2 ε . \begin{aligned}
\mathcal{L}F_0(i\xi) &= \int_{-\infty}^{\infty} e^{-i\xi y}F_0(y)\,\mathrm{d}y = \frac{4}{\pi}\lim_{\varepsilon\to+0}\sum_{m=0}^{\infty}(-1)^m(2m+1)\int_{\varepsilon}^{\infty}e^{-(i\xi+(2m+1)^2)y}\,\mathrm{d}y \\
&= \frac{4}{\pi}\lim_{\varepsilon\to+0}e^{-i\xi\varepsilon}\sum_{m=0}^{\infty}(-1)^m\frac{2m+1}{i\xi+(2m+1)^2}e^{-(2m+1)^2\varepsilon}.
\end{aligned} L F 0 ( i ξ ) = ∫ − ∞ ∞ e − i ξ y F 0 ( y ) d y = π 4 ε → + 0 lim m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) ∫ ε ∞ e − ( i ξ + ( 2 m + 1 ) 2 ) y d y = π 4 ε → + 0 lim e − i ξ ε m = 0 ∑ ∞ ( − 1 ) m i ξ + ( 2 m + 1 ) 2 2 m + 1 e − ( 2 m + 1 ) 2 ε . We now justify the interchange of the limit and the infinite sum in the last line. Using the identity
2 m + 1 i ξ + ( 2 m + 1 ) 2 = 1 2 m + 1 − i ξ ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) , \frac{2m+1}{i\xi+(2m+1)^2}=\frac{1}{2m+1}-\frac{i\xi}{(2m+1)(i\xi+(2m+1)^2)}, i ξ + ( 2 m + 1 ) 2 2 m + 1 = 2 m + 1 1 − ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) i ξ , it suffices to consider separately
lim ε → + 0 ∑ m = 0 ∞ ( − 1 ) m 1 2 m + 1 e − ( 2 m + 1 ) 2 ε \lim_{\varepsilon\to+0}\sum_{m=0}^{\infty}(-1)^m\frac{1}{2m+1}e^{-(2m+1)^2\varepsilon} ε → + 0 lim m = 0 ∑ ∞ ( − 1 ) m 2 m + 1 1 e − ( 2 m + 1 ) 2 ε and
lim ε → + 0 ∑ m = 0 ∞ ( − 1 ) m i ξ ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) e − ( 2 m + 1 ) 2 ε . \lim_{\varepsilon\to+0}\sum_{m=0}^{\infty}(-1)^m\frac{i\xi}{(2m+1)(i\xi+(2m+1)^2)}e^{-(2m+1)^2\varepsilon}. ε → + 0 lim m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) i ξ e − ( 2 m + 1 ) 2 ε . For the second expression, the limit and the infinite sum may be interchanged by the Weierstrass M M M ε ≥ 0 \varepsilon\ge0 ε ≥ 0
∣ i ξ ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) e − ( 2 m + 1 ) 2 ε ∣ ≤ ∣ ξ ∣ ( 2 m + 1 ) 3 , \left|\frac{i\xi}{(2m+1)(i\xi+(2m+1)^2)}e^{-(2m+1)^2\varepsilon}\right|\leq\frac{|\xi|}{(2m+1)^3}, ( 2 m + 1 ) ( i ξ + ( 2 m + 1 ) 2 ) i ξ e − ( 2 m + 1 ) 2 ε ≤ ( 2 m + 1 ) 3 ∣ ξ ∣ , and
∑ m = 0 ∞ ∣ ξ ∣ ( 2 m + 1 ) 3 < ∞ . \sum_{m=0}^{\infty}\frac{|\xi|}{(2m+1)^3}<\infty. m = 0 ∑ ∞ ( 2 m + 1 ) 3 ∣ ξ ∣ < ∞. For the first expression, define
b m ( ε ) : = 1 2 m + 1 e − ( 2 m + 1 ) 2 ε . b_m(\varepsilon):=\frac{1}{2m+1}e^{-(2m+1)^2\varepsilon}. b m ( ε ) := 2 m + 1 1 e − ( 2 m + 1 ) 2 ε . For each ε ≥ 0 \varepsilon\geq0 ε ≥ 0 { b m ( ε ) } m ≥ 0 \{b_m(\varepsilon)\}_{m\geq0} { b m ( ε ) } m ≥ 0 m m m
∣ ∑ m = M ∞ ( − 1 ) m b m ( ε ) ∣ ≤ b M ( ε ) ≤ 1 2 M + 1 → 0 as M → ∞ . \left|\sum_{m=M}^{\infty}(-1)^m b_m(\varepsilon)\right|\leq b_M(\varepsilon)\leq\frac{1}{2M+1}\to0\quad\text{as }M\to\infty. m = M ∑ ∞ ( − 1 ) m b m ( ε ) ≤ b M ( ε ) ≤ 2 M + 1 1 → 0 as M → ∞. Therefore, the first series converges uniformly in ε ≥ 0 \varepsilon\geq0 ε ≥ 0
L F 0 ( i ξ ) = 4 π ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 i ξ + ( 2 m + 1 ) 2 . \mathcal{L}F_0(i\xi)=\frac{4}{\pi}\sum_{m=0}^{\infty}(-1)^m\frac{2m+1}{i\xi+(2m+1)^2}. L F 0 ( i ξ ) = π 4 m = 0 ∑ ∞ ( − 1 ) m i ξ + ( 2 m + 1 ) 2 2 m + 1 . By the partial fraction expansion of 1 / cosh 1/\cosh 1/ cosh
L F 0 ( i ξ ) = 4 π ∑ m = 0 ∞ ( − 1 ) m 2 m + 1 i ξ + ( 2 m + 1 ) 2 = 1 cosh ( π 2 i ξ ) . \mathcal{L}F_0(i\xi)=\frac{4}{\pi}\sum_{m=0}^{\infty}(-1)^m\frac{2m+1}{i\xi+(2m+1)^2}=\frac{1}{\cosh\left(\frac{\pi}{2}\sqrt{i\xi}\right)}. L F 0 ( i ξ ) = π 4 m = 0 ∑ ∞ ( − 1 ) m i ξ + ( 2 m + 1 ) 2 2 m + 1 = cosh ( 2 π i ξ ) 1 . On the other hand, we have
∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s = ∑ m = 0 ∞ ∫ ( 2 m + 1 ) 2 ∞ ( 1 i ξ + s − 1 s + i ξ s 2 ) d s = ∑ m = 0 ∞ [ log ( 1 + i ξ s ) − i ξ s ] ( 2 m + 1 ) 2 ∞ = − ∑ m = 0 ∞ ( i ξ ( 2 m + 1 ) 2 − log ( 1 + i ξ ( 2 m + 1 ) 2 ) ) = i π 2 8 ξ − ∑ m = 0 ∞ log ( 1 + i ξ ( 2 m + 1 ) 2 ) . \begin{aligned}
\int_{-\infty}^{\infty} K(i\xi,s)\varphi_0(s)\,ds=\sum_{m=0}^{\infty}\int_{(2m+1)^2}^{\infty}\left(\frac{1}{i\xi+s}-\frac{1}{s}+\frac{i\xi}{s^2}\right)\,ds \\
&=\sum_{m=0}^{\infty}\left[\log\left(1+\frac{i\xi}{s}\right)-\frac{i\xi}{s}\right]_{(2m+1)^2}^{\infty} \\
&=-\sum_{m=0}^{\infty}\left(\frac{i\xi}{(2m+1)^2}-\log\left(1+\frac{i\xi}{(2m+1)^2}\right)\right) \\
&=i\frac{\pi^2}{8}\xi-\sum_{m=0}^{\infty}\log\left(1+\frac{i\xi}{(2m+1)^2}\right).
\end{aligned} ∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s = m = 0 ∑ ∞ ∫ ( 2 m + 1 ) 2 ∞ ( i ξ + s 1 − s 1 + s 2 i ξ ) d s = m = 0 ∑ ∞ [ log ( 1 + s i ξ ) − s i ξ ] ( 2 m + 1 ) 2 ∞ = − m = 0 ∑ ∞ ( ( 2 m + 1 ) 2 i ξ − log ( 1 + ( 2 m + 1 ) 2 i ξ ) ) = i 8 π 2 ξ − m = 0 ∑ ∞ log ( 1 + ( 2 m + 1 ) 2 i ξ ) . The first equality follows from Fubini’s theorem. Indeed, the estimate ∣ K ( i ξ , s ) ∣ ≤ ξ 2 / s 3 |K(i\xi,s)|\leq\xi^2/s^3 ∣ K ( i ξ , s ) ∣ ≤ ξ 2 / s 3 K ( i ξ , s ) φ 0 ( s ) K(i\xi,s)\varphi_0(s) K ( i ξ , s ) φ 0 ( s ) cosh \cosh cosh
exp ( − i π 2 8 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s ) = ∏ m = 0 ∞ ( 1 + i ξ ( 2 m + 1 ) 2 ) − 1 = 1 cosh ( π 2 i ξ ) = L F 0 ( i ξ ) . \exp\left(-i\frac{\pi^2}{8}\xi+\int_{-\infty}^{\infty}K(i\xi,s)\varphi_0(s)\,ds\right)=\prod_{m=0}^{\infty}\left(1+\frac{i\xi}{(2m+1)^2}\right)^{-1}=\frac{1}{\cosh\left(\frac{\pi}{2}\sqrt{i\xi}\right)}=\mathcal{L}F_0(i\xi). exp ( − i 8 π 2 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 0 ( s ) d s ) = m = 0 ∏ ∞ ( 1 + ( 2 m + 1 ) 2 i ξ ) − 1 = cosh ( 2 π i ξ ) 1 = L F 0 ( i ξ ) . Therefore, we get the representation (2.1). Condition (c) in Theorem 2.2 follows from cosh ( π 2 i ξ ) ≠ 0 \cosh\left(\frac{\pi}{2}\sqrt{i\xi}\right)\neq0 cosh ( 2 π i ξ ) = 0 ξ ∈ R \xi\in\mathbb{R} ξ ∈ R
Lemma 2.6. For each s ∈ C s\in\mathbb{C} s ∈ C
cosh ( s π ) cosh ( 2 s π ) = 2 2 π ∑ m = 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ 2 m + 1 ( 4 s ) 2 + ( 2 m + 1 ) 2 . \frac{\cosh( s\pi)}{\cosh(2s\pi)}=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^{\lfloor m/2\rfloor}\frac{2m+1}{(4s)^2+(2m+1)^2}. cosh ( 2 s π ) cosh ( s π ) = π 2 2 m = 0 ∑ ∞ ( − 1 ) ⌊ m /2 ⌋ ( 4 s ) 2 + ( 2 m + 1 ) 2 2 m + 1 . Proof. Using the exponential representation of cosh \cosh cosh
cosh ( s ) cosh ( 2 s ) = 1 2 2 ( 1 cosh ( s + π i / 4 ) + 1 cosh ( s − π i / 4 ) ) . \frac{\cosh(s)}{\cosh(2s)}=\frac{1}{2\sqrt{2}}\left(\frac{1}{\cosh(s+\pi i/4)}+\frac{1}{\cosh(s-\pi i/4)}\right). cosh ( 2 s ) cosh ( s ) = 2 2 1 ( cosh ( s + π i /4 ) 1 + cosh ( s − π i /4 ) 1 ) . Thus, by the partial fraction expansion of 1 / cosh 1/\cosh 1/ cosh
cosh ( π s ) cosh ( 2 π s ) = 2 π ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) ( 1 ( 2 s + i / 2 ) 2 + ( 2 m + 1 ) 2 + 1 ( 2 s − i / 2 ) 2 + ( 2 m + 1 ) 2 ) . \frac{\cosh(\pi s)}{\cosh(2\pi s)}=\frac{\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^m(2m+1)\left(\frac{1}{(2s+i/2)^2+(2m+1)^2}+\frac{1}{(2s-i/2)^2+(2m+1)^2}\right). cosh ( 2 π s ) cosh ( π s ) = π 2 m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) ( ( 2 s + i /2 ) 2 + ( 2 m + 1 ) 2 1 + ( 2 s − i /2 ) 2 + ( 2 m + 1 ) 2 1 ) . Using the identity
a ( 1 ( x + i / 2 ) 2 + a 2 + 1 ( x − i / 2 ) 2 + a 2 ) = 2 ( 2 a − 1 ( 2 x ) 2 + ( 2 a − 1 ) 2 + 2 a + 1 ( 2 x ) 2 + ( 2 a + 1 ) 2 ) , a\left(\frac{1}{(x+i/2)^2+a^2}+\frac{1}{(x-i/2)^2+a^2}\right)=2\left(\frac{2a-1}{(2x)^2+(2a-1)^2}+\frac{2a+1}{(2x)^2+(2a+1)^2}\right), a ( ( x + i /2 ) 2 + a 2 1 + ( x − i /2 ) 2 + a 2 1 ) = 2 ( ( 2 x ) 2 + ( 2 a − 1 ) 2 2 a − 1 + ( 2 x ) 2 + ( 2 a + 1 ) 2 2 a + 1 ) , which can be verified by putting both sides over a common denominator, we obtain
cosh ( π s ) cosh ( 2 π s ) = 2 2 π ∑ m = 0 ∞ ( − 1 ) m ( 4 m + 1 ( 4 s ) 2 + ( 4 m + 1 ) 2 + 4 m + 3 ( 4 s ) 2 + ( 4 m + 3 ) 2 ) = 2 2 π ∑ m = 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ 2 m + 1 ( 4 s ) 2 + ( 2 m + 1 ) 2 . \begin{aligned}
\frac{\cosh(\pi s)}{\cosh(2\pi s)}=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^m\left(\frac{4m+1}{(4s)^2+(4m+1)^2}+\frac{4m+3}{(4s)^2+(4m+3)^2}\right) \\
&=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^{\lfloor m/2\rfloor}\frac{2m+1}{(4s)^2+(2m+1)^2}.
\end{aligned} cosh ( 2 π s ) cosh ( π s ) = π 2 2 m = 0 ∑ ∞ ( − 1 ) m ( ( 4 s ) 2 + ( 4 m + 1 ) 2 4 m + 1 + ( 4 s ) 2 + ( 4 m + 3 ) 2 4 m + 3 ) = π 2 2 m = 0 ∑ ∞ ( − 1 ) ⌊ m /2 ⌋ ( 4 s ) 2 + ( 2 m + 1 ) 2 2 m + 1 . This completes the proof. □
Lemma 2.7. Define
φ 1 / 2 ( s ) = ∑ m = 0 ∞ 1 [ ( 2 m + 1 ) 2 , ∞ ) ( s ) − ∑ m = 0 ∞ 1 [ 4 ( 2 m + 1 ) 2 , ∞ ) ( s ) . \varphi_{1/2}(s)=\sum_{m=0}^{\infty}\mathbf{1}_{[(2m+1)^2,\infty)}(s)-\sum_{m=0}^{\infty}\mathbf{1}_{[4(2m+1)^2,\infty)}(s). φ 1/2 ( s ) = m = 0 ∑ ∞ 1 [( 2 m + 1 ) 2 , ∞ ) ( s ) − m = 0 ∑ ∞ 1 [ 4 ( 2 m + 1 ) 2 , ∞ ) ( s ) . Then, F 1 / 2 F_{1/2} F 1/2 φ 1 / 2 \varphi_{1/2} φ 1/2 Theorem 2.2 , and we have
L F 1 / 2 ( i ξ ) = exp ( − i 3 π 2 32 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 1 / 2 ( s ) d s ) . \mathcal{L}F_{1/2}(i\xi)=\exp\left(-i\frac{3\pi^2}{32}\xi+\int_{-\infty}^{\infty}K(i\xi,s)\varphi_{1/2}(s)\,ds\right). L F 1/2 ( i ξ ) = exp ( − i 32 3 π 2 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 1/2 ( s ) d s ) . Consequently, F 1 / 2 F_{1/2} F 1/2 F 1 / 2 F_{1/2} F 1/2
Proof. Since the proof is similar to that of Lemma 2.5 , we omit the details and only sketch the argument. The interchange of the limits is justified, by Lemma 2.6 , we obtain
L F 1 / 2 ( i ξ ) = ∫ − ∞ ∞ e − i ξ y F 1 / 2 ( y ) d y = 2 2 π ∑ m = 0 ∞ ∫ 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ ( 2 m + 1 ) e − ( i ξ + ( 2 m + 1 ) 2 ) y d y = 2 2 π ∑ m = 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ 2 m + 1 i ξ + ( 2 m + 1 ) 2 = cosh ( i ξ π / 4 ) cosh ( i ξ π / 2 ) . \begin{aligned}
\mathcal{L}F_{1/2}(i\xi)=\int_{-\infty}^{\infty}e^{-i\xi y}F_{1/2}(y)\,\mathrm{d}y=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}\int_{0}^{\infty}(-1)^{\lfloor m/2\rfloor}(2m+1)e^{-(i\xi+(2m+1)^2)y}\,\mathrm{d}y \\
&=\frac{2\sqrt{2}}{\pi}\sum_{m=0}^{\infty}(-1)^{\lfloor m/2\rfloor}\frac{2m+1}{i\xi+(2m+1)^2} \\
&=\frac{\cosh(\sqrt{i\xi}\pi/4)}{\cosh(\sqrt{i\xi}\pi/2)}.
\end{aligned} L F 1/2 ( i ξ ) = ∫ − ∞ ∞ e − i ξ y F 1/2 ( y ) d y = π 2 2 m = 0 ∑ ∞ ∫ 0 ∞ ( − 1 ) ⌊ m /2 ⌋ ( 2 m + 1 ) e − ( i ξ + ( 2 m + 1 ) 2 ) y d y = π 2 2 m = 0 ∑ ∞ ( − 1 ) ⌊ m /2 ⌋ i ξ + ( 2 m + 1 ) 2 2 m + 1 = cosh ( i ξ π /2 ) cosh ( i ξ π /4 ) . On the other hand, we have
∫ − ∞ ∞ K ( i ξ , s ) φ 1 / 2 ( s ) d s = i 3 π 2 32 ξ + ∑ m = 0 ∞ log ( 1 + i ξ 4 ( 2 m + 1 ) 2 1 + i ξ ( 2 m + 1 ) 2 ) . \int_{-\infty}^{\infty}K(i\xi,s)\varphi_{1/2}(s)\,\mathrm{d}s=i\frac{3\pi^2}{32}\xi+\sum_{m=0}^{\infty}\log\left(\frac{1+\frac{i\xi}{4(2m+1)^2}}{1+\frac{i\xi}{(2m+1)^2}}\right). ∫ − ∞ ∞ K ( i ξ , s ) φ 1/2 ( s ) d s = i 32 3 π 2 ξ + m = 0 ∑ ∞ log ( 1 + ( 2 m + 1 ) 2 i ξ 1 + 4 ( 2 m + 1 ) 2 i ξ ) . Thus, by the Weierstrass product for cosh, we have
exp ( − i 3 π 2 32 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 1 / 2 ( s ) d s ) = ∏ m = 0 ∞ ( 1 + i ξ 4 ( 2 m + 1 ) 2 1 + i ξ ( 2 m + 1 ) 2 ) = cosh ( i ξ π / 4 ) cosh ( i ξ π / 2 ) = L F 1 / 2 ( i ξ ) . \exp\left(-i\frac{3\pi^2}{32}\xi+\int_{-\infty}^{\infty}K(i\xi,s)\varphi_{1/2}(s)\,\mathrm{d}s\right)=\prod_{m=0}^{\infty}\left(\frac{1+\frac{i\xi}{4(2m+1)^2}}{1+\frac{i\xi}{(2m+1)^2}}\right)=\frac{\cosh(\sqrt{i\xi}\pi/4)}{\cosh(\sqrt{i\xi}\pi/2)}=\mathcal{L}F_{1/2}(i\xi). exp ( − i 32 3 π 2 ξ + ∫ − ∞ ∞ K ( i ξ , s ) φ 1/2 ( s ) d s ) = m = 0 ∏ ∞ ( 1 + ( 2 m + 1 ) 2 i ξ 1 + 4 ( 2 m + 1 ) 2 i ξ ) = cosh ( i ξ π /2 ) cosh ( i ξ π /4 ) = L F 1/2 ( i ξ ) . Therefore, we get the representation (2.1).
Theorem 1.1 (Restated). For each integer n > 0 n>0 n > 0 G { 2 } n G_{\{2\}^n} G { 2 } n n n n { i y ∣ y ∈ R > 0 } \{iy\mid y\in\mathbb{R}_{>0}\} { i y ∣ y ∈ R > 0 } { 1 / 2 + i y ∣ y ∈ R > 0 } \{1/2+iy\mid y\in\mathbb{R}_{>0}\} { 1/2 + i y ∣ y ∈ R > 0 }
Proof. Here, we prove only that G { 2 } n G_{\{2\}^n} G { 2 } n n n n { τ ∈ H ∣ Re ( τ ) = 1 / 2 } \{\tau\in\mathbb{H}\mid\operatorname{Re}(\tau)=1/2\} { τ ∈ H ∣ Re ( τ ) = 1/2 } F 0 ( n ) F_0^{(n)} F 0 ( n ) n n n ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
Since F 0 F_0 F 0 F 0 ( n ) F_0^{(n)} F 0 ( n ) n n n ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) F 0 ( n ) F_0^{(n)} F 0 ( n ) n n n ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) F 0 ( n ) F_0^{(n)} F 0 ( n ) n n n F 0 ( n ) F_0^{(n)} F 0 ( n ) 0 0 0 x → 0 x\to0 x → 0 x → ∞ x\to\infty x → ∞ F 0 ( n + 1 ) F_0^{(n+1)} F 0 ( n + 1 ) n + 2 n+2 n + 2
We now use the fact that, on ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) F 0 F_0 F 0 F 0 ( n + 1 ) F_0^{(n+1)} F 0 ( n + 1 ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
0 < x 1 < ⋯ < x n + 2 0<x_1<\cdots<x_{n+2} 0 < x 1 < ⋯ < x n + 2 be distinct zeros of F 0 ( n + 1 ) F_0^{(n+1)} F 0 ( n + 1 ) F 0 ( n + 1 ) F_0^{(n+1)} F 0 ( n + 1 ) 0 0 0 x → 0 x\to0 x → 0 x → ∞ x\to\infty x → ∞
( 0 , x 1 ) , ( x 1 , x 2 ) , … , ( x n + 1 , x n + 2 ) , ( x n + 2 , ∞ ) (0,x_1),\quad(x_1,x_2),\quad\ldots,\quad(x_{n+1},x_{n+2}),\quad(x_{n+2},\infty) ( 0 , x 1 ) , ( x 1 , x 2 ) , … , ( x n + 1 , x n + 2 ) , ( x n + 2 , ∞ ) contains a nonzero local extremum of F 0 ( n + 1 ) F_0^{(n+1)} F 0 ( n + 1 ) F 0 ( n + 2 ) F_0^{(n+2)} F 0 ( n + 2 ) F 0 ( n + 2 ) F_0^{(n+2)} F 0 ( n + 2 ) n + 3 n+3 n + 3 F 0 F_0 F 0 F 0 ( n ) F_0^{(n)} F 0 ( n ) n n n ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
It follows from Theorem 2.3 and Lemmas 2.5 and Lemma 2.7 that there exist constants M 0 , M 1 / 2 > 0 M_0,M_{1/2}>0 M 0 , M 1/2 > 0 n = 1 , 2 , … n=1,2,\ldots n = 1 , 2 , … F 0 ( n ) F_0^{(n)} F 0 ( n ) F 1 / 2 ( n ) F_{1/2}^{(n)} F 1/2 ( n ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) ( 0 , n M 0 ] (0,nM_0] ( 0 , n M 0 ] ( 0 , n M 1 / 2 ] (0,nM_{1/2}] ( 0 , n M 1/2 ] r r r y r ( ∗ , n ) / n n ≥ r y_r^{(\ast,n)}/n_{n\ge r} y r ( ∗ , n ) / n n ≥ r
Finally, we apply Theorem 1.1 to investigate the number of sign changes in the Fourier coefficients of G { 2 } n G_{\{2\}^n} G { 2 } n
We say that a sequence ( a n ) n ∈ Z (a_n)_{n\in\mathbb{Z}} ( a n ) n ∈ Z n n n n + 1 n+1 n + 1 m m m n 1 < n 2 < ⋯ < n m n_1<n_2<\cdots<n_m n 1 < n 2 < ⋯ < n m Z \mathbb{Z} Z a n j a n j + 1 < 0 a_{n_j}a_{n_{j+1}}<0 a n j a n j + 1 < 0 j = 1 , … , m − 1 j=1,\ldots,m-1 j = 1 , … , m − 1
Conjecture 2.8. Write the Fourier expansion of G { 2 } n G_{\{2\}n} G { 2 } n
G { 2 } n ( τ ) = ∑ m = 0 ∞ a m q m . G_{\{2\}n}(\tau)=\sum_{m=0}^{\infty}a_mq^m. G { 2 } n ( τ ) = m = 0 ∑ ∞ a m q m . Then, a m ≠ 0 a_m\ne0 a m = 0 m ≥ 0 m\ge0 m ≥ 0 ( a m ) m ≥ 0 (a_m)_{m\ge0} ( a m ) m ≥ 0 n n n
Lemma 2.9. If the sequence ( a m ) m ≥ 0 (a_m)_{m\ge0} ( a m ) m ≥ 0 n n n
f ( x ) = ∑ m = 0 ∞ a m x m f(x)=\sum_{m=0}^{\infty}a_mx^m f ( x ) = m = 0 ∑ ∞ a m x m has at most n n n
Proof. We proceed by induction on n n n n = 0 n=0 n = 0 n = μ − 1 n=\mu-1 n = μ − 1 ( a m ) m ≥ 0 (a_m)_{m\ge0} ( a m ) m ≥ 0 μ \mu μ a 0 ≠ 0 a_0\ne0 a 0 = 0 M M M
g ( x ) = x M + 1 d d x ( x − M f ( x ) ) = x M + 1 d d x ( x − M ∑ m = 0 ∞ a m x m ) = ∑ m = 0 ∞ b m x m . g(x)=x^{M+1}\frac{d}{dx}\left(x^{-M}f(x)\right)=x^{M+1}\frac{d}{dx}\left(x^{-M}\sum_{m=0}^{\infty}a_mx^m\right)=\sum_{m=0}^{\infty}b_mx^m. g ( x ) = x M + 1 d x d ( x − M f ( x ) ) = x M + 1 d x d ( x − M m = 0 ∑ ∞ a m x m ) = m = 0 ∑ ∞ b m x m . where b m = ( m − M ) a m b_m=(m-M)a_m b m = ( m − M ) a m ( b m ) m ≥ 0 (b_m)_{m\ge0} ( b m ) m ≥ 0 μ − 1 \mu-1 μ − 1 g g g μ − 1 \mu-1 μ − 1 x M + 1 > 0 x^{M+1}>0 x M + 1 > 0 x > 0 x>0 x > 0
d d x ( x − M f ( x ) ) \frac{d}{dx}\left(x^{-M}f(x)\right) d x d ( x − M f ( x ) ) also has at most μ − 1 \mu-1 μ − 1 x − M f ( x ) x^{-M}f(x) x − M f ( x ) μ \mu μ x − M ≠ 0 x^{-M}\ne0 x − M = 0 x > 0 x>0 x > 0 f ( x ) f(x) f ( x )
Corollary 2.10. Let ( a m ) m ≥ 0 (a_m)_{m\ge0} ( a m ) m ≥ 0 Conjecture 2.8 . Then, the sequence changes sign at least n n n
Proof. This follows immediately from Theorem 1.1 and Lemma 2.9 . □
Asymptotic of zeros
In this section, we give the proof of Theorems 1.3 and 1.4.
Lemma 3.1. Let x > 1 x>1 x > 1 ( a m ) m ≥ 0 (a_m)_{m\ge0} ( a m ) m ≥ 0 ( b m ) m ≥ 0 (b_m)_{m\ge0} ( b m ) m ≥ 0 n ≥ 1 n\ge1 n ≥ 1
A n = ∑ m = 0 ∞ a m x n b m , A_n=\sum_{m=0}^{\infty}a_mx^{nb_m}, A n = m = 0 ∑ ∞ a m x n b m , and assume that the series converges absolutely when n = 1 n=1 n = 1 k k k a k ≠ 0 a_k\ne0 a k = 0 b k > b m b_k>b_m b k > b m m ≠ k m\ne k m = k a m ≠ 0 a_m\ne0 a m = 0 n n n A n ≠ 0 A_n\ne0 A n = 0 sgn A n = sgn a k \operatorname{sgn}A_n=\operatorname{sgn}a_k sgn A n = sgn a k
Proof. Dividing A n A_n A n x n b k x^{nb_k} x n b k
x − n b k A n = a k + ∑ m ≥ 0 m ≠ k a m x − n ( b k − b m ) . x^{-nb_k}A_n=a_k+\sum_{\substack{m\ge0\\m\ne k}}a_mx^{-n(b_k-b_m)}. x − n b k A n = a k + m ≥ 0 m = k ∑ a m x − n ( b k − b m ) . For every m ≠ k m\ne k m = k a m ≠ 0 a_m\ne0 a m = 0 b k − b m > 0 b_k-b_m>0 b k − b m > 0
a m x − n ( b k − b m ) ⟶ 0 ( n → ∞ ) . a_mx^{-n(b_k-b_m)}\longrightarrow0\qquad(n\to\infty). a m x − n ( b k − b m ) ⟶ 0 ( n → ∞ ) . Moreover, for $n\ge 1,
∣ a m x − n ( b k − b m ) ∣ ≤ ∣ a m ∣ x − ( b k − b m ) = x − b k ∣ a m ∣ x b m . \left|a_mx^{-n(b_k-b_m)}\right|\le|a_m|x^{-(b_k-b_m)}=x^{-b_k}|a_m|x^{b_m}. a m x − n ( b k − b m ) ≤ ∣ a m ∣ x − ( b k − b m ) = x − b k ∣ a m ∣ x b m . Since
∑ m = 0 ∞ ∣ a m ∣ x b m < ∞ , \sum_{m=0}^{\infty}|a_m|x^{b_m}<\infty, m = 0 ∑ ∞ ∣ a m ∣ x b m < ∞ , the dominated convergence theorem yields
lim n → ∞ x − n b k A n = a k . \lim_{n\to\infty} x^{-nb_k}A_n=a_k. n → ∞ lim x − n b k A n = a k . Since x n b k > 0 x^{nb_k}>0 x n b k > 0 n n n A n ≠ 0 A_n\ne0 A n = 0 sgn A n = sgn a k \operatorname{sgn} A_n=\operatorname{sgn} a_k sgn A n = sgn a k
Theorem 1.3 (Restated on the imaginary axis). Write the zeros of G { 2 } ( n ) ( τ ) G_{\{2\}}^{(n)}(\tau) G { 2 } ( n ) ( τ ) i y 1 ( n ) , i y 2 ( n ) , … , i y n ( n ) iy_1^{(n)},iy_2^{(n)},\ldots,iy_n^{(n)} i y 1 ( n ) , i y 2 ( n ) , … , i y n ( n ) y 1 ( n ) > y 2 ( n ) > ⋯ > y n ( n ) > 0 y_1^{(n)}>y_2^{(n)}>\cdots>y_n^{(n)}>0 y 1 ( n ) > y 2 ( n ) > ⋯ > y n ( n ) > 0 r > 0 r>0 r > 0
lim n → ∞ ( y r ( n ) − ( n + 1 2 ) 1 π r log ( 2 r + 1 2 r − 1 ) ) = 0. \lim_{n\to\infty}\left(y_r^{(n)}-\left(n+\frac{1}{2}\right)\frac{1}{\pi r}\log\left(\frac{2r+1}{2r-1}\right)\right)=0. n → ∞ lim ( y r ( n ) − ( n + 2 1 ) π r 1 log ( 2 r − 1 2 r + 1 ) ) = 0. Proof. Let y ~ r ( n ) : = y r ( n ) / n \widetilde y_r^{(n)}:=y_r^{(n)}/n y r ( n ) := y r ( n ) / n (5) , y ~ r ( n ) \widetilde y_r^{(n)} y r ( n ) r r r
f n : R > 0 → R ; f n ( y ) : = ( π 4 ) n ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) 2 n + 1 e − π n y ( 2 m + 1 ) 2 / 4 . f_n:\mathbb{R}_{>0}\to\mathbb{R};\quad f_n(y):=\left(\frac{\pi}{4}\right)^n\sum_{m=0}^{\infty}(-1)^m(2m+1)^{2n+1}e^{-\pi n y(2m+1)^2/4}. f n : R > 0 → R ; f n ( y ) := ( 4 π ) n m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) 2 n + 1 e − π n y ( 2 m + 1 ) 2 /4 . Thus it suffices to show that
y ~ r ( n ) = ( 1 + 1 2 n ) 1 π r log ( 2 r + 1 2 r − 1 ) + o ( 1 n ) . \widetilde y_r^{(n)}=\left(1+\frac{1}{2n}\right)\frac{1}{\pi r}\log\left(\frac{2r+1}{2r-1}\right)+o\left(\frac{1}{n}\right). y r ( n ) = ( 1 + 2 n 1 ) π r 1 log ( 2 r − 1 2 r + 1 ) + o ( n 1 ) . Write λ m = π 4 ( 2 m + 1 ) 2 \lambda_m=\frac{\pi}{4}(2m+1)^2 λ m = 4 π ( 2 m + 1 ) 2 Φ m ( y ) = log λ m − λ m y \Phi_m(y)=\log\lambda_m-\lambda_m y Φ m ( y ) = log λ m − λ m y
f n ( y ) = ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) λ m n e − n y λ m = ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) e n Φ m ( y ) . f_n(y)=\sum_{m=0}^{\infty}(-1)^m(2m+1)\lambda_m^n e^{-ny\lambda_m}=\sum_{m=0}^{\infty}(-1)^m(2m+1)e^{n\Phi_m(y)}. f n ( y ) = m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) λ m n e − n y λ m = m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) e n Φ m ( y ) . For each integer r > 0 r>0 r > 0 x r ∈ R > 0 x_r\in\mathbb{R}_{>0} x r ∈ R > 0
Φ r − 1 ( x r ) = Φ r ( x r ) . \Phi_{r-1}(x_r)=\Phi_r(x_r). Φ r − 1 ( x r ) = Φ r ( x r ) . Equivalently, log ( λ r / λ r − 1 ) = ( λ r − λ r − 1 ) x r \log(\lambda_r/\lambda_{r-1})=(\lambda_r-\lambda_{r-1})x_r log ( λ r / λ r − 1 ) = ( λ r − λ r − 1 ) x r
x r = 1 π r log ( 2 r + 1 2 r − 1 ) . x_r=\frac{1}{\pi r}\log\left(\frac{2r+1}{2r-1}\right). x r = π r 1 log ( 2 r − 1 2 r + 1 ) . Note that x 1 > x 2 > ⋯ > 0 x_1>x_2>\cdots>0 x 1 > x 2 > ⋯ > 0
Φ r ( x ) − Φ r − 1 ( x ) = Φ r ( x ) − Φ r − 1 ( x ) − Φ r ( x r ) + Φ r − 1 ( x r ) = − λ r x + λ r − 1 x + λ r x r − λ r − 1 x r = ( λ r − λ r − 1 ) ( x r − x ) = 2 π r ( x r − x ) . (6) \begin{aligned}
\Phi_r(x)-\Phi_{r-1}(x)
&=\Phi_r(x)-\Phi_{r-1}(x)-\Phi_r(x_r)+\Phi_{r-1}(x_r)\\
&=-\lambda_r x+\lambda_{r-1}x+\lambda_r x_r-\lambda_{r-1}x_r\\
&=(\lambda_r-\lambda_{r-1})(x_r-x)=2\pi r(x_r-x). \tag*{(6)}
\end{aligned} Φ r ( x ) − Φ r − 1 ( x ) = Φ r ( x ) − Φ r − 1 ( x ) − Φ r ( x r ) + Φ r − 1 ( x r ) = − λ r x + λ r − 1 x + λ r x r − λ r − 1 x r = ( λ r − λ r − 1 ) ( x r − x ) = 2 π r ( x r − x ) . (6) Thus, for x r < y < x r − 1 x_r<y<x_{r-1} x r < y < x r − 1
Φ 0 ( y ) < ⋯ < Φ r − 2 ( y ) < Φ r − 1 ( y ) > Φ r ( y ) > Φ r + 1 ( y ) > ⋯ \Phi_0(y)<\cdots<\Phi_{r-2}(y)<\Phi_{r-1}(y)>\Phi_r(y)>\Phi_{r+1}(y)>\cdots Φ 0 ( y ) < ⋯ < Φ r − 2 ( y ) < Φ r − 1 ( y ) > Φ r ( y ) > Φ r + 1 ( y ) > ⋯ and
Φ 0 ( x r ) < ⋯ < Φ r − 2 ( x r ) < Φ r − 1 ( x r ) = Φ r ( x r ) > Φ r + 1 ( x r ) > ⋯ . \Phi_0(x_r)<\cdots<\Phi_{r-2}(x_r)<\Phi_{r-1}(x_r)=\Phi_r(x_r)>\Phi_{r+1}(x_r)>\cdots. Φ 0 ( x r ) < ⋯ < Φ r − 2 ( x r ) < Φ r − 1 ( x r ) = Φ r ( x r ) > Φ r + 1 ( x r ) > ⋯ . Let M r = Φ r − 1 ( x r ) = Φ r ( x r ) M_r=\Phi_{r-1}(x_r)=\Phi_r(x_r) M r = Φ r − 1 ( x r ) = Φ r ( x r )
F n , r ( t ) : = e − n M r f n ( x r + t n ) = ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) e n ( Φ m ( x r ) − M r ) e − λ m t . F_{n,r}(t):=e^{-nM_r}f_n\left(x_r+\frac{t}{n}\right)=\sum_{m=0}^{\infty}(-1)^m(2m+1)e^{n(\Phi_m(x_r)-M_r)}e^{-\lambda_m t}. F n , r ( t ) := e − n M r f n ( x r + n t ) = m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) e n ( Φ m ( x r ) − M r ) e − λ m t . Since Φ m ( x r ) − M r = 0 \Phi_m(x_r)-M_r=0 Φ m ( x r ) − M r = 0 m = r − 1 , r m=r-1,r m = r − 1 , r Φ m ( x r ) − M r < 0 \Phi_m(x_r)-M_r<0 Φ m ( x r ) − M r < 0 m ≠ r − 1 , r m\ne r-1,r m = r − 1 , r F n , r ( t ) F_{n,r}(t) F n , r ( t ) R > 0 \mathbb{R}_{>0} R > 0
F ~ r ( t ) = ( − 1 ) r ( ( 2 r + 1 ) e − λ r t − ( 2 r − 1 ) e − λ r − 1 t ) \widetilde F_r(t)=(-1)^r\left((2r+1)e^{-\lambda_r t}-(2r-1)e^{-\lambda_{r-1}t}\right) F r ( t ) = ( − 1 ) r ( ( 2 r + 1 ) e − λ r t − ( 2 r − 1 ) e − λ r − 1 t ) as n → ∞ n\to\infty n → ∞ F ~ r ( t ) \widetilde F_r(t) F r ( t )
t r = 1 2 π r log ( 2 r + 1 2 r − 1 ) = x r 2 . t_r=\frac{1}{2\pi r}\log\left(\frac{2r+1}{2r-1}\right)=\frac{x_r}{2}. t r = 2 π r 1 log ( 2 r − 1 2 r + 1 ) = 2 x r . and this zero is simple. Therefore, for sufficiently small ε > 0 \varepsilon> 0 ε > 0 n n n F n , r ( t ) F_{n,r}(t) F n , r ( t ) ( t r − ε , t r + ε ) (t_r-\varepsilon,t_r+\varepsilon) ( t r − ε , t r + ε ) t n , r t_{n,r} t n , r x n , r = x r + t n , r / n x_{n,r}=x_r+t_{n,r}/n x n , r = x r + t n , r / n f n ( y ) f_n(y) f n ( y )
lim n → ∞ n ( x n , r − x r ) = lim n → ∞ t n , r = t r \lim_{n\to\infty} n(x_{n,r}-x_r)=\lim_{n\to\infty}t_{n,r}=t_r n → ∞ lim n ( x n , r − x r ) = n → ∞ lim t n , r = t r and t r = x r / 2 t_r=x_r/2 t r = x r /2
x n , r = x r + x r 2 n + o ( 1 n ) . x_{n,r}=x_r+\frac{x_r}{2n}+o\left(\frac{1}{n}\right). x n , r = x r + 2 n x r + o ( n 1 ) . Finally, we prove that y ~ r ( n ) = x n , r \widetilde{y}^{(n)}_r=x_{n,r} y r ( n ) = x n , r n n n x n , r x_{n,r} x n , r r r r ε > 0 \varepsilon>0 ε > 0
U i = ( x i − ε , x i + ε ) , i = 1 , … , r U_i=(x_i-\varepsilon,x_i+\varepsilon),\quad i=1,\ldots,r U i = ( x i − ε , x i + ε ) , i = 1 , … , r are pairwise disjoint. By the above argument, for all sufficiently large n n n f n ( y ) f_n(y) f n ( y ) x n , j x_{n,j} x n , j U j U_j U j
I ⊂ ( x r − ε , ∞ ) ∖ ⋃ i = 1 r U i , I\subset(x_r-\varepsilon,\infty)\setminus\bigcup_{i=1}^{r}U_i, I ⊂ ( x r − ε , ∞ ) ∖ i = 1 ⋃ r U i , there exists an integer m m m 1 ≤ m ≤ r 1\leq m\leq r 1 ≤ m ≤ r
Φ 0 ( y ) < ⋯ < Φ m − 2 ( y ) < Φ m − 1 ( y ) > Φ m ( y ) > Φ m + 1 ( y ) > ⋯ \Phi_0(y)<\cdots<\Phi_{m-2}(y)<\Phi_{m-1}(y)>\Phi_m(y)>\Phi_{m+1}(y)>\cdots Φ 0 ( y ) < ⋯ < Φ m − 2 ( y ) < Φ m − 1 ( y ) > Φ m ( y ) > Φ m + 1 ( y ) > ⋯ for all y ∈ I y\in I y ∈ I Lemma 3.1 , f n f_n f n I I I n n n
y ~ r ( n ) = ( 1 + 1 2 n ) x r + o ( 1 n ) . \widetilde{y}^{(n)}_r=\left(1+\frac{1}{2n}\right)x_r+o\left(\frac{1}{n}\right). y r ( n ) = ( 1 + 2 n 1 ) x r + o ( n 1 ) . This completes the proof.
Theorem 1.3 (Restated on the half-line 1 2 + i R > 0 \frac{1}{2}+i\mathbb{R}_{>0} 2 1 + i R > 0 Write the zeros of G { 2 } n ( τ ) G_{\{2\}^n}(\tau) G { 2 } n ( τ ) { τ ∈ H ∣ Re ( τ ) = 1 / 2 } \{\tau\in\mathbb{H}\mid\operatorname{Re}(\tau)=1/2\} { τ ∈ H ∣ Re ( τ ) = 1/2 } 1 2 + i y 1 ( n ) , 1 2 + i y 2 ( n ) , … , 1 2 + i y n ( n ) \frac{1}{2}+iy^{(n)}_1,\frac{1}{2}+iy^{(n)}_2,\ldots,\frac{1}{2}+iy^{(n)}_n 2 1 + i y 1 ( n ) , 2 1 + i y 2 ( n ) , … , 2 1 + i y n ( n ) y 1 ( n ) > y 2 ( n ) > ⋯ > y n ( n ) > 0 y^{(n)}_1>y^{(n)}_2>\cdots>y^{(n)}_n>0 y 1 ( n ) > y 2 ( n ) > ⋯ > y n ( n ) > 0 r > 0 r>0 r > 0
lim n → ∞ ( y r ( n ) − ( n + 1 2 ) 1 2 π r log ( 4 r + 1 4 r − 1 ) ) = 0. \lim_{n\to\infty}\left(y^{(n)}_r-\left(n+\frac{1}{2}\right)\frac{1}{2\pi r}\log\left(\frac{4r+1}{4r-1}\right)\right)=0. n → ∞ lim ( y r ( n ) − ( n + 2 1 ) 2 π r 1 log ( 4 r − 1 4 r + 1 ) ) = 0. Proof. The proof is exactly the same as above, with f n f_n f n
g n : R > 0 → R ; g n ( y ) : = ( π 4 ) n ∑ m = 0 ∞ ( − 1 ) ⌊ m / 2 ⌋ ( 2 m + 1 ) 2 n + 1 e − π n y ( 2 m + 1 ) 2 / 4 . g_n:\mathbb{R}_{>0}\to\mathbb{R};\quad g_n(y):=\left(\frac{\pi}{4}\right)^n\sum_{m=0}^{\infty}(-1)^{\lfloor m/2\rfloor}(2m+1)^{2n+1}e^{-\pi n y(2m+1)^2/4}. g n : R > 0 → R ; g n ( y ) := ( 4 π ) n m = 0 ∑ ∞ ( − 1 ) ⌊ m /2 ⌋ ( 2 m + 1 ) 2 n + 1 e − π n y ( 2 m + 1 ) 2 /4 . Note that the only difference is the sign pattern of the terms in the infinite sum.
Theorem 1.4 (Restated). For any integers r > 0 r>0 r > 0 j j j { z r , j ( n ) } n ≥ r \{z^{(n)}_{r,j}\}_{n\geq r} { z r , j ( n ) } n ≥ r G { 2 } n G_{\{2\}^n} G { 2 } n
lim n → ∞ ( z r , j ( n ) − ( j r + i ( n + 1 2 ) 1 π r log ( 2 r + 1 2 r − 1 ) ) ) = 0. \lim_{n\to\infty}\left(z^{(n)}_{r,j}-\left(\frac{j}{r}+i\left(n+\frac{1}{2}\right)\frac{1}{\pi r}\log\left(\frac{2r+1}{2r-1}\right)\right)\right)=0. n → ∞ lim ( z r , j ( n ) − ( r j + i ( n + 2 1 ) π r 1 log ( 2 r − 1 2 r + 1 ) ) ) = 0. Proof. If we carry out the argument in the first half of the proof of Theorem 1.3 without restricting to the imaginary axis, then the function f n f_n f n
h n : H → C ; h n ( τ ) = ( π 4 ) n ∑ m = 0 ∞ ( − 1 ) m ( 2 m + 1 ) 2 n + 1 e n Φ m ( y ) + i λ m x , τ = x + i y h_n:\mathbb{H}\to\mathbb{C};\quad h_n(\tau)=\left(\frac{\pi}{4}\right)^n\sum_{m=0}^{\infty}(-1)^m(2m+1)^{2n+1}e^{n\Phi_m(y)+i\lambda_m x},\quad\tau=x+iy h n : H → C ; h n ( τ ) = ( 4 π ) n m = 0 ∑ ∞ ( − 1 ) m ( 2 m + 1 ) 2 n + 1 e n Φ m ( y ) + i λ m x , τ = x + i y Accordingly, for each r > 0 r>0 r > 0
H n , r ( x , t ) = e − n M r h n ( x + i ( x r + t n ) ) , H_{n,r}(x,t)=e^{-nM_r}h_n\left(x+i\left(x_r+\frac{t}{n}\right)\right), H n , r ( x , t ) = e − n M r h n ( x + i ( x r + n t ) ) , we find that H n , r ( x , t ) H_{n,r}(x,t) H n , r ( x , t )
H ~ r ( x , t ) = ( − 1 ) r ( ( 2 r + 1 ) e − λ r t + i λ r x − ( 2 r − 1 ) e − λ r − 1 t + i λ r − 1 x ) . \widetilde{H}_r(x,t)=(-1)^r\left((2r+1)e^{-\lambda_r t+i\lambda_r x}-(2r-1)e^{-\lambda_{r-1}t+i\lambda_{r-1}x}\right). H r ( x , t ) = ( − 1 ) r ( ( 2 r + 1 ) e − λ r t + i λ r x − ( 2 r − 1 ) e − λ r − 1 t + i λ r − 1 x ) . The zeros of H ~ r \widetilde{H}_r H r
x = j r and t = 1 2 π r log ( 2 r + 1 2 r − 1 ) . x=\frac{j}{r}\quad\text{and}\quad t=\frac{1}{2\pi r}\log\left(\frac{2r+1}{2r-1}\right). x = r j and t = 2 π r 1 log ( 2 r − 1 2 r + 1 ) . The same argument as in the proof of Theorem 1.3 then yields the desired sequence of zeros of G { 2 } n ( τ ) G_{\{2\}n}(\tau) G { 2 } n ( τ )
Global properties of zeros
In this section, we give the proof of Theorems 1.5, 1.6 and 1.8. Let E k ( τ ) : = 1 ζ ( k ) G k ( τ ) E_k(\tau):=\frac{1}{\zeta(k)}G_k(\tau) E k ( τ ) := ζ ( k ) 1 G k ( τ ) k ≥ 2 k\geq2 k ≥ 2 E k E_k E k E k E_k E k G k G_k G k
ζ ( k ) = − 1 2 ( 2 π i ) k B k k ! , \zeta(k)=-\frac{1}{2}\frac{(2\pi i)^kB_k}{k!}, ζ ( k ) = − 2 1 k ! ( 2 π i ) k B k , B k B_k B k k k k k > 0 k>0 k > 0 G k / ( 2 π i ) k G_k/(2\pi i)^k G k / ( 2 π i ) k E k E_k E k 2 π i 2\pi i 2 π i
Simplicity of zeros
Theorem 4.1 (Corollary 2 of Gun–Oesterlé [12 ] ). Let P P P Q [ X , Y , Z ] \mathbb{Q}[X,Y,Z] Q [ X , Y , Z ] a a a ψ P = P ( E 2 , E 4 , E 6 ) \psi_P=P(E_2,E_4,E_6) ψ P = P ( E 2 , E 4 , E 6 ) H \mathbb{H} H e e e R R R P P P ψ R ( a ) = 0 \psi_R(a)=0 ψ R ( a ) = 0 ψ R = R ( E 2 , E 4 , E 6 ) \psi_R=R(E_2,E_4,E_6) ψ R = R ( E 2 , E 4 , E 6 ) ψ R \psi_R ψ R a a a R R R P P P v R ( P ) v_R(P) v R ( P )
Theorem 1.5 (Restated). For each integer n > 0 n>0 n > 0 G { 2 } n G_{\{2\}n} G { 2 } n
Proof. Let P n ( X , Y , Z ) ∈ Q [ X , Y , Z ] P_n(X,Y,Z)\in\mathbb{Q}[X,Y,Z] P n ( X , Y , Z ) ∈ Q [ X , Y , Z ] G { 2 } n = P n ( G 2 , G 4 , G 6 ) G_{\{2\}n}=P_n(G_2,G_4,G_6) G { 2 } n = P n ( G 2 , G 4 , G 6 ) a ∈ H a\in\mathbb{H} a ∈ H G { 2 } n G_{\{2\}n} G { 2 } n e ≥ 2 e\geq2 e ≥ 2 R R R P n P_n P n
R ( G 2 , G 4 , G 6 ) ( a ) = 0 and v R ( P n ) = e ≥ 2. R(G_2,G_4,G_6)(a)=0\quad\text{and}\quad v_R(P_n)=e\geq2. R ( G 2 , G 4 , G 6 ) ( a ) = 0 and v R ( P n ) = e ≥ 2. Let S = P n / R e S=P_n/R^e S = P n / R e
∂ R ∂ X ≠ 0. \frac{\partial R}{\partial X}\ne0. ∂ X ∂ R = 0. Then, we have
∂ P n ∂ X = e R e − 1 S ∂ R ∂ X + R e ∂ S ∂ X . (7) \frac{\partial P_n}{\partial X}=eR^{e-1}S\frac{\partial R}{\partial X}+R^e\frac{\partial S}{\partial X}. \tag*{(7)} ∂ X ∂ P n = e R e − 1 S ∂ X ∂ R + R e ∂ X ∂ S . (7) Since R R R ∂ R ∂ X ≠ 0 \frac{\partial R}{\partial X}\ne0 ∂ X ∂ R = 0
v R ( ∂ P n ∂ X ) = e − 1. v_R\left(\frac{\partial P_n}{\partial X}\right)=e-1. v R ( ∂ X ∂ P n ) = e − 1. On the other hand, from (2) , we have
∂ P n ∂ X = P n − 1 . (8) \frac{\partial P_n}{\partial X}=P_{n-1}. \tag*{(8)} ∂ X ∂ P n = P n − 1 . (8) Thus, a a a G { 2 } n − 1 G_{\{2\}n-1} G { 2 } n − 1 e − 1 e-1 e − 1 a a a D ( G { 2 } n − 1 ) D(G_{\{2\}n-1}) D ( G { 2 } n − 1 ) e − 2 e-2 e − 2
D ( G { 2 } n − 1 ( τ ) ) = 8 n − 1 π 2 n − 2 ( 2 n − 1 ) ! D n ( η ( τ ) 3 ) η ( τ ) 3 − 8 n − 1 π 2 n − 2 ( 2 n − 1 ) ! D n − 1 ( η ( τ ) 3 ) η ( τ ) 3 D ( η ( τ ) 3 ) η ( τ ) 3 = 1 4 π 2 ( n ( 2 n + 1 ) G { 2 } n ( τ ) − 3 G 2 ( τ ) G { 2 } n − 1 ( τ ) ) . \begin{aligned}
D(G_{\{2\}n-1}(\tau))=\frac{8^{n-1}\pi^{2n-2}}{(2n-1)!}\frac{D^n(\eta(\tau)^3)}{\eta(\tau)^3}-\frac{8^{n-1}\pi^{2n-2}}{(2n-1)!}\frac{D^{n-1}(\eta(\tau)^3)}{\eta(\tau)^3}\frac{D(\eta(\tau)^3)}{\eta(\tau)^3} \\
&=\frac{1}{4\pi^2}\left(n(2n+1)G_{\{2\}n}(\tau)-3G_2(\tau)G_{\{2\}n-1}(\tau)\right).
\end{aligned} D ( G { 2 } n − 1 ( τ )) = ( 2 n − 1 )! 8 n − 1 π 2 n − 2 η ( τ ) 3 D n ( η ( τ ) 3 ) − ( 2 n − 1 )! 8 n − 1 π 2 n − 2 η ( τ ) 3 D n − 1 ( η ( τ ) 3 ) η ( τ ) 3 D ( η ( τ ) 3 ) = 4 π 2 1 ( n ( 2 n + 1 ) G { 2 } n ( τ ) − 3 G 2 ( τ ) G { 2 } n − 1 ( τ ) ) . Since a a a G 2 G { 2 } n − 1 G_2G_{\{2\}^{n-1}} G 2 G { 2 } n − 1 e − 1 e-1 e − 1 D ( G { 2 } n − 1 ) D(G_{\{2\}^{n-1}}) D ( G { 2 } n − 1 ) e − 2 e-2 e − 2 a a a G { 2 } n G_{\{2\}^n} G { 2 } n e − 2 e-2 e − 2 a a a G { 2 } n G_{\{2\}^n} G { 2 } n e e e
∂ R ∂ X = 0. \frac{\partial R}{\partial X}=0. ∂ X ∂ R = 0. Then, from (7) , R R R
∂ j P n ∂ X j ( j = 0 , 1 , 2 , … ) . \frac{\partial^j P_n}{\partial X^j}\qquad(j=0,1,2,\ldots). ∂ X j ∂ j P n ( j = 0 , 1 , 2 , … ) . Therefore, from (8) , R R R
∂ n P n ∂ X n = P 0 = 1. \frac{\partial^n P_n}{\partial X^n}=P_0=1. ∂ X n ∂ n P n = P 0 = 1. Hence R R R R ( G 2 , G 4 , G 6 ) ( a ) = 0 R(G_2,G_4,G_6)(a)=0 R ( G 2 , G 4 , G 6 ) ( a ) = 0
SL2(Z)-inequivalence of zeros
El Basraoui and Sebbar [8 ] showed that E 2 E_2 E 2 S L 2 ( Z ) SL_2(\mathbb{Z}) S L 2 ( Z ) [21 ] proved the analogous result for quasimodular forms of depth one. A similar statement holds for quasimodular forms of maximal depth.
Theorem 1.6 (Restated). Let f f f k k k k / 2 k/2 k /2 f f f f f f S L 2 ( Z ) SL_2(\mathbb{Z}) S L 2 ( Z ) G { 2 } n G_{\{2\}^n} G { 2 } n n > 0 n>0 n > 0
Proof. We prove this by following the same strategy as in the proof of Theorem 3.5 in [8 ] . From the quasimodular transformation, for τ ∈ H \tau\in\mathbb{H} τ ∈ H g = ( a b c d ) ∈ S L 2 ( Z ) g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}) g = ( a c b d ) ∈ S L 2 ( Z )
1 ( c τ + d ) k f ( g τ ) = ∑ r = 0 k / 2 ( c c τ + d ) r f r ( τ ) , where 0 ≠ f k / 2 ∈ C . (9) \frac{1}{(c\tau+d)^k}f(g\tau)=\sum_{r=0}^{k/2}\left(\frac{c}{c\tau+d}\right)^r f_r(\tau),\qquad\text{where }0\ne f_{k/2}\in\mathbb{C}. \tag*{(9)} ( c τ + d ) k 1 f ( g τ ) = r = 0 ∑ k /2 ( c τ + d c ) r f r ( τ ) , where 0 = f k /2 ∈ C . (9) Therefore, if the equation
Ψ ( v , τ ) : = ∑ r = 0 k / 2 ( τ + v ) k / 2 − r f r ( τ ) = 0 \Psi(v,\tau):=\sum_{r=0}^{k/2}(\tau+v)^{k/2-r}f_r(\tau)=0 Ψ ( v , τ ) := r = 0 ∑ k /2 ( τ + v ) k /2 − r f r ( τ ) = 0 admits a solution ( v , τ ) ∈ Q × H (v,\tau)\in\mathbb{Q}\times\mathbb{H} ( v , τ ) ∈ Q × H g τ g\tau g τ f ( τ ) f(\tau) f ( τ ) g = ( a b c d ) ∈ S L 2 ( Z ) g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb{Z}) g = ( a c b d ) ∈ S L 2 ( Z ) c ≠ 0 c\ne0 c = 0 v = d / c v=d/c v = d / c τ 0 \tau_0 τ 0 f ( τ ) f(\tau) f ( τ ) g 0 = ( a 0 b 0 c 0 d 0 ) ∈ S L 2 ( Z ) g_0=\begin{pmatrix}a_0&b_0\\c_0&d_0\end{pmatrix}\in SL_2(\mathbb{Z}) g 0 = ( a 0 c 0 b 0 d 0 ) ∈ S L 2 ( Z ) c 0 ≠ 0 c_0\ne0 c 0 = 0 ( d 0 / c 0 , g 0 − 1 τ 0 ) (d_0/c_0,g_0^{-1}\tau_0) ( d 0 / c 0 , g 0 − 1 τ 0 ) Ψ = 0 \Psi=0 Ψ = 0 (9) , multiplying by ( τ + d / c ) k / 2 (\tau+d/c)^{k/2} ( τ + d / c ) k /2 τ \tau τ
c − k / 2 f ′ ( g τ ) ( c τ + d ) k / 2 + 2 − c 1 − k / 2 k 2 f ( g τ ) ( c τ + d ) k / 2 + 1 = d d τ ∑ r = 0 k / 2 ( τ + d c ) k / 2 − r f r ( τ ) . c^{-k/2}\frac{f'(g\tau)}{(c\tau+d)^{k/2+2}}-c^{1-k/2}\frac{k}{2}\frac{f(g\tau)}{(c\tau+d)^{k/2+1}}=\frac{d}{d\tau}\sum_{r=0}^{k/2}\left(\tau+\frac{d}{c}\right)^{k/2-r}f_r(\tau). c − k /2 ( c τ + d ) k /2 + 2 f ′ ( g τ ) − c 1 − k /2 2 k ( c τ + d ) k /2 + 1 f ( g τ ) = d τ d r = 0 ∑ k /2 ( τ + c d ) k /2 − r f r ( τ ) . Thus, since τ 0 \tau_0 τ 0 f f f
d Ψ d τ ( d 0 c 0 , g 0 − 1 τ 0 ) ≠ 0. \frac{d\Psi}{d\tau}\left(\frac{d_0}{c_0},g_0^{-1}\tau_0\right)\ne0. d τ d Ψ ( c 0 d 0 , g 0 − 1 τ 0 ) = 0. Therefore, by the implicit function theorem, there exist a neighborhood V ⊂ R V\subset\mathbb{R} V ⊂ R d 0 / c 0 d_0/c_0 d 0 / c 0 φ : V → H \varphi:V\to\mathbb{H} φ : V → H
φ ( d 0 / c 0 ) = g 0 − 1 τ 0 and Ψ ( v , φ ( v ) ) = 0. \varphi(d_0/c_0)=g_0^{-1}\tau_0\qquad\text{and}\qquad\Psi(v,\varphi(v))=0. φ ( d 0 / c 0 ) = g 0 − 1 τ 0 and Ψ ( v , φ ( v )) = 0. We claim that φ \varphi φ V V V φ \varphi φ Ψ ( v , g 0 − 1 τ 0 ) = 0 \Psi(v,g_0^{-1}\tau_0)=0 Ψ ( v , g 0 − 1 τ 0 ) = 0 v ∈ V v\in V v ∈ V Ψ ( X − g 0 − 1 τ 0 , g 0 − 1 τ 0 ) \Psi(X-g_0^{-1}\tau_0,g_0^{-1}\tau_0) Ψ ( X − g 0 − 1 τ 0 , g 0 − 1 τ 0 ) X X X f k / 2 ≠ 0 f_{k/2}\ne0 f k /2 = 0 V ∩ Q V\cap\mathbb{Q} V ∩ Q φ \varphi φ { g 0 φ ( v ) ∣ v ∈ V ∩ Q } \{g_0\varphi(v)\mid v\in V\cap\mathbb{Q}\} { g 0 φ ( v ) ∣ v ∈ V ∩ Q } V V V g 0 φ ( V ) g_0\varphi(V) g 0 φ ( V ) ⋃ γ : f i n i t e γ F \bigcup_{\gamma:\mathrm{finite}}\gamma\mathcal{F} ⋃ γ : finite γ F f f f S L 2 ( Z ) SL_2(\mathbb{Z}) S L 2 ( Z ) T T T
Transcendence of zeros
Theorem 4.2 (Chudnovsky[7 ] , generalized by Nesterenko[23 ] ). For each τ ∈ H \tau\in\mathbb{H} τ ∈ H we have
trdeg Q { E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ ) } ≥ 2. \operatorname{trdeg}_{\mathbb{Q}}\{E_2(\tau), E_4(\tau), E_6(\tau)\} \ge2. trdeg Q { E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ )} ≥ 2. Theorem 1.8 (Restated). Let f ∈ Q [ E 2 , E 4 , E 6 ] f \in\mathbb{Q}[E_2,E_4,E_6] f ∈ Q [ E 2 , E 4 , E 6 ] k k k k / 2 k/2 k /2 f f f H \mathbb{H} H n > 0 n > 0 n > 0 G { 2 } n G_{\{2\}n} G { 2 } n
Proof. Let τ 0 ∈ H \tau_0 \in\mathbb{H} τ 0 ∈ H f f f τ 0 \tau_0 τ 0 j ( τ 0 ) j(\tau_0) j ( τ 0 ) [28 Proposition II.2.1] . By the classical identity
j ( τ ) = 1728 E 4 ( τ ) 3 E 4 ( τ ) 3 − E 6 ( τ ) 2 , (10) j(\tau) = 1728\frac{E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2}, \tag*{(10)} j ( τ ) = 1728 E 4 ( τ ) 3 − E 6 ( τ ) 2 E 4 ( τ ) 3 , (10) we have
trdeg Q Q ( E 4 ( τ 0 ) , E 6 ( τ 0 ) ) ≤ 1. \operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}\bigl(E_4(\tau_0), E_6(\tau_0)\bigr) \le1. trdeg Q Q ( E 4 ( τ 0 ) , E 6 ( τ 0 ) ) ≤ 1. Since f f f k k k k / 2 k/2 k /2 P ( X , Y , Z ) ∈ Q [ X , Y , Z ] P(X,Y,Z) \in\mathbb{Q}[X,Y,Z] P ( X , Y , Z ) ∈ Q [ X , Y , Z ]
f ( τ ) = P ( E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ ) ) , f(\tau) = P(E_2(\tau), E_4(\tau), E_6(\tau)), f ( τ ) = P ( E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ )) , and the coefficient of X k / 2 X^{k/2} X k /2 P P P Q \mathbb{Q} Q P ( X , E 4 ( τ 0 ) , E 6 ( τ 0 ) ) P(X,E_4(\tau_0),E_6(\tau_0)) P ( X , E 4 ( τ 0 ) , E 6 ( τ 0 )) X X X f ( τ 0 ) = 0 f(\tau_0)=0 f ( τ 0 ) = 0 E 2 ( τ 0 ) E_2(\tau_0) E 2 ( τ 0 ) P ( X , E 4 ( τ 0 ) , E 6 ( τ 0 ) ) = 0 P(X,E_4(\tau_0),E_6(\tau_0))=0 P ( X , E 4 ( τ 0 ) , E 6 ( τ 0 )) = 0
trdeg Q Q ( E 2 ( τ 0 ) , E 4 ( τ 0 ) , E 6 ( τ 0 ) ) ≤ 1 , \operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}\bigl(E_2(\tau_0), E_4(\tau_0), E_6(\tau_0)\bigr) \le1, trdeg Q Q ( E 2 ( τ 0 ) , E 4 ( τ 0 ) , E 6 ( τ 0 ) ) ≤ 1 , which contradicts Theorem 4.2 . Hence τ 0 \tau_0 τ 0
Remark. [6 Corollary 1.10] states that if τ \tau τ j ( τ ) j(\tau) j ( τ ) τ \tau τ G 2 n G_{2n} G 2 n τ \tau τ E 2 E_2 E 2 [6 Corollary 1.10] would imply that j ( τ ) j(\tau) j ( τ ) E 2 ( τ ) = 0 E_2(\tau)=0 E 2 ( τ ) = 0 (10) that
trdeg Q Q ( E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ ) ) ≤ 1 , \operatorname{trdeg}_{\mathbb{Q}}\mathbb{Q}\bigl(E_2(\tau), E_4(\tau), E_6(\tau)\bigr) \le1, trdeg Q Q ( E 2 ( τ ) , E 4 ( τ ) , E 6 ( τ ) ) ≤ 1 , which contradicts Theorem 4.2 . This remark is due to an observation by Professor Abdellah Sebbar.
Theorem 4.3 (Gelfond[9 ] –Schneider[26 ] ). If a a a and b b b are algebraic numbers with a ≠ 0 , 1 a \ne0,1 a = 0 , 1 and b b b irrational, then the value a b a^b a b is transcendental.
Corollary 4.4. If a a a is an algebraic real number, then the value 1 π log a \frac{1}{\pi}\log a π 1 log a is transcendental. In particular, the limiting values for the zeros appearing in Theorem 1.3,
1 π κ r log ( 2 κ r + 1 2 κ r − 1 ) , κ ∈ { 1 , 2 } , \frac{1}{\pi\kappa r}\log\left(\frac{2\kappa r+1}{2\kappa r-1}\right), \qquad\kappa\in\{1,2\}, π κ r 1 log ( 2 κ r − 1 2 κ r + 1 ) , κ ∈ { 1 , 2 } , are transcendental.
Proof. Suppose that x = 1 π log a x=\frac{1}{\pi}\log a x = π 1 log a − i x -ix − i x
( − 1 ) − i x = e ( − i x ) π i = e log a = a (-1)^{-ix}=e^{(-ix)\pi i}=e^{\log a}=a ( − 1 ) − i x = e ( − i x ) π i = e l o g a = a is algebraic. This contradicts Theorem 4.3 . Therefore, x x x
Zeros inside Ford circle
In this section, we prove Theorem 1.9. Our argument follows a strategy similar to that used in [15 ] .
Theorem 1.9 (Restated). Each Ford circle contains exactly two distinct (and simple) zeros of G 2 , 2 G_{2,2} G 2 , 2
Lemma 5.1. Let τ ∈ H \tau\in\mathbb{H} τ ∈ H g = ( a b c d ) ∈ S L 2 ( Z ) g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in\mathrm{SL}_{2}(\mathbb{Z}) g = ( a c b d ) ∈ SL 2 ( Z ) c ≠ 0 c \ne0 c = 0 G 2 , 2 ( g τ ) = 0 G_{2,2}(g\tau)=0 G 2 , 2 ( g τ ) = 0
τ − π i G 2 ( τ ) + G 4 ( τ ) = − d c or τ − π i G 2 ( τ ) − G 4 ( τ ) = − d c . \tau-\frac{\pi i}{G_{2}(\tau)+\sqrt{G_{4}(\tau)}}=-\frac{d}{c}
\quad\text{or}\quad
\tau-\frac{\pi i}{G_{2}(\tau)-\sqrt{G_{4}(\tau)}}=-\frac{d}{c}. τ − G 2 ( τ ) + G 4 ( τ ) π i = − c d or τ − G 2 ( τ ) − G 4 ( τ ) π i = − c d . Proof. From the transformation law of G 2 G_{2} G 2
( G 2 ∣ 2 g ) ( τ ) = G 2 ( τ ) − c π i c τ + d (G_{2}|_{2}g)(\tau)=G_{2}(\tau)-\frac{c\pi i}{c\tau+d} ( G 2 ∣ 2 g ) ( τ ) = G 2 ( τ ) − c τ + d c π i and (1) , we obtain
( G { 2 } n ∣ 2 n g ) ( τ ) = ∑ r = 0 n 1 ( n − r ) ! ( − π i τ + d / c ) n − r G { 2 } r ( τ ) . (G_{\{2\}^{n}}|_{2n}g)(\tau)=\sum_{r=0}^{n}\frac{1}{(n-r)!}\left(-\frac{\pi i}{\tau+d/c}\right)^{n-r}G_{\{2\}^{r}}(\tau). ( G { 2 } n ∣ 2 n g ) ( τ ) = r = 0 ∑ n ( n − r )! 1 ( − τ + d / c π i ) n − r G { 2 } r ( τ ) . This implies that G 2 , 2 ( g τ ) = 0 G_{2,2}(g\tau)=0 G 2 , 2 ( g τ ) = 0
0 = G 2 , 2 ( τ ) + X G 2 ( τ ) + 1 2 X 2 , X = − π i τ + d / c . 0=G_{2,2}(\tau)+XG_{2}(\tau)+\frac{1}{2}X^{2},\qquad X=-\frac{\pi i}{\tau+d/c}. 0 = G 2 , 2 ( τ ) + X G 2 ( τ ) + 2 1 X 2 , X = − τ + d / c π i . Using the identity ( G 2 ) 2 = 2 G 2 , 2 + G 4 (G_{2})^{2}=2G_{2,2}+G_{4} ( G 2 ) 2 = 2 G 2 , 2 + G 4
X = − G 2 ( τ ) + G 4 ( τ ) or X = − G 2 ( τ ) − G 4 ( τ ) . X=-G_{2}(\tau)+\sqrt{G_{4}(\tau)}
\quad\text{or}\quad
X=-G_{2}(\tau)-\sqrt{G_{4}(\tau)}. X = − G 2 ( τ ) + G 4 ( τ ) or X = − G 2 ( τ ) − G 4 ( τ ) . Equivalently,
τ − π i G 2 ( τ ) + G 4 ( τ ) = − d c or τ − π i G 2 ( τ ) − G 4 ( τ ) = − d c . \tau-\frac{\pi i}{G_{2}(\tau)+\sqrt{G_{4}(\tau)}}=-\frac{d}{c}
\quad\text{or}\quad
\tau-\frac{\pi i}{G_{2}(\tau)-\sqrt{G_{4}(\tau)}}=-\frac{d}{c}. τ − G 2 ( τ ) + G 4 ( τ ) π i = − c d or τ − G 2 ( τ ) − G 4 ( τ ) π i = − c d . This completes the proof □
For simplicity, let h ± ( τ ) : = G 2 ( τ ) ± G 4 ( τ ) h_{\pm}(\tau):=G_{2}(\tau)\pm\sqrt{G_{4}(\tau)} h ± ( τ ) := G 2 ( τ ) ± G 4 ( τ ) g ± ( τ ) : = τ − π i G 2 ( τ ) ± G 4 ( τ ) = τ − π i h ± ( τ ) g_{\pm}(\tau):=\tau-\frac{\pi i}{G_{2}(\tau)\pm\sqrt{G_{4}(\tau)}}=\tau-\frac{\pi i}{h_{\pm}(\tau)} g ± ( τ ) := τ − G 2 ( τ ) ± G 4 ( τ ) π i = τ − h ± ( τ ) π i τ = x + i y \tau=x+iy τ = x + i y g ± ( τ ) = U ± ( x , y ) + i V ± ( x , y ) = U ± ( τ ) + i V ± ( τ ) g_{\pm}(\tau)=U_{\pm}(x,y)+iV_{\pm}(x,y)=U_{\pm}(\tau)+iV_{\pm}(\tau) g ± ( τ ) = U ± ( x , y ) + i V ± ( x , y ) = U ± ( τ ) + i V ± ( τ ) C ± : = ζ ( 2 ) ± ζ ( 4 ) C_{\pm}:=\zeta(2)\pm\sqrt{\zeta(4)} C ± := ζ ( 2 ) ± ζ ( 4 ) E 4 ( τ ) \sqrt{E_{4}(\tau)} E 4 ( τ ) G 4 ( τ ) \sqrt{G_{4}(\tau)} G 4 ( τ ) Im τ ≥ 1 \operatorname{Im}\tau\ge1 Im τ ≥ 1 E 4 ( τ ) \sqrt{E_{4}(\tau)} E 4 ( τ ) (11) below gives α < 1 \alpha<1 α < 1
Lemma 5.2. For each τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H
∣ E 2 ( τ ) − 1 ∣ ≤ 24 r ( 1 − r ) 3 . |E_{2}(\tau)-1|\le\frac{24r}{(1-r)^{3}}. ∣ E 2 ( τ ) − 1∣ ≤ ( 1 − r ) 3 24 r . Moreover, for each τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ≥ 1 y\ge1 y ≥ 1
∣ E 4 ( τ ) − 1 ∣ ≤ 1 − 1 − α , |\sqrt{E_{4}(\tau)}-1|\le1-\sqrt{1-\alpha}, ∣ E 4 ( τ ) − 1∣ ≤ 1 − 1 − α , where α ( y ) : = α : = 240 r ( r 2 + 4 r + 1 ) ( 1 − r ) 5 \alpha(y):=\alpha:=\frac{240r(r^{2}+4r+1)}{(1-r)^{5}} α ( y ) := α := ( 1 − r ) 5 240 r ( r 2 + 4 r + 1 ) r ( y ) : = r : = ∣ q ∣ r(y):=r:=|q| r ( y ) := r := ∣ q ∣
Proof. The first inequality was obtained in [15 Lemma 2.3] . Therefore, we only prove the second inequality. From the identity E 4 ( τ ) = 1 + 240 ∑ n ≥ 1 n 3 q n 1 − q n E_{4}(\tau)=1+240\sum_{n\ge1}\frac{n^{3}q^{n}}{1-q^{n}} E 4 ( τ ) = 1 + 240 ∑ n ≥ 1 1 − q n n 3 q n
∣ E 4 ( τ ) − 1 ∣ ≤ 240 ∑ n ≥ 1 n 3 r n 1 − r n . |E_{4}(\tau)-1|\le240\sum_{n\ge1}\frac{n^{3}r^{n}}{1-r^{n}}. ∣ E 4 ( τ ) − 1∣ ≤ 240 n ≥ 1 ∑ 1 − r n n 3 r n . On the other hand, applying r d d r r\frac{d}{dr} r d r d ∑ n ≥ 0 r n = 1 1 − r \sum_{n\ge0}r^{n}=\frac{1}{1-r} ∑ n ≥ 0 r n = 1 − r 1
∑ n ≥ 0 n 3 r n = r ( r 2 + 4 r + 1 ) ( 1 − r ) 4 . \sum_{n\ge0}n^{3}r^{n}=\frac{r(r^{2}+4r+1)}{(1-r)^{4}}. n ≥ 0 ∑ n 3 r n = ( 1 − r ) 4 r ( r 2 + 4 r + 1 ) . Therefore,
∣ E 4 ( τ ) − 1 ∣ ≤ 240 ∑ n ≥ 1 n 3 r n 1 − r n ≤ 240 1 − r ∑ n ≥ 1 n 3 r n = 240 r ( r 2 + 4 r + 1 ) ( 1 − r ) 5 = α . (11) \lvert E_4(\tau)-1\rvert\le240\sum_{n\ge1}\frac{n^3r^n}{1-r^n}\le\frac{240}{1-r}\sum_{n\ge1}n^3r^n=\frac{240r(r^2+4r+1)}{(1-r)^5}=\alpha. \tag*{(11)} ∣ E 4 ( τ ) − 1 ∣ ≤ 240 n ≥ 1 ∑ 1 − r n n 3 r n ≤ 1 − r 240 n ≥ 1 ∑ n 3 r n = ( 1 − r ) 5 240 r ( r 2 + 4 r + 1 ) = α . (11) Note that 0 < α < 1 0<\alpha<1 0 < α < 1 y ≥ 1 y\ge1 y ≥ 1 α ( y ) \alpha(y) α ( y ) y y y α ( 1 ) < 1 \alpha(1)<1 α ( 1 ) < 1 ∣ E 4 ( τ ) − 1 ∣ \lvert\sqrt{E_4(\tau)}-1\rvert ∣ E 4 ( τ ) − 1 ∣
Re ( E 4 ( τ ) ) 2 − Im ( E 4 ( τ ) ) 2 = Re ( E 4 ( τ ) ) , \operatorname{Re}(\sqrt{E_4(\tau)})^2-\operatorname{Im}(\sqrt{E_4(\tau)})^2=\operatorname{Re}(E_4(\tau)), Re ( E 4 ( τ ) ) 2 − Im ( E 4 ( τ ) ) 2 = Re ( E 4 ( τ )) , Re ( E 4 ( τ ) ) 2 + Im ( E 4 ( τ ) ) 2 = ∣ E 4 ( τ ) ∣ , \operatorname{Re}(\sqrt{E_4(\tau)})^2+\operatorname{Im}(\sqrt{E_4(\tau)})^2=\lvert E_4(\tau)\rvert, Re ( E 4 ( τ ) ) 2 + Im ( E 4 ( τ ) ) 2 = ∣ E 4 ( τ )∣ , together with Re ( E 4 ( τ ) ) > 0 \operatorname{Re}(\sqrt{E_4(\tau)})>0 Re ( E 4 ( τ ) ) > 0 (11) , we obtain
Re ( E 4 ( τ ) ) = ∣ E 4 ( τ ) ∣ + Re ( E 4 ( τ ) ) 2 ≥ Re ( E 4 ( τ ) ) ≥ 1 − α . \operatorname{Re}(\sqrt{E_4(\tau)})=\sqrt{\frac{\lvert E_4(\tau)\rvert+\operatorname{Re}(E_4(\tau))}{2}}\ge\sqrt{\operatorname{Re}(E_4(\tau))}\ge\sqrt{1-\alpha}. Re ( E 4 ( τ ) ) = 2 ∣ E 4 ( τ )∣ + Re ( E 4 ( τ )) ≥ Re ( E 4 ( τ )) ≥ 1 − α . Thus, we obtain
∣ E 4 ( τ ) + 1 ∣ ≥ 1 + Re ( E 4 ( τ ) ) ≥ 1 + 1 − α . \lvert\sqrt{E_4(\tau)}+1\rvert\ge1+\operatorname{Re}(\sqrt{E_4(\tau)})\ge1+\sqrt{1-\alpha}. ∣ E 4 ( τ ) + 1 ∣ ≥ 1 + Re ( E 4 ( τ ) ) ≥ 1 + 1 − α . As a consequence, we obtain
∣ E 4 ( τ ) − 1 ∣ = ∣ E 4 ( τ ) − 1 ∣ ∣ E 4 ( τ ) + 1 ∣ ≤ α 1 + 1 − α = 1 − 1 − α . \lvert\sqrt{E_4(\tau)}-1\rvert=\frac{\lvert E_4(\tau)-1\rvert}{\lvert\sqrt{E_4(\tau)}+1\rvert}\le\frac{\alpha}{1+\sqrt{1-\alpha}}=1-\sqrt{1-\alpha}. ∣ E 4 ( τ ) − 1 ∣ = ∣ E 4 ( τ ) + 1 ∣ ∣ E 4 ( τ ) − 1 ∣ ≤ 1 + 1 − α α = 1 − 1 − α . By Lemma 5.2 , we have
∣ h ± ( τ ) − C ± ∣ ≤ ζ ( 2 ) 24 r ( 1 − r ) 3 + ζ ( 4 ) ( 1 − 1 − α ) = : M ( y ) . (12) \lvert h_{\pm}(\tau)-C_{\pm}\rvert\le\zeta(2)\frac{24r}{(1-r)^3}+\sqrt{\zeta(4)}\left(1-\sqrt{1-\alpha}\right)=:M(y). \tag*{(12)} ∣ h ± ( τ ) − C ± ∣ ≤ ζ ( 2 ) ( 1 − r ) 3 24 r + ζ ( 4 ) ( 1 − 1 − α ) =: M ( y ) . (12) Note that both 24 r ( 1 − r ) 3 \frac{24r}{(1-r)^3} ( 1 − r ) 3 24 r 1 − 1 − α 1-\sqrt{1-\alpha} 1 − 1 − α y y y M ( y ) M(y) M ( y ) y y y y = 1 y=1 y = 1 M ( 1 ) < C − < C + M(1)<C_-<C_+ M ( 1 ) < C − < C + M ( y ) ≤ M ( 1 ) < C − < C + M(y)\le M(1)<C_-<C_+ M ( y ) ≤ M ( 1 ) < C − < C + y ≥ 1 y\ge1 y ≥ 1
Lemma 5.3. For each τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ≥ 1.04 y\ge1.04 y ≥ 1.04
∣ g + ′ ( τ ) − 1 ∣ < 1. \lvert g_+'(\tau)-1\rvert<1. ∣ g + ′ ( τ ) − 1 ∣ < 1. Proof. From g + ( τ ) = τ − π i h + ( τ ) g_+(\tau)=\tau-\frac{\pi i}{h_+(\tau)} g + ( τ ) = τ − h + ( τ ) π i
∣ g + ′ ( τ ) − 1 ∣ = π ∣ h + ′ ( τ ) ∣ ∣ h + ( τ ) 2 ∣ . \lvert g_+'(\tau)-1\rvert=\frac{\pi\lvert h_+'(\tau)\rvert}{\lvert h_+(\tau)^2\rvert}. ∣ g + ′ ( τ ) − 1 ∣ = ∣ h + ( τ ) 2 ∣ π ∣ h + ′ ( τ )∣ . Thus, we need to compute a lower bound for ∣ h + ( τ ) ∣ \lvert h_+(\tau)\rvert ∣ h + ( τ )∣ ∣ h + ′ ( τ ) ∣ \lvert h_+'(\tau)\rvert ∣ h + ′ ( τ )∣ Im ( τ ) ≥ 1.04 \operatorname{Im}(\tau)\ge1.04 Im ( τ ) ≥ 1.04 (12) , we obtain
∣ h + ( τ ) ∣ ≥ C + − M ( y ) . \lvert h_+(\tau)\rvert\ge C_+-M(y). ∣ h + ( τ )∣ ≥ C + − M ( y ) . Next, from h + ( τ ) = G 2 ( τ ) + G 4 ( τ ) h_+(\tau)=G_2(\tau)+\sqrt{G_4(\tau)} h + ( τ ) = G 2 ( τ ) + G 4 ( τ )
∣ h + ′ ( τ ) ∣ ≤ ζ ( 2 ) ∣ E 2 ′ ( τ ) ∣ + ζ ( 4 ) 2 ∣ E 4 ′ ( τ ) ∣ ∣ E 4 ( τ ) ∣ . (13) \lvert h_+'(\tau)\rvert\le\zeta(2)\lvert E_2'(\tau)\rvert+\frac{\sqrt{\zeta(4)}}{2}\frac{\lvert E_4'(\tau)\rvert}{\lvert\sqrt{E_4(\tau)}\rvert}. \tag*{(13)} ∣ h + ′ ( τ )∣ ≤ ζ ( 2 ) ∣ E 2 ′ ( τ )∣ + 2 ζ ( 4 ) ∣ E 4 ( τ ) ∣ ∣ E 4 ′ ( τ )∣ . (13) Using the same argument as (11) yields
∣ E 2 ′ ( τ ) ∣ ≤ 48 π r ( 1 + r ) ( 1 − r ) 5 and ∣ E 4 ′ ( τ ) ∣ ≤ 480 π r ( 1 + 11 r + 11 r 2 + r 3 ) ( 1 − r ) 7 . \lvert E_2'(\tau)\rvert\le48\pi\frac{r(1+r)}{(1-r)^5}
\quad\text{and}\quad
\lvert E_4'(\tau)\rvert\le480\pi\frac{r(1+11r+11r^2+r^3)}{(1-r)^7}. ∣ E 2 ′ ( τ )∣ ≤ 48 π ( 1 − r ) 5 r ( 1 + r ) and ∣ E 4 ′ ( τ )∣ ≤ 480 π ( 1 − r ) 7 r ( 1 + 11 r + 11 r 2 + r 3 ) . Moreover, by (11) , we have ∣ E 4 ( τ ) ∣ ≥ 1 − α \lvert\sqrt{E_4(\tau)}\rvert\ge\sqrt{1-\alpha} ∣ E 4 ( τ ) ∣ ≥ 1 − α
∣ h + ′ ( τ ) ∣ ≤ ζ ( 2 ) 48 π r ( 1 + r ) ( 1 − r ) 5 + 240 π ζ ( 4 ) r ( 1 + 11 r + 11 r 2 + r 3 ) 1 − α ( 1 − r ) 7 . \lvert h_+'(\tau)\rvert\le\zeta(2)48\pi\frac{r(1+r)}{(1-r)^5}+240\pi\sqrt{\zeta(4)}\frac{r(1+11r+11r^2+r^3)}{\sqrt{1-\alpha}(1-r)^7}. ∣ h + ′ ( τ )∣ ≤ ζ ( 2 ) 48 π ( 1 − r ) 5 r ( 1 + r ) + 240 π ζ ( 4 ) 1 − α ( 1 − r ) 7 r ( 1 + 11 r + 11 r 2 + r 3 ) . Finally, we have
∣ g + ′ ( τ ) − 1 ∣ = π ∣ h + ′ ( τ ) ∣ ∣ h + ( τ ) 2 ∣ ≤ π ζ ( 2 ) 48 π r ( 1 + r ) ( 1 − r ) 5 + 240 π ζ ( 4 ) r ( 1 + 11 r + 11 r 2 + r 3 ) 1 − α ( 1 − r ) 7 ( C + − M ( y ) ) 2 . \lvert g_+'(\tau)-1\rvert=\pi\frac{\lvert h_+'(\tau)\rvert}{\lvert h_+(\tau)^2\rvert}\le\pi\frac{\zeta(2)48\pi\frac{r(1+r)}{(1-r)^5}+240\pi\sqrt{\zeta(4)}\frac{r(1+11r+11r^2+r^3)}{\sqrt{1-\alpha}(1-r)^7}}{(C_+-M(y))^2}. ∣ g + ′ ( τ ) − 1 ∣ = π ∣ h + ( τ ) 2 ∣ ∣ h + ′ ( τ )∣ ≤ π ( C + − M ( y ) ) 2 ζ ( 2 ) 48 π ( 1 − r ) 5 r ( 1 + r ) + 240 π ζ ( 4 ) 1 − α ( 1 − r ) 7 r ( 1 + 11 r + 11 r 2 + r 3 ) . ::: Since the denominator of the fraction on the right-hand side is increasing in y y y y y y y y y y = 1.04 y=1.04 y = 1.04 y ≥ 1.04 y \ge1.04 y ≥ 1.04
Lemma 5.4. For each τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ≥ 1.04 y\ge1.04 y ≥ 1.04 Re g + ′ ( τ ) = ∂ y V + ( x , y ) > 0 \operatorname{Re} g_+^{\prime}(\tau)=\partial_y V_+(x,y)>0 Re g + ′ ( τ ) = ∂ y V + ( x , y ) > 0
Proof. Recall that, by the Cauchy–Riemann equations, Re g + ′ ( τ ) = ∂ y V + ( x , y ) \operatorname{Re} g_+^{\prime}(\tau)=\partial_y V_+(x,y) Re g + ′ ( τ ) = ∂ y V + ( x , y ) Lemma 5.3 , we have
∣ Re g + ′ ( τ ) − 1 ∣ < 1 , \left|\operatorname{Re} g_+^{\prime}(\tau)-1\right|<1, Re g + ′ ( τ ) − 1 < 1 , which implies Re g + ′ ( τ ) > 0 \operatorname{Re} g_+^{\prime}(\tau)>0 Re g + ′ ( τ ) > 0
Lemma 5.5. For each τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ∈ [ 1 , 1.04 ] y\in[1,1.04] y ∈ [ 1 , 1.04 ]
V + ( x , y ) < 0. V_+(x,y)<0. V + ( x , y ) < 0. Proof. By definition of V + V_+ V +
V + ( x , y ) = y − Re ( π h + ( τ ) ) . V_+(x,y)=y-\operatorname{Re}\left(\frac{\pi}{h_+(\tau)}\right). V + ( x , y ) = y − Re ( h + ( τ ) π ) . We first claim
Re ( 1 h + ) ≥ 1 C + + M ( y ) . \operatorname{Re}\left(\frac{1}{h_+}\right)\ge\frac{1}{C_++M(y)}. Re ( h + 1 ) ≥ C + + M ( y ) 1 . To prove this, we let h + ( τ ) = a + b i h_+(\tau)=a+bi h + ( τ ) = a + bi a , b ∈ R a,b\in\mathbb{R} a , b ∈ R Re ( 1 h + ) = a a 2 + b 2 \operatorname{Re}\left(\frac{1}{h_+}\right)=\frac{a}{a^2+b^2} Re ( h + 1 ) = a 2 + b 2 a
a ( C + + M ( y ) ) ≥ a 2 + b 2 . a(C_++M(y))\ge a^2+b^2. a ( C + + M ( y )) ≥ a 2 + b 2 . On the other hand, by (12) we have
( a − C + ) 2 + b 2 ≤ M ( y ) 2 . (a-C_+)^2+b^2\le M(y)^2. ( a − C + ) 2 + b 2 ≤ M ( y ) 2 . Therefore, it suffices to show that
M ( y ) 2 − C + 2 + 2 a C + ≤ a C + + a M ( y ) . M(y)^2-C_+^2+2aC_+\le aC_++aM(y). M ( y ) 2 − C + 2 + 2 a C + ≤ a C + + a M ( y ) . Or equivalently,
( C + − M ( y ) ) ( C + + M ( y ) − a ) ≥ 0 , (C_+-M(y))(C_++M(y)-a)\ge0, ( C + − M ( y )) ( C + + M ( y ) − a ) ≥ 0 , which follows from the facts that M ( y ) < C + M(y)<C_+ M ( y ) < C + (12) . Finally, we obtain an upper bound for V + ( τ ) V_+(\tau) V + ( τ )
V + ( τ ) ≤ y − π C + + M ( y ) . V_+(\tau)\le y-\frac{\pi}{C_++M(y)}. V + ( τ ) ≤ y − C + + M ( y ) π . Note that M ( y ) M(y) M ( y ) [ 1 , 1.04 ] [1,1.04] [ 1 , 1.04 ] [ 1 , 1.02 ] [1,1.02] [ 1 , 1.02 ] [ 1.02 , 1.04 ] [1.02,1.04] [ 1.02 , 1.04 ]
1.02 − π C + + M ( 1 ) and 1.04 − π C + + M ( 1.02 ) , 1.02-\frac{\pi}{C_++M(1)}
\qquad\text{and}\qquad
1.04-\frac{\pi}{C_++M(1.02)}, 1.02 − C + + M ( 1 ) π and 1.04 − C + + M ( 1.02 ) π , respectively. Since both are negative, we conclude that V + ( τ ) < 0 V_+(\tau)<0 V + ( τ ) < 0 Im ( τ ) ∈ [ 1 , 1.04 ] \operatorname{Im}(\tau)\in[1,1.04] Im ( τ ) ∈ [ 1 , 1.04 ]
The corresponding argument for g − g_- g −
Lemma 5.6. The following are true.
For τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ≥ 1.46 y\ge1.46 y ≥ 1.46 ∣ g − ( τ ) − 1 ∣ < 1 \lvert g_-(\tau)-1\rvert<1 ∣ g − ( τ ) − 1 ∣ < 1
For τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ≥ 1.46 y\ge1.46 y ≥ 1.46 Re g − ( τ ) = ∂ y V − ( x , y ) > 0 \operatorname{Re}g_-(\tau)=\partial_yV_-(x,y)>0 Re g − ( τ ) = ∂ y V − ( x , y ) > 0
For τ = x + i y ∈ H \tau=x+iy\in\mathbb{H} τ = x + i y ∈ H y ∈ [ 1 , 1.46 ] y\in[1,1.46] y ∈ [ 1 , 1.46 ] V − ( x , y ) < 0 V_-(x,y)<0 V − ( x , y ) < 0
As before, we only treat the case of g + g_+ g + g − g_- g − V + ( x , y ) V_+(x,y) V + ( x , y )
lim y → ∞ V + ( x , y ) = + ∞ . \lim_{y\to\infty}V_+(x,y)=+\infty. y → ∞ lim V + ( x , y ) = + ∞. Therefore, by Lemma 5.5 and the intermediate value theorem, for each x ∈ R x\in\mathbb{R} x ∈ R y ∈ ( 1.04 , ∞ ) y\in(1.04,\infty) y ∈ ( 1.04 , ∞ ) V + ( x + i y ) = 0 V_+(x+iy)=0 V + ( x + i y ) = 0 Lemma 5.4 ensures that it is unique. Thus, we have a function ϕ + : R → ( 1.04 , ∞ ) \phi_+:\mathbb{R}\to(1.04,\infty) ϕ + : R → ( 1.04 , ∞ ) ϕ + ( x ) = y \phi_+(x)=y ϕ + ( x ) = y V + ( x + i y ) = 0 V_+(x+iy)=0 V + ( x + i y ) = 0
A + : = { τ ∈ H ∣ Im τ ≥ 1 , ∣ Re τ ∣ ≤ 1 / 2 , V + ( τ ) = 0 } = { x + i ϕ + ( x ) ∣ x ∈ [ − 1 / 2 , 1 / 2 ] } . A_+:=\{\tau\in\mathbb{H}\mid\operatorname{Im}\tau\ge1,\ |\operatorname{Re}\tau|\le1/2,\ V_+(\tau)=0\}=\{x+i\phi_+(x)\mid x\in[-1/2,1/2]\}. A + := { τ ∈ H ∣ Im τ ≥ 1 , ∣ Re τ ∣ ≤ 1/2 , V + ( τ ) = 0 } = { x + i ϕ + ( x ) ∣ x ∈ [ − 1/2 , 1/2 ]} . Furthermore, by Lemma 5.4 and the implicit function theorem, y = ϕ + ( x ) y=\phi_+(x) y = ϕ + ( x ) R \mathbb{R} R ϕ + ′ ( x ) = − ∂ x V + ∂ y V + \phi_+'(x)=-\frac{\partial_xV_+}{\partial_yV_+} ϕ + ′ ( x ) = − ∂ y V + ∂ x V +
d d x ( U + ( x , ϕ + ( x ) ) ) = ( ∂ x U + ) ( x , ϕ + ( x ) ) + ϕ + ′ ( x ) ( ∂ y U + ) ( x , ϕ + ( x ) ) = ∂ x U + − ∂ y U + ∂ x V + ∂ y V + = ∂ x U + ∂ y V + − ∂ y U + ∂ x V + ∂ y V + = ( ∂ y V + ) 2 + ( ∂ x V + ) 2 ∂ y V + > 0 \begin{aligned}
\frac{d}{dx}\left(U_+(x,\phi_+(x))\right)&=(\partial_xU_+)(x,\phi_+(x))+\phi_+'(x)(\partial_yU_+)(x,\phi_+(x))\\
&=\partial_xU_+-\partial_yU_+\frac{\partial_xV_+}{\partial_yV_+}\\
&=\frac{\partial_xU_+\partial_yV_+-\partial_yU_+\partial_xV_+}{\partial_yV_+}\\
&=\frac{(\partial_yV_+)^2+(\partial_xV_+)^2}{\partial_yV_+}>0
\end{aligned} d x d ( U + ( x , ϕ + ( x )) ) = ( ∂ x U + ) ( x , ϕ + ( x )) + ϕ + ′ ( x ) ( ∂ y U + ) ( x , ϕ + ( x )) = ∂ x U + − ∂ y U + ∂ y V + ∂ x V + = ∂ y V + ∂ x U + ∂ y V + − ∂ y U + ∂ x V + = ∂ y V + ( ∂ y V + ) 2 + ( ∂ x V + ) 2 > 0 for any x ∈ R x\in\mathbb{R} x ∈ R U + U_+ U + A + A_+ A + x x x [ − 1 / 2 , 1 / 2 ] [-1/2,1/2] [ − 1/2 , 1/2 ] A − A_- A − ϕ − \phi_- ϕ −
Lemma 5.7. We have g + ( A + ) = [ − 1 / 2 , 1 / 2 ] g_+(A_+)=[-1/2,1/2] g + ( A + ) = [ − 1/2 , 1/2 ] g − ( A − ) = [ − 1 / 2 , 1 / 2 ] g_-(A_-)=[-1/2,1/2] g − ( A − ) = [ − 1/2 , 1/2 ]
Proof. We only treat the case of g + g_+ g + g − g_- g − A + A_+ A + g + ( τ ) = U + ( τ ) g_+(\tau)=U_+(\tau) g + ( τ ) = U + ( τ ) τ ∈ A + \tau\in A_+ τ ∈ A + U + U_+ U + A + A_+ A +
U + ( − 1 / 2 , ϕ + ( − 1 / 2 ) ) = − 1 / 2 , U + ( 1 / 2 , ϕ + ( 1 / 2 ) ) = 1 / 2. (14) U_+(-1/2,\phi_+(-1/2))=-1/2,\qquad U_+(1/2,\phi_+(1/2))=1/2. \tag*{(14)} U + ( − 1/2 , ϕ + ( − 1/2 )) = − 1/2 , U + ( 1/2 , ϕ + ( 1/2 )) = 1/2. (14) Since G 2 G_2 G 2 G 4 G_4 G 4 T = ( 1 1 0 1 ) T=\begin{pmatrix}1&1\\0&1\end{pmatrix} T = ( 1 0 1 1 ) g + ( τ + 1 ) = g + ( τ ) + 1 g_+(\tau+1)=g_+(\tau)+1 g + ( τ + 1 ) = g + ( τ ) + 1 V + ( x + 1 , ϕ + ( x ) ) = V + ( x , ϕ + ( x ) ) = 0 V_+(x+1,\phi_+(x))=V_+(x,\phi_+(x))=0 V + ( x + 1 , ϕ + ( x )) = V + ( x , ϕ + ( x )) = 0 x x x y > 1.04 y>1.04 y > 1.04 V + ( x , y ) = 0 V_+(x,y)=0 V + ( x , y ) = 0 ϕ + ( x + 1 ) = ϕ + ( x ) \phi_+(x+1)=\phi_+(x) ϕ + ( x + 1 ) = ϕ + ( x ) g + ( τ + 1 ) = g + ( τ ) + 1 g_+(\tau+1)=g_+(\tau)+1 g + ( τ + 1 ) = g + ( τ ) + 1
U + ( − 1 2 , ϕ + ( − 1 2 ) ) + 1 = U + ( 1 2 , ϕ + ( 1 2 ) ) . (15) U_+\left(-\frac{1}{2},\phi_+\left(-\frac{1}{2}\right)\right)+1=U_+\left(\frac{1}{2},\phi_+\left(\frac{1}{2}\right)\right). \tag*{(15)} U + ( − 2 1 , ϕ + ( − 2 1 ) ) + 1 = U + ( 2 1 , ϕ + ( 2 1 ) ) . (15) On the other hand, by our choice of the branch of G 4 \sqrt{G_4} G 4 g + ( − τ ‾ ) = − g + ( τ ) ‾ g_+(-\overline{\tau})=-\overline{g_+(\tau)} g + ( − τ ) = − g + ( τ ) τ = 1 2 + ϕ + ( 1 2 ) i \tau=\frac{1}{2}+\phi_+\left(\frac{1}{2}\right)i τ = 2 1 + ϕ + ( 2 1 ) i
U + ( − 1 2 , ϕ + ( − 1 2 ) ) = U + ( − 1 2 , ϕ + ( 1 2 ) ) = − U + ( 1 2 , ϕ + ( 1 2 ) ) . (16) U_+\left(-\frac{1}{2},\phi_+\left(-\frac{1}{2}\right)\right)=U_+\left(-\frac{1}{2},\phi_+\left(\frac{1}{2}\right)\right)=-U_+\left(\frac{1}{2},\phi_+\left(\frac{1}{2}\right)\right). \tag*{(16)} U + ( − 2 1 , ϕ + ( − 2 1 ) ) = U + ( − 2 1 , ϕ + ( 2 1 ) ) = − U + ( 2 1 , ϕ + ( 2 1 ) ) . (16) Combining (15) and (16) , we obtain (14) . □
By Lemma 5.7 and the identity g ± ( τ + 1 ) = g ± ( τ ) + 1 g_\pm(\tau+1)=g_\pm(\tau)+1 g ± ( τ + 1 ) = g ± ( τ ) + 1
Theorem 5.8. For each α ∈ R \alpha\in\mathbb{R} α ∈ R τ ∈ H \tau\in\mathbb{H} τ ∈ H g + ( τ ) = α g_+(\tau)=\alpha g + ( τ ) = α Im ( τ ) > 1 \operatorname{Im}(\tau)>1 Im ( τ ) > 1 n ∈ Z n\in\mathbb{Z} n ∈ Z α ∈ [ − 1 / 2 + n , 1 / 2 + n ] \alpha\in[-1/2+n,1/2+n] α ∈ [ − 1/2 + n , 1/2 + n ] Re ( τ ) ∈ [ − 1 / 2 + n , 1 / 2 + n ] \operatorname{Re}(\tau)\in[-1/2+n,1/2+n] Re ( τ ) ∈ [ − 1/2 + n , 1/2 + n ] g − g_- g −
Lemma 5.9. Let τ + \tau_+ τ + τ − \tau_- τ − g + ( τ ) = α g_+(\tau)=\alpha g + ( τ ) = α g − ( τ ) = α g_-(\tau)=\alpha g − ( τ ) = α α ∈ R \alpha\in\mathbb{R} α ∈ R Im ( τ + ) > 1 \operatorname{Im}(\tau_+)>1 Im ( τ + ) > 1 Im ( τ − ) > 1 \operatorname{Im}(\tau_-)>1 Im ( τ − ) > 1 τ + ≠ τ − \tau_+\ne\tau_- τ + = τ −
Proof. Suppose that τ 0 : = τ + = τ − \tau_0:=\tau_+=\tau_- τ 0 := τ + = τ −
τ 0 − π i G 2 ( τ 0 ) + G 4 ( τ 0 ) = τ 0 − π i G 2 ( τ 0 ) − G 4 ( τ 0 ) . \tau_0-\frac{\pi i}{G_2(\tau_0)+\sqrt{G_4(\tau_0)}}=\tau_0-\frac{\pi i}{G_2(\tau_0)-\sqrt{G_4(\tau_0)}}. τ 0 − G 2 ( τ 0 ) + G 4 ( τ 0 ) π i = τ 0 − G 2 ( τ 0 ) − G 4 ( τ 0 ) π i . It follows that G 4 ( τ 0 ) = 0 G_4(\tau_0)=0 G 4 ( τ 0 ) = 0 [25 ] that G 4 ( τ ) G_4(\tau) G 4 ( τ ) Im ( τ ) > 1 \operatorname{Im}(\tau)>1 Im ( τ ) > 1
By Lemmas 5.1 and 5.9 and Theorem 5.8 , we have the following.
Theorem 5.10. For each g = ( a b c d ) ∈ S L 2 ( Z ) g=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb{Z}) g = ( a c b d ) ∈ SL 2 ( Z ) c ≠ 0 c\ne0 c = 0 G 2 , 2 ( g τ ) = 0 G_{2,2}(g\tau)=0 G 2 , 2 ( g τ ) = 0 { τ ∈ H ∣ Im ( τ ) > 1 } \{\tau\in\mathbb{H}\mid\operatorname{Im}(\tau)>1\} { τ ∈ H ∣ Im ( τ ) > 1 }
Since the interior of the Ford circle corresponding to a / c a/c a / c { τ ∈ H ∣ Im ( τ ) > 1 } \{\tau\in\mathbb{H} \mid\operatorname{Im}(\tau) > 1\} { τ ∈ H ∣ Im ( τ ) > 1 } g = ( a b c d ) ∈ SL 2 ( Z ) g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in\operatorname{SL}_{2}(\mathbb{Z}) g = ( a c b d ) ∈ SL 2 ( Z ) Theorem 5.10 implies that G 2 , 2 G_{2,2} G 2 , 2 a / c a/c a / c Theorem 1.5 , we obtain the desired result. This completes the proof of Theorem 1.9.
Graduate School of Mathematics, Kyushu University, Motooka 744, Nishi-ku, Fukuoka 819-0395 Japan
Email address: kkina@math.kyushu-u.ac.jp
KNU Research Institute of Mathematical Sciences, Kangwon National University, Chuncheon 24341, Republic of Korea
Email address: gshin@kangwon.ac.kr
