On a partial theta function and its spectrum
OOn a partial theta function and its spectrum
Vladimir Petrov KostovUniversit´e de Nice, Laboratoire de Math´ematiques, Parc Valrose,06108 Nice Cedex 2, France, e-mail: [email protected]
Abstract
The bivariate series θ ( q, x ) := (cid:80) ∞ j =0 q j ( j +1) / x j defines a partial theta function . For fixed q ( | q | < θ ( q, . ) is an entire function. For q ∈ ( − ,
0) the function θ ( q, . ) has infinitelymany negative and infinitely many positive real zeros. There exists a sequence { ¯ q j } of valuesof q tending to − + such that θ (¯ q k , . ) has a double real zero ¯ y k (the rest of its real zerosbeing simple). For k odd (resp. for k even) θ (¯ q k , . ) has a local minimum at ¯ y k and ¯ y k is therightmost of the real negative zeros of θ (¯ q k , . ) (resp. θ (¯ q k , . ) has a local maximum at ¯ y k andfor k sufficiently large ¯ y k is the second from the left of the real negative zeros of θ (¯ q k , . )).For k sufficiently large one has − < ¯ q k +1 < ¯ q k <
0. One has ¯ q k = 1 − ( π/ k ) + o (1 /k ) and | ¯ y k | → e π/ = 4 . . . . . AMS classification:
Keywords: partial theta function; spectrum
The bivariate series θ ( q, x ) := (cid:80) ∞ j =0 q j ( j +1) / x j defines an entire function in x for every fixed q from the open unit disk. This function is called a partial theta function because θ ( q , x/q ) = (cid:80) ∞ j =0 q j x j whereas the Jacobi theta function is defined by the same series, but when summationis performed from −∞ to ∞ (i.e. when summation is not partial).There are several domains in which the function θ finds applications: in the theory of (mock)modular forms (see [3]), in statistical physics and combinatorics (see [18]), in asymptotic analysis(see [2]) and in Ramanujan type q -series (see [19]). Recently it has been considered in the contextof problems about hyperbolic polynomials (i.e. real polynomials having all their zeros real, see[4], [16], [5], [15], [6], [13] and [9]). These problems have been studied earlier by Hardy, Petrovitchand Hutchinson (see [4], [5] and [16]). For more information about θ , see also [1].For q ∈ C , | q | ≤ . θ ( q, . ) has no multiple zeros, see [11]. For q ∈ [0 ,
1) thefunction θ has been studied in [13], [8], [9] and [10]. The results are summarized in the followingtheorem: Theorem 1. (1) For any q ∈ (0 , the function θ ( q, . ) has infinitely many negative zeros.(2) There exists a sequence of values of q (denoted by < ˜ q < ˜ q < · · · ) tending to − suchthat θ (˜ q k , . ) has a double negative zero y k which is the rightmost of its real zeros and which is alocal minimum of θ ( q, . ) . One has ˜ q = 0 . . . . .(3) For the remaining values of q ∈ [0 , the function θ ( q, . ) has no multiple real zero.(4) For q ∈ (˜ q k , ˜ q k +1 ) (we set ˜ q = 0 ) the function θ ( q, . ) has exactly k complex conjugatepairs of zeros counted with multiplicity.(5) One has ˜ q k = 1 − ( π/ k ) + o (1 /k ) and y k → − e π = − . . . . . a r X i v : . [ m a t h . C A ] A p r efinition 2. A value of q ∈ C , | q | <
1, is said to belong to the spectrum of θ if θ ( q, . ) has amultiple zero.In the present paper we consider the function θ in the case when q ∈ ( − , q ∈ [0 ,
1) one can notice the following fact. For q ∈ ( − ,
0] set v := − q . Then θ ( q, x ) = θ ( − v, x ) = θ ( v , − x /v ) − vxθ ( v , − vx ) . (1)We prove the analog of the above theorem. The following three theorems are proved respectivelyin Sections 3, 4 and 5. Theorem 3.
For any q ∈ ( − , the function θ ( q, . ) has infinitely many negative and infinitelymany positive real zeros. Theorem 4. (1) There exists a sequence of values of q (denoted by ¯ q j ) tending to − + suchthat θ (¯ q k , . ) has a double real zero ¯ y k (the rest of its real zeros being simple). For the remainingvalues of q ∈ ( − , the function θ ( q, . ) has no multiple real zero.(2) For k odd (resp. for k even) one has ¯ y k < , θ (¯ q k , . ) has a local minimum at ¯ y k and ¯ y k is the rightmost of the real negative zeros of θ (¯ q k , . ) (resp. ¯ y k > , θ (¯ q k , . ) has a local maximumat ¯ y k and for k sufficiently large ¯ y k is the leftmost but one (second from the left) of the realnegative zeros of θ (¯ q k , . ) ).(3) For k sufficiently large one has − < ¯ q k +1 < ¯ q k < .(4) For k sufficiently large and for q ∈ (¯ q k +1 , ¯ q k ) the function θ ( q, . ) has exactly k complexconjugate pairs of zeros counted with multiplicity. Remark 5.
Numerical experience confirms the conjecture that parts (2), (3) and (4) of thetheorem are true for any k ∈ N . Proposition 14 clarifies part (2) of the theorem. Theorem 6.
One has ¯ q k = 1 − ( π/ k ) + o (1 /k ) and | ¯ y k | → e π/ = 4 . . . . . Remarks 7. (1) Theorems 1 and 4 do not tell whether there are values of q ∈ ( − ,
1) for which θ ( q, . ) has a multiple complex conjugate pair of zeros.(2) It would be interesting to know whether the sequences { y k } and { ¯ y k − } are monotonedecreasing and { ¯ y k } is monotone increasing. This is true for at least the five first terms of eachsequence.(3) It would be interesting to know whether there are complex non-real values of q of theopen unit disk belonging to the spectrum of θ and (as suggested by A. Sokal) whether | ˜ q | is thesmallest of the moduli of the spectral values.(4) The following statement is formulated and proved in [7]: The sum of the series (cid:80) ∞ j =0 q j ( j +1) / x j (considered for q ∈ (0 , and x ∈ C ) tends to / (1 − x ) (for x fixed and as q → − ) exactly when x belongs to the interior of the closed Jordan curve { e | s | + is , s ∈ [ − π, π ] } . This statement and equation (1) imply that as q → − + , θ ( q, x ) → (1 − x ) / (1 + x ) for x ∈ ( − e π/ , e π/ ), e π/ = 4 . . . . . Notice that the radius of convergence of the Taylorseries at 0 of the function (1 − x ) / (1 + x ) equals 1.(5) On Fig. 1 and 2 we show the graphs of θ (¯ q k , . ) for k = 1, . . . , 8. The ones for k = 1, 2,5 and 6 are shown in black, the others are drawn in grey. One can notice by looking at Fig. 2that for x ∈ [ − . , .
5] the graphs of θ (¯ q k , . ) for k ≥ − x ) / (1 + x ).(6) The approximative values of ¯ q k and ¯ y k for k = 1, 2, . . . − ¯ q k . . . . y k − .
991 2 . − .
621 3 . k − ¯ q k . . . . y k − .
908 3 . − .
08 4 . θ This section contains properties of the function θ , known or proved in [9]. When a property isvalid for all q from the unit disk or for all q ∈ ( − , θ ( q, x ). When a property holdstrue only for q ∈ [0 ,
1) or only for q ∈ ( − , θ ( v, x ) or θ ( − v, x ), where v ∈ [0 , Theorem 8. (1) The function θ satisfies the following functional equation: θ ( q, x ) = 1 + qxθ ( q, qx ) (2) and the following differential equation: q∂θ/∂q = 2 x∂θ/∂x + x ∂ θ/∂x = x∂ ( xθ ) /∂x (3) (2) For k ∈ N one has θ ( v, − v − k ) ∈ (0 , v k ) .(3) In the following two situations the two conditions sgn ( θ ( v, − v − k − / )) = ( − k and | θ ( v, − v − k − / ) | > hold true:(i) For k ∈ N and v > small enough;(ii) For any v ∈ (0 , fixed and for k ∈ N large enough. The real entire function ψ ( z ) is said to belong to the Laguerre-P´olya class L − P if it can berepresented as ψ ( x ) = cx m e − αx + βx ω (cid:89) k =1 (1 + x/x k ) e − x/x k , where ω is a natural number or infinity, c , β and x k are real, α ≥ m is a nonnegativeinteger and (cid:80) x − k < ∞ . Similarly, the real entire function ψ ∗ ( x ) is a function of type I in theLaguerre-P´olya class, written ψ ∗ ∈ L − PI , if ψ ∗ ( x ) or ψ ∗ ( − x ) can be represented in the form ψ ∗ ( x ) = cx m e σx ω (cid:89) k =1 (1 + x/x k ) , (4)where c and σ are real, σ ≥ m is a nonnegative integer, x k >
0, and (cid:80) /x k < ∞ . It is clearthat L − PI ⊂ L − P . The functions in
L − P , and only these, are uniform limits, on compactsubsets of C , of hyperbolic polynomials (see, for example, Levin [14, Chapter 8]). Similarly, ψ ∈ L − PI if and only if ψ is a uniform limit on the compact sets of the complex plane ofpolynomials whose zeros are real and are either all positive, or all negative. Thus, the classes L − P and
L − PI are closed under differentiation; that is, if ψ ∈ L − P , then ψ ( ν ) ∈ L − P forevery ν ∈ N and similarly, if ψ ∈ L − PI , then ψ ( ν ) ∈ L − PI . P´olya and Schur [17] provedthat if ψ ( x ) = ∞ (cid:88) k =0 γ k x k k ! (5)3igure 1: The graphs of the functions θ (¯ q k , . ) for k = 1, 2, 3 and 4.4igure 2: The graphs of the functions θ (¯ q k , . ) for k = 5, 6, 7 and 8.5elongs to L − P and its Maclaurin coefficients γ k = ψ ( k ) (0) are all nonnegative, then ψ ∈L − PI .The following theorem is the basic result contained in [12]: Theorem 9. (1) For any fixed q ∈ C ∗ , | q | < , and for k sufficiently large, the function θ ( q, . ) has a zero ζ k close to − q − k (in the sense that | ζ k + q − k | → as k → ∞ ). These are all butfinitely-many of the zeros of θ .(2) For any q ∈ C ∗ , | q | < , one has θ ( q, x ) = (cid:81) k (1 + x/x k ) , where − x k are the zeros of θ counted with multiplicity.(3) For q ∈ (˜ q j , ˜ q j +1 ] the function θ ( q, . ) is a product of a degree j real polynomial with-out real roots and a function of the Laguerre-P´olya class L − PI . Their respective forms are (cid:81) jk =1 (1 + x/η k ) and (cid:81) k (1 + x/ξ k ) , where − η k and − ξ k are the complex and the real zeros of θ counted with multiplicity.(4) For any fixed q ∈ C ∗ , | q | < , the function θ ( q, . ) has at most finitely-many multiplezeros.(5) For any q ∈ ( − , the function θ ( q, . ) is a product of the form R ( q, . )Λ( q, . ) , where R = (cid:81) jk =1 (1 + x/ ˜ η k ) is a real polynomial with constant term and without real zeros and Λ = (cid:81) k (1+ x/ ˜ ξ k ) , ˜ ξ k ∈ R ∗ , is a function of the Laguerre-P´olya class L − P . One has ˜ ξ k ˜ ξ k +1 < .The sequence {| ˜ ξ k |} is monotone increasing for k large enough. One can use equation (1). By part (3) of Theorem 8 with v for v , for each v ∈ (0 ,
1) fixed andfor k large enough, if − x /v = − ( v ) − k − / (i.e. if | x | = v − k − / ), then | θ ( v , − x /v ) | > θ ( v , − x /v )) = ( − k . At the same time part (2) of Theorem 8 implies thatfor − vx = − ( v ) − k (i.e. again for | x | = v − k − / ) one has θ ( v , − vx ) ∈ (0 , v k ) hence | vxθ ( v , − vx ) | < v k +1 / <
1. This means that for v ∈ (0 ,
1) fixed and for k large enoughthe equality sgn( θ ( v , − x /v )) = ( − k holds, i.e. there is a zero of θ on each interval of theform ( − v − k − / , − v − k +1 / ) and ( v − k +1 / , v − k − / ). θ The present subsection contains some preliminary information about the zeros of θ . Lemma 10.
For q ∈ [ − . , all zeros of θ ( q, . ) are real and distinct.Proof. Indeed, it is shown in [11] that for q ∈ C , | q | ≤ . θ are of the form − q − j ∆ j , ∆ j ∈ [0 . , . | q | ≤ . q is real, then as all coefficients of θ are real, each of its zeros is eitherreal or belongs to a complex conjugate pair. The moduli of the zeros being distinct the zerosare all real and distinct. Notation 11.
We denote by 0 < x < x < · · · the positive and by · · · < x < x < θ . For q ∈ [ − . ,
0) this notation is in line with the fact that x j is close to − q − j . 6 emarks 12. (1) The quantities ∆ j are constructed in [11] as convergent Taylor series in q ofthe form 1 + O ( q ).(2) For q ∈ ( − ,
0) the function θ ( q, . ) has no zeros in [ − , θ = 1 + q x + q x + · · · + qx (1 + q x + q x + · · · ) . Each of the two series is sign-alternating, with positive first term and with decreasing moduli ofits terms for q ∈ ( − , x ∈ [ − , qx >
0, one has θ > Lemma 13.
For q ∈ [ − . , the real zeros of θ and their products with q are arranged onthe real line as shown on Fig. 3.Proof. The lemma follows from equation (2). Indeed,0 = θ ( q, x k +2 ) = 1 + qx k +2 θ ( q, qx k +2 ) , x k +2 < q < θ ( q, qx k +2 ) < . For small values of q the quantity qx k +2 is close to − q − k − (see Lemma 10 and part (1) ofRemarks 12), i.e. close to x k +1 and as θ ( q, qx k +2 ) <
0, one must have x k +1 < qx k +2 < x k +3 .By continuity these inequalities hold for all q < x k +1 < x k +3 are realand distinct.In the same way one can justify the disposition of the other points of the form qx j w.r.t. thepoints x j . x4k+4x qxx4k+6qx4k+7 qx4k+5 x 4k+34k+14k+3x 4k+2 01 qx 4k+2 qx 4k+4 x 4k+54k+6qx Figure 3: The real zeros of θ for q ∈ ( − , Proposition 14.
The function θ ( q, . ) with q ∈ ( − , has a zero in the interval (0 , − /q ) . Moreprecisely, one has θ ( q, − /q ) < .Proof. Setting as above v = − q one gets θ ( − v, /v ) = − v Φ( v ) + v Ψ( v ) , where Φ( v ) = ∞ (cid:88) j =0 ( − j v j +3 j , Ψ( v ) = ∞ (cid:88) j =0 ( − j v j +5 j . Further we use some results of [9]. Consider the functions ϕ k ( τ ) := (cid:80) ∞ j =0 ( − j τ kj + j ( j − / = θ ( τ, − τ k − ) and ξ k ( τ ) := 1 / (1 + τ k ), τ ∈ [0 , k > roposition 15. (1) The functions ϕ k are real analytic on [0 , ; when k > , then ϕ k < ξ k ;when k > , then ϕ k > ξ k − ; one has lim τ → − ϕ k ( τ ) = 1 / .(2) Consider the function ϕ k as a function of the two variables ( k, τ ) . One has ∂ϕ k /∂k > for k > .(3) For q ∈ (0 , , x ∈ ( − q − , ∞ ) one has ∂θ/∂x > .(4) For q ∈ (0 , , x ∈ ( − q − / , ∞ ) one has θ > / . Before proving Proposition 15 we finish the proof of Proposition 14. Part (2) of Proposition 15implies ϕ / ( v ) > ϕ / ( v ). As − v Φ( v ) = ϕ / ( v ) − − v Ψ( v ) = ϕ / ( v ) −
1, this meansthat θ ( − v, /v ) <
0, i.e. θ ( q, − /q ) < Proof of Proposition 15:
Part (1) of the proposition is proved in [9].To prove part (2) observe that ∂ϕ k /∂k = ( − log τ ) (cid:80) ∞ j =1 ( − j − jτ kj + j ( j − / = ( − log τ ) τ k (cid:80) ∞ j =1 ( ϕ k + j − τ k + j ϕ k + j +1 )As ϕ k + j − τ k + j ϕ k + j +1 = 2 ϕ k + j − ϕ k + j ( τ ) > ξ k + j − ( τ ) ≥ /
2, each difference ϕ k + j − τ k + j ϕ k + j +1 is positive on [0 , − log τ and τ k are alsopositive.For x ∈ ( − q − ,
0] part (3) follows from part (2) and from ϕ k ( τ ) = θ ( τ, − τ k − ). Indeed, onecan represent x as − τ k − for some k >
0; for fixed τ , as x increases, k also increases. One has0 < ∂ϕ k /∂k = ( − log τ ) ∂θ/∂x | x = − τ k − and − log τ > . For x > θ ( v, x ) considered as a series in x being positive.For x ∈ ( − q − / ,
0] part (4) follows from part (3). Indeed, consider the function ψ :=1 + 2 (cid:80) ∞ j =1 ( − j τ j , τ ∈ [0 , τ → − , see [8]. As 0 < ψ ( τ / ) / ϕ / ( τ ) − /
2, for k ≥ / θ ( τ, − τ k − ) = ϕ k ( τ ) ≥ ϕ / ( τ ) > / . For x = 0 one has θ = 1 hence for x > Lemma 16.
For any ε > sufficiently small there exists δ > such that for q ∈ ( − , − δ ] the function θ ( q, . ) has a single real zero in the interval ( − e π/ + ε, e π/ − ε ) . This zero is simpleand positive. θ coalesce? Further we describe the way multiple zeros are formed when q decreases from 0 to − Definition 17.
We say that the phenomenon A happens before the phenomenon B if A takesplace for q = q , B takes place for q = q and − < q < q <
0. By phenomena we mean thatcertain zeros of θ or another function coalesce. Notation 18.
We denote by x j ≺ x k the following statement: The zeros x k and x k +2 of θ cancoalesce only after x j and x j +2 have coalesced. emma 19. One has x k +2 ≺ x k +3 , x k +2 ≺ x k +1 , x k +3 ≺ x k +4 and x k +5 ≺ x k +4 , k ∈ N ∪ .Proof. The statements follow respectively from qx k +5 < x k +4 < x k +2 < qx k +3 , x k +1 < qx k +2 < qx k +4 < x k +3 ,qx k +4 < x k +3 < x k +5 < qx k +6 and x k +6 < qx k +7 < qx k +5 < x k +4 . Lemma 20.
One has x k +2 ≺ x k +6 , k ∈ N ∪ .Proof. Indeed, equation (2) implies the following one: θ ( q, x ) = 1 + qx + q x + q x + q x θ ( q, q x ) . (6)For x = x k +2 /q one obtains θ ( q, x k +2 /q ) = 1 + x k +2 /q + x k +2 /q + x k +2 /q = (1 + x k +2 /q ) + ( x k +2 /q + x k +2 /q ) . Each of the two sums is negative due to q ∈ ( − , x k +2 < − | q | one has x k +2 /q ∈ ( x k +8 , x k +6 ) because x j = − q − j (1 + O ( q )) and θ ( q, x k +2 /q ) <
0. By continuity this holds true for all q ∈ ( − ,
0) for which the zeros x k +8 , x k +6 , x k +4 and x k +2 are real and distinct. Hence if x k +2 and x k +4 have not coalesced, then x k +6 and x k +8 are real and distinct. Remarks 21. (1) Recall that for q ∈ ( − ,
0) we set v := − q and that equation (1) holds true.(2) Equation (3) implies that the values of θ at its local maxima decrease and its values atlocal minima increase as q decreases in ( − , ∂θ/∂x = 0,so 2 q∂θ/∂q = x ∂ θ/∂x . At a minimum one has ∂ θ/∂x ≥
0, so ∂θ/∂q ≤ θ increases as q decreases; similarly for a maximum.(3) In particular, this means that θ can only lose real zeros, but not acquire such as q decreases on ( − , θ depend continuously on q . If at some point of thereal axis a new zero of even multiplicity appears, then it cannot be a maximum because thecritical value must decrease and it cannot be a minimum because its value must increase.(4) To treat the cases of odd multiplicities of the zeros it suffices to differentiate both sidesof equation (3) w.r.t. x . For example, a simple zero x of θ cannot become a triple one because2 q∂/∂q ( ∂θ/∂x ) = 2 ∂θ/∂x + 4 x∂ θ/∂x + x ∂ θ/∂x which means that as ∂θ/∂x | x = x = ∂ θ/∂x | x = x = 0, then either ∂ θ/∂x | x = x > x one has ∂θ/∂x ≥ ∂/∂q ( ∂θ/∂x ) <
0, or ∂ θ/∂x | x = x < ∂θ/∂x ≤ ∂/∂q ( ∂θ/∂x ) > x ), so in both cases the triple zerobifurcates into a simple one and a complex pair as q decreases. The case of a zero of multiplicity2 m + 1, m ∈ N , is treated by analogy.(5) In equation (1) the first argument (i.e. v ) of both functions θ ( v , − x /v ) and θ ( v , − vx )is the same, so when one of them has a double zero, then they both have double zeros. If thedouble zero of the first one is at x = a , then the one of the second is at x = a/v .9 − −+ + ++ Figure 4: The graphs of ψ (solid line) and ψ (dashed line). Proposition 22.
For any k ∈ N ∪ there exists q ∗ k ∈ ( − , such that for q = q ∗ k the zeros x k +2 and x k +4 coalesce. Notation 23.
We denote by ψ and ψ the functions θ ( v , − x /v ) and − vxθ ( v , − vx ). By y ± k and z ± k we denote their real zeros for v ∈ (0 , . k ∈ N , y k > y − k < z k > z − k <
0. We set z = 0. Proof of Proposition 22:
On Fig. 4 we show for v ∈ (0 , . ψ and ψ (drawn in solid and dashed line respectively) look like, the former being even and thelatter odd, see part (1) of Remarks 21. The signs + and − indicate places, where it is certainthat their sum θ ( − v, x ) is positive or negative. It is positive (resp. negative) if both functionsare of this sign. It is positive near the origin because ψ | x =0 > ψ | x =0 = 0.For values of v close to 0 the zeros y ± k of ψ (resp. the zeros z ± k of ψ /x ) are close tothe numbers ± v − (4 k − / (resp. ± v − (4 k +1) / ), k ∈ N , see part (1) of Remarks 12. From theseremarks follows also that for small positive values of v the zeros of θ ( − v, x ) are close to thenumbers − v − k , k ∈ N . Hence on the negative half-axis x one obtains the following arrangementof these numbers: · · · < − v − / < − v − / < − v − < − v − / < − v − / < − v − < − v − / < .y − z − x y − z − x y − Hence for v ∈ (0 , . − there is exactlyone zero of θ ( − v, . ).As v increases from 0 to 1, it takes countably-many values at each of which one of thefunctions ψ and ψ (in turn) has a double zero and when the value is passed, this double zerobecomes a conjugate pair, see Theorem 1. Hence on the negative real half-line the regions,where the corresponding function ψ or ψ is negative, disappear one by one starting from theright. As two consecutive changes of sign of θ ( − v, . ) are lost, θ ( − v, . ) has a couple of consecutivereal negative zeros replaced by a complex conjugate pair. The quantity of these losses beingcountable implies the proposition. Proposition 24.
Any interval of the form [ γ, , γ ∈ ( − , , contains at most finitely manyspectral values of q . roof. Indeed, if γ ≥ − . q , see Lemma 10; allzeros of θ are real and distinct and the graphs of the functions ψ , look as shown on Fig. 4.Suppose that γ < − . q decreases in [ γ, ψ , changes only on some closed interval containing0, but outside it the zones marked by + and − continue to exist (but their borders changecontinuously) and the simple zeros of θ that are to be found between two such consecutive zonesof opposite signs are still to be found there. Besides, no new real zeros appear, see part (3) ofRemarks 21.Therefore there exists s ∈ N such that when q decreases from 0 to γ , the zeros x s , x s +1 , x s +2 , . . . remain simple and depend continuously on q . Hence only the rest of the zeros (i.e. x , . . . , x s − ) can participate in the bifurcations. Proposition 25.
For q ∈ ( − , the function θ ( q, . ) can have only simple and double real zeros.Positive double zeros are local maxima and negative double zeros are local minima.Proof. Equality (2) implies the following one: θ ( q, x/q ) = 1+ x/q +( x /q ) θ ( q, x ). Set x = x k +2 .Hence x k +2 < −
1, see part (2) of Remarks 12. As θ ( q, x k +2 ) = 0 and 1 + x k +2 /q >
0, thisimplies θ ( q, x k +2 /q ) >
0. In the same way θ ( q, x k +4 /q ) > q close to 0 the numbers x k +2 /q and x k +4 /q are close respectively to x k +4 and x k +6 ,see part (1) of Remarks 12. Hence for such values of q one has x k +6 < x k +4 /q < x k +2 /q < x k +4 . (7)This string of inequalities holds true (by continuity) for q belonging to any interval of the form( a, a ∈ ( − , q of which the zeros x k +6 , x k +4 and x k +2 are real and distinct.Equation (7) implies that x k +6 and x k +4 cannot coalesce if x k +4 and x k +2 are real (butnot necessarily distinct). Hence when for the first time negative zeros of θ coalesce, this happenswith exactly two zeros, and the double zero is a local minimum of θ .For positive zeros one obtains in the same way the string of inequalities x k +5 < x k +3 /q < x k +5 /q < x k +7 . Indeed, one has 1 + x k +3 /q < x k +3 > q ∈ (0 , θ ( q, x k +3 /q ) < θ ( q, x k +5 /q ) <
0. Thus x k +5 and x k +7 cannotcoalesce if x k +3 and x k +5 are real (but not necessarily distinct). Hence when for the first timepositive zeros of θ coalesce, this happens with exactly two zeros, and the double zero is a localmaximum of θ .After a confluence of zeros takes place, one can give new indices to the remaining real zerosso that the indices of consecutive zeros differ by 2 ( x s +2 < x s < < x s +1 < x s +3 ).After this for the next confluence the reasoning is the same. Lemma 26.
For k ∈ N sufficiently large one has x k +3 ≺ x k +6 .Proof. Suppose that x k +3 ≥
3. Then θ ( q, x k +3 /q ) = 1 + x k +3 /q + x k +3 /q + ( x k +3 /q ) θ ( q, x k +3 ) = 1 + x k +3 /q + x k +3 /q . For x k +3 ≥ q ∈ ( − ,
0) the right-hand side is negative. In the same way θ ( q, x k +5 /q ) <
0. We prove below that 11 k +8 < x k +5 /q < x k +3 /q < x k +6 . (8)Hence the zeros x k +8 and x k +6 cannot coalesce before x k +5 and x k +3 have coalesced. Toprove the string of inequalities (8) observe that for q close to 0 the numbers x k +8 and x k +5 /q are close to one another ( x k +3 /q and x k +6 as well, see part (1) of Remarks 12) which implies(8). By continuity, as long as x k +3 ≥ q ∈ ( − , q not necessarily close to 0.The result of V. Katsnelson (see part (4) of Remarks 7) implies that there exists a ∈ ( − , q ∈ ( − , a ] the function θ ( q, x ) has no zeros in [ − ,
3] except the one which issimple and close to 1 (see Proposition 14). Hence for q ∈ ( − , a ] the condition x k +3 ≥ x k +3 has been real and simple for q ∈ ( a, q ∈ [ a, θ have coalesced, and only finitely many have belonged to theinterval [ − ,
3] for some value of q , see Proposition 24. Therefore there exists k ∈ N such thatfor k ≥ k one has x k +3 ≥ Proposition 25 and Remarks 21 show that θ ( q, . ) has no real zero of multiplicity higher that 2.Lemma 20 implies the string of inequalities − < · · · < ¯ q l +2 < ¯ q l < · · · <
0. For k sufficientlylarge one has − < · · · < ¯ q k +1 < ¯ q k < · · · <
0. This follows from Lemmas 19 and 26. It resultsfrom Proposition 24 and from the above inequalities that the set of spectral values has − Recall that ψ = θ ( v , − x /v ) and ψ = − vxθ ( v , − vx ), see Notation 23. Recall that thespectral values ˜ q k of q for θ ( q, x ), q ∈ (0 ,
1) satisfy the asymptotic relation ˜ q k = 1 − π/ k + o (1 /k ).Hence the values of v for which the function θ ( v , x ) has a double zero are of the form˜ v k = (˜ q k ) / = 1 − π/ k + o (1 /k )and the functions ψ , have double zeros for v = ˜ v k .Consider three consecutive values of k the first of which is odd – k , k + 1 and k + 2. Set v := ˜ v k . Denote by a < b < ψ , | v =˜ v k . These zerosare local minima and on the whole interval [ a, b ] one has θ >
0. The values of θ at local minimaincrease (see part (2) of Remarks 21), therefore the double zero of θ (¯ q k , . ) is obtained for some | q | < | ˜ v k | , i.e. before the functions ψ , | v =˜ v k have double zeros. This follows from equality (1)in which both summands in the right-hand side have local minima (recall that as k is odd, thedouble zero of θ is negative, so x < | ¯ q k | < | ˜ v k | = 1 − π/ k + o (1 /k ) . (9)In the same way | ¯ q k +2 | < | ˜ v k +2 | = 1 − π/ k + 2) + o (1 / ( k + 2)) . (10)12n the case of k + 1 the function ψ | v =˜ v k has a local minimum while ψ | v =˜ v k has a localmaximum (because k + 1 is even, the double zero of θ is positive, so x > x > θ has a local maximum and as the values of θ at local maxima decrease (see part (2)of Remarks 21), the double zero of θ (¯ q k +1 , . ) is obtained for some | q | > | ˜ v k +1 | , i.e. after thefunctions ψ , | v =˜ v k have double zeros. Therefore | ¯ q k +1 | > | ˜ v k +1 | = 1 − π/ k + 1) + o (1 / ( k + 1)) . (11)When k is sufficiently large one has | ¯ q k | < | ¯ q k +1 | < | ¯ q k +2 | (this follows from part (3) ofTheorem 4). Using equations (10) and (11) one gets1 − π/ k + 1) + o (1 / ( k + 1)) < | ¯ q k +1 | < | ¯ q k +2 | < − π/ k + 2) + o (1 / ( k + 2)) . Hence | ¯ q k +1 | = 1 − π/ k + 1) + o (1 / ( k + 1)) and | ¯ q k +2 | = 1 − π/ k + 2) + o (1 / ( k + 2)).This implies the statement of Theorem 6. References [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