On the double zeros of a partial theta function
aa r X i v : . [ m a t h . C A ] A p r On the double zeros of a partial theta function
Vladimir Petrov KostovUniversit´e de Nice, Laboratoire de Math´ematiques, Parc Valrose,06108 Nice Cedex 2, France, e-mail: [email protected]
Abstract
The series θ ( q, x ) := P ∞ j =0 q j ( j +1) / x j converges for q ∈ [0 , x ∈ R , and defines a partialtheta function . For any fixed q ∈ (0 ,
1) it has infinitely many negative zeros. For q taking oneof the spectral values ˜ q , ˜ q , . . . (where 0 . . . . = ˜ q < ˜ q < · · · <
1, lim j →∞ ˜ q j = 1)the function θ ( q, . ) has a double zero y j which is the rightmost of its real zeros (the rest ofthem being simple). For q = ˜ q j the partial theta function has no multiple real zeros. Weprove that ˜ q j = 1 − π/ j + (log j ) / j + O (1 /j ) and y j = − e π e − (log j ) / j + O (1 /j ) . AMS classification:
Key words: partial theta function; spectrum; asymptotics
Consider the bivariate series θ ( q, x ) := P ∞ j =0 q j ( j +1) / x j . For each fixed q of the open unit diskit defines an entire function in x called a partial theta function . This terminology is explainedby the fact that the Jacobi theta function is the sum of the series P ∞ j = −∞ q j x j and one has θ ( q , x/q ) = P ∞ j =0 q j x j . There are different domains in which the function θ finds applications:asymptotic analysis ([2]), statistical physics and combinatorics ([16]), Ramanujan type q -series([17]) and the theory of (mock) modular forms ([3]). See also [1] for more information aboutthis function.The function θ satisfies the following functional equation: θ ( q, x ) = 1 + qxθ ( q, qx ) (1)In what follows we consider q as a parameter and x as a variable. We treat only the case q ∈ (0 , x ∈ R . For each fixed q the function θ ( q, . ) has infinitely many negative zeros. (Ithas no positive zeros because its Taylor coefficients are all positive.) There exists a sequence ofvalues ˜ q j of q (called spectral values ) such that 0 . . . . = ˜ q < ˜ q < · · · (where ˜ q j → − as j → ∞ ) for which and only for which the function θ ( q, . ) has a multiple real zero y j (see [12]and [9]). This zero is negative, of multiplicity 2 and is the rightmost of its real zeros. The restof them are simple. The function θ (˜ q j , . ) has a local minimum at y j . As q increases and passesfrom ˜ q − j to ˜ q + j , the rightmost two real zeros coalesce and give birth to a complex conjugate pair.The double zero of θ (˜ q , . ) equals − . . . . . The spectral value ˜ q is of interest inthe context of a problem due to Hardy, Petrovitch and Hutchinson, see [4], [14], [5], [13], [6]and [12]. The following asymptotic formula and limit are proved in [10]:˜ q j = 1 − π/ j + o (1 /j ) , lim j →∞ y j = − e π = − . . . . (2)1n the present paper we make this result more precise: Theorem 1.
The following asymptotic estimates hold true: ˜ q j = 1 − π/ j + (log j ) / j + b/j + o (1 /j ) y j = − e π e − (log j ) / j + α/j + o (1 /j ) , (3) where b ∈ [1 . . . . , . . . . ] and α = − π/ − b + π / hence α ∈ [ − . . . . , − . . . . ] . The first several numbers y j form a monotone decreasing sequence. We list the first five ofthem: − . . . . , − . . . . , − . . . . , − . . . . , − . . . . . The theorem implies that for j large enough the sequence must also be decreasing and gives anidea about the rate with which the sequences { ˜ q j } and { y j } tend to their limit values. Acknowledgement.
The author has discussed partial theta functions with B.Z. Shapiroduring his visits to the University of Stockholm and with A. Sokal by e-mail. V. Katsnelson hassent to the author an earlier version of his paper [7]. To all of them the author expresses hismost sincere gratitude.
For q ∈ (0 , ˜ q ) the function θ ( q, . ) has only simple negative zeros which we denote by ξ j , where · · · < ξ < ξ < q ∈ (0 , θ ( q, x ) = ∞ Y j =1 (1 − x/ξ j ) (4)For q ∈ [˜ q ,
1) the indexation of the zeros is such that zeros change continuously as q varies.For q ∈ (0 , ˜ q ) all derivatives ( ∂ k θ/∂x k )( q, . ) have only simple negative zeros. For k = 1 thismeans that the numbers t s and w s , where the function θ ( q, . ) has respectively local minima andmaxima, satisfy the string of inequalities · · · < t s +1 < ξ s +1 < w s < ξ s < t s < ξ s − < · · · < . (5)The above inequalities hold true for any q ∈ (0 ,
1) whenever ξ s − is real negative (which impliesthat this is also the case of ξ j for j > s − Lemma 2.
Suppose that q ∈ (˜ q j , ˜ q j +1 ] (we set ˜ q = 0 ). Then for s ≥ j + 1 one has t s +1 ≤ w s /q and w s ≤ t s /q .Proof. Equality (1) implies( ∂θ/∂x )( q, x ) = xq ( ∂θ/∂x )( q, qx ) + θ ( q, qx ) . (6)When qx = t s and s ≥ j + 1, then θ ( q, t s ) ≤
0, ( ∂θ/∂x )( q, t s ) = 0 and ( ∂θ/∂x )( q, t s /q ) ≤ t s /q , i. e. w s ≤ t s /q .2n the same way if one sets qx = w s , one gets θ ( q, w s ) ≥
0, ( ∂θ/∂x )( q, w s ) = 0 and( ∂θ/∂x )( q, w s /q ) ≥ t s +1 ≤ w s /q . Notation 3. (1) In what follows we are using the numbers u s := − q − s +1 / and v s := − q − s − / .(2) We denote by ˜ r s the solution to the equation θ ( q, u s ) = 0 and we set z s := − (˜ r s ) − s +1 / . Remark 4.
For s sufficiently large the equation θ ( q, u s ) = 0 has a unique solution. This followsfrom part (2) of Lemma 15.It is shown in [9] that · · · < ξ < u < ξ < − q − < ξ /q < v < ξ /q < − q − < ξ < u < ξ < − q − < . (7)These inequalities hold true for q > q ∈ (0 ,
1) if the index j of ξ j is sufficiently large.Comparing the inequalities (5) and (7) we see that ξ s +1 < w s , v s < ξ s and ξ s < t s , u s <ξ s − . In this sense we say that a number u s (resp. v s ) corresponds to a local minimum (resp.maximum) of θ ( q, . ).We prove the following theorems respectively in Sections 5 and 3: Theorem 5.
The following asymptotic estimates hold true: ˜ r j = 1 − π/ j + (log j ) / j + b ∗ /j + o (1 /j ) z j = − e π e − (log j ) / j + α ∗ /j + o (1 /j ) , (8) where b ∗ ∈ [1 . . . . , . . . . ] and α ∗ = − π/ − b ∗ + π / hence α ∗ ∈ [ − . . . . , − . . . . ] . Theorem 6.
For j sufficiently large one has < ˜ r j ≤ ˜ q j ≤ ˜ r j +1 < . Theorem 1 follows from the above two theorems. Indeed, as1 − π/ j + (log j ) / j + O (1 /j )) = ˜ r j ≤ ˜ q j and1 − π/ j + 1) + (log( j + 1)) / j + 1) + O (1 / ( j + 1) ) = ˜ r j +1 ≥ ˜ q j one deduces immediately the equality ˜ q j = 1 − π/ j + o (1 /j ) ( A ). It is also clear that ˜ r j +1 =1 − π/ j + (log j ) / j + O (1 /j ).To obtain an estimate of the term O (1 /j ) recall that˜ r j = 1 − π/ j + (log j ) / j + b ∗ /j + o (1 /j ) hence˜ r j +1 = 1 − π/ j + (log j ) / j + ( b ∗ + π/ /j + o (1 /j )(we use the equality 1 / ( j +1) = 1 /j − /j ( j +1)). This implies ˜ q j = 1 − π/ j +(log j ) / j + b/j + o (1 /j ), where b ∈ [ b ∗ , b ∗ + π/
2] hence b ∈ [1 . . . . , . . . . ]. The quantities α and α ∗ are expressed by similar formulas via b and b ∗ , see Theorems 1 and 5. This gives theclosed intervals to which α and α ∗ belong.Section 4 contains properties of the function ψ used in the proofs. At first reading one canread only the statements of Theorem 10 and Proposition 11 from that section.3 Proof of Theorem 6
We prove first the inequality ˜ r j ≤ ˜ q j . When q increases and becomes equal to a spectral value˜ q j , then two negative zeros of θ coalesce. The corresponding double zero of θ (˜ q j , . ) is a localminimum. It equals t j . Hence for some value of q not greater than ˜ q j one has θ ( q, u j ) = 0. Thisvalue is ˜ r j , see Notation 3.To prove the inequality ˜ q j ≤ ˜ r j +1 ( ∗ ) we use a result due to V. Katsnelson, see [7]: The sum of the series P ∞ j =0 q j ( j +1) / x j (considered for q ∈ (0 , and x complex) tends to / (1 − x ) (for x fixed and as q → − ) exactly when x belongs to the interior of the closed Jordancurve { e | s | + is , s ∈ [ − π, π ] } . Hence in particular θ ( q, x ) converges to 1 / (1 − x ) as q → − for each fixed x ∈ ( − e π , j sufficiently large one has y j < −
23 (because the function 1 / (1 − x ) has no zeroson ( −∞ ,
0] and 23 < e π ). Proposition 7.
For j sufficiently large one has θ (˜ q j , v j ) > / . Before proving the proposition we deduce the inequality ( ∗ ) from it. Recall that ˜ q j ≥ ˜ q > . θ (˜ q j , u j +1 ) = 1 + ˜ q j u j +1 θ (˜ q j , v j ) < − . × × (1 /
3) = − . < . Hence for q = ˜ q j the value of θ ( q, u j +1 ) is still negative, i. e. one has θ ( q, u j +1 ) = 0 for somevalue q > ˜ q j . Proof of Proposition 7.
We deduce the proposition from the following two lemmas:
Lemma 8.
Suppose that the quantity v j is computed for q equal to the spectral value ˜ q j . Set Ξ j := ( − ˜ q − jj − v j (˜ q j )) / ( − ˜ q − j + ˜ q − j − ) . Then lim j →∞ Ξ j = 1 / . The next lemma considers certain points of the graph of θ (˜ q j , . ). Recall that θ (˜ q j , w j ) = 1because for q = ˜ q j one has θ (˜ q j , t j ) = 0, w j = t j / ˜ q j and by equation (1) θ (˜ q j , w j ) = 1 +˜ q j w j θ (˜ q j , t j ) = 1. Lemma 9.
The point ( v j , θ (˜ q j , v j )) lies above or on the straight line passing through the twopoints ( − ˜ q − jj , θ (˜ q j , − ˜ q − jj )) and ( w j , θ (˜ q j , w j )) = ( w j , . The two lemmas imply that for j sufficiently large the following inequality holds true: θ (˜ q j , v j ) − θ (˜ q j , − ˜ q − jj ) ≥ Ξ j ( θ (˜ q j , w j ) − θ (˜ q j , − ˜ q − jj )) > (1 / − θ (˜ q j , − ˜ q − jj )) . Hence θ (˜ q j , v j ) > / / θ (˜ q j , − ˜ q − jj ). It is shown in [9] (see Proposition 9 there) that for q ∈ (0 ,
1) one has θ ( q, − q − s ) ∈ (0 , q s ), s ∈ N . Hence θ (˜ q j , v j ) > / Proof of Lemma 8.
It is clear that ( − ˜ q − jj − v j (˜ q j )) / ( − ˜ q − jj + ˜ q − j − j ) = p ˜ q j (1 − p ˜ q j ) / (1 − ˜ q j ) = p ˜ q j / (1 + p ˜ q j ). As j → ∞ one has ˜ q j → / Proof of Lemma 9.
We are going to prove a more general statement from which the lemmafollows. Suppose that w s ≤ x < x < x ≤ − q − s < − < s ∈ N . Set θ ( q, x ) = C , θ ( q, x ) = B , θ ( q, x ) = A . Suppose that A > B > C > A ≥
1. We use the letters A , B and C also to denote the points of the graph of θ ( q, . ) with coordinates ( x , C ), ( x , B ) and( x , A ). 4 e prove that the point B is above or on the straight line AC . Indeed, suppose that the point B is below the straight line AC . Then B − C | x − x | < A − B | x − x | (9)Consider the points ( x /q, C ′ ), ( x /q, B ′ ) and ( x /q, A ′ ) of the graph of θ ( q, . ). Equation (1)implies that C ′ = 1 + x C , B ′ = 1 + x B , A ′ = 1 + x A .
In the same way for the points ( x /q , C ′′ ), ( x /q , B ′′ ) and ( x /q , A ′′ ) of the graph of θ ( q, . )one gets C ′′ = 1 + x q C ′ B ′′ = 1 + x q B A ′′ = 1 + x q A ′ = 1 + x q + x q C = 1 + x q + x q B = 1 + x q + x q A . (10)It is clear that x /q < x /q < x /q ≤ − q − s − . By Lemma 2 one has w s +1 ≤ t s +1 /q ≤ w s /q ≤ x /q ≤ − q − s − . The inequalities
A > B > C > x < x < x < − A ′ < B ′ < C ′ , see (10). As A ≥ x < −
1, one has A ′ <
0. Therefore A ′′ = 1 + ( x /q ) A ′ > w s +1 ≤ x /q < x /q < x /q ≤ − q − s − that B ′′ > C ′′ >
0. (Indeed, θ ( q, x ) > x ∈ ( − q − s − , − q − s ), see [9].) If B ′ < C ′ <
0, then x < x < x < − A ′′ > B ′′ > C ′′ , see (10). If B ′ < ≤ C ′ , then again by (10) one gets A ′′ > B ′′ > C ′′ .If 0 ≤ B ′ < C ′ , then one obtains A ′′ > B ′′ and A ′′ > C ′′ . If B ′′ ≤ C ′′ , then the point B ′′ lies below the straight line A ′′ C ′′ . In this case one can find a point ( x /q , B ∗′′ ) of the graph of θ ( q, . ) such that w s +1 ≤ x /q < x /q < x /q < x /q ≤ − q − s − < − < , A ′′ > B ∗′′ > C ′′ > B ∗′′ lies below the straight line A ′′ B ′′ .Suppose that A ′′ > B ′′ > C ′′ . We show that the point B ′′ is below the straight line A ′′ C ′′ .This is equivalent to proving that( x /q ) + ( x /q ) B − ( x /q ) − ( x /q ) C | x − x | /q < ( x /q ) + ( x /q ) A − ( x /q ) − ( x /q ) B | x − x | /q or to proving the inequality B ( x /x ) | x − x | − C | x − x | − A ( x /x ) | x − x | < . (11)Inequality (9) can be given another presentation: B | x − x | − C | x − x | − A | x − x | < . (12)One can notice that inequality (11) (which we want to prove) is the sum of inequality (12)(which is true) and the inequality 5 (cid:18) x x − (cid:19) | x − x | − A (cid:18) x x − (cid:19) | x − x | < . (13)So if we show that inequality (13) is true, then this will imply that inequality (11) is also true.Recall that x < x < x < A > B > C >
0. Hence inequality (13) is equivalent to B ( | x | + | x | ) − A ( | x | + | x | ) < B ∗′′ = B ′′ and x = x .For s ∈ N we define in the same way the three points ( x /q s , C (2 s ) ), ( x s /q s , B ∗ (2 s ) ) and( x /q s , A (2 s ) ) by the condition that they belong to the graph of θ ( q, . ), w s ≤ x /q s < x s /q s
Figure 1: Part of the graph of a partial theta function and the points A , B , C and C ′′ .This implies that the graph of θ ( q, . ) has on each interval ( − q − s − N , − q − s ), N ∈ N , at least O (4 N ) inflection points, twice as much as O (2 N ), the one that should be. (This contradictionproves the lemma.) Indeed, on Fig. 1 we show part of the graph of θ ( q, . ) as it should look like(the sinusoidal curve) and the points C , B , A and C ′′ . If the point B is below the straight line AC , then the change of convexity requires two more inflection points between a local minimumof θ ( q, . ) and the local maximum to its left. 6 The function ψ In the present section we consider the function ψ ( q ) := 1 + 2 P ∞ j =1 ( − j q j . It is real-analyticon ( − , Theorem 10. (1) By the Jacobi triple product identity the function ψ can be expressed as follows(see [15], Chapter 1, Problem 56): ψ ( q ) = ∞ Y j =1 − q j q j (14) (2) The function ψ is decreasing, i.e. ψ ′ < for all q ∈ ( − , .(3) For the endpoints of its interval of definition one has the limits lim q → − ψ ( q ) = 0 , lim q →− + ψ ( q ) = + ∞ .(4) The function ψ is flat at , i.e. for any l ∈ N , ψ ( q ) = o (( q − l ) as q → − .(5) The function ψ is convex, i.e. ψ ′′ ≥ for all q ∈ ( − , , with equality only for q = 0 .(6) Consider the function τ ( q ) := ( q −
1) log ψ ( q ) . It is increasing on (0 , and lim q → − τ ( q ) = π / . This implies that for any ε > there exists δ > such that e π q − < ψ ( q ) ≤ e π − ε q − for q ∈ (1 − δ, .(7) As q → − + , the growth rate of the function ψ satisfies the conditions ψ ( q ) = o (( q +1) − ) and ( q + 1) α /ψ ( q ) = o (1) for any α ∈ ( − , . Set D := 1 / π / . . . . . Property (6) can be further detailed: Proposition 11.
For q close to the function τ is of the form τ = π / / − q ) log(1 − q ) − K (1 − q ) + o (1 − q ) with K ∈ [ D, D + 1 /
12] = [2 . . . . , . . . . ] . Hence ψ = e τq − = e K (1 + o (1)) (1 − q ) − e π q − .Proof. The logarithm of the j th factor of the right-hand side of formula (14) equalslog 1 − q j q j = ( − (cid:18) q j + q j q j · · · (cid:19) . (15)This means that log ψ ( q ) := ( − ∞ X j =1 (cid:18) q j + q j q j · · · (cid:19) = ( − ∞ X k =0 q k +1 (2 k + 1)(1 − q k +1 ) . (16)Hence τ ( q ) := ( q −
1) log ψ ( q ) = 2 P ∞ k =0 ζ k ( q ), where ζ k ( q ) := q k +1 (2 k + 1)(1 + q + q + · · · + q k ) = q k +1 (1 − q )(2 k + 1)(1 − q k +1 ) . emma 12. For q ∈ (0 , the following inequalities hold true: q k +1 / (2 k + 1) ≤ ζ k ( q ) ≤ q k +1 / (2 k + 1) with equalities only for q = 1 .Proof of Lemma 12. The inequalities result from 1 + q + · · · + q k ≤ k + 1 and q k + j + q k − j ≥ q k hence 1 + q + · · · + q k ≥ (2 k + 1) q k (with equalities only for q = 1).The above lemma gives the idea to compare the function τ (for q close to 1) with the function h ( q ) := 2 P ∞ k =0 q k +1 / (2 k + 1) . The lemma implies the following result: h ( q ) /q ≤ τ ( q ) ≤ h ( q ) , ( τ ( q ) = h ( q )) ⇐⇒ ( q = 1) (17)Our next step is to compare the asymptotic expansions of the functions τ and h close to 1: Lemma 13.
For q close to the following equality holds: h ( q ) = π / / − q ) log(1 − q ) − D (1 − q ) + O ((1 − q ) log(1 − q )) (18) Proof of Lemma 13.
Notice first that lim q → − τ ( q ) = h (1) = π / h = h + h , where h := 2 ∞ X k =0 q k +1 / (2 k + 1)(2 k + 2) and h := 2 ∞ X k =0 q k +1 / (2 k + 1) (2 k + 2) . Equation (15) implies log((1 + q ) / (1 − q )) = 2 P ∞ k =0 q k +1 / (2 k + 1). Integrating both sides ofthis equality yields(1 + q ) log(1 + q ) + (1 − q ) log(1 − q ) = 2 ∞ X k =0 q k +2 / (2 k + 1)(2 k + 2) . Thus h = (1 + q / ) log(1 + q / ) + (1 − q / ) log(1 − q / ). The first summand is real analyticin a neighbourhood of 1 and equals 2 log 2 − (1 / − q ) + O ((1 − q ) ). The secondone is equal to[ (1 − q ) / (1 + q / ) ] (log(1 − q ) − log(1 + q / )) =(1 − q ) [ (1 / O (1 − q )) log(1 − q ) − (log(1 + q / )) / (1 + q / ) ] =(1 / − q ) log(1 − q ) − (1 / − q ) + O ((1 − q ) log(1 − q )) . About the function h one can notice that there exist the limits lim q → − h and lim q → − h ′ (thelatter equals π / Lemma 14.
For q ∈ (0 , it is true that h ( q ) − τ ( q ) ≤ (1 − q ) / . (19)Proposition 11 results from the last two lemmas.8 roof of Lemma 14. To prove formula (19) set R := 1 / (2 k + 1) (1 + q + · · · + q k ) and S l :=1 + q + · · · + q l . Hence12 k + 1 (cid:18) q k +1 k + 1 − q k +1 q + · · · + q k (cid:19) = ( q k +1 (1 + q + · · · + q k ) − (2 k + 1) q k +1 ) R = k − X j =0 ( q k +1+ j + q k +1 − j − q k +1 ) R = k − X j =0 q k +1+ j (1 − q k − j ) R = q k +1 (1 − q ) k − X j =0 q j (1 + q + · · · + q k − j − ) R = q k +1 (1 − q ) k − X j =0 q j k − j − X ν =0 q ν S k − j − − ν ! R . (20)The sums S l enjoy the following property:( l − S l ≥ ( l + 1) qS l − . (21)Indeed, this is equivalent to ( l − q l ) ≥ qS l − . The last inequality follows from 1 + q r ≥ q + q r − (i.e. (1 − q )(1 − q r − ) ≥
0) applied for suitable choices of the exponent r . Equation(21) implies the next property (whenever the indices are meaningful):( l − ν + 1) S l ≥ ( l + 1) q ν S l − ν . (22)Using equation (22) one can notice that the right-hand side of (20) is not larger than q k +1 (1 − q ) k − X j =0 q j k − j − X ν =0 k − j − − ν k − j − S k − j − ! R ≤ q k +1 (1 − q ) k − X j =0 k − j − X ν =0 (cid:18) k − j − − ν k − j − (cid:19) (cid:18) k − j − k − (cid:19) S k − ! R = q k +1 (1 − q ) k − X j =0 k − j − X ν =0 k − j − − ν k − S k − R = q k +1 (1 − q ) k − X j =0 ( k − j ) k − S k − R = q k +1 (1 − q ) k ( k + 1)(2 k + 1)6(2 k − k + 1) S k − S k ≤ q k (1 − q ) k ( k + 1)6(2 k + 1) . At the last line we used property (21) with l = 2 k . The last fraction is less than 1 /
24. Hence h ( q ) − τ ( q ) ≤ (1 − q ) P ∞ k =0 q k /
12 = (1 − q ) / Proof of Theorem 5
We follow the same path of reasoning as the one used in [10]. In this section we use the resultsof [10] and [8]. Set λ s ( q ) := ∞ X j =2 s ( − j q j / , χ s ( q ) := λ s ( q ) /q s = ∞ X j =2 s ( − j q j / − s = ∞ X j =0 ( − j q ( j +4 js ) / . The equation θ ( q, − q − s +1 / ) = 0 is equivalent to (see [10]) ψ ( q / ) = λ s ( q ) = q s χ s ( q ) . (23)The following lemma is also proved in [10]: Lemma 15. (1) One has lim q → − λ s ( q ) = lim q → − χ s ( q ) = 1 / .(2) For s ∈ N sufficiently large the graphs of the functions ψ ( q / ) and λ s ( q ) (considered for q ∈ [0 , ) intersect at exactly one point belonging to (0 , and at .(3) For q ∈ [0 , the inequality λ s ( q ) ≥ λ s +1 ( q ) holds true with equality for q = 0 and q = 1 .(4) For q ∈ [0 , one has / ≤ χ s ( q ) ≤ . Part (2) of the lemma implies that for each s sufficiently large the number ˜ r s is correctlydefined. Part (3) implies that the numbers ˜ r s form an increasing sequence. Indeed, this followsfrom ψ ( q / ) being a decreasing function, see part (2) of Theorem 10.Recall that the constant K was introduced by Proposition 11. Set q := ˜ r s = 1 − h s /s .Consider the equalities (23). The left-hand side is representable in the form e K (1 + o (1)) [ (1 − q ) − / / (1 + √ q ) − / ] e π ( √ q +1) / q − , see Proposition 11. Hence log ψ ( q / ) is of the form( π / − s/h s )(2 − (1 / h s /s ) + o ( h s /s )) − (1 /
2) log( h s /s ) + (1 /
2) log 2 + K + o (1)= − ( π / s/h s ) + (1 /
2) log s − (1 /
2) log h s + L + o (1) , where L := K + (1 /
2) log 2 + π /
8. The right-hand side of (23) equals (1 − h s /s ) s χ s (1 − h s /s ).Hence its logarithm is of the form(2 s ) log(1 − h s /s ) + log( χ s (1 − h s /s )) = − (2 s )( h s /s + h s / s + O (1 /s )) − log 2 + o (1)= − sh s − ( h s ) − log 2 + o (1) . (we use χ s = 1 / o (1), see part (1) of Lemma 15; hence log h s = log( π/
2) + o (1)). Set h s := π/ d s . Hence d s = o (1), see equality ( A ) after Theorem 6, and − ( π / s/ ( π/ d s )) + (1 /
2) log s − (1 /
2) log( π/
2) + L + o (1)= − s ( π/ d s ) − ( π/ d s ) − log 2 + o (1)10r equivalently − ( π / s + ( π/ d s )((log s ) / − (log( π/ / L ) + o (1)= − s ( π/ d s ) − ( π/ d s ) − ( π/ d s ) log 2 + o (1) (24)The terms − π / s cancel. Hence(( π/ d s ) /
2) log s = − sπd s − s ( d s ) + O (1)i.e. d s = − ((log s ) / s )(1 + o (1)). Set d s := − (log s ) / s + g s . Using equation (24) one gets − ( π / s + ( π/ − (log s ) / s + g s )((log s ) / − (log( π/ / L ) + o (1)= − s ( π/ − (log s ) / s + g s ) − ( π/ − (log s ) / s + g s ) − ( π/ − (log s ) / s + g s ) log 2(25)To find the main asymptotic term in the expansion of g s we have to leave only the linear termsin g s and the terms independent of g s (because g s = o ( g s )). The left-hand side of equation (25)takes the form: − ( π / s + ( π/ s ) − ( π/
4) log( π/
2) + ( π/ L + g s ((log s ) / o (1)) + o (1) . The right-hand side equals − s ( π/ + π (log s ) / − sπg s − π / − (3 π / g s − ( π/
2) log 2 − g s log 2 + o (1) . The terms s and log s cancel. The remaining terms give the equality(((log s ) / o (1)) + 2 sπ + O (1)) g s = ( π/
4) log( π/ − ( π/ L − π / − ( π/
2) log 2 + o (1) . Hence g s = (1 /s )( M + o (1)), where M := (log( π/ / − L/ − π / − (log 2) / π/ / − L/ − π / . Now recall that L = K + (1 /
2) log 2 + π / K ∈ [2 . . . . , . . . . ] . Hence M = − b ∗ = (log( π/ / − π / − K/
4, so b ∗ ∈ [1 . . . . , . . . . ].Thus we have proved the first of formulas (8). To prove the second one it suffices to noticethat z s = − (˜ r s ) − s +1 / = − (1 − π/ s + (log s ) / s + b ∗ /s + · · · ) − s +1 / . Set Φ := π/ s − (log s ) / s − b ∗ /s + · · · . Hence z s = − e ( − s +1 /
2) log(1 − Φ) = − e ( − s +1 / − Φ − Φ / −··· ) = − e π e − (log s ) / s + α ∗ /s + ··· with α ∗ = − π/ − b ∗ + π / ∈ [ − . . . . , − . . . . ].11 eferences [1] G. E. Andrews, B. C. Berndt, Ramanujan’s lost notebook. Part II. Springer, NY, 2009.[2] B. C. Berndt, B. Kim, Asymptotic expansions of certain partial theta functions. Proc.Amer. Math. Soc. 139 (2011), no. 11, 3779–3788.[3] K. Bringmann, A. Folsom, R. C. Rhoades, Partial theta functions and mock mod-ular forms as q -hypergeometric series, Ramanujan J. 29 (2012), no. 1-3, 295–310,http://arxiv.org/abs/1109.6560[4] G. H. Hardy, On the zeros of a class of integral functions, Messenger of Mathematics, 34(1904), 97–101.[5] J. I. Hutchinson, On a remarkable class of entire functions, Trans. Amer. Math. Soc. 25(1923), pp. 325–332.[6] O.M. Katkova, T. Lobova and A.M. Vishnyakova, On power series having sections withonly real zeros. Comput. Methods Funct. Theory 3 (2003), no. 2, 425–441.[7] B. Katsnelson, On summation of the Taylor series for the function 1 / (1 − zz