Asymptotic expansion of Fourier coefficients of reciprocals of Eisenstein series
aa r X i v : . [ m a t h . N T ] J a n ASYMPTOTIC EXPANSION OF FOURIER COEFFICIENTS OFRECIPROCALS OF EISENSTEIN SERIES
BERNHARD HEIM AND MARKUS NEUHAUSER
Abstract.
In this paper we give a classification of the asymptotic expansionof the q -expansion of reciprocals of Eisenstein series E k of weight k for themodular group SL ( Z ). For k ≥
12 even, this extends results of Hardy andRamanujan, and Berndt, Bialek and Yee, utilizing the Circle Method on the onehand, and results of Petersson, and Bringmann and Kane, developing a theory ofmeromorphic Poincar´e series on the other. We follow a uniform approach, basedon the zeros of the Eisenstein series with the largest imaginary part. Thesespecial zeros provide information on the singularities of the Fourier expansionof 1 /E k ( z ) with respect to q = e πiz .2010 Mathematics Subject Classification.
Primary 11F30, 11M36, 26C10; Secondary 05A16,11B37.
Key words and phrases.
Eisenstein series, Fourier coefficients, meromorphic modular forms,polynomials, Ramanujan, recurrence relations. Introduction
In this paper we provide a new approach to determine the main asymptoticgrowth terms in the Fourier expansion of the reciprocals 1 /E k of Eisenstein seriesof weight k . We refer to [BFOR17], Chapter 15 for a very good introduction intothe topic.Eisenstein series are defined by(1) E k ( z ) := 1 − kB k ∞ X n =1 σ k − ( n ) q n ( q := e πiz ) . They are modular forms [O03] on the upper half of the complex plane H . Thealgebra of modular forms with respect to the modular group SL ( Z ) is generatedby E and E . As usual B k denotes the k -th Bernoulli number and σ ℓ ( n ) := P d | n d ℓ .Hardy and Ramanujan [HR18B] launched in their last joint paper, the study ofcoefficients of meromorphic modular forms with a simple pole in the standard fun-damental domain F . They demonstrated that, similar to their famous asymptoticformula for the partition numbers(2) p ( n ) ∼ n √ e π √ n , ∞ X n =0 p ( n ) q n := q η ( z ) , which had been given birth to the Circle Method [HR18A], formulas for the co-efficients of reciprocals of modular forms can be obtained. The reciprocal of theDedekind η -function is a weakly modular form of weight − / H .Hardy and Ramanujan focused on the reciprocal of the Eisenstein series E .They proved an explicit formula for the coefficients. Shortly afterwards, in aletter to Hardy, Ramanujan stated several formulas of the same type, includingthe q -expansion of 1 /E . No proofs were given.Bialek in his Ph.D. thesis, written under the guidance of Berndt [BB05], andfinally Berndt, Bialek and Yee [BBY02] have proven the claims in the letter ofRamanujan by extending the methods applied in [HR18B].We illustrate the case k = 4. Following Ramanujan, we frequently put E k ( q z ) := E k ( z ) for q = q z := e πiz . Let ρ be the unique zero of E in F . Let λ run over theintegers of the form 3 a Q rℓ =1 p a ℓ ℓ , where a = 0 or 1. Here, p ℓ is a prime of the form6 m + 1, and a j ∈ N . Then [BB05]:(3) β ( n ) = ( − n ( q ρ ) X ( λ ) X ( c,d ) h ( c,d ) ( n ) λ e πn √ λ . Here, ( c, d ) runs over distinct solutions to λ = c − cd + d , such that integers a, b exist solving ad − bc = 1. Let h (1 , ( n ) := 1, h (2 , ( n ) := ( − n , and for λ ≥ h ( c,d ) ( n ) := 2 cos ( ad + bc − ac − bd + λ ) π nλ − c √ d − c !! . For the definition of distinct we refer to [BB05], Section 3. From the explicitformula (3) one observes that the main asymptotic growth comes from ( c, d ) =
SYMPTOTIC EXPANSION 3 (1 , β ( n ) ∼ ( − n E ( ρ ) e πn √ , (5) β ( n ) ∼ E ( i ) e πn , (6)where P ∞ n =0 β k ( n ) q n := E k ( z ) . We added the asymptotic (6), which can be ob-tained in a similar way.Petersson [P50] offered an alternative approach to study the q -expansion ofmeromorphic modular forms. He defined Poincar´e series with poles at arbitrarypoints in H and of arbitrary order, to provide a basis for the underlying vec-tor spaces. Recently, Bringmann and Kane [BK17] have generalized Petersson’smethod. They have also recorded several important examples.In this paper we study the asymptotic expansions for all reciprocals of Eisen-stein series. Instead of proving first an explicit formula and then detecting themain growth terms, we provide a direct approach. This is based on the distribu-tion of the zeros in the standard fundamental domain with the largest imaginarypart.Before we state our results, we want to point out as a warning that the limitsas n → ∞ for β ( n ) /β ( n + 1) and β ( n ) /β ( n + 1) exist, but that this is nottrue for all k as indicated in Table 1. n β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ − . · − . · − . · − . · − − . · − . · − − . · − . · − − . · − . · − . · − . · − − . · − . · − − . · − . · − − . · − . · − . · − . · − − . · − . · − − . · − . · − − . · − . · − . · − . · − ... ... ... ... ...19 − . · − . · − . · − . · − − . · − . · − − . · − . · − Table 1.
Quotients of successive coefficients of 1 / E k for k ∈{ , , , } Results
The constants in the asymptotic expansion of β k ( n ), the coefficients of the q -expansion of the reciprocal of E k , involve the Ramanujan Θ-operator [R16,BKO04] induced by residue calculation. The differential operator Θ := q dd q acts BERNHARD HEIM AND MARKUS NEUHAUSER on formal power series by:(7) Θ ∞ X n = h a ( n ) q n ! := ∞ X n = h n a ( n ) q n . Let E ( q ) := 1 − P ∞ n =1 σ ( n ) q n . Ramanujan observed that(8) Θ(E ) = (E E − E ) / ) = (E E − E ) / . Our first results give an explicit interpretation of the data presented in Table 1for k = 6 and k = 14. Theorem 1.
Let k ≥ and k ≡ be an integer. Then / E k has a q -expansion with radius q i = e − π : (9) 1E k ( q ) = ∞ X n =0 β k ( n ) q n . The coefficients β k ( n ) are non-zero and have the asymptotic expansion (10) β k ( n ) ∼ − k )( q i ) q − ni . The number q i = e − π ≈ . · − is transcendental. It is well-knownthat the so-called Gel ′ fond constant e π is transcendental. This was first proven byGel ′ fond in 1929. It can also be deduced from the Gel ′ fond–Schneider Theorem,which solved Hilbert’s seventh problem [W08]. We refer to a result by Nesterenko(also [W08], Section 5.6). Let z ∈ H . Then at least already three of the fournumbers(11) q z , E ( q z ) , E ( q z ) , and E ( q z )are algebraically independent. Since E ( q ρ ) = E ( q i ) = 0, we obtain that q i , E ( q i )and q ρ , E ( q ρ ) are transcendental.Moreover, Θ(E k )( q i ) for k = 6 , ,
14 can be explicitly expressed by Γ( ) and π . For example,(12) Θ(E )( q i ) = −
12 E ( q i ) , where E ( q i ) = 3 Γ( ) (2 π ) . We can also extract the numbers q i and E ( q i ) from the coefficients. Corollary 1.
Let k ≥ and k ≡ . Then lim n →∞ β k ( n ) β k ( n + 1) = q i , (13) lim n →∞ β ( n ) β ( n ) = lim n →∞ β ( n ) β ( n ) = E ( q i ) . (14)Hardy and Ramanujan stated lower and upper bounds at the end of their initialwork [HR18B] on the coefficients of the reciprocal of 1 /E . We generalize theiridea to all cases k ≡ k = 2 and also improve their result inthe original case k = 6. SYMPTOTIC EXPANSION 5
Theorem 2.
Let k ≡ and k a positive integer. Let x := kB k . Then wehave for all n ∈ N (15) (cid:16) x + √ ∆ k (cid:17) n +1 − (cid:16) x −√ ∆ k (cid:17) n +1 √ ∆ k ≤ β k ( n ) with ∆ k = x + 4 (cid:0) k − + 1 (cid:1) x and (16) β k ( n ) ≤ (cid:16) x − b k −√ D k (cid:17) (cid:16) b k + √ D k (cid:17) n + (cid:16) b k + √ D k − x (cid:17) (cid:16) b k −√ D k (cid:17) n √ D k with b k = x + a k , c k = (cid:0) k − + 1 − a k (cid:1) x , and D k = b k + 4 c k for all k where a = p / and a k = k − +12 k − +1 for k ≥ . The case k ≡ k , we cannot expectthat the limit as n → ∞ of β k ( n ) /β k ( n + 1) exists, since we have two poles on thecircle of convergence. But for k = 4 and k = 8 there is still only one pole. Proposition 1.
Let q ρ = e πρ = − e − π √ . Let m ∈ N . Then the coefficients β ,m ( n ) of the m th power of E − i. e. (17) ∞ X n =0 β ,m ( n ) q n := (cid:18) ( q ) (cid:19) m satisfy for all m : (18) lim n →∞ β ,m ( n ) β ,m ( n + 1) = q ρ . Remarks. a) For small weights the following identities exist:(19) E = E , E = E · E and E = E · E . b) Let the principal part of E − m at the pole q ρ be given by(20) m X k =1 λ m,k ( q − q ρ ) k , then λ m,m = res q ρ (cid:0) E − (cid:1) m . It would be interesting to get explicit formulas for all λ m,k , 1 ≤ k ≤ m . Especially for the case m = 2.c) We have res q ρ (E − ) = − q ρ E ( q ρ ) .We know that β ( n ) and β ( n ) are non-zero for all n ∈ N [HN20B]. Weprovide new proof of the asymptotic expansion for k = 4. This is the main termof a formula first conjectured by Ramanujan and proven about 80 years later byBialek [BB05]. For the case k = 8, we also refer to [BK17]. Theorem 3.
We have ( − n β ( n ) ∈ N for all n ∈ N . Further, we have theasymptotic expansion (21) β ( n ) ∼ − )( q ρ ) q − nρ , BERNHARD HEIM AND MARKUS NEUHAUSER where
Θ(E )( q ρ ) = − E ( q ρ ) / . F. K. C. Rankin and H. P. F. Swinnerton-Dyer [RS70] have proven that all thezeros of E k ( z ) in the standard fundamental domain F are in C = { z ∈ F : | z | =1 } ⊂ F . We recall the following basic facts [O03]. The modular group Γ := SL ( Z )operates on the complex upper half plane H , denoted by γ ( z ), where γ ∈ Γ and z ∈ H . The standard fundamental domain F is given by F = { z ∈ H : | z | ≥ ≤ Re ( z ) ≤ / } ∪{ z ∈ H : | z | > − / < Re ( z ) < } . Proposition 2 (Rankin, Swinnerton-Dyer [RS70]) . Let k ≥ be an even integer.Let z k be the zero of E k with the largest imaginary part. Then (22) z = z = ρ and z k = i for k ≡ . All other zeros satisfy z k ∈ C \ { i, ρ } . Only for k = 8 the zero z k is not simple. Further, from [RS70] and Kohnen [K04] we obtain
Corollary 2.
Let k ≥ and k ≡ . Let k = 12 N + s for s ∈ { , , } .Then z k = e πi ϕ , where ϕ ∈ (cid:0) N − N , (cid:1) . Theorem 4.
Let k be a positive integer. Let k ≥ and k ≡ . Then / E k has a q -expansion with radius | q z k | , where z k is the zero of E k with the largestimaginary part. Then (23) β k ( n ) q nz k + 1Θ(E k )( q z k ) + 1Θ(E k )( q z k ) (cid:18) q z k q z k (cid:19) n constitutes a zero sequence. The expression(24) 1Θ(E k )( q z k ) + 1Θ(E k )( q z k ) (cid:18) q z k q z k (cid:19) n is bounded. But this is not sufficient to obtain an asymptotic expansion. Nev-ertheless we have discovered a new property of the coefficients of 1 / E k for k ≡ Theorem 5.
Let k ≡ and k ≥ . Then there exists a subsequence { n t } ∞ t =1 of { n } ∞ n =1 such that (25) lim t →∞ β k ( n t ) − q − n t z k (cid:16) k )( q zk ) + k )( q zk ) (cid:16) q zk q zk (cid:17) n t (cid:17) = 1 . The statement of this theorem is equivalent to(26) lim t →∞ β k ( n t ) − (cid:18) q − ntzk Θ(E k )( q zk ) (cid:19) = 1 . We further have the following properties.
Theorem 6.
Let k be a positive integer. Let k ≥ and k ≡ . SYMPTOTIC EXPANSION 7 a) Let A k ( n ) denote the number of changes of sign in the sequence { β k ( m ) } nm =0 and let z k = x k + i y k ∈ F be the zero of E k with the largest imaginary part.Then lim n →∞ A k ( n ) n = 2 x k . b) Let B k ( n ) be the number of non-zero coefficients among the n coefficients { β k ( m ) } n − m =0 . Then lim sup n →∞ nB k ( n ) ≤ . We end this section with a considerably surprising result.
Corollary 3.
For large weights k divisible by , the coefficients of / E k ( q ) satisfy (27) lim ℓ →∞ lim n →∞ A ℓ ( n ) n = 0 . Proofs
Proof of Corollary 1, Proposition 1, and Theorem 1.
We first recall aresult from complex analysis. Let f ( q ) = P ∞ n =0 a ( n ) q n be a power series regularat q = 0 with finite radius of convergence. Assume that there is only one singularpoint q on the circle of convergence. Let q be a pole. Then it is known ([PS78],Part 3) that(28) lim n →∞ a ( n ) a ( n + 1) = q . This follows from the Laurent expansion of f ( q ), which has a finite principal part.Let E k ( q ) have exactly one zero q ∈ B (0) with absolute value smaller thanall other zeros. Then we obtain the property (28) for the coefficients of 1 / E k .Note that every zero of a modular form has one representative in the fundamentaldomain F .The zeros of E k are controlled by a theorem by Rankin and Swinnerton-Dyer([RS70], see also Section 2). They proved that every zero in F has absolute value1. Further, let k be a positive, even integer and k ≥
4. Let k = 12 N + s , where s ∈ { , , , , , } . Then E k has N simple zeros in C \ { i, ρ } . Additionally wehave simple zeros ρ for s = 4 and i for s = 6. Further, E k has the double zero ρ for s = 8, the simple zeros i and ρ for s = 10, and the simple zero i and the doublezero ρ for s = 14. Further, let z k be the zero of E k with the largest imaginarypart. Note that(29) z ′ k := J ( z k ) = (cid:18) −
11 0 (cid:19) z k and z k have the same imaginary part. Note that J ( i ) = i and J ( ρ ) = ρ −
1. Thus,1 / E k has exactly one pole on the radius of convergence iff z k = i or z k = ρ . Proof of Corollary 1.
From the theorem of Rankin and Swinnerton-Dyer we ob-tain that for k ≡ z k = i and q i = e − π . This gives a first proofof Corollary 1 (13). Corollary 1 (13) also follows directly from Theorem 1. Thequotients for small k converge very quickly. We refer to Table 1 and Table 2. BERNHARD HEIM AND MARKUS NEUHAUSER n β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ β ( n ) β ( n +1) ≈ − . · − . · − . · − . · − . · − − . · − . · − − . · − . · − − . · − − . · − . · − . · − . · − . · − − . · − . · − − . · − . · − . · − − . · − . · − . · − . · − − . · − Table 2.
Quotients of successive coefficients of 1 / E k for k ∈ { , , , , } .Since Θ(E )( q i ) = − E ( q i ) and Θ(E )( q i ) = − E ( q i ) , the second part of theCorollary also follows from Theorem 1 and (19). An approximate numerical valueof E ( q i ) can be read off Table 3. The theorem by Nesterenko implies that thisnumber is transcendental, since E ( q i ) = 0. (cid:3) n β ( n ) β ( n ) ≈ . . . . . · · ·
80 1 . . . Table 3.
Quotients of β ( n ) and β ( n ).Note that for each integer ℓ ≥
2, the limit as n → ∞ of β ℓ − ( n ) β ℓ +2 ( n ) exists, but it isgenerally not equal to E ( q i ). Proof of Proposition 1.
Since (1 / E ) m has only the pole q ρ on the circle of con-vergence, again we have formula (28), which proves the proposition. (cid:3) Proof of Theorem 1.
Let w be any complex number. Let B r ( w ) = { z ∈ C : | z − w | < r } be the open ball with radius r around w . We denote the closureby B r ( w ) and its boundary by ∂B r ( w ). Let k ≡ k has thespecial property that restricted to B | q i | (0) it has exactly one zero at q i , which isalso simple. This implies that the Taylor series expansion of the reciprocal of E k has radius of convergence | q i | and only a simple pole at q i :(30) 1E k ( q ) = ∞ X n =0 β k ( n ) q n ( | q | < | q i | ) . Note that subtracting the principal part at q i provides a new Taylor series expan-sion with a larger radius of convergence:(31) 1E k ( q ) − res q i (1 / E k ) q − q i = ∞ X n =0 b ( n ) q n . SYMPTOTIC EXPANSION 9
This implies that b ( n ) q ni constitutes a zero sequence. Here, res q i (1 / E k ) denotesthe residue at the pole q i . We obtain that(32) q n +1 i β k ( n ) + res q i (1 / E k )constitutes a zero sequence. By a standard argument, we obtain that(33) res q i (1 / E k ) = 1 dd q E k ( q i ) . Finally, we obtain the asymptotic behavior(34) β k ( n ) ∼ − k )( q i ) q − ni . (cid:3) Proof of Theorem 2.
We use the following easy to prove lemmata.
Lemma 1. σ ℓ ( n ) < ℓℓ − n ℓ for ℓ > and σ ( n ) ≤ (1 + ln n ) n .Proof. σ ℓ ( n ) ≤ (cid:0) R n t − ℓ d t (cid:1) n ℓ < ℓℓ − n ℓ for ℓ > ≤ (1 + ln n ) n for ℓ =1. (cid:3) Lemma 2.
For ℓ ≥ holds ℓ q − ℓ − ℓ > . .Proof. Considering ℓ as a real variable ≥
5, we obtain the following logarithmicderivative dd ℓ ℓ ln (cid:18) − ℓ − ℓ (cid:19) = − ℓ ln (cid:18) − ℓ − ℓ (cid:19) + 1 ℓ − ℓ − ℓ (cid:18) − − ℓ ln 31 + 3 − ℓ + 2 − ℓ ln 21 + 2 − ℓ (cid:19) > − ln 33 ℓ +1 + ln 22 ℓ +1 > − ln 33 ℓ + ln 22 ℓ +1 > ℓ ≥
5. Therefore, the values of theoriginal sequence are increasing and we take the smallest value for ℓ = 5. (cid:3) Proof of Theorem 2.
With ε k ( n ) = kB k σ k − ( n ) we obtain E k ( z ) = 1 − ∞ X n =1 ε k ( n ) q n . Let 1 / (1 − ε k (1) q − ε k (2) q ) = P ∞ n =0 α k ( n ) q n . The α k ( n ) fulfill the recur-rence relation α k ( n ) = ε k (1) α k ( n −
1) + ε k (2) α k ( n −
2) for n ≥
2. Obviously, α k (0) = β k (0), α k (1) = β k (1), and by induction α k ( n ) = ε k (1) α k ( n −
1) + ε k (2) α k ( n − ≤ P nj =1 ε k ( j ) β k ( n − j ) = β k ( n ) using the power series expan-sion of 1 / E k .For the upper bound let a = p / k ≥ a k = ε k (3) ε k (2) = σ k − (3) σ k − (2) = k − +12 k − +1 . For all k ≡ b k = a k + ε k (1), c k = ε k (2) − a k ε k (1), and − b k q − c k q − a k q = 1 − P ∞ n =1 δ k ( n ) q n . Therefore, δ k (1) = b k − a k = ε k (1), δ k (2) = c k + a k δ k (1) = ε k (2), and δ k ( n ) = a k δ k ( n −
1) for n ≥
3. Therefore δ k ( n ) = ε k (2) a n − k . (1) First, let k = 2. Then δ ( n ) = 72 (7 / ( n − / . For n ∈ { , , , } we ob-tain 24 σ ( n ) ≤ δ ( n ). Using Lemma 1 we obtain ε ( n ) ≤
24 (1 + ln n ) n .For n = 7 we obtain 24 · (1 + ln 7) · < <
72 (7 / (7 − / and for n ≥ n +1)1+ln n n +1 n ≤ (cid:18) ln ( ) n (cid:19) < . < p /
3. Therefore, ε ( n ) ≤ δ ( n ).(2) Now, let k ≥ δ k ( n ) = ε k (3) a n − k = 2 kB k (cid:0) k − + 1 (cid:1) (cid:18) (cid:19) k − − k − k ! n − . Using Lemma 1 we obtain σ k − ( n ) < k − k − n k − . Since k ≥ k − k − ≤ = 1 + < (cid:0) (cid:1) ≤ (cid:0) (cid:1) k − . Therefore k − r B k k ε k ( n ) = k − p σ k − ( n ) < k − r k − k − n k − < n. Using Lemma 2 implies k − q B k k δ k ( n ) > . (cid:0) (cid:1) n − . Now n < . (cid:0) (cid:1) n − for n ≥ . < .
47 for n = 4 and nn − < for n > ε k ( n ) = δ k ( n ) for n ∈ { , } and ε k ( n ) ≤ δ k ( n ) for all n ≥ − a k q − b k q − c k q = P ∞ n =0 γ k ( n ) q n . Then β k ( n ) = γ k ( n ) for n ∈ { , } andby induction γ k ( n ) = P nj =1 δ k ( j ) γ k ( n − j ) ≥ P nj =1 ε k ( j ) β k ( n − j ) = β k ( n ) for n ≥ α k ( n ) ≤ β k ( n ) ≤ γ k ( n ) for all n ≥
1. From the generatingfunctions we can now determine formulas for α k ( n ) and γ k ( n ). The characteristicequation for α k ( n ) is λ k − ε k (1) λ k − ε k (2) = 0. Let ∆ k = ε k (1) + 4 ε k (2) = (cid:16) kB k (cid:17) + kB k (cid:0) k − + 1 (cid:1) . Then λ k, ± = (cid:0) ε k (1) ± √ ∆ k (cid:1) . We obtain (cid:18) L k, + L k, − (cid:19) = (cid:18) λ k, + λ k, − (cid:19) − (cid:18) ε k (1) (cid:19) = λ k, + − λ k, − (cid:18) ε k (1) − λ k, − λ k, + − ε k (1) (cid:19) = √ ∆ k (cid:18) λ k, + − λ k, − (cid:19) .Therefore, α k ( n ) = L k, + λ nk, + + L k, − λ nk, − = λ n +1 k, + − λ n +1 k, − √ ∆ k for all n .The characteristic equation for γ k ( n ) is µ k − b k µ k − c k = 0. Let D k = b k + 4 c k .Then µ k, ± = (cid:0) b k ± √ D k (cid:1) , (cid:18) M k, + M k, − (cid:19) = (cid:18) µ k, + µ k, − (cid:19) − (cid:18) ε k (1) (cid:19) = 1 √ D k (cid:18) ε k (1) − µ k, − µ k, + − ε k (1) (cid:19) , and γ k ( n ) = M k, + µ nk, + + M k, − µ nk, − . (cid:3) Example (Slight improvement of [HR18B]) . Let k = 6. Then α ( n ) = 1 √
504 + √ ! n +1 − − √ ! n +1 ≈ . (cid:0) . n +1 − ( − . n +1 (cid:1) . SYMPTOTIC EXPANSION 11 n n +1 − ( − n +1 α ( n ) β ( n ) γ ( n ) · . n +21( − n . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · . · Table 4.
Improvement of upper and lower bounds (approxima-tion) for β ( n ).With x = B = 504, a = , b = , c = , D = and √ D ≈ .
59 we obtain µ , ± = b ±√ D , M , + = 1 √ D (cid:18) x − b − √ D (cid:19) , M , − = 1 √ D (cid:18) b + √ D − x (cid:19) . By (16) this finally yields γ ( n ) = M , + µ n , + + M , − µ n , − ≈ . · . n + 31 . · ( − . n . . The second and last column in Table 4 are the lower and upper bounds from[HR18B].3.3.
Proof of Theorem 3.
For the special case of k = 4 we refer to a resultof [HN20B]. We have proven that ( − n β ( n ) ∈ N for all n ∈ N (see also[AKN97], last section, for an announcement of the result of strict sign changes).We are mainly interested in the implication β ( n ) = 0. Proof of Theorem 3.
Let k = 4. Then z = ρ and J ( z ) = ρ −
1. This impliesthat 1 / E ( q ) = P ∞ n =0 β ( n ) q n has | q ρ | as the radius of convergence. Further, theonly singularity on the circle of convergence is given by the pole q ρ . Now we canproceed as in the proof of Theorem 1 and obtain the asymptotic expansion of β ( n ). Here we use the fact that res q ρ E − is equal to(35) q ρ Θ (E ) ( q ρ ) = − q ρ E ( q ρ ) . (cid:3) Proof of Theorem 4 and Theorem 5.
Proof of Theorem 4.
Let k ≡ E k which contribute to poles on the circle of convergence of the power series(36) 1E k ( q ) = ∞ X n =0 β k ( n ) q n . Let k ≥
12 then Proposition 2 and Corollary 2 imply that there are exactly twosingularities provided by the two poles at q z k and q z k . This implies that theradius of convergence is equal to | q z k | . Here we also used the well-known fact,that the imaginary part of γ ( z ), when γ is in the modular group and z in thefundamental domain, does not increase. Next we consider the Laurent expansionof 1 / E k ( q ) around q z k . We subtract the principal part from 1 / E k ( q ) and obtain aholomorphic function at q z k . We iterate this procedure and consider the Laurentexpansion around the other pole q z k and subtract again the principal part. Notethat we have poles of order one. This implies that(37) 1E k ( q ) − res q zk E − k q − q z k − res q zk E − k q − q z k has a holomorphic expansion P ∞ n =0 b ( n ) q n , with a radius of convergence largerthan | q z k | = | q z k | . This implies that b ( n ) q nz k and b ( n ) q nz k constitute zero sequences.The residue values can be expressed by Θ(E k ) evaluated at the poles. This leads toan expression which allows in the final formula the number q − nz k to appear insteadof q − ( n +1) z k . See also the proof of Theorem 1. By the identity principle b ( n ) is equalto(38) β k ( n ) + 1Θ (E k ) ( q z k ) q − nz k + 1Θ (E k ) ( q z k ) q − nz k . This implies that ∞ X n =0 (cid:18) β k ( n ) + 1Θ (E k ) ( q z k ) q − nz k + 1Θ (E k ) ( q z k ) q − nz k (cid:19) q n = ∞ X n =0 b ( n ) q nz k (cid:18) qq z k (cid:19) n for q ∈ C and | q | < | q z k | . Let w = q/q z k . Then ∞ X n =0 (cid:18) β k ( n ) q nz k + 1Θ (E k ) ( q z k ) + 1Θ (E k ) ( q z k ) (cid:18) q z k q z k (cid:19) n (cid:19) w n = ∞ X n =0 b ( n ) q nz k w n . In the final step we compare the coefficients with respect to w n and use the identityprinciple for regular power series. Since b ( n ) q nz k constitutes a zero sequence, theclaim of the theorem follows. (cid:3) Proof of Theorem 5.
Let k ≡ k ≥
12. Let z k = x k + iy k be thezero of E k in F with the largest imaginary part. Then z k = i, ρ . This impliesby results by Kanou [K00] and Kohnen [K03] that z k is transcendental. Since wehave chosen z k on the circle of unity, we can conclude that x k and y k are alsotranscendental. By a well-known result by Kronecker [K84], since x k is irrational,the orbit(39) O k := (cid:26)(cid:18) q z k q z k (cid:19) n : n ∈ N (cid:27) SYMPTOTIC EXPANSION 13 is dense in { w = e πiα : α ∈ [0 , } . Let C k := 1 / Θ(E k )( q z k ). Since(40) C k = 1 / Θ(E k )( q z k ) , for the closure of the set, we obtain(41) D k := (cid:26) k )( q z k ) + 1Θ(E k )( q z k ) (cid:18) q z k q z k (cid:19) n : n ∈ N (cid:27) a circle with center C k and radius | C k | :(42) ∂B | C k | ( C k ) = n z ∈ C : | z − C k | = | C k | o . We note that 0 and 2 C k are not elements of D k . Let d k ∈ ∂B | C k | ( C k ) \ { } . Thenthere exists a subsequence { n t } ∞ t =1 of { n } ∞ n =1 such that,(43) lim t →∞ k )( q z k ) + 1Θ(E k )( q z k ) (cid:18) q z k q z k (cid:19) n t = d k Combining this result with Theorem 4 proves the claim. (cid:3)
Proof of Theorem 6 and Corollary 3.
We recall a result from complexanalysis. P´olya and Szeg˝o recorded the following beautiful property ([PS78], PartThree, Chapter 5). Let f ( x ) = P ∞ n =0 a ( n ) x n be a power series with radius ofconvergence 0 < r < ∞ and real coefficients. We assume that we have onlytwo singularities on the circle of convergence and that these two singularities arepoles: x = re iα and x = re − iα with 0 < α < π . Let A ( n ) denote the number ofchanges of sign in the sequence { a ( m ) } nm =0 . Then lim n →∞ A ( n ) n = απ . The numberof changes of sign in a sequence of real numbers is given by the sign changes ofthe sequence, when all zeros are removed. Results in this direction had also beengiven by K¨onig [K75] in 1875. Proof of Theorem 6, part a).
Let k ≡ / E k ( q ) = P ∞ n =0 β k ( n ) q n has a radius of convergence | q z k | , where z k = x k + iy k is the zero of E k with thelargest imaginary part with 0 < x k < /
2. We stated already that q z k and q z k arethe single two singularities on the circle of convergence. Note that q z k = r k · e πix k ,where r k = e − πy k = | q z k | . Further, q z k = r k · e − πix k . Thus all assumptions arefulfilled to apply the above cited result for A ( n ) = A k ( n ) and α = 2 x k . (cid:3) Example . We have z ≈ . . i . See Table 5 for values A ( n ) /n .We also recall another interesting result stated in [PS78] (Part Three, Chapter5). Let f ( x ) = P ∞ n =0 a ( n ) x n be a power series with finite positive radius ofconvergence. We assume that there are only poles on the circle of convergence.Let B ( n ) be the number of non-zero coefficients among the first n coefficients { a ( m ) } n − m =0 . Then the number of poles is not smaller than(44) lim sup n →∞ nB ( n ) . Proof of Theorem 6, part b).
The number of poles is 2. Thus, by the result above,two is an upper bound for the term (44), which completes the proof. (cid:3) n A ( n ) n ≈ A (10 n )10 n ≈ A (100 n )100 n ≈ . . . . . . . . . . . . . . . . . . . . . . . . . . . Table 5.
Portion of sign changes for k = 16. Example . We have B ( n ) = B ( n ) = B ( n ) = n for n ≤ Proof of Corollary 3.
From Theorem 6 we obtain(45) lim n →∞ A ℓ ℓ = 2 x ℓ , where x ℓ is the real part of the zero of E ℓ with the largest imaginary part.Finally, from Corollary 2 the claim follows, since x ℓ tends to zero. (cid:3) Acknowledgments.
To be entered later.
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