# Solutions of a Class of Multiplicatively Advanced Differential Equations II: Fourier Transforms

Authors: David W. Pravica; Njinasoa Randriampiry; Michael J. Spurr
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

For a wide class of solutions to multiplicatively advanced differential equations (MADEs), a comprehensive set of relations is established between their Fourier transforms and Jacobi theta functions. In demonstrating this set of relations, the current study forges a systematic connection between the theory of MADEs and that of special functions. In a large subset of the general case, we introduce a new family of Schwartz wavelet MADE solutionsWμ,λt for μ and λ rational with λ>0. These Wμ,λt have all moments vanishing and have a Fourier transform related to theta functions. For low parameter values derived from λ, the connection of the Wμ,λt to the theory of wavelet frames is begun. For a second set of low parameter values derived from λ, the notion of a canonical extension is introduced. A number of examples are discussed. The study of convergence of the MADE solution to the solution of its analogous ODE is begun via an in depth analysis of a normalized example W−4/3,1/3t/W−4/3,1/30. A useful set of generalized q-Wallis formulas are developed that play a key role in this study of convergence.


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## Body

## 1. Introduction

This paper expands the study of a class of solutions of multiplicatively advanced differential equations (MADEs) by determining the relationship of their Fourier transforms to Jacobi theta functions. The class of solutions under consideration consists of the Dirichlet-like [1] series:
(1)fμ,λt≡∑m=−∞∞−1me−qmtqmm−μ/λ,fort≥0,whereμ∈ℚ,λ∈ℚ+andq>1.Each of thefμ,λt in (1) satisfies the MADE.
(2)fμ,λδt=−1γ+δqγγ+μ/λfμ,λqγt,where λ/2=γ/δ with γ/δ in a reduced form and γ,δ∈ℕ; see [2]. Note that (2) is multiplicatively advanced in that the argument qγt is an advancing of the parameter t by qγ>1, as q is taken to be greater than 1 and γ∈ℕ.In general, as is shown in [2], there are nonunique ways to extend the fμ,λt from 0≤t<∞ to the negative reals, obtaining Fμ,λt on −∞<t<∞ with Fμ,λt also satisfying the MADE (2). To overcome this issue of nonuniqueness, we first extend fμ,λt (perhaps discontinuously) to the negative reals by defining f~μ,λt to be.
(3)f~μ,λt≡fμ,λt,t≥0,0,t<0.Then, we utilize onlyf~μ,λt to obtain a new function Wμ,λt in C∞ℝ which is naturally generated by fμ,λt, and we further observe that for low parameters (namely, 1≤δ≤3 and δ=0modβ) Wμ,λt is indeed an extension of fμ,λt satisfying (2). Here, β is obtained as μ+1/2=α/β with α/β in a reduced form.In attempting to compute the Fourier transform(4)Ff~μ,λtx=12π∫−∞∞e−ixtf~μ,λtdt,of f~μ,λt, we are led, instead, into discovering a relation between a weighted average of the Ff~μ,λtx of form
(5)1M∑ℓ=0M−1ωℓp1Ff~μ,λtxωℓp2,where M,p1,p2 depend on μ,λ and ω is an Mth root of unity to a similar weighted average of z3p3/θQ;z3M.(6)1Dγ∑κ=0Dγ−1ω~κz3p3θQ;ω~κz3M,where M,D,γ,p3 are determined by μ,λ and ω~ is a Dγth root of unity, while θ is the Jacobi theta function given by (10) below, z3 is a scaled Dγth root of ix, and Q=q2/λ. See equations (98) and (99) of Theorem 13 for further details.One then applies the inverse Fourier transform to (5) to obtain the new functions
(7)Wμ,λt=F−1∑ℓ=0M−1ωℓp1Ff~μ,λtxωℓp2t,which are central to this study. These new Wμ,λt comprehensively generalize each of the main examples we have previously studied, including Kq,t in [3] and qCost,qSint in [4, 2]. For certain low values of δ, we show that the Wμ,λt give unique, canonical extensions of the associated fμ,λt which satisfy the MADE (2). In general, for higher values of δ, the Wμ,λt are not extensions of fμ,λt; however, they are all Schwartz wavelets with connections not only to wavelet theory but also to special function theory in that FWμ,λtx is expressed in terms of (6) and thus in terms of the Jacobi theta function. See (98) and (99) of Theorem 13 for specifics.We next determine a simple criteria for theWμ,λt to not identically vanish, whereby the relationship of their Fourier transforms to the theta function remains substantive, as described in Theorem 23. This relationship is seen to greatly extend and generalize each of the special cases of Fourier transform computations that have been computed in all of our previous collected work, including [2–6]. Furthermore, even in those cases where Wμ,λt vanishes identically, we are still able to provide a (more technical) relation between the Fourier transform of f~μ,λt and the Jacobi theta function in Theorem 28. We then begin applying the above discoveries to wavelet theory, producing numerous examples of new Schwartz wavelets generating frames for L2ℝ.Furthermore, considering the MADE (2) to be a perturbation of the classical ODE
(8)fδt=−1γ+δft,with q>1 in (2) considered to be the perturbation parameter as q⟶1+, we then initiate a study of convergence of the normalized solution of (2) to the classical solution of (8). In particular, we prove convergence for the normalized W−4/3,1/3t to the classical solution of (8) with convergence being uniform on any compact set of ℝ; see Figure 1. We also exhibit graphical evidence for such convergence of other normalized Wμ,λt.Figure 1
Plot ofy=cost (blue dots) approached by y=W−4/3,1/3t/W−4/3,1/30 for: q=1.3 (red dashed), q=1.1 (solid black). Convergence is uniform on compact subsets of ℝ as q⟶1+.The convergence seen in Figure1 mirrors earlier convergence results [4] we have been previously able to obtain in the canonical extensions for the special cases: (1) μ=0;λ=1 which normalizes to qCost discussed in Section 5; and (2) μ=1;λ=1 which normalizes to qSint, also discussed in Section 5. The convergence of qCost to cost and the convergence of qSint to sint are illustrated in Figure 2.Figure 2
(a)y=cosx (solid black) approached by y=qCost for q=1.35 (dotted-dash blue), q=1.15 (dashed red). (b) y=sinx (solid black) approached by y=qSint for q=1.35 (dotted-dash blue), q=1.15 (dashed red).
(a)(b)We conclude the paper with a set of generalizations of Wallis’ formula forπ/2 that we call generalized q-Wallis formulas, and we demonstrate their utility in the study of convergence of normalized solutions of MADEs to their classical analogue ODEs.We mention that the current work falls in the area of functional differential equations of multiplicatively advanced type. Studies in functional differential equations include for instance [7–9]. More precisely, the current work falls under the area of q-difference differential equations, where the multiplicative advancement yt⟶yqt is seen as a dilation that is denoted σqyt=yqt. There is a robust study within the area of q-difference differential equations with dilations involving q>1. This is highlighted by works of L. Di Vizio [10–12]; C. Hardouin [11]; T. Dreyfus [13, 14]; A. Lastra [14], [15–20], [21–23]; S. Malek [14], [15–20], [21–23], [24–27]; J. Sanz [21–23]; H. Tahara [28]; and C. Zhang [12, 29], along with further references by these researchers and others. Also, for good background references to the current work, consult [2–6, 30–34] (especially [2, 4]). These last references also exhibit a number of various applications of global solutions of MADEs.


### 1.1. Preliminaries and Salient Properties of the Jacobi Theta Function

We shall need to extend the definition offμ,λt to the case that the argument is complex and lying in the right half plane.Definition 1.
Letq>1,μ,λ∈ℚ, with λ>0. Then for t≥0 and the function fμ,λt given by (1), one defines for Rz≥0 (that is, for the real part of z∈ℂ nonnegative)
(9)fμ,λz≡∑m=−∞∞−1me−qmzqmm−μ/λ,which is analytic for Rz>0.
Next, recall that forq>1 the Jacobi theta function is given by
(10)θq;u=∑n=−∞∞unqnn−1/2=μq∏n=0∞1+uqn1+1uqn+1,where
(11)μq=∏n=0∞1−1qn+1.Two properties of the Jacobi theta function of interest are that(12)forallp∈ℤθq;qpu=qpp+1/2upθq;u,uθq;u−1=θq;u,which are proven in [5] and [6], respectively. From the product formula in (10), one sees that
(13)θq;u=0⟺u=−qpforsomep∈ℤ.As indicated earlier, the Jacobi theta function plays a major role in this study in the computation of Fourier transforms.


## 1.1. Preliminaries and Salient Properties of the Jacobi Theta Function

We shall need to extend the definition offμ,λt to the case that the argument is complex and lying in the right half plane.Definition 1.
Letq>1,μ,λ∈ℚ, with λ>0. Then for t≥0 and the function fμ,λt given by (1), one defines for Rz≥0 (that is, for the real part of z∈ℂ nonnegative)
(9)fμ,λz≡∑m=−∞∞−1me−qmzqmm−μ/λ,which is analytic for Rz>0.
Next, recall that forq>1 the Jacobi theta function is given by
(10)θq;u=∑n=−∞∞unqnn−1/2=μq∏n=0∞1+uqn1+1uqn+1,where
(11)μq=∏n=0∞1−1qn+1.Two properties of the Jacobi theta function of interest are that(12)forallp∈ℤθq;qpu=qpp+1/2upθq;u,uθq;u−1=θq;u,which are proven in [5] and [6], respectively. From the product formula in (10), one sees that
(13)θq;u=0⟺u=−qpforsomep∈ℤ.As indicated earlier, the Jacobi theta function plays a major role in this study in the computation of Fourier transforms.


## 2. Proof of the Relation of Fourier Transforms to Jacobi Theta Functions

We proceed immediately to the computation of Fourier transforms. Forμ,λ∈ℝ with λ>0 and given fμ,λt is as in (1), we define
(14)f~μ,λt≡fμ,λt=∑k=−∞∞−1ke−qkt/qkk−μ/λ,fort≥0,0,fort<0.We now restrictμ,λ∈ℚ to be rational with λ>0, and let x∈ℝ. One then has the following computation of the Fourier transform Ff~μ,λtx of f~μ,λt:
(15)Ff~μ,λtx=12π∫−∞∞e−ixtf~μ,λtdt=12π∫0∞e−ixt∑k=−∞∞−1ke−qktqkk−μ/λdt=12π∑k=−∞∞−1k1qkk−μ/λ∫0∞e−ixt−qktdt=12π∑k=−∞∞−1k1qkk−μ/λe−ixt−qkt−ix−qk0∞=12π∑k=−∞∞−1kqkk−μ/λ1ix+qk=12π∑k=−∞∞−1kqkk+1/λqkμ+1/λix+qk=12π∑k=−∞∞−1kq2/λkk+1/2q2/λkμ+1/2ix+q2/λkλ/2=12π∑k=−∞∞−1kQkk+1/2Qkμ+1/2ix+Qkλ/2(16)=12π∑k=−∞∞−1kQkk+1/2Qkα/βix+Qkγ/δ,where in (15) and (16) Q≡q2/λ and for conciseness in moving from (15) to (16), one has μ+1/2≡α/β and λ/2≡γ/δ, where α/β and γ/δ in ℚ are taken to be in a reduced form with α∈ℤ and β,γ,δ∈ℕ. We extend Ff~μ,λtx to the complex plane by setting
(17)Ff~μ,λtζ≡12π∑k=−∞∞−1kQkk+1/2Qkα/βiζ+Qkγ/δ,for ζ∈ℂ\S where S=iQkγ/δ∣k∈ℤ denotes the set of ζ where the denominator in (17) vanishes. Note that Ff~μ,λtζ is defined at ζ=0 by virtue of the quadratic exponent kk+1/2 of Q in the denominator counteracting any growth of the linear exponent kα/β−γ/δ of Q in the numerator of (17). Also for ℂ∗=ℂ\0, note that, if U is any open region with compact closure U¯⊂ℂ∗\S, one has that for ζ∈U the distance dζ from ζ to S is positive, as the only cluster point of S is 0. Furthermore, the distance dU¯,S from U¯ to S is also positive. Hence, we have
(18)12π∑k=−∞∞−1kQkk+1/2Qkα/βiζ+Qkγ/δ≤12π∑k=−∞∞1Qkk+1/2Qkα/βdζ≤12π∑k=−∞∞1Qkk+1/2Qkα/βdU¯,S<∞.Hence, the truncated sums(19)12π∑k=−NN−1kQkk+1/2Qkα/βiζ+Qkγ/δ,which are analytic on U, approach Ff~μ,λtζ uniformly on U as N⟶∞. Thus, Ff~μ,λtζ is analytic on U [35] and therefore analytic on ℂ∗\S.We have seen in [2] that the “alternating Q-combinatoric” −1k/Qkk+1/2 in (16) can be given by the residue of 1/uθQ;u at a simple pole u=−Qk under a computation of an appropriate contour integral about a region containing u=−Qk. Here, θQ;u is the Jacobi theta function given by (10). Observe that the term Qkα/β/ix+Qkγ/δ in (16) would then be obtained from evaluation of −uα/β/ix+−uγ/δ at u=−Qk. Therefore, as a starting point, we would be interested in integrating the expression
(20)1uθQ;u−uα/βz+−uγ/δdu,around an appropriate closed contour Γ in the complex u-plane, where we set z=ix later. However, since there is in general a multivalued issue with expression (20) if α/β or γ/δ are not integers, we set u=vM in (20), where M>0 is the least integer such that Mα/β and Mγ/δ are both integers, and we integrate
(21)∫Γ1vMθQ;vM−vMα/βz+−vMγ/δMvM−1dv=M∫Γ1vθQ;vMvαM/βeiπα/βz+vγM/δeiπγ/δdv.Note that, sinceα and β have no common factors and since Mα/β∈ℤ, one has that β divides M. Similarly, since γ and δ have no common factors and since Mγ/δ∈ℕ, δ also divides M. Since M is the least such integer, M is the least common multiple of β and δ. Thus,
(22)forα∈ℤ,β,γ,δ∈ℕwithμ+12=αβ,λ2=γδinreducedform,we set
(23)M=lcmβ,δ,B=Mβ,D=Mδ.We then divide (21) by M and integrate the following simplified version of (21)
(24)∫Γ1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv,around a closed contourΓ in the complex v-plane, where the exponents M,Bα, and Dγ are now integers (avoiding any multivalued issue in a would-be contour integral involving (20) by instead using the integration in (24)). The contour Γ will later be taken to be the oriented boundary of an annulus centered at the origin. This key step in avoiding multivalued issues in moving away from (20) to the integral in (24) allows us to overcome the limiting assumptions in earlier work (that μ is odd and λ is even in Theorem 6.3 of [2] or that μ is an integer and λ is 1 in Theorem 6.5 of [2]) to now handle the general case in this study (where μ and λ are allowed to be rational, with λ>0).In anticipation of a residue computation of the expression (24), we begin by examining the product representation of the Jacobi theta function θQ;u in (10) and removing one appropriate factor from the product corresponding to the vanishing of θQ;u when u=−Qk. That is, note that from (10), one has that for k≥0(25)θQ;u=μQ∏n=0∞1+uQn1+1uQn+1=1+uQkμQ∏n=0,n≠k∞1+uQn∏n=0∞1+1uQn+1(26)=1+uQkθk∣Q;u=Qk+uQkθk∣Q;u,where for k≥0, the expression θk∣Q;u in (26) is defined by the bracketed expression in (25), namely,
(27)θk∣Q;u≡μQ∏n=0,n≠k∞1+uQn∏n=0∞1+1uQn+1.Similarly, fork<0 one has that
(28)θQ;u=μQ∏n=0∞1+uQn1+1uQn+1=1+1uQkμQ∏n=0∞1+uQn∏n=0,n≠k−1∞1+1uQn+1(29)=1+1uQkθk∣Q;u=u+Qkuθk∣Q;u,where for k<0, the expression θk∣Q;u in (29) is defined by the bracketed expression in (28), namely,
(30)θk∣Q;u≡μQ∏n=0∞1+uQn∏n=0,n≠k−1∞1+1uQn+1.Thus, via (27) and (30), the expression θk∣Q;u is defined for each k∈ℤ.We pause the discussion on representingθQ;u in terms of θk∣Q;u in order to record a series of useful computational lemmas. The first such lemma evaluates θk∣Q;−Qk as an “alternating combinatoric.”Lemma 2.
Fork≥0 and θk∣Q;u as in (26), one has
(31)θk∣Q;−Qk=μQ∏n=0,n≠k∞1−QkQn∏n=0∞1−1Qk+n+1=−1kμQ3Qkk+1/2.
And fork<0 and θk∣Q;u as in (29), one has
(32)θk∣Q;−Qk=μQ∏n=0∞1−QkQn∏n=0,n≠k−1∞1−1Qk+n+1=−−1kμQ3Qkk+1/2.Proof.
The proof is given in Lemma 5.1 of [2].The second lemma will provide a structure for the proof of the third lemma, and it will be utilized in a subsequent residue computation.Lemma 3.
For an integerM≥2, let ω=e2πi/M be an Mth root of unity. Then,
(33)xM−bM=x−b∑j=0M−1xjbM−1−j=x−b∏p=1M−1x−ωpb.
Hence,(34)∑j=0M−1xjbM−1−j=∏p=1M−1x−ωpb.Proof.
Upon expansion of the middle expression in (33), the left-most equality is self-evident. The right-most equality in (33) follows from the fact that for each p∈0,1,⋯,M−1 one has ωpb is a root of xM−bM, and hence, x−ωpb is a factor. To obtain (34), one divides the right two expressions in (33) by x−b. The lemma is now proven.The third lemma will simplify the computation in (50) below.Lemma 4.
For an integerM≥2 let ω=e2πi/M be an Mth root of unity. One has
(35)∏p=1M−11−ωp=M.Proof.
Setx=1=b in (34) to obtain (35). The lemma is shown.We record three further lemmas on the behavior of roots of unity for later computational use.Lemma 5.
Letω=e2πi/M be an Mth root of unity, and let ϕ=ωp for some p∈ℤ have order N. Then,
(36)∑ℓ=0N−1ϕℓ=1,ifN=1,0,ifN≥2,Inparticular,∑ℓ=0M−1ωℓ=1,ifM=1,0,ifM≥2.Proof.
IfN=1, then ϕ0=1. If N≥2, then let σ=∑ℓ=0N−1ϕℓ. Observe that since ϕN=ϕ0 we have ϕσ=σ which gives 0=1−ϕσ. Since 1−ϕ≠0, we conclude σ=0. In particular, if p=1, then N=M and the second equality in (36) holds. The lemma is demonstrated.The next lemma generalizes the previous lemma.Lemma 6.
Letω=e2πi/M be an Mth root of unity. Let p∈ℤ be fixed. Then,
(37)∑ℓ=0M−1ωℓp=M,ifp=0modM,0,ifp≠0modM..Proof.
Ifp is a multiple of M then ωℓp=1 and then ∑ℓ=0M−1ωℓp=∑ℓ=0M−11=M. If p is not divisible by M, then ωp≠1 is a root of unity with order, say, N>1, with NT=M. Then,
(38)∑ℓ=0M−1ωℓp=∑ℓ=0M−1ωpℓ=∑ℓ=0N−1ωpℓ+∑ℓ=N2N−1ωpℓ+⋯∑ℓ=j−1NjN−1ωpℓ+⋯+∑ℓ=T−1NTN−1ωpℓ(39)=0,where the vanishing in (39) follows from the vanishing of each summand
(40)∑ℓ=j−1NjN−1ωpℓ=ωpj−1N∑ℓ=0N−1ωpℓ,in (38), which in turn follows from an application of Lemma 5. This proves the lemma.The following is a refinement of Lemma6.Lemma 7.
Letω=e2πi/M be an Mth root of unity, with M=ab, where a,b∈ℕ. Let p∈ℤ be fixed. Then if a=1, one has
(41)∑ℓ=0a−1ωℓp=1,forallp,and if a>1, one has
(42)∑ℓ=0a−1ωℓp=a,ifp=0modM,0,ifp≠0modMandp=0modb,ωap−1ωp−1≠0,ifp≠0modMandp≠0modb..Proof.
Letσ=∑ℓ=0a−1ωℓp. Then, if a=1, one has σ=ω0p=1p=1, giving (41). If a>1, one has
(43)ωpσ=∑ℓ=0a−1ωℓ+1p=ωap+∑ℓ=0a−1ωℓp−ω0p=ωap+σ−1.
Hence,(44)ωp−1σ=ωap−1.
Ifp=0modM, σ=∑ℓ=0a−1ωℓp=∑ℓ=0a−11=a, giving the first case in (42). If p≠0modM and p=0modb, then ωa is a bth root of unity and the right hand side of (44) vanishes while ωp−1≠0, giving σ=0 for the second case. Finally, if p≠0modM and p≠0modb, then ωa is a bth root of unity and the right hand side of (44) does not vanish while ωp−1≠0, resulting in σ=ωap−1/ωp−1≠0. This shows the third case and finishes the proof. Note that the third case is the only case with p≠0modb, because in the first case if p=0modM with M=ab, then p=0modb. This gives the lemma.We return now to the discussion in (25)–(30). Let k≥0. In (25) and (26), we let u=vM and ω=e2πi/M be the Mth root of unity; one then has that, for each ℓ with 0≤ℓ≤M−1,
(45)θQ;vM=μQ∏n=0∞1+vMQn1+1vMQn+1=1+vMQkμQ∏n=0,n≠k∞1+vMQn∏n=0∞1+1vMQn+1(46)=1+vMQkθk∣Q;vM=Qk+vMQkθk∣Q;vM(47)=∏j=0M−1v−ωjeiπ/MQk/MQkθk∣Q;vM=v−ωℓeiπ/MQk/MQk∏j=0,j≠ℓM−1v−ωjeiπ/MQk/Mθk∣Q;vM,where (27) was used to move from (45) to (46) and (33) in Lemma 3 was used to move from (46) to (47). Thus, for k≥0, the residue of f1v=1/vθQ;vM at vk,ℓ=ωℓeiπ/MQk/M is given by
(48)Resf1,vk,ℓ=Qkωℓeiπ/MQk/M1∏j=0,j≠ℓM−1ωℓeiπ/MQk/M−ωjeiπ/MQk/M⋅1θk∣Q;ωℓeiπ/MQk/MM(49)=Qkωℓeiπ/MQk/MM1∏j=0,j≠ℓM−11−ωj−ℓθk∣Q;−Qk(50)=Qk−Qk1Mθk∣Q;−Qk=1−11M−1kμQ3Qkk+1/2(51)=−1k+1μQ3Qkk+1/21M,where we have factored out ωℓeiπ/MQk/M from each factor in ∏j=0,j≠ℓM−1ωℓeiπ/MQk/M−ωjeiπ/MQk/M in the denominator of (48) to obtain (49); and the first equality in (50) follows from Lemma 4 if M≥2 (and is automatic if M=1); and the second equality in (50) follows from (31) of Lemma 2.Letk<0. In (28) and (29), we let u=vM and ω=e2πi/M; one then has that, for each ℓ with 0≤ℓ≤M−1,
(52)θQ;vM=μQ∏n=0∞1+vMQn1+1vMQn+1=1+1vMQkμQ∏n=0∞1+vMQn∏n=0,n≠k−1∞1+1vMQn+1(53)=1+1vMQkθk∣Q;vM=vM+QkvMθk∣Q;vM(54)=∏j=0M−1v−ωjeiπ/MQk/MvMθk∣Q;vM=v−ωℓeiπ/MQk/MvM∏j=0,j≠ℓM−1v−ωjeiπ/MQk/Mθk∣Q;vM,where (30) was used to move from (52) to (53) and (33) in Lemma 3 was used to move from (53) to (54). Thus, for k<0, the residue of f1v=1/vθQ;vM at vk,ℓ=ωℓeiπ/MQk/M is given by
(55)Resf1,vk,ℓ=ωℓeiπ/MQk/MMωℓeiπ/MQk/M1∏j=0,j≠ℓM−1ωℓeiπ/MQk/M−ωjeiπ/MQk/M⋅1θk∣Q;ωℓeiπ/MQk/MM(56)=−Qkωℓeiπ/MQk/MM1∏j=0,j≠ℓM−11−ωj−ℓθk∣Q;−Qk(57)=−Qk−Qk1Mθk∣Q;−Qk=111M−−1kμQ3Qkk+1/2(58)=−1k+1μQ3Qkk+1/21M,where we have factored out ωℓeiπ/MQk/M from each factor in ∏j=0,j≠ℓM−1ωℓeiπ/MQk/M−ωjeiπ/MQk/M in the denominator of (55) to obtain (56); and the first equality in (57) follows from Lemma 4 if M≥2 (and is automatic if M=1); and the second equality in (57) follows from (32) of Lemma 2.Note that the form of the final expression in (51) agrees with the form of the final expression in (58). Hence, we have that for all k∈ℤ the residue of f1v=1/vθQ;vM at vk,ℓ=ωℓeiπ/MQk/M is given by
(59)Resf1,vk,ℓ=−1k+1μQ3Qkk+1/21M.The previous discussion allows us to reach the following conclusion:Proposition 8.
Letα,β,γ,δ,M,B and D be as in (22) and (23). Let k∈ℤ, ω=e2πi/M, and ℓ satisfying 0≤ℓ≤M−1 all be fixed. Let z∈ℂ\−ωℓ+1Qk/MDγ be fixed. Then the residue of
(60)f2v=1vθQ;vMvBαeiπα/βz+vDγeiπγ/δ
atvk,ℓ=ωℓeiπ/MQk/M is given by
(61)Resf2,vk,ℓ=−1k+1μQ3Qkk+1/21Mωℓeiπ/MQk/MBαeiπα/βz+ωℓeiπ/MQk/MDγeiπγ/δ(62)=−1k+1μQ3Qkk+1/21Mωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ.Proof.
Referring to (24) and (59) and the discussion above, equality in (61) follows immediately, once one determines that
(63)vBαeiπα/βz+vDγeiπγ/δis analytic in v at vk,ℓ=ωℓeiπ/MQk/M. Thus, we must require that z+ωℓeiπ/MQk/MDγeiπγ/δ≠0, which is equivalent to z≠−ωℓ+1Qk/MDγ, as seen in the next sentence. Equality in (62) follows directly from the facts that eiπα/β=eiπ/MMα/β=eiπ/MBα and eiπγ/δ=eiπ/MMγ/δ=eiπ/MDγ. The proposition is now shown.Note that whenℓ=M−1 and z=ix in (62), the resulting expression matches the kth summand in (16) up to the constant factors 1/2π and −1/MμQ3.Having found the residues atωℓeiπ/MQk/M, the next proposition allows for the determination of the residues of f2v in (60) at the roots of z+vDγeiπγ/δ. First, observe that z+vDγeiπγ/δ=0 precisely when vDγ=eiπe−iπγ/δz, namely, when v=ω~κeiπ/Dγe−iπγ/δDγz0 where z0 is a Dγth root of z, ω~=e2πi/Dγ is the Dγth root of unity, and 0≤κ≤Dγ−1.Proposition 9.
Letz∈ℂ∗\S~ where S~≡−ωj+1Qk/MDγ∣k∈ℤ,0≤j≤M−1. For 0≤κ≤Dγ−1, ω~=e2πi/Dγ, z0 a fixed Dγth root of z, and z1=eiπ/Dγe−iπγ/δDγz0, the residue of
(64)f2v=1vθQ;vMvBαeiπα/βz+vDγeiπγ/δ
atvκ=ω~κeiπ/Dγe−iπγ/δDγz0=ω~κz1 is given by
(65)Resf2,vκ=1Dγ1−zω~κz1Bαeiπα/βθQ;ω~κz1M.Proof.
Equation (65) holds for Dγ=1 upon setting v=−e−iπγ/δz in the expression vBαeiπα/β/vθQ;vMeiπγ/δ. Observe that for Dγ≥2 the expression z+vDγeiπγ/δ can be factored as follows:
(66)z+vDγeiπγ/δ=eiπγ/δvDγ−eiπe−iπγ/δz(67)=eiπγ/δ∏j=0Dγ−1v−ω~jeiπ/Dγe−iπγ/δDγz0(68)=eiπγ/δ∏j=0Dγ−1v−ω~jz1(69)=v−ω~κz1eiπγ/δ∏j=0,j≠κDγ−1v−ω~jz1,where one uses (33) to move from (66) to (67) and one uses the definition z1=eiπ/Dγe−iπγ/δDγz0 to simplify (67) into (68). Notice next that evaluation of the bracketed expression in (69) at ω~κz1 yields
(70)eiπγ/δ∏j=0,j≠κDγ−1ω~κz1−ω~jz1=eiπγ/δω~κz1Dγ−1∏j=0,j≠κDγ−11−ω~j−κ=eiπγ/δω~κz1Dγ−1Dγ,where equality in (70) follows from (35) in Lemma 4.To compute the residue off2v in (64) at v=vκ=ω~κz1 one observes
(71)Resf2,vκ=1ω~κz1θQ;ω~κz1Mω~κz1Bαeiπα/βeiπγ/δ∏j=0,j≠κDγ−1ω~κz1−ω~jz1(72)=1ω~κz1θQ;ω~κz1Mω~κz1Bαeiπα/βeiπγ/δω~κz1Dγ−1Dγ(73)=1eiπγ/δω~κz1DγθQ;ω~κz1Mω~κz1Bαeiπα/βDγ(74)=1−zθQ;ω~κz1Mω~κz1Bαeiπα/βDγ.where (71) follows from (69) and (72) follows from (70), while (73) follows from a consolidation of the factors ω~κz1 and (74) follows from the facts that ω~ is a Dγth root of unity and z1Dγ=−e−πiγ/δz. Of course, we must have that 1/vθQ;vM is analytic at v=vκ=ω~κz1, and hence, ω~κz1M≠−Qk for any k∈ℤ, which is seen to be equivalent to z≠−ωj+1Qk/MDγ for any j=0,⋯,M−1, and any k∈ℤ. Furthermore, we must have ω~κz1≠0, which is equivalent to z≠0. Thus, we require that z∈ℂ∗\S~, as hypothesized. The proposition is now proven.Lemma 10.
LetM=lcmβ,δ, and let N∈ℕ. Set ΓN=CN−cN=∂AN be the positively oriented boundary of the annular region AN in ℂ enclosed by the circular paths CN=QN+1/M+QN/M/2eiϕ and cN=Q−N−1/M+Q−N/M/2eiϕ, where ϕ increases from 0 to 2π. Let z∈ℂ∗\S~, where S~≡−ωj+1Qk/MDγ∣k∈ℤ,0≤j≤M−1 and ω=e2πi/M.
Letv=ω~κz1 for 0≤κ≤Dγ−1 denote the roots of z+vDγeiπα/β=0, where ω~=e2πi/Dγ, z0 is a Dγth root of z, and z1=eiπ/Dγe−iπγ/δDγz0. Choose N sufficiently large so that for 0≤κ≤Dγ−1 one has ω~κz1∈IntAN, the interior of AN. Then, for f2v as in (62), one has
(75)∫ΓNf2vdv=∫ΓN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=2πi1μQ31M∑ℓ=0M−1∑k=−NN−1k+1Qkk+1/2ωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ(76)+2πieiπα/β−z1Dγ∑κ=0Dγ−1ω~κz1BαθQ;ω~κz1M.Proof.
The integral overΓN yields 2πi times the enclosed residues, which occur at v=vκ=ω~κz1 for 0≤κ≤Dγ−1 in IntAN the interior of AN, as well as at the zeroes of θQ;vM in AN, which by construction of ΓN are v=vk,ℓ=ωℓeiπ/MQk/M for −N≤k≤N and 0≤ℓ≤M−1. So the residue theorem gives
(77)∫ΓN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=2πi∑ℓ=0M−1∑k=−NNResf2,vk,ℓ+2πi∑κ=0Dγ−1Resf2,vκ(78)=2πi1μQ31M∑ℓ=0M−1∑k=−NN−1k+1Qkk+1/2ωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ(79)+2πieiπα/β−z1Dγ∑κ=0Dγ−1ω~κz1BαθQ;ω~κz1M,where Resf2,vk,ℓ in (77) has been replaced by (62) to obtain the summation in (78) and where Resf2,vκ in (77) has been replaced by (65) to obtain the summation in (79). The expressions in (78) and (79) now give (75) and (76), and the lemma is proven.Lemma 11.
Let the assumptions in the first paragraph of Lemma10 hold. Then,
(80)limN⟶∞∫CN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0,(81)limN⟶∞∫cN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0,
and hence,(82)limN⟶∞∫ΓN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0,(83)limN⟶∞∫ΓN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0.Proof.
We show (80) and (81), which implies immediately that (82) and (83) hold. Now, the vanishing of each of the limits in (80) and (81) holds because θQ;vM grows sufficiently rapidly as N approaches infinity for v∈CN or v∈cN. This rapid growth will follow directly from the identity (12). Designate C0=w∣w=Q1/M+1/2eiϕ,ϕ∈0,2π as a reference circle of radius ρ≔Q1/M+1/2>1 which by construction satisfies that θQ;wM never vanishes. This nonvanishing is the result of the fact that 1<ρ=w<Q1/M, whereby 1<wM<Q, along with the fact that all zeros of θQ;vM occur for vM=−Qk for some k∈ℤ, from (13). By continuity of θQ;wM on the compact set C0, there exist constants b and B, each depending on Q, such that for all w∈C0(84)0<b≤θQ;wM≤B<∞.Note thatv∈CN implies ∃w∈C0 with v=QN/Mw, and by (12), one has
(85)θQ;vM=θQ;QN/MwM=θQ;QNwM=QNN+1/2wNMθQ;wM,with θQ;vM=QNN+1/2ρNMθQ;wM≥QNN+1/2ρNMb. Then for N0 such that QN0/MρDγ−z>0 and all N≥N0, one has
(86)∫CN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=∫CN1θQ;vMvBαz+vDγeiπγ/δdvv≤1QNN+1/2ρNMbQN/MρBαQN/MρDγ−z2π,which approaches 0 as N approaches infinity, as QNN+1/2 is the dominant term. Similarly, v∈cN implies ∃w∈C0 with v=Q−N−1/Mw, and by (12), one has
(87)θQ;vM=θQ;Q−N−1/MwM=θQ;Q−N−1wM=Q−N−1−N/2wM−N−1θQ;wM,with θQ;vM=QNN+1/2ρM−N−1θQ;wM≥QNN+1/2ρM−N−1b.Thus, forN1 such that z−Q−N1−1/MρDγ>0 and all N≥N1, one has
(88)∫cN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=∫cN1θQ;vMvBαz+vDγeiπγ/δdvv≤1QNN+1/2ρM−N−1bQ−N−1/MρBαz−Q−N−1/MρDγ2π,which also vanishes as N approaches infinity, as QNN+1/2 is the dominant term. The lemma is now shown.With Lemma11 in mind, we record the following corollary which will be utilized later.Corollary 12.
LetS≡Sϕ1,ϕ2=v∈ℂ∗∣ϕ1≤argv≤ϕ2∪0 be the sector in ℂ emanating from the origin with argument falling in the interval ϕ1,ϕ2 with ϕ2−ϕ1≤2π. Let sN≡S∩cN and SN≡S∩CN, where z,CN, and cN are as in Lemma 10. Then
(89)limN⟶∞∫SN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0,(90)limN⟶∞∫sN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv=0.Proof.
One has(91)0≤∫SN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv(92)≤∫SN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv(93)≤∫CN1vθQ;vMvBαeiπα/βz+vDγeiπγ/δdv.Since the limit asN approaches infinity of (93) vanishes, by (80), the vanishing in (89) holds. The vanishing in (90) holds by replacing SN with sN and CN with cN in (91)–(93), taking limits as N approaches infinity, and relying on (81). This proves the corollary.We have now arrived at a preliminary version of the main theorem of this study. It is “preliminary” in that in certain cases each side of the main equality (98) and (99) in Theorem 13 immediately below reduces to an identically 0 function, and it will be necessary later to give nonidentically vanishing criteria in our main result Theorem 23, where we will also relate the expression in (99) to the Fourier transform of a function naturally generated by fμ,λt. First, we pause to catalogue the current set of results.Theorem 13.
Forq>1, let fμ,λt be as in (1) and f~μ,λt be as in (14), respectively, where we assume μ∈ℚ and λ∈ℚ+. Let z∈ℂ∗\S~ where S~≡−ωj+1Qk/MDγ∣k∈ℤ,0≤j≤M−1. Then,
(94)1μQ31M∑ℓ=0M−1∑k=−∞∞−1kQkk+1/2ωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ(95)=eiπα/β−z1Dγ∑κ=0Dγ−1ω~κz1BαθQ;ω~κz1M,where μQ is given by (11), where θQ;z is the Jacobi theta function, where
(96)αβ=μ+12γδ=λ2M=lcmβ,δQ=q2/λ=qδ/γB=MβD=Mδω=e2πi/Mω~=e2πi/Dγ,
and whereα∈ℤ and β,γ,δ∈ℕ with α/β and γ/δ in reduced form. M is taken to be the least common multiple of β and δ. Also, for z0 any fixed Dγth root of z, one has that z1 in (95) is given by
(97)z1=eiπ/Dγe−iπγ/δDγz0,wherebyz1Dγ=−e−πiγ/δz.
Settingz=ix (94) for x∈ℝ and requiring that δ≠0mod4 give the following relation of the weighted average of the rotations of the Fourier transforms Ff~μ,λtx with the average of the rotations of z3/θQ;z3M:
(98)1MFf~μ,λtx+∑ℓ=0M−2ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ(99)=μQ3eiπα/β2π1−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M.
Here,Ff~μ,λtx/ωℓ+1Dγ is given by (17). Also, for z2 any fixed Dγth root of ix, one has that z3 in (99) is given by
(100)z3=eiπ/Dγe−iπγ/δDγz2,wherebyz3Dγ=−e−πiγ/δix.Proof.
Integrating the integrand (64) over the oriented boundary ΓN of the annular region AN as in Lemma 10 gives expressions (75) and (76). Taking the limit of (75) and (76) as N approaches infinity, and relying on Lemma 11 gives
(101)0=2πi1μQ31M∑ℓ=0M−1∑k=−∞∞−1k+1Qkk+1/2ωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ(102)+2πieiπα/β−z1Dγ∑κ=0Dγ−1ω~κz1BαθQ;ω~κz1M,where z1 is given by (97). Dividing (101) and (102) by 2πi and moving the double summation to the left side of the equality give (94) and (95). Now, multiplying (94) and (95) by μQ3/2π gives
(103)1M∑ℓ=0M−112π∑k=−∞∞−1kQkk+1/2ωℓ+1Qk/MBαz+ωℓ+1Qk/MDγ(104)=μQ32πeiπα/β−z1Dγ∑κ=0Dγ−1ω~κz1BαθQ;ω~κz1M.Settingz=ix in (103) and (104) gives that (97) becomes (100), and we then also replace z1 in (104) by z3 to obtain (105)–(107) below. Factoring out the powers of ωℓ+1 from (103) now yields
(105)1M∑ℓ=0M−1ωℓ+1Bα112π∑k=−∞∞−1kQkk+1/2Qk/MBαix+ωℓ+1Qk/MDγ(106)=1M∑ℓ=0M−1ωℓ+1Bαωℓ+1Dγ12π∑k=−∞∞−1kQkk+1/2Qk/MBαix/ωℓ+1Dγ+Qk/MDγ(107)=μQ32πeiπα/β−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M.One now recognizes that in (106)
(108)12π∑k=−∞∞−1kQkk+1/2Qk/MBαix/ωℓ+1Dγ+Qk/MDγ=12π∑k=−∞∞−1kQkk+1/2Qk/MM/βαix/ωℓ+1Dγ+Qk/MM/δγ=12π∑k=−∞∞−1kQkk+1/2Qkα/βix/ωℓ+1Dγ+Qkγ/δ(109)=Ff~μ,λtx/ωℓ+1Dγ,where (109) follows from (17).Relying on (109) one sees that (106)-(107) becomes
(110)1M∑ℓ=0M−1ωℓ+1Bα−DγFf~μ,λtx/ωℓ+1Dγ(111)=μQ32πeiπα/β−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M,which is equivalent to (98) and (99) when one observes that for ℓ=M−1 in (110) one has ωℓ+1=ωM=1.At this point, we have shown (98) and (99) only for x≠0, so we next handle the case that x=0. Notice that (108) and (109) are defined at x=0 via
(112)Ff~μ,λtx/ωℓ+1Dγx=0=12π∑k=−∞∞−1kQkk+1/2Qkα/β0+Qkγ/δ(113)=12π∑k=−∞∞−Qα/β−γ/δ−1kQkk−1/2(114)=12πθQ;−Qα/β−γ/δ−1,where (114) follows from (10). Now, at x=0, from (114) one observes that (110) and (111) still hold in the sense that it becomes
(115)1M∑ℓ=0M−1ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγx=0=1M∑ℓ=0M−1ωℓ+1Bα−Dγ12πθQ;−Qα/β−γ/δ−1(116)=0=limx⟶0μQ32πeiπα/β−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M,where the vanishing of (115) follows from the vanishing of (159) which is proven in the paragraph containing (164) in Proposition 20 below (with n set to 0) and the vanishing of the limit in (116) follows from Corollary 32 below. Finally the requirement that δ≠0mod4 is equivalent to the condition that z=ix does not belong to S~ for any x∈ℝ by Lemma 14 below. The theorem is now proven.The following lemma gives a simple characterization for having somex with ix∈S~ (which is to be avoided in Theorem 13).Lemma 14.
In the setting of Theorem13, with the notation as in (96), one has
(117)∃x∈ℝsuchthatix∈S~=−ωj+1Qk/MDγ∣j,k∈ℤ(118)⟺δ=0mod4.Proof.
Observe that the existence of anx∈ℝ with ix∈S~ is equivalent to the existence of a j∈ℤ with either i=ωj+1Dγ (in which case x=−Qk/MDγ for some k∈ℤ) or i=−ωj+1Dγ (in which case x=Qk/MDγ for some k∈ℤ).
Sinceω=exp2πi/M, one has
(119)∃x∈ℝwithix∈S~⟺∃j∈ℤwith±i=ωj+1Dγ⟺∃j∈ℤ,m∈1,3,andn∈ℤwitheπim/2e2πin=e2πij+1Dγ/M(120)⟺∃j∈ℤ,m∈1,3,andn∈ℤwithm2+2n=2j+1γδ(121)⟺∃j∈ℤ,m∈1,3,andn∈ℤwithδm+4n=4j+1γ,where (120) follows from the fact that D=M/δ. Now from (121), one has that δ must be divisible by 4. Thus,
(122)∃x∈ℝwithix∈S~⇒δ=0mod4.
Conversely, ifδ=0mod4, then ∃p∈ℕ with δ=4p. Since γ and δ have no common factors, one concludes that γ is not divisible by 2 or 4. Thus, γ=m+4n for some m∈1,3 and some n∈ℤ. Hence,
(123)4pγ=δγ=δm+4n.
Settingj=p−1 gives that (121) holds. We conclude
(124)δ=0mod4⇒∃x∈ℝwithix∈S~.
Equations (122) and (124) now give the equivalence (117) and (118), and the lemma is proven.


## 3. The Functions Naturally Generated by thefμ,λt

In this section, we again assume the notation of the previous section. In particular, the notation in (96) holds. Namely, μ,λ∈ℚ are rational with λ>0, with x,t∈ℝ. Also μ+1/2=α/β and λ/2=γ/δ with α∈ℤ and β,γ,δ∈ℕ, where both α/β and γ/δ are in a reduced form. Finally, M=lcmβ,δ is the least common multiple of β and δ, B=M/β, D=M/δ, and ω=e2πi/M. Let Q=q2/λ, and let 0≤ℓ≤M−1. With this notation in mind, we are able to make the following definitions.Definition 15.
Assume thatδ≠0mod4. Then by Lemma 14, one sees from (117), (118), and (119) that ∄j∈ℤwith±i=ωj+1Dγ. Hence, the real part Rωℓ+1Dγ≠0. For the real part Rωℓ+1Dγ>0 define
(125)f~μ,λωℓ+1Dγt≡+1∑k∈ℤ−1kexp−ωℓ+1Qk/MDγtqkk−μ/λ=+1fμ,λωℓ+1Dγt,fort≥0,0,fort<0,and for real part Rωℓ+1Dγ<0 define
(126)f~μ,λ≤ωℓ+1Dγt≡0,fort>0,−1∑k∈ℤ−1kexp−ωℓ+1Qk/MDγtqkk−μ/λ=−1fμ,λωℓ+1Dγt,fort≤0,while for Rωℓ+1Dγ<0 we also define
(127)f~μ,λωℓ+1Dγt≡χ−∞,0tf~μ,λ≤ωℓ+1Dγt(128)=0,fort≥0,−1fμ,λωℓ+1Dγt,fort<0,where χ−∞,0t is the characteristic function of the interval −∞,0. We emphasize the +1 coefficient in (125) versus the −1 coefficient in (126) and (128).Note that forRωℓ+1Dγ<0(129)limt⟶0−f~μ,λωℓ+1Dγt=f~μ,λ≤0=−fμ,λ0,and for Rωℓ+1Dγ>0(130)limt⟶0+f~μ,λωℓ+1Dγt=f~μ,λ0=fμ,λ0.Definition 16.
Assumeδ≠0mod4. The function naturally generated by fμ,λt is given by
(131)Wμ,λt≡∑ℓ=0M−1ωℓ+1Bαf~μ,λωℓ+1Dγt(132)=∑−ωℓ+1Bαf~μ,λωℓ+1Dγt+∑+ωℓ+1Bαf~μ,λωℓ+1Dγt,where the leftmost summation over the negative index − in (132) stands for summation over the indices ℓ with Rωℓ+1Dγ<0 and the rightmost summation over the positive index + in (125) stands for summation over the indices ℓ with Rωℓ+1Dγ>0.While theWμ,λt are constructed here in Definition 16, they were originally obtained as the inverse Fourier transform of the expression in (98) in Theorem 13 (scaled by M). In this sense, the Wμ,λt are natural. We point out that the Wμ,λt may vanish identically, as discussed in Proposition 21 below. However, there are a wealth of Wμ,λt that are not vanishing identically, as characterized in Proposition 21. In any case, the Wμ,λt have useful properties that are now recorded as a series of propositions. The first observation is that Wμ,λt is real valued.Proposition 17.
For allt∈ℝ, Wμ,λt∈ℝ.Proof.
Observe that forωℓ+1Dγ=Rωℓ+1Dγ+iIωℓ+1Dγ one has Conjωℓ+1Dγ=Rωℓ+1Dγ−iIωℓ+1Dγ, and hence, conjugation preserves the sign of Rωℓ+1Dγ. Thus, conjugation also preserves the sets −≡ℓ∣Rωℓ+1Dγ<0 and +≡ℓ∣Rωℓ+1Dγ>0. This allows us to conclude that
(133)Conjf~μ,λωℓ+1Dγt=f~μ,λω¯ℓ+1Dγt
independently of the−1 factor in (128) versus the +1 factor in (125), where ω¯=Conjω=e−2πi/M. Hence, from (132), one has
(134)ConjWμ,λt=∑−ω¯ℓ+1Bαf~μ,λω¯ℓ+1Dγt+∑+ω¯ℓ+1Bαf~μ,λω¯ℓ+1Dγt=∑ℓ=0M−1ω¯ℓ+1Bαf~μ,λω¯ℓ+1Dγt=∑ℓ=0M−1ωℓ+1Bαf~μ,λωℓ+1Dγt,where the last equation in (134) follows since conjugation permutes each of the sets ωℓ∣0≤ℓ≤M−1, ωℓ∣ℓ∈−, and ωℓ∣ℓ∈+. The proposition now follows.The second result gives smoothness.Proposition 18.
Wμ,λt is C∞ at t=0, and hence, Wμ,λt is in C∞ℝ. Furthermore, Wμ,λt is Schwartz.Proof.
Away fromt=0, Wμ,λt is C∞ via (125)-(126). From (132), note that for t<0, one has Wμ,λt=∑−ωℓ+1Bαf~μ,λωℓ+1Dγt and for t≥0 one has Wμ,λt=∑+ωℓ+1Bαf~μ,λωℓ+1Dγt. Smoothness will follow at t=0, by showing that limt⟶0−Wμ,λnt=limt⟶0+Wμ,λnt for all n∈ℕ0. From (125) and (126), one has
(135)limt⟶0−Wμ,λnt=limt⟶0−∑−ωℓ+1Bαdndtnf~μ,λωℓ+1Dγt=limt⟶0−∑−ωℓ+1Bα−1nωℓ+1Dγnf~μ+nλ,λωℓ+1Dγt=∑−ωℓ+1Bα−1nωℓ+1Dγnf~μ+nλ,λ≤0(136)=∑−ωℓ+1Bα−1nωℓ+1Dγn−1fμ+nλ,λ0,where (135) follows directly from differentiating (126) and (136) also follows from (126). Also,
(137)limt⟶0+Wμ,λnt=limt⟶0+∑+ωℓ+1Bαdndtnf~μ,λωℓ+1Dγt=limt⟶0+∑+ωℓ+1Bα−1nωℓ+1Dγnf~μ+nλ,λωℓ+1Dγt=∑+ωℓ+1Bα−1nωℓ+1Dγnf~μ+nλ,λ0(138)=∑+ωℓ+1Bα−1nωℓ+1Dγnfμ+nλ,λ0,where (135) follows directly from differentiating (125) and (138) follows from (1). Equality of (136) with (138) would be equivalent with
(139)0=∑−ωℓ+1Bαωℓ+1Dγn+∑+ωℓ+1Bαωℓ+1Dγnfμ+nλ,λ0.Now ifBα+Dγn≠0modM, then
(140)∑−ωℓ+1Bαωℓ+1Dγn+∑+ωℓ+1Bαωℓ+1Dγn=0from Lemma 6. Hence (139) holds. On the other hand, if Bα+Dγn=0modM, then
(141)∑−ωℓ+1Bαωℓ+1Dγn+∑+ωℓ+1Bαωℓ+1Dγn=M≠0from Lemma 6. In this case, one has that for some k∈ℤ(142)Mk=Bα+Dγn=Mαβ+γδn⇔k=αβ+γδn∈ℤ(143)⇔k−1=αβ−1+γδn∈ℤ,where we have used (96) to rewrite B=M/β,D=M/δ in the second equality of (142). Now from Lemma 2.4 of [2], one has fμ+nλ0=θq2/λ;−qμ+nλ−1/λ=θQ;−Qμ+nλ−1/2, where from (96) we have used q2/λ=Q. From (13), it follows that θQ;−Qμ+nλ−1/2=0 if and only if there is a k∈ℤ with k=μ+nλ−1/2. Now,
(144)k=μ+nλ−12⇔k=μ+12−1+nλ2⇔k=αβ−1+γδn∈ℤ,where (96) was used to rewrite μ+1/2=α/β and λ/2=γ/δ. Observing equality of the rightmost expression in (144) with the rightmost expression (143), one has that in (139)
(145)∑ℓ=0M−1ωℓ+1Bαωℓ+1Dγn≠0preciselywhenfμ+nλ,λ0=0.Thus, in all cases, (139) holds, whence equality of (136) and (138) holds. We conclude that Wμ,λt is C∞ at t=0 and thus on ℝ.Finally, the fact thatWμ,λt is Schwartz follows from (1) the fact that it is in C∞ℝ and (2) from the fact that for t sufficiently large one has Wμ,λt≤K1t−K2lnt+K3 for constants K1,K2>0 and K3∈ℝ, which in turn follows from the expressions (125) and (126) and from Proposition 8.1 of [2]. The proposition is now proven.One consequence of Proposition18 is that Wμ,λt has a Fourier transform which is Schwartz. The following proposition allows us to observe that Wμ,λt was defined so that its Fourier transform would be given by (98) (up to the constant factor 1/M).Proposition 19.
The Fourier transform ofWμ,λt is given by
(146)FWμ,λtx=Ff~μ,λtx+∑ℓ=0M−2ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ.Proof.
From (131), one has
(147)FWμ,λx=∑ℓ=0M−1ωℓ+1BαFf~μ,λωℓ+1Dγtx.
We next evaluateFf~μ,λωℓ+1Dγtx in (147). Observe that for Rωℓ+1Dγ>0 one has
(148)Ff~μ,λωℓ+1Dγtx=12π∫0∞e−ixt∑k=−∞∞−1kexp−ωℓ+1Qk/MDγtqkk−μ/λdt=12π∑k=−∞∞−1kqkk−μ/λ∫0∞exp−ixt−ωℓ+1Qk/MDγtdt=12π∑k=−∞∞−1kqkk−μ/λexp−ixt−ωℓ+1Qk/MDγt−ix−ωℓ+1Qk/MDγ0∞=12π∑k=−∞∞−1kqkk−μ/λ1ix+ωℓ+1Qk/MDγ.
Similarly, forRωℓ+1Dγ<0, one has
(149)Ff~μ,λωℓ+1Dγtx=12π∫−∞0e−ixt−1∑k=−∞∞−1kexp−ωℓ+1Qk/MDγtqkk−μ/λdt=12π∑k=−∞∞−1kqkk−μ/λ∫−∞0−1exp−ixt−ωℓ+1Qk/MDγtdt=12π∑k=−∞∞−1kqkk−μ/λ−1exp−ixt−ωℓ+1Qk/MDγt−ix−ωℓ+1Qk/MDγ−∞0=12π∑k=−∞∞−1kqkk−μ/λ1ix+ωℓ+1Qk/MDγ.
From the matching forms of (148) and (149), one sees that (147) becomes
(150)FWμ,λtx=∑ℓ=0M−1ωℓ+1Bα12π∑k=−∞∞−1kqkk−μ/λ1ix+ωℓ+1Qk/MDγ.
Noticing that(151)−1kqkk−μ/λ=−1kqkμ+1/λqkk+1/λ=−1kq2/λkμ+1/2q2/λkk+1/2=−1kQkμ+1/2Qkk+1/2=−1kQk/MMα/βQkk+1/2=−1kQk/MBαQkk+1/2
reexpresses (150) as
(152)FWμ,λtx=∑ℓ=0M−1ωℓ+1Bα12π∑k=−∞∞−1kQk/MBαQkk+1/21ix+ωℓ+1Qk/MDγ(153)=∑ℓ=0M−1ωℓ+1Bαωℓ+1Dγ12π∑k=−∞∞−1kQkk+1/2Qk/MBαix/ωℓ+1Dγ+Qk/MDγ=∑ℓ=0M−1ωℓ+1Bα−DγFf~μ,λtx/ωℓ+1Dγ,where (153) follows from (17). Now, (153) gives (146) and the proposition is proven.Another property ofWμ,λt is that all of its moments vanish.Proposition 20.
All moments ofWμ,λt vanish. That is,
(154)∫−∞∞tnWμ,λtdt=0,∀n∈ℕ∪0.
Equivalently, all the derivatives ofFWμ,λtx satisfy
(155)dndxnFWμ,λtxx=0=0.Proof.
We proceed by showing (155). Differentiating (152) yields
(156)dndxnFWμ,λtx=−inn!∑ℓ=0M−1ωℓ+1Bα2π∑k=−∞∞−1kQkk+1/2Qk/MBαix+ωℓ+1Qk/MDγn+1.
Evaluating atx=0 gives
(157)dndxnFWμ,λtxx=0=−inn!∑ℓ=0M−1ωℓ+1Bα2π∑k=−∞∞−1kQkk+1/2Qk/MBαωℓ+1Qk/MDγn+1=−inn!2π∑ℓ=0M−1ωℓ+1Bαωℓ+1Dγn+1∑k=−∞∞−1kQkk+1/2Qk/MBα−Dγn+1(158)=−inn!2π∑ℓ=0M−1ωℓ+1Bα−Dγn+1θQ;−Qα/β−n+1γ/δ−1(159)=−inn!2π∑ℓ=0M−1ωℓ+1Bα−Dγn+1θQ;−Qα/β−n+1γ/δ−1.
Here, movement from (157) to (158) is justified as follows:
(160)∑k=−∞∞−1kQkk+1/2Qk/MBα−Dγn+1=∑k=−∞∞−1kQkk+1/2Qk/MMα/β−γn+1/δ(161)=∑k=−∞∞−1kQkk+1/2Qα/β−γn+1/δk(162)=∑k=−∞∞−1kQkk−1/2Qα/β−γn+1/δ−1k(163)=θQ;−Qα/β−n+1γ/δ−1,where (160) follows from B=M/β,D=M/δ as in (96); (161) follows from cancelling out M; (162) follows from multiplying up and down by Q−k; and (163) follows from (10).Examining (159), one sees that if Bα−Dγn+1≠0modM in (159), then by Lemma 6, one has ∑ℓ=0M−1ωℓ+1Bα−Dγn+1=0 and (159) vanishes. On the other hand, if Bα−Dγn+1=0modM, then there is a k∈ℤ with
(164)Mk=Bα−Dγn+1=Mαβ−γδn+1⟺k=αβ−γδn+1⟺k−1=αβ−γδn+1−1∈ℤ,whence θQ;−Qα/β−n+1γ/δ−1=0 by (13). Thus, (159) again vanishes. Since (159) now vanishes in all cases, we have vanishing of every derivative of FWμ,λtx at x=0, giving (155). Thus, (154) now holds, and we have vanishing of all moments. This completes the proof of the proposition.Notice the similarity of the argument in Proposition20 with that of Proposition 18. In each case, a factor formed from the sum of powers of roots of unity fails to vanish precisely when the remaining theta function factor vanishes.The next proposition gives a simple characterization forWμ,λt to vanish (and to not vanish) identically.Proposition 21.
For all notation as in (96), the following equivalences hold:
(165)δ≠0modβ⇔Wμ,λt≡0⇔1MFWμ,λtx≡0(166)⇔1M∑ℓ=0M−1ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ≡0(167)⇔μQ3eiπα/β2π1−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M≡0.
Thus,δ=0modβ occurs precisely when Wμ,λt does not vanish identically, which in turn occurs precisely when the identity (98) and (99) in Theorem 13 is not an identically zero tautology.Proof.
The right-most equivalence in (165) holds by linearity and injectivity of the Fourier transform; and the equivalences in (166) and (167) hold by Proposition 19 and Theorem 13, respectively. It remains to show the following equivalence:
(168)δ≠0modβ⇔thevanishingof167.
Observe that in the argument of the theta function in (167), the expression ω~κM=ω~kM holds precisely when ∃jwithe2πij/Dω~κ=ω~k (and then k=κ+jγ). One might expect that the previous statement would be that ∃Jwithe2πiJ/Mω~κ=ω~k; however, e2πiJ/M=e2πiJ/Dδ=e2πiJγ/Dγδ is an integral power of ω~=e2πi/Dγ precisely when J=jδ is a multiple of δ, resulting in canceling of δ terms.Next, summing over indices with like values ofω~κM first gives
(169)∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M=∑κ=0γ−1∑j=0D−1e2πij/Dω~κz3BαθQ;e2πij/Dω~κz3M=∑κ=0γ−1∑j=0D−1e2πij/DBαω~κz3BαθQ;ω~κz3M=∑κ=0γ−1∑j=0D−1e2πij/DBαω~κz3BαθQ;ω~κz3M(170)=∑j=0D−1e2πj/DBα∑κ=0γ−1ω~κz3BαθQ;ω~κz3M.By Proposition22 below, the expression ∑κ=0γ−1ω~κz3Bα/θQ;ω~κz3M in (170) does not vanish identically. Thus, (169) and (170) vanish identically if and only if ∑j=0D−1e2πj/DBα vanishes. However, this summation of Dth roots of unity to the Bα power vanishes precisely when Bα≠0modD by Lemma 6. Thus, Bα≠0modD is equivalent to identically vanishing in (169) and (170) and therefore equivalent to identically vanishing in (167). Now,
(171)Bα≠0modD⟺∄nsuchthatBα=Dn⟺∄nsuchthatMβα=Mδn⟺∄nsuchthatαδβ=n⇔δ≠0modβ,where the last equivalence holds from the fact that α/β is in a reduced form. Thus, the leftmost equivalence in (165) holds. From this left-most equivalence, one deduces that Wμ,λt does not vanish identically precisely when δ=0modβ. The proposition is now proven.The following proposition, utilized in Proposition21, relies on properties of the Jacobi theta function to obtain a nonidentically vanishing condition.Proposition 22.
The function(172)∑κ=0γ−1ω~κzBαθQ;ω~κzMis not identically 0 in the argument z∈ℂ. Hence,
(173)∑κ=0γ−1ω~κz3BαθQ;ω~κz3Mgiven in (170) is not identically 0 in z3, where z3 is as in (100) and (184).Proof.
Ifγ=1, then (172) becomes zBα/θQ;zM, which does not vanish identically. For instance when z=1, it becomes 1/θQ;1≠0. If γ>1, we show that the function
(174)∑κ=0γ−1ω~κzBαθQ;ω~κzM=∑κ=0γ−1ω~κzBα∏0≤j≠κγ−1θQ;ω~jzM∏j=0γ−1θQ;ω~jzM
is also not identically0. This in turn is equivalent to the numerator in the right hand side of (174), namely,
(175)∑κ=0γ−1ω~κzBα∏0≤j≠κγ−1θQ;ω~jzM=zBαθQ;ω~zMθQ;ω~2zM⋯θQ;ω~γ−1zM(176)+θQ;zM∑κ=1γ−1ω~κzBα∏1≤j≠κγ−1θQ;ω~jzM,
not being identically0 in z. Setting z=−Q1/M in (175) and (176) gives that
(177)θQ;−Q1/MM=θQ;−Q=0and then only the summand in (175) survives:
(178)∑κ=0γ−1ω~κ−Q1/MBα∏0≤j≠κγ−1θQ;ω~jM−Q=−Q1/MBαθQ;ω~M−QθQ;ω~2M−Q⋯θQ;ω~γ−1M−Q≠0,where the nonvanishing in (178) is obtained from (13) along with the fact that
(179)−Qω~jM=−Qe2πij/DγM=−Qe2πijδ/γdoes not lie on the negative real axis for j=1,⋯,γ−1. To see this latter point, if jδ/γ=n∈ℤ, one has jδ=γn, from which one has that j is divisible by γ (as γ/δ is assumed to be in reduced form). However, each j=1,⋯,γ−1 is not divisible by γ. Thus, (172) does not vanish identically. By the identity theorem, (172) does not vanish on any subset of ℂ having a limit point. One concludes that (173) is not identically 0 in z3, as the set of z3 as in (184) ranges over a set with limit point as x varies in ℝ. The proposition is now shown.In light of Proposition21 and Lemma 14, we refine and sharpen Theorem 13 by
(1)
removing all identically0=0 tautologies in (98) and (99) via making the additional assumption that δ=0modβ(2)
guaranteeing that the expressions in (98) and (99) are well-defined via making the easily checked assumption that δ≠0mod4(3)
incorporatingWμ,λt into the theorem while including the additional properties of Wμ,λt garnered from Propositions 17–20In doing so, we arrive at the main theorem of this study.Theorem 23.
Letq>1, and for t≥0, let fμ,λt be defined as in (1), with μ∈ℚ,λ∈ℚ+. For t∈ℝ, let Wμ,λt, as in Definition 16, be the function naturally generated by fμ,λt. Let the notation of (96) hold. In particular, let μ+1/2=α/β and λ/2=γ/δ be in a reduced form; let δ≠0mod4 with δ=0modβ (which gives M=δ and D=1); and let ω=e2πi/δ with ω~=e2πi/γ. Then, Wμ,λt is a real valued Schwartz wavelet with all moments vanishing with Fourier transform given by FWμ,λtx satisfying
(180)1δFWμ,λtx(181)=1δFf~μ,λtx+∑ℓ=0δ−2ωℓ+1Bα−γFf~μ,λtxωℓ+1γ(182)=μQ3eiπα/β2π1−ix1γ∑κ=0γ−1ω~κz3BαθQ;ω~κz3δ,where θQ;z is the Jacobi theta function, where
(183)Q=qδ/γ,B=δβ,ω=e2πi/δ,ω~=e2πi/γ,
and where forz2 any fixed γth root of ix, one has that z3 in (182) is given by
(184)z3=eiπ/γe−iπ/δz2,z3γ=−e−πiγ/δix.
Also,Ff~μ,λtx/ωℓ+1Dγ is given by (17). Furthermore, given fμ,λt, one has that Wμ,λt is uniquely defined by Definition 16, and it satisfies the same multiplicatively advanced differential equation (MADE) on ℝ as does fμ,λt on 0,∞, namely,
(185)Wμ,λδt=−1γ+δqγγ+μ/λWμ,λqγt.Proof.
Wμ,λt is real valued and Schwartz via Propositions 17 and 18. From Proposition 19, one has that equality in (187) below holds, and, from equation (98) and (99) in Theorem 13, one has that equality in (188) below holds.
(186)1MFWμ,λtx(187)=1MFf~μ,λtx+∑ℓ=0M−2ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ(188)=μQ3eiπα/β2π1−ix1Dγ∑κ=0Dγ−1ω~κz3BαθQ;ω~κz3M.
From Lemma14, the hypothesis that δ≠0mod4 is equivalent to the condition that (186)–(188) are defined for all x∈ℝ. By Proposition 21, Wμ,λt is identically 0 in t precisely when δ≠0modβ; thus, the hypothesis that δ=0modβ in the theorem gives precisely all the nonidentically zero examples of (186)–(188). Furthermore, from the hypothesis that δ=0modβ, one has that M=lcmβ,δ=δ, from which we conclude that D=M/δ=δ/δ=1. Setting M=δ and D=1 in (186)–(188) yields equations (180)–(182). Similarly, setting M=δ and D=1 in (96) (respectively (100)) yields (183) (respectively (184)). Next, all moments vanish by Proposition 20. It remains to show that Wμ,λt is a wavelet solving the MADE (185). The wavelet property follows from the following three criteria:
(1)
Wμ,λt∈L1ℝ∩L2ℝ∩L∞ℝ because Wμ,λt is Schwartz(2)
∫−∞∞Wμ,λtdt=0 because each moment, including the 0th-moment, of Wμ,λt vanishes(3)
∫−∞∞FWμ,λtx2/xdx<∞ because FWμ,λtx is Schwartz and decays rapidly in the tails and because FWμ,λtx vanishes to infinite order at x=0 by Proposition 20Now, from [2], fμ,λt on 0,∞ satisfies the MADE
(189)fμ,λδt=−1γ+δqγγ+μ/λfμ,λqγt.This follows, since(190)λ2=γδ,whenceλ=2γδ.SettingA=γ and L=δ in equations (16)–(18) in Theorem 2.2 of [2] gives the MADE (189). The MADE for Wμ,λt now follows from (189) as Wμ,λt is a linear combination of expressions involving fμ,λt (by Definition 16 and by (125)–(127)). This proves the theorem.Remark 24.
While the assumption thatδ=0modβ in Theorem 23 may at first seem to be limiting, in fact, there remains a wealth of nonidentically-zero cases (with δ=0modβ). The examples below in Sections 5 and 6 demonstrate that Theorem 23 is quite general in nature. For now, observe that if δ is a multiple of β then lcmβ,δ=δ=M. Hence, if M=δ=∏j=1Jpjnj is the prime factorization of M, the nonidentically zero cases for Wμ,λt and its Fourier transform are handled by those β=∏j=1Jpjkj with 0≤kj≤nj for all j∈1,⋯,J and with the parameters μ and λ satisfying μ+1/2=α/β and λ/2=γ/δ in a reduced form.Remark 25.
We also point out that even whenδ≠0modβ, we are able to relate the Fourier transform of fμ,λt to the Jacobi theta function. However, the Fourier relation in this setting comes from an analogue of Wμ,λt that is generically noncontinuous at t=0 and is not a wavelet. See Theorem 28 and the related discussion below.Remark 26.
In a development similar to (169)-(170), it is also possible to show that
(191)∑ℓ=0M−1ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ(192)=∑j=0D−1e2πj/DBα∑ℓ=0δ−1ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ.However, the development in (169) and (170) is preferred in order to harness properties of the Jacobi theta function to gain a nonidentically vanishing criterion, as in Proposition 22. Nonetheless, in comparing the expression in (170) with that in (192), observing that they share a common factor ∑j=0D−1e2πj/DBα, and contemplating equation (98) - (99), one is led to consider the relation between
(193)∑κ=0γ−1ω~κz3BαθQ;ω~κz3Mand∑ℓ=0δ−1ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ,including whenδ≠0modβ. This will be done in Theorem 28 below, where it will allow for the recovery of Fourier transform information for fμ,λt in the cases that δ≠0modβ.Remark 27.
At this juncture, we pause to convey a surprising consequence of expressing the Fourier transform ofWμ,λ in terms of Jacobi theta functions, as in Theorem 23. Knowledge of the Fourier transform of Wμ,λ in terms of the Jacobi theta function allows for the proof of nonvanishing results in the cases of γ=1,2 in Corollary 34 and Proposition 36 below. This pair of nonvanishing results serves to underpin a main connection between the solutions of MADEs of type Wμ,λ to the theory of wavelet frames in harmonic analysis. That is, in these cases, Wμ,λ is a mother wavelet generating a wavelet frame for the square integrable functions L2ℝ, as is demonstrated in Theorem 37 below. Thus, the current study lays the foundation for a set of strong connections between solutions of MADEs, special function theory, and the theory of wavelet frames. Next, we recover information on the Fourier transform of f~μ,λt when δ≠0modβ in the second main theorem of this work, again relating the Fourier transform to the Jacobi theta function.Theorem 28.
Forq>1, let fμ,λt be as in (1) and f~μ,λt be as in (14), respectively, where we assume μ∈ℚ and λ∈ℚ+, with notation as in (96), in particular μ+1/2=α/β and λ/2=γ/δ are in reduced form. Let θQ;z be the Jacobi theta function. Let δ≠0modβ and let δ≠0mod4. Then,
(194)1μQ31δ∑ℓ=−1δ−2∑k=−∞∞−1kQkk+1/2ωℓ+1Qk/MBαix+ωℓ+1Qk/MDγ(195)=eiπα/β−ix1γ∑κ=κ1κ1+γ−1ω~κz3BαθQ;ω~κz3M(196)+De2πiαδ/β−12πi∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv.
Here, forz2 a fixed Dγth root of ix, one has that z3 in (195) (and (201) below) is given by
(197)z3=eiπ/Dγe−iπγ/δDγz2,wherebyz3Dγ=−e−πiγ/δix.
Also,R1 is the ray emanating from the origin given by (210) and (206), and
(198)κ1≡minκ∣argω~κz3>−πM.
Furthermore,μQ is given by (11), with
(199)ω=e2πi/Mandω~=e2πi/Dγ,where M is taken to be the least common multiple of β and δ.
One also has the following relation of the weighted partial average of the rotations of the Fourier transformsFf~μ,λtx with a partial average of rotations of z3/θQ;z3M:
(200)1δFf~μ,λtx+∑ℓ=0δ−2ωℓ+1Bα−DγFf~μ,λtxωℓ+1Dγ(201)=μQ3eiπα/β2π1−ix1γ∑κ=κ1κ1+γ−1ω~κz3BαθQ;ω~κz3M(202)+μQ3De2πiαδ/β−12π3/2i∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv.
Again,z3 in (201) is given by (197). The expression in (202) is referred to as the defect.Proof.
Instead of integrating the integrandf2v in (64) over the oriented boundary ΓN of the annulus AN as in Theorems 13 and 23, we shall integrate f2v over the oriented boundary Γ~N of the intersection of the sector S=Sϕ1,ϕ2=v∈ℂ∗∣ϕ1≤argv≤ϕ2∪0 with the annulus AN, where ϕ1 is chosen as in (206) below and ϕ2−ϕ1=2π/D. Thus, Γ~N=R1,N+SN−R2,N−sN. Here, sN≡S∩cN and SN≡S∩CN, where as before, CN=eiϕQN+1/M+QN/M/2 and cN=eiϕQ−N−1/M+Q−N/M/2, but in the current setting, one has SN and sN that ϕ increases from ϕ1 to ϕ2. Also, R1,N and R2,N are portions of two rays, as given by
(203)R1,N=v=ρeiϕ1Q−N−1/M+Q−N/M2≤ρ≤QN+1/M+QN/M2,(204)R2,N=v=ρeiϕ2Q−N−1/M+Q−N/M2≤ρ≤QN+1/M+QN/M2.Referring to (117) and (118) in Lemma 14, one has
(205)δ≠0mod4⟺∀x∈ℝix∉S~=−ωj+1Qk/MDγj,k∈ℤ,and this in turn is equivalent to the fact that for all x∈ℝ the roots (in v) of ix+vDγeiπγ/δ=0 do not lie among the zeroes of θQ;vM by the remarks at the ends of Propositions 8 and 9. Hence, for all x∈ℝ and each associated z3 as in (197) and for each integer κ, one has ω~κz3 which does not fall among the zeroes of θQ;vM. Namely, for all x∈ℝ and each associated z3, and for all values of κ one has ω~κz3∉eπi/MωℓQk/M∣ℓ,k∈ℤ. In particular, setting ℓ=−1, then, for all x∈ℝ and each associated z3 and for all values of κ and n, one has that argω~κz3≠−π/M+2πn. We now choose
(206)ϕ1≡−π/M−ε,ϕ2≡ϕ1+2π/D,where ε<2π/M is chosen sufficiently small so that none of the ω~κz3 falls in the sector S−π/M−ε,−π/M. Then, the corresponding sector S=Sϕ1,ϕ2 gives the intersection S∩AN which has oriented boundary Γ~N=R1,N+SN−R2,N−sN as describe above. Define κ1≡minκ∣argω~κz3>−π/M. Choose N sufficiently large such that all ω~κz3 are contained in AN. Then integrating f2v in (64) over the contour Γ~N gives 2πi times the enclosed residues. Namely,
(207)∫Γ~Nf2vdv=∫Γ~N1vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=2πi∑ℓ=−1δ−2∑k=−NNResf2,ωℓeiπ/MQk/M+2πi∑κ=κ1κ1+γ−1Resf2,ω~κz3(208)=2πi1μQ31M∑ℓ=−1δ−2∑k=−NN−1k+1Qkk+1/2ωℓ+1Qk/MBαix+ωℓ+1Qk/MDγ(209)+2πieiπα/β−ix1Dγ∑κ=κ1κ1+γ−1ω~κz3BαθQ;ω~κz3M,where the evaluation of the enclosed residues in (207) occurs via Propositions 8 and 9, similar to the analogous computation in the setting of Theorem 13. Also, comparing (208) with (75), notice the summation (over ℓ) in (208) has M/D=δ terms (as compared to M terms in (75)) as the sector S=Sϕ1,ϕ2 sweeps through an argument of 2π/D in (208) as opposed to sweeping through an argument of 2π in (75). The choice of ϕ1=−π/M−ε is made in order to have ℓ start at −1 in (208), whereby the first term (at ℓ=−1) in the double summation in (208) will have ωℓ+1=ω0=1, matching the expression for Ff~μ,λx in (16) after letting N approach infinity. Again, comparing (209) with (76), notice the summation (over κ) in (209) has Dγ/D=γ terms (as compared to Dγ terms in (76)) as the sector S=Sϕ1,ϕ2 sweeps through an argument of 2π/D in (209) as opposed to sweeping through an argument of 2π in (76). The choice of κ1 is made to give ω~κ1z3 as the first (v) root of ix+vDγeπiγ/δ=0 falling in Sϕ1,ϕ2∩AN, that is with smallest argument such that ϕ1<argω~κz3<ϕ2.Now, defining the rays(210)R1=v=ρeiϕ1∣0≤ρ<∞,(211)R2=v=ρeiϕ2∣0≤ρ<∞,one has that integration along Rm,N approaches integration along Rm for m=1,2 as N⟶∞. Hence, one has the following limit:
(212)limN⟶∞∫Γ~Nf2vdv=limN⟶∞∫Γ~N1vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=limN⟶∞∫R1,N+SN−R2,N−sN1vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=limN⟶∞∫R1,N1vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv(213)−limN⟶∞∫R2,N1vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv(214)−∫R21vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv(215)−∫R11ve2πi/DθQ;ve2πi/DMve2πi/DBαeiπα/βix+ve2πi/DDγeiπγ/δe2πi/Ddv(216)=∫R1vBαeiπα/βvθQ;vMix+vDγeiπγ/δ−ve2πi/DBαeiπα/βvθQ;vMix+vDγeiπγ/δdv(217)=1−e2πiαδ/β∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv,where in moving from (212) to (213), one can drop the integrals over SN and sN in the limit by Corollary 12; in moving from (213) to (214), we have relied on the remarks following (211); in moving from (214) to (215) we used R2=e2πi/DR1; in moving from (215) to (216), we have simplified powers of e2πi/D by utilizing M/D=δ and Dγ/D=γ; and in moving from (216) to (217), we have used B=M/β and D=M/δ. Relying on (217) and taking the limit of (207)–(209) as N approaches infinity yield
(218)1−e2πiαδ/β∫R11vθQ;vMvBαeiπα/βix+vDγeiπγ/δdv=2πi1μQ31M∑ℓ=−1δ−2∑k=−∞∞−1k+1Qkk+1/2ωℓ+1Qk/MBαix+ωℓ+1Qk/MDγ(219)+2πieiπα/β−ix1Dγ∑κ=κ1κ1+γ−1ω~κz3BαθQ;ω~κz3M.Rearranging terms in (218) and (219) and multiplying through by D/2πi give (194)–(196). Referring to (16), multiplying (194)-(196) by μQ3/2π, and factoring out the powers ωℓ+1Bα from the numerator and ωℓ+1Dγ from the denominator of (194) give (200)–(202). The theorem is now demonstrated.Remark 29.
Whenx=0, we will consider the expression (202) (the integral together with its coefficient) to be the defect from having the expression in (200) vanish at x=0, as the expression in (201) vanishes as x approaches 0 by Corollary 32 below. As a consequence, one concludes that if the defect (202) (when x=0) does not vanish, then expression (200) cannot be the Fourier transform of a wavelet. Observe that when δ≠0modβ, the term e2πiαδ/β−1 in (202) does not vanish, and vanishing of the defect when x=0 is equivalent to the vanishing of the integral expression in (202).Remark 30.
If one were to allow for the caseδ=0modβ in Theorem 28, one would then have M=lcmβ,δ=δ and D=M/δ=1 in this case. Furthermore, in this case, Theorem 28 would reduce to Theorem 23, as the coefficient would be e2πiαδ/β−1=1−1=0 in (202). Thus, if one were to remove the hypothesis that δ≠0modβ in Theorem 28, Theorem 23 could be subsumed into Theorem 28. We choose to keep the cases δ=0modβ and δ≠0modβ separate in order to highlight the special properties of Wμ,λt when δ=0modβ.The following proposition gives a pair of flatness conditions, and it forms the technical basis for the proof of Corollary32.Proposition 31.
Letq>1 and ϕ be fixed with −π<ϕ<π. Then for 0<r<∞ and for any power p∈ℝ, one has
(220)limr⟶0rpθq;reϕ=0=limr⟶∞rpθq;reϕ.Due to its length, we leave the proof of Proposition31 until the end of this section, where the interested reader can peruse it. Instead, we immediately proceed to Corollary 32 and its proof, as Corollary 32 is used throughout this study.Corollary 32.
Letδ≠0mod4, and let z3 be as in (184) and (197). Namely, z3=eiπ/Dγe−iπγ/δDγz2, where z2 is any Dγth root of ix. As a consequence, z3Dγ=−e−πiγ/δix. Let ω~=e2πi/Dγ. Then for κ=1,⋯,Dγ(221)limx⟶01ixω~κz3BαθQ;ω~κz3M=0.Proof.
Note that−eπiγ/δz3Dγ=ix. Hence,
(222)1ixω~κz3BαθQ;ω~κz3M=1−eπiγ/δz3Dγω~κz3BαθQ;ω~κz3M=−e−πiγ/δω~κDγω~κz3Bα−DγθQ;ω~κz3M.
One concludes that(223)1ixω~κz3BαθQ;ω~κz3M=z3MBα−Dγ/MθQ;ω~κz3M.
Applying Proposition31 with p=Bα−Dγ/M as x⟶0 gives (221). Note that the assumption that δ≠0mod4 assures that ω~κz3M does not fall along the negative real axis. This gives the corollary.We finish this section with the proof of Proposition31.Proof of Proposition31. Set r=qτ, where τ ranges from −∞ to ∞. Then, from (10), one has
(224)θq;reiϕ2=θq;qτeiϕ2=θq;qτeiϕθq;qτe−iϕ=μq∏n=0∞1+qτeiϕqn1+1qτeiϕqn+1⋅μq∏n=0∞1+qτe−iϕqn1+1qτe−iϕqn+1=μq2∏n=0∞1+2qτcosϕqn+q2τq2n∏n=0∞1+2cosϕqτqn+1+1q2τq2n+1=μq2∏n=0∞qτ−n+cosϕ2+sin2ϕ⋅∏n=0∞q−τ−n−1+cosϕ2+sin2ϕ(225)=μq2JτJ−τ−1,where
(226)Jτ≡∏n=0∞qτ−n+cosϕ2+sin2ϕ.From (224) and (10), one has that, for cosϕ≥0,
(227)θq;qτeiϕ2≥μq2∏n=0∞1+q2τq2n∏n=0∞1+1q2τq2n+1=μq2μq2θq2;q2τ=μq2μq2∑m∈ℤq2τmq2mm−1/2(228)>μq2μq2q2τmq2mm−1/2,where the inequality in (228) holds for all m∈ℤ. Hence, inverting (227) and (228), taking square roots, and multiplying by rp=qτp, one obtains that for each m∈ℤ(229)rpθq;reiϕ=qτpθq;qτeiϕ≤μq2μqqmm−1/2qp−mτ.To handle the limit asτ⟶+∞, pick m>p. To handle the limit as τ⟶−∞, pick m<p. Thus, (220) holds when cosϕ≥0.The remaining case is−1<cosϕ<0. Here, the possibility that cosϕ=−1 is excluded, since if cosϕ were to equal −1 then θq;reiϕ=θq;−r, which has an infinite number of zeroes (for r=qj with j∈ℤ). It will be shown below (see (239) through (241)) that we have the following two limits:
(230)limτ⟶∞J−τ−1=1=limτ⟶−∞Jτ.It will also be shown (see (242) and (259) below) that the following rate of growth holds:
(231)Jτ≈q2ττ−τ2+τ=qτ2+1+2τ−ττasτ⟶∞,where τ denotes the greatest integer function evaluated at τ. Hence, from (231), we have the related rate of growth
(232)J−τ−1≈q2−τ−1−τ−1−−τ−12+−τ−1=qτ2+−1+2τ−ττ−2τ−τasτ⟶−∞.For the time being, we assume (230)–(232) and use (225) to show that (220) holds as r⟶∞ (equivalently as τ⟶∞) via
(233)rpθq;reiϕ=qτpθq;qτeiϕ=qpτμqJτJ−τ−1≈qpτqτ+1+2τ−ττ/2=qpτ−τ+τ−τ+1+2τ−ττ/2(234)=q2p−τ−1−2τ−ττ/2qpτ−τ⟶0asτ⟶∞,where one moves to (233) via (230) and (231). The vanishing in (234) is seen from the fact that when τ>2p−1 the exponent 2p−τ−1−2τ−ττ/2 in (234) is increasingly negative as τ⟶∞, while pτ−τ remains bounded.Similarly, we use (225) to show next that (220) holds as r⟶0 (equivalently as τ⟶−∞). In this setting,
(235)rpθq;reiϕ=qτpθq;qτeiϕ=qpτμqJτJ−τ−1≈q−p−τqτ−1+2τ−ττ/2−τ−τ(236)=q−pτ−τ+τ−τ−1+2τ−ττ/2+τ−τ(237)=q−2p+τ−1+2τ−ττ/2+−p+1τ−τ⟶0asτ⟶−∞.where one moves to (235) and (236) via (230) and (232). The vanishing in (237) is seen from the fact that when τ>1−2p the exponent −2p+τ−1+2τ−ττ/2 in (237) is increasingly negative as τ⟶−∞, while −p+1τ−τ remains bounded.All that remains is to show (230)–(232). We first show limτ⟶∞J−τ−1=1, from which it will immediately follow that limτ⟶−∞Jτ=1. Note that
(238)J−τ−1=∏n=0∞q−τ−n−1+cosϕ2+sin2ϕ=∏n=0∞1q2τ+n+1+2cosϕqτ+n+1+1=∏n=0∞1q2τ−τ+τ+n+1+2cosϕqτ−τ+τ+n+1+1=∏k=τ+1∞1q2τ−τ+2k+2cosϕqτ−τ+k+1,where the reindexing k=τ+n+1 occurs in moving to (238). For τ satisfying τ>logq2 and −1<cosϕ<0, one traps J−τ−1 in (238) by
(239)exp∑k=τ+1∞−4qk<∏k=τ+1∞−2qk+1<J−τ−1<∏k=τ+1∞1q2k+1<exp∑k=τ+1∞1q2k,where we have further assumed that τ is sufficiently large so that ln1−2/qτ+1>−4/qτ+1, (that is 1−2/qτ+1 falls in the interval x∈ℝ∣lnx>2x−1⊃2/3,1≠∅). Since
(240)limτ⟶∞exp∑k=τ+1∞−4qk=limτ⟶∞exp−4qτ+1qq−1=1,(241)limτ⟶∞exp∑k=τ+1∞1q2k=limτ⟶∞exp1q2τ+1q2q2−1=1hold, the trapping in (239) gives that limτ⟶∞J−τ−1=1. An immediate consequence is that limτ⟶−∞Jτ=1, giving (230). We conclude that there is an N1 so that for τ>N1 one has
(242)12≤J−τ−1≤32.We next show (231). From (226), observe that for τ≥1 one has
(243)Jτ=∏n=0∞qτ−n+cosϕ2+sin2ϕ=∏n=0τ−1qτ−n+cosϕ2+sin2ϕ⋅∏n=τ∞qτ−n+cosϕ2+sin2ϕ=∏n=0τ−1qτ−n+cosϕ2+sin2ϕ⋅∏n=τ∞qτ−τ−n−τ+cosϕ2+sin2ϕ(244)=∏n=0τ−1qτ−n+cosϕ2+sin2ϕ⋅∏k=0∞qτ−τ−k+cosϕ2+sin2ϕ(245)=∏n=0τ−1qτ−n+cosϕ2+sin2ϕJτ−τ=∏n=0τ−1q2τ−n∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2Jτ−τ=q2ττq−2ττ−1/2∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2Jτ−τ(246)=q2ττ−τ2+τ∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2Jτ−τ,where a reindexing by k=n−τ moves one from (243) to (244); one obtains (245) from (244) by (226); and one factors out a q2τ−n from each product in the left most product expression in (245) to proceed forward.Now,Jτ is nonvanishing for all τ, as θq;reiϕ2=θq;qτeiϕ2 has vanishing points if and only if ϕ=±π, up to full revolutions, and τ∈ℤ (by (10)). Thus, the compact interval 0,1 under J maps to a compact interval M1,M2 (with 0<M1≤M2<∞). Since τ−τ∈0,1, one has
(247)0<M1≤Jτ−τ≤M2<∞,and Jτ−τ is bounded for all τ.It remains to show that∏n=0τ−11+cosϕ/qτ−n2+sinϕ/qτ−n2 in (246) is bounded as τ approaches infinity (again, in the setting that −1<cosϕ<0). An upper bound C2 independent of τ follows via
(248)∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2=∏n=0τ−11+cosϕqτ−τ+τ−n2+sinϕqτ−τ+τ−n2(249)=∏k=1τ1+cosϕqτ−τ+k2+sinϕqτ−τ+k2(250)=∏k=1τ1+2cosϕqτ−τ+k+1q2τ−τ+2k<∏k=1τ1+1q2k≤exp∑k=1τ1q2k(251)≤exp∑k=1∞1q2k=exp1q2−1≡C2,where the reindexing k=τ−n is used to obtain (249).Starting with (248)–(249) to obtain (252), a lower bound C1 independent of τ for τ≥N (with N determined below) follows via
(252)∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2=∏k=1τ1+cosϕqτ−τ+k2+sinϕqτ−τ+k2(253)>∏k=1τ1+cosϕqk2+sinϕq1+k2=∏k=1τ1+2cosϕqk+cos2ϕq2k+sin2ϕq2k+1(254)=∏k=1N−11+2cosϕqk+cos2ϕq2k+sin2ϕq2k+1⋅∏k=Nτ1+2cosϕqk+cos2ϕq2k+sin2ϕq2k+1(255)>∏k=1N−11+2cosϕqk+cos2ϕq2k∏k=Nτ1+2cosϕqk(256)>∏k=1N−11+cosϕqk2exp∑k=Nτ4cosϕqk(257)>1+cosϕq2N−1exp4cosϕqNqq−1≡C1,where N appearing in (254) is chosen sufficiently large so that ln1+2cosϕ/qN>4cosϕ/qN (that is, 1+2cosϕ/qN falls in the interval x∈ℝ∣lnx>2x−1⊃2/3,1≠∅). This justifies movement from (255) to (256).Hence, we have from (251) and (257) that
(258)0<C1<∏n=0τ−11+cosϕqτ−n2+sinϕqτ−n2<C2.From (242), (247), and (258) applied to (225) and (246), one obtains
(259)0<μq2q2ττ−τ2+τC1M112<μq2JτJ−τ−1=θq;qτeiϕ2≤μq2q2ττ−τ2+τC2M232<∞,for τ>maxN1,N, where q>1 is fixed. Thus, (231) is demonstrated. As a consequence, we also have that (232) now holds. The proposition is proven.


## 4. Connection to the Theory of Wavelet Frames

We have seen in the previous section that expressing the Fourier transform ofWμ,λt in terms of Jacobi theta functions, as in Theorem 23, provides a strong connection to the theory of special functions. However, more can be concluded. Namely, from the relation of the Wμ,λt to the Jacobi theta function, we also demonstrate in this section the connection to the theory of wavelets and wavelet frames. In particular, for low values of γ, we establish that each Wμ,λt as in Theorem 23 is a Schwartz mother wavelet for a wavelet frame generating all of L2ℝ.Recall thatft is a wavelet if it belongs to L1ℝ∩L2ℝ∩L∞ℝ, has first moment ∫−∞∞ftdt=0, and satisfies the admissibility condition that ∫−∞∞Fftx2/xdx<∞. Furthermore, such a wavelet ft is a mother wavelet for a frame of form
(260)Sf;a0,b0=a0n/2fa0nt+mb0/fn,m∈ℤ,ifSf;a0,b0 generates L2ℝ, where a0>1 is the scale factor, b0>0 is the translation parameter, and f=f2 is the norm of f in L2ℝ. One defines the diagonal term G0fx by
(261)G0fx=1f2∑n=−∞∞Ffta0nx2and the off-diagonal term G1fx by
(262)G1fx=1f2∑j∈ℤ∑k∈ℤ\0Ffta0jx⋅Ffta0jx+2πk/b0,which together give the frame condition
(263)0<inf1≤x≤a0G0fx−G1fx≤sup1≤x≤a0G0fx−G1fx<∞,sufficient for Sf;a0,b0 in (260) to be a frame. As is shown in [36] and [37], for b0>0 sufficiently small, (263) is in turn implied by the conditions (264) immediately below:
(264)0<inf1≤x≤a0G0fxand∃C>0withFftx≤Cx1+x23/2.In Proposition36 below, we see that there are natural scale factors a0 that allow us to compute G0fx for the wavelet ft=Wμ,λt for low values of γ, where knowledge of properties of Jacobi theta functions lets us compute G0Wμ,λx exactly, which in turn will establish the nonvanishing of G0Wμ,λx in the left hand criteria of (264).But first, we need to obtain an even stronger nonvanishing analogue of Proposition22 in the setting of Theorem 23 (in particular, with δ=0modβ) under the additional assumption that γ=1,2.Proposition 33.
Under the notation and assumptions of Theorem23 (in particular δ=0modβ and M=δ), along with the additional assumption that γ=1,2, one has that (172) becomes
(265)∑κ=0γ−1ω~κzBαθQ;ω~κzδ,which never vanishes for z∈ℂ∗ if γ=1 and for γ=2 and z∈ℂ∗ (265) vanishes precisely when
(266)z2δ=−Q4p+1+2j,forsomep∈ℤwithj=0ifBαiseven,z2δ=−Q4p+1+2j,forsomep∈ℤwithj=1ifBαisodd.
Equivalently, (265) vanishes precisely when z=eπi/2δe2πiℓ/2δQ4p+1+2j/2δ for some p,ℓ∈ℤ, where j=0 if Bα is even, and j=1 if Bα is odd.Proof.
Ifγ=1, then ω~=e2πi/γ=e2πi/1=1 and (265) becomes zBα/θQ;zδ, which never vanishes for z∈ℂ∗. If γ=2, then ω~=e2πi/γ=e2πi/2=−1 and (265) becomes
(267)∑κ=0γ−1ω~κzBαθQ;ω~κzδ=zBαθQ;zδ+−zBαθQ;−zδ=zBα1BαθQ;−zδ+−1BαθQ;zδθQ;zδθQ;−zδ,where equality in (267) follows from the fact that δ must be odd, as γ/δ is in reduced form and γ=2. Now from (10) the numerator in the bracketed expression in (267) becomes
(268)1BαθQ;−zδ+−1BαθQ;zδ(269)=∑n=−∞∞−zδnQnn−1/2+−1BαzδnQnn−1/2(270)=∑n=−∞∞−zδn+−1BαzδnQnn−1/2(271)=∑k=−∞∞2zδ2kQ2k2k−1/2,ifBαiseven,∑k=−∞∞−2zδ2k+1Q2k+12k/2,ifBαisodd,where
(1)
ifBα is even in (270) the n= odd cases cancel, leaving the n=2k in the Bα even case of (271)(2)
ifBα is odd in (270) the n= even cases cancel, leaving the n=2k+1 in the Bα odd case of (271)
Now, in theBα even case,
(272)∑k=−∞∞2zδ2kQ2k2k−1/2=∑k=−∞∞2z2δkQ−kQ4k2−2k/2Q−2k/2=2∑k=−∞∞z2δ/QkQ4kk−1/2=2θQ4;z2δQ,where the last equality in (272) follows from (10). And in the Bα odd case,
(273)∑k=−∞∞−2zδ2k+1Q2k+12k/2=∑k=−∞∞−2zδz2δkQ−3kQ4k2+2k/2Q−6k/2=−2zδ∑k=−∞∞z2δ/Q3kQ4kk−1/2=−2zδθQ4;z2δQ3,where the last equality in (273) again follows from (10). Thus, (268)–(271) reduces to
(274)1BαθQ;−zδ+−1BαθQ;zδ=2θQ4;z2δQ,ifBαiseven,−2zδθQ4;z2δQ3,ifBαisodd.
Thus, (267) reduces to
(275)∑κ=0γ−1ω~κzBαθQ;ω~κzδ=zBαθQ;zδ+−zBαθQ;−zδ=2zBαθQ4;z2δ/QθQ;zδθQ;−zδ,ifBαiseven−2zBα+δθQ4;z2δ/Q3θQ;zδθQ;−zδ,ifBαisodd(276)=−1j2zBα+jδθQ4;z2δ/Q1+2jθQ;zδθQ;−zδ,wherej=0,ifBαiseven,j=1,ifBαisodd.
Then, (276) vanishes for z∈ℂ∗ precisely when θQ4;z2δ/Q1+2j=0, which by (10) occurs precisely when z2δ/Q1+2j=−Q4p for some p∈ℤ. This last statement is equivalent to (266). Equivalently, we have such vanishing when z=−11/2δQ4p+1+2j/2δ=eπi/2δe2πiℓ/2δQ4p+1+2j/2δ for some p,ℓ∈ℤ. This gives the proposition.Corollary 34.
In the setting of, and under the assumptions of, Theorem23, with δ=0modβ and δ≠0mod4, with γ=1 or γ=2, one has that
(277)∑κ=0γ−1ω~κz3BαθQ;ω~κz3δ≠0,forallx∈ℝ∗=ℝ\0,where z3=eπi/γe−πi/δz2 and z2 is any fixed γth root of ix.Proof.
By Proposition33, when γ=1, the nonvanishing of (277) holds automatically for x≠0. Also by Proposition 33, equation (266), when γ=2, the vanishing of (277) holds if and only if for
(278)z3=eπi/γe−πi/δeπi/2γeiν/γx1/γ(279)=eπi/2e−πi/δeπi/4eiν/2x1/2,
one hasz32δ=−Q4p+1+2j for some p∈ℤ and for j=0 if Bα is even or for j=1 if Bα is odd, where in (278)–(279) one has ν=0 if x>0 and ν=π if x<0. From (279) and the fact that δ is odd (since γ=2 is even), one sees that
(280)z32δ=ei3πδ/2eiνδxδ=±i±1xδ=±ixδ≠−Q4p+1+2j,
for any values ofp and j, because ±ixδ is imaginary while −Q4p+1+2j is real. Thus, (277) never vanishes for x∈ℝ∗ when γ=2, and the corollary is proven.Remark 35.
Although, forγ=1,2 by Corollary 34 one has (277) never vanishes for x∈ℝ∗, one has that (277) vanishes to infinite order at x=0 by Corollary 32.Proposition 36.
Letq>1. Under the assumptions and notation of Theorem 23, with Wμ,λt as in Definition 16, for γ=1 and a0=q>1 with Q=q2/λ=qδ/γ=qδ whereby q=Q1/δ=a0, one has that
(281)G0Wμ,λx=1Wμ,λ2θQ2;Q2Bα/δQ2+2/δz32δx2δδμQ3eiπα/β2π1−ixz3xBαθQ;z3δx2(282)=1Wμ,λ2θQ2;Q2Bα/δQ2+2/δz32δx2δFWμ,λtx2,which is C∞ and never vanishes for x∈ℝ∗, in particular for x∈1,q=1,a0.
And forγ=2 and a0=q2>1 with Q=q2/λ=qδ/γ=qδ/2 whereby q2=Q4/δ=a0, one has that
(283)G0Wμ,λx=1Wμ,λ2θQ8;Q4Bα/δQ6+8/δz3x4δ⋅δμQ3eiπα/β22π−1j−ix2z3xBα+jδθQ4;z32δx/Q1+2jθQ;z3δxθQ;−z3δx2(284)=1Wμ,λ2θQ8;Q4Bα/δQ6+8/δz3x4δFWμ,λtx2,where in (283) j=0 if Bα is even and j=1 if Bα is odd. Expressions (283)–(284) are C∞ and never vanish for x∈ℝ∗, in particular for x∈1,q2=1,a0.Proof.
First, we handle theγ=1 case with a0=q=Q1/δ. Under this assumption, along with the assumptions of Theorem 23, one has that (180)–(182) become
(285)FWμ,λtx=C^1−ixz3xBαθQ;z3δx,whereC^=δμQ3eiπα/β2π.
From (184), one has z3Qpx=Qp/γz3x=Qpz3x for p∈ℝ, whence z3δQn/δx=Qnz3δx for n∈ℤ in the theta function expression in (285). One concludes that
(286)FWμ,λtQn/δx=C^1−iQn/δxz3Qn/δxBαθQ;z3δQn/δx=C^1−iQn/δxQnBα/δz3xBαθQ;Qnz3δx=C^1−iQn/δxQnBα/δz3xBαQnn+1/2z3δxnθQ;z3δx,where equality in (286) follows from (12). From (286) and the fact that a0=q=Q1/δ, we have (288) below:
(287)G0Wμ,λtx=1Wμ,λ2∑n=−∞∞FWμ,λtQn/δx2(288)=C∧2Wμ,λ2∑n=−∞∞Q2nBα/δQ2n/δQ2nn+1/2z3δx2n1−ixz3xBαθQ;z3δx2=1Wμ,λ2C∧−ixz3xBαθQ;z3δx2⋅∑n=−∞∞Q2Bα/δQ−2/δz3x−2δnQ2nn+1/2Q2−2n/2Q2−2n/2(289)=1Wμ,λ2C∧−ixz3xBαθQ;z3δx2∑n=−∞∞Q2Bα/δQ−2/δz3x−2δQ−2nQ2nn−1/2(290)=1Wμ,λ2C∧−ixz3xBαθQ;z3δx2θQ2;Q2Bα/δQ−2/δz3x−2δQ−2,where the move from (289) to (290) is justified by (10). Now (290) is equal to (281), which by (285), is equal to (282). From its definition in (287), G0Wμ,λtx is invariant under multiplication of the argument x by Q1/δ=qδ/γ1/δ=q1/γ=q. This q-invariance can also be checked via properties of the theta functions in expression (281). From (277) in Corollary 34 coupled with (285), one has nonvanishing of FWμ,λtx for x∈ℝ∗. Since θQ2;z only vanishes for z=−Q2k for k∈ℤ by (10), we have that θQ2;Q2Bα/δQ−2/δz3x−2δQ−2 never vanishes for any x∈ℝ∗ as its argument is positive. From these two nonvanishing results applied to (282) one has that G0Wμ,λtx is C∞ and never vanishes for x∈ℝ∗. The γ=1 case is now shown.We turn next to theγ=2 case, with a0=q2=Q4/δ. In this setting, from Theorem 23 one has that (180)–(182) becomes
(291)FWμ,λtx=C^21−ixz3xBαθQ;z3δx+−z3xBαθQ;−z3δx(292)=C^21−ix−1j2z3Bα+jδxθQ4;z32δx/Q1+2jθQ;zδxθQ;−z3δx,(293)wherej=0,ifBαiseven,j=1,ifBαisodd,(294)whereC^=δμQ3eiπα/β2π,and where (292) and (293) follow from (275) and (276). From (184), one has z3Qpx=Qp/γz3x=Qp/2z3x for p∈ℝ, whence z3δQ4n/δx=Q2nz3δx and z32δQ4n/δx=Q4nz32δx for n∈ℤ in the theta function expressions in (292). From (292), along with the fact that with a0=q2=Q4/δ, one then has that
(295)FWμ,λtQ4n/δx=C^21−iQ4n/δx−1j2z3Bα+jδQ4n/δxθQ4;z32δQ4n/δx/Q1+2jθQ;zδQ4n/δxθQ;−z3δQ4n/δx=C^2−1j2Q2n/δz3xBα+jδ−iQ4n/δxθQ4;Q4nz32δx/Q1+2jθQ;Q2nzδxθQ;−Q2nz3δx=C^2−1j2Q2n/δz3xBα+jδ−iQ4n/δx⋅Q4nn+1/2z32δx/Q1+2jnθQ4;z32δx/Q1+2jQ2n2n+1/2z2nδxθQ;zδxQ2n2n+1/2−1nz2nδxθQ;−z3δx=Q−4/δQ2Bα+jδ/δQ−1−2j−1z3−2δxnQ2n2(296)⋅C^21−ix−1j2z3Bα+jδxθQ4;z32δx/Q1+2jθQ;zδxθQ;−z3δx(297)=Q−4/δQ2Bα+jδ/δQ−1−2j−1z3−2δxQ−2nQ4nn−1/2FWμ,λtx,where j is as in (293); (12) was used to move from (295) to the subsequent line; all terms involving a power of n have been factored out in (296); and in moving to (297), the expressions involving powers of n have been multiplied by Q−2n/Q−2n while the bracketed expression in (296) was recognized as the Fourier transform of Wμ,λt via (292). One then uses (297) to compute the diagonal term G0Wμ,λtx as follows:
(298)G0Wμ,λtx=1Wμ,λ2∑n=−∞∞FWμ,λtQ4n/δx2=1Wμ,λ2FWμ,λtx2(299)⋅∑n=−∞∞Q−8/δQ4Bα+jδ/δQ−2−4jz3x−4δQ−4nQ8nn−1/2(300)=1Wμ,λ2FWμ,λtx2θQ8;Q4Bα/δQ6+8/δz3x4δ,where (297) is used to move from (298) to (299) (after taking absolute value and squaring) and (10) is used to move from (299) to (300). Finally, (300) is seen to be equivalent to (284), which is in turn equivalent to (283). From its definition in (298), G0Wμ,λtx is invariant under multiplication of the argument x by Q4/δ=q4δ/γ1/δ=q4/γ=q4/2=q2. This q2-invariance can also be checked via properties of the theta functions in expression (300). From (277) in Corollary 34 coupled with (291), one has non-vanishing of FWμ,λtx for x∈ℝ∗. Since θQ8;z only vanishes for z=−Q8k for k∈ℤ by (10), we have that θQ8;Q4Bα/δQ−6−8/δz3x−4δ never vanishes for any x∈ℝ∗ as its argument is positive. From these two nonvanishing results applied to (300), one has that G0Wμ,λtx is C∞ and never vanishes for x∈ℝ∗. The γ=2 case is now shown, and the proposition is proven.At this point, at least for low values ofγ=1,2, we are prepared to explicitly make the connection to wavelet frame theory that follows from knowledge of the Fourier transform of Wμ,λt in terms of the theta function.Theorem 37.
Letq>1. Under the assumptions and notation of Theorem 23, in particular δ=0modβ, and with Wμ,λt as in Definition 16, one has, for γ=1 with a0=q, and also for γ=2 with a0=q2, that for b0>0 sufficiently small
(301)Wμ,λtisamotherwaveletforaframeSWμ,λ;a0,b0generatingL2ℝ.Proof.
The theorem will follow by establishing (264) above. Now, G0Wμ,λx≠0 for 1≤x≤a0 by Proposition 36. Next, from Theorem 23, we have that Wμ,λt is a Schwartz wavelet with all moments vanishing. Since all moments vanish, FWμ,λtx is flat at x=0. Since Wμ,λt is Schwartz, FWμ,λtx is also Schwartz and therefore decays faster than 1/xp for any p∈ℕ for x near ±∞. Hence, for choice of C sufficiently large, one has FWμ,λtx≤Cx/1+x23/2. Thus, (264) is satisfied, and the theorem is demonstrated.


## 5. Canonical Extensions

From Theorem 3.2 in [2], a given fμ,λt on 0,∞ has, in general, infinitely many Schwartz wavelet extensions Fμ,λt to all of ℝ with Fμ,λt satisfying the same MADE as fμ,λt. In this section, we shall demonstrate, in the setting of Theorem 23 for low values of M=δ=1,2,3, that there is a natural uniquely determined extension Fμ,λt of fμ,λt to ℝ, namely, the canonical extension Wμ,λt of fμ,λt given as follows. We remark that for these values 1,2,3 one has δ≠0mod4 as in Theorem 23 automatically.Definition 38.
Under the assumptions and notation of Theorem23 (in particular δ=0modβ), with M=δ=1,2,3 (and δ=0modβ), the canonical extension of fμ,λt from 0,∞ to all of ℝ is defined to be Wμ,λt, where Wμ,λt is the function naturally generated by fμ,λt as given by Definition 16, with Wμ,λt0,∞=fμ,λt.We clarify the above definition by emphasizing that canonical extensions offμ,λt as the μ,λ vary (as given in Definition 38) form a strict subset of the set of functions naturally generated by fμ,λt (as in Definition 16) . This follows from the fact that, for δ>3, in a large number of cases one has that Wμ,λt≠fμ,λt for t≥0 and is therefore not an extension (see Propositions 42 and 43 below). We next show that for δ=1,2,3,Wμ,λt is indeed an extension of fμ,λt.Proposition 39.
Under the assumptions and notation of Theorem23, for δ=1,2,3, with δ=0modβ one has that Wμ,λt0,∞=fμ,λt for t≥0, where fμ,λt is given by (1).Proof.
We handle each value ofδ separately. First, let δ=1. In this setting, one has ω=e2πi/M=e2πi/δ=e2πi/1=1. From (131) and (132), one has that
(302)Wμ,λt=∑ℓ=0δ−1ωℓ+1Bαf~μ,λωℓ+1Dγt=∑ℓ=001Bαf~μ,λ1Dγt=f~μ,λt,
and hence,Wμ,λt0,∞=fμ,λt on 0,∞ holds for δ=1 because f~μ,λt=fμ,λt for t≥0. This last equality follows from (125) in conjunction with (1) after noting that Q=qδ/γ and so Qk/MDγ=qδ/γk/d1γ=qk. We remark that, from CaseM=1 in Section 6.1 below, the δ=1 Case under consideration here consists of precisely the fμ,λt that are flat at the origin and the canonical extension is fμ,λt extended to be 0 on the negative real axis, equivalently the canonical extension is f~μ,λt=Wμ,λt.Next, letδ=2. Since γ/δ=γ/2 is in reduced form, one concludes that γ=2k+1 is odd. Now, ω=e2πi/M=e2πi/δ=e2πi/2=−1. From (131)-(132), one has that
(303)Wμ,λt=∑ℓ=0δ−1ωℓ+1Bαf~μ,λωℓ+1Dγt=∑ℓ=02−1−1ℓ+1Bαf~μ,λ−1ℓ+112k+1t=−1Bαf~μ,λ−1t+1Bαf~μ,λ1t(304)=1Bαfμ,λt,fort≥0,1Bα−1fμ,λ−1t,fort<0,where one moves from (303) to (304) via (125) and (128). From (304), one sees that Wμ,λt0,∞=fμ,λt when δ=2. Also note from (304) that the canonical extension Wμ,λt is an even function if Bα is odd, and it is an odd function if Bα is even when δ=2.Finally, letδ=3. Since γ/δ=γ/3 is in reduced form, one concludes that γ=3p+j for p∈ℕ0 with j=1,2. In this setting, ω=e2πi/M=e2πi/δ=e2πi/3 which gives that
(305)ωℓ+1Dγ=ωℓ+113p+j=e2πi/33p+jℓ+1(306)=e2πij/3ℓ+1=e2πij/3,ifℓ=0,e4πij/3,ifℓ=1.1,ifℓ=2.For each case thatj=1,2 one has that Re2πij/3<0 and Re4πij/3<0. From (306), ωℓ+1Dγ has negative real part for ℓ=0,1 and ωℓ+1Dγ=1 for ℓ=2. Then for t≥0 and for ℓ=0,1, (128) gives that f~μ,λωℓ+1Dγt=0. Combining these results with (131) and (132) and (125) gives for t≥0 that
(307)Wμ,λt=∑ℓ=0δ−1ωℓ+1Bαf~μ,λωℓ+1Dγt=∑ℓ=02ωℓ+1Bαf~μ,λωℓ+1Dγt(308)=0+0+ω2+1Bαf~μ,λω2+1Dγt=f~μ,λt=fμ,λt.Hence, in each of the casesδ=1,2,3 one has Wμ,λt0,∞=fμ,λt, and the proposition is proven.In the next theorem, we assume all of the hypotheses of both Theorems13 and 23 to determine a family of fμ,λt with canonical extensions having optimal properties.Theorem 40.
Letq>1. Assume that all the hypotheses and the notation of Theorems 13 and 23 hold. Then for
(309)γ,δsatisfyingγ∈1,2andδ∈1,2,3,
equivalently, for(310)μ,λ∈ℤ,1or2ℤ+1,2or2ℤ+1,4or2ℤ+1+2j/3,2/3or2ℤ+1+2j/3,4/3forj=0,1,2,the associated fμ,λt have canonical extensions Wμ,λt which
(1)
are Schwartz wavelets onℝ(2)
are mother wavelets for a frameSWμ,λ;a0,b0 generating L2ℝ, as in (301), where b0 is sufficiently small, and a0=q>1 when γ=1 and a0=q2>1 when γ=2(3)
have all moments vanishing(4)
have Fourier transforms given by (180)-(182) which relate the canonical extension of fμ,λt to the Jacobi theta function(5)
satisfy the MADE (185)(6)
have nonvanishing diagonal frame termsG0Wμ,λx on 1,a0: given by (281)-(282) for γ=1 and a0=q, and given by (283)-(284) for γ=2 and a0=q2, each expressible in terms of the Jacobi theta functionProof.
Sinceδ∈1,2,3 one has Wμ,λt is a canonical extension of fμ,λt to ℝ by Proposition 39 and Definition 38. Properties (1), (3), (4), and (5) follow from Theorem 23. Property (2) follows from Theorem 37. Since γ∈1,2, Property (6) follows from Proposition 36. It remains to show the equivalence of (297) with (310).We first show (297) ⇒ (310). Assuming (297), one has λ/2=γ/δ∈1/1,2/1,1/2,1/3,2/3 where the case that γ/δ=2/2 is ruled out for not being in reduced form. From this, one concludes λ∈1,2,4,2/3,4/3. Since δ=0modβ, one concludes that β=1 when δ=1; β=1,2 when δ=2; and β=1,3 when δ=3.Whenβ=1, which happens for each value of λ, one has
(311)μ+12=αβ=α1fromwhichμ=2α−1isodd.Thus, theβ=1 case gives that μ,λ falls in one of 2ℤ+1,1, 2ℤ+1,2, 2ℤ+1,4, 2ℤ+1,2/3, or 2ℤ+1,4/3.Whenβ=2, which happens only when δ=2 (for δ∈1,2,3), equivalently, when λ/2=γ/δ=1/2, equivalently, when λ=1, one has
(312)μ+12=αβ=2k+12,whereαmustbeodd,asα2isinareducedform.Hence,μ=2k is even when β=2, and in this case, μ,λ falls in 2ℤ,1.Whenβ=3, which happens only when δ=3 (for δ∈1,2,3), equivalently, when λ/2=γ/δ=γ/3, equivalently, when λ=2/3,4/3, one has
(313)μ+12=αβ=3k+j3,withj=1,2sinceα3isinareducedform.Hence,μ=2k−1+2j/3 for j=1,2 when β=3, and in this case, μ,λ falls in 2ℤ+1+2j/3,2/3 or 2ℤ+1+2j/3,4/3.All of the above casesβ=1,2,3 combine to require that μ,λ satisfies (310).Conversely, we show (310) ⇒ (309). So assume (310). Within this setting, examine the case that μ=2k−1 is odd, with λ∈1,2,4,2/3,4/3. Then,
(314)μ+12=2k−1+12=k=k1=αβinareducedform,from which we deduce thatβ=1 and γ/δ=λ/2∈1/2,1,2,1/3,2/3. This implies δ=1,2,or3, none of which is ruled out as each such δ is divisible by β=1. Furthermore, from the possible λ/2, one sees that γ∈1,2.Next, examine the case thatμ=2k is even with λ=1. Then, μ+1=2k+1. Hence,
(315)μ+12=2k+12=αβwhichisinreducedform.One concludes thatβ=2 and λ/2=γ/δ∈1/2, in a reduced form. Thus, δ=2 (which is divisible by β=2), and γ=1.Finally, examine the case thatμ=2k−1+2j/3 with j=1,2 and λ∈2/3,4/3. Then,
(316)μ+12=3k+j3=αβwhichisinreducedformasj=1,2.One concludes thatβ=3 which does divide δ for λ/2=γ/δ∈1/3,2/3, all in a reduced form. Thus, δ=3 and γ=1,2.All cases combine to giveγ∈1,2 and δ∈1,2,3, which is (309). Thus, (309) ⟺ (310), and the theorem is now proven.Selected examples of canonical extensions are now provided.[Exampleμ=−1odd and λ=2] In this case,
(317)f−1,2t=∑k=−∞∞−1kexp−qktqkk+1/2,fort≥0.One sees thatμ+1/2=0/2=0/1=α/β and λ/2=2/2=1/1=γ/δ. Thus, α=0 and γ=1, while β=δ=1, consistent with Examples [M=1] in Section 6.1 and [M=δ=1, μ=2N+1, and λ=2n] in Section 6.2 below, where the current case that μ=−1 and λ=2 is seen to be flat at the origin. One extends f−1,2t to be identically 0 for t<0. This yields f~−1,2t, which equals the canonical extension W−1,2t by (302) as δ=1. This particular canonical extension was first introduced in [3] as the function Kt. From Theorem 40, one has that Kt satisfies properties (1) through (6), including that Kt is a Schwartz wavelet with all moments vanishing, satisfying the MADE K′t=Kqt, generating a frame for L2ℝ, and having Fourier transform iμq3/2πxθq;ix, as was previously proven in [3] but is now seen as a special case of Theorem 40.[Exampleμ=0even and λ=1] In this case,
(318)f0,1t=∑k=−∞∞−1kexp−qktqk2,fort≥0.One sees thatμ+1/2=1/2=α/β and λ/2=1/2=γ/δ. Thus, α=γ=1, while β=δ=2, consistent with Examples [M=δ=p,β=p] (with the prime p taken to be 2) in Section 6.1 and [M=δ=2, β=2] in Section 6.2 below. Also, we record that B=M/β=δ/β=2/2=1. Since δ=2, f0,1t has canonical extension W0,1 given by (304). That is,
(319)W0,1t=1Bαf0,1t,fort≥0−1Bα−1f0,1−1t,fort<0=11⋅1f0,1t,fort≥0−11⋅1−1f0,1−1t,fort<0=f0,1t=∑k=−∞∞−1kexp−qktqk2,which is the extension of f0,1 to be an even function. This canonical extension W0,1t was first introduced in [4] and denoted there by f0t. From Theorem 40, one has that f0t satisfies properties (1) through (6), including that f0t is a Schwartz wavelet with all moments vanishing, satisfying the MADE f0″t=−qf0qt, generating a frame for L2ℝ, and having Fourier transform 2μq23/2πθq2;x2, as was previously proven in [4] but is now seen as a special case of Theorem 40. If one normalizes f0t by f00, one obtains qCost≡f0t/f00, which converges uniformly [4] to cost on compact subsets of ℝ as q⟶1+, as is illustrated in Figure 2(a). Furthermore, qCost satisfies properties (1) through (6), including satisfying the same MADE above (by linearity) and having Fourier transform 2μq23/2πf00θq2;x2. See [4] and [6] for further details.[Exampleμ=1odd and λ=1] In this case,
(320)f1,1t=∑k=−∞∞−1kexp−qktqkk−1,fort≥0.One sees thatμ+1/2=1+1/2=1/1=α/β and λ/2=1/2=γ/δ. Thus, α=β=γ=1, while δ=2, consistent with Examples [M=δ=p,β=1] in Section 6.1 (with the prime p taken to be 2) and [M=δ=2, β=1] in Section 6.2 below. Also, we record that B=M/β=δ/β=2/1=2. Since δ=2, f1,1t has canonical extension W1,1 given by (290). That is,
(321)W1,1t=1Bαf1,1t,fort≥0−1Bα−1f1,1−1t,fort<0=12⋅1f1,1t,fort≥0−12⋅1−1f1,1−1t,fort<0=∑k=−∞∞−1kexp−qktqkk−1,fort≥0,−1∑k=−∞∞−1kexp−qktqkk−1,fort<0,which is the extension of f1,1 to be an odd function. This canonical extension W1,1t was first introduced in [4] and denoted there by f1t. From Theorem 40, one has that f1t satisfies properties (1) through (6), including that f1t is a Schwartz wavelet with all moments vanishing, satisfying the MADE f1′′t=−q2f1qt, generating a frame for L2ℝ, and having Fourier transform 2μq23−ix/2πθq2;x2, as was previously proven in [4] but is now seen as a special case of Theorem 40. If one normalizes f1t by f00, one obtains qSint≡f1t/f00, which converges uniformly to sint [4] on compact subsets of ℝ as q⟶1+, as is illustrated in Figure 2(b). Furthermore, qSint satisfies properties (1) through (6), including satisfying the same MADE above (by linearity) and having Fourier transform 2μq23−ix/2πf00θq2;x2. Also, qSin′t=q⋅qCosqt while qCos′t=−qSint. See [4] and [6] for further details.While the above examples are consistent with earlier examples in our previous work, the examples for(322)μ,λ∈2ℤ+1,4or2ℤ+1+,or2ℤ+1+2j3,,forj=0,1,2are all not previously seen. To illustrate an example from (322), we choose μ,λ∈2ℤ+1+2j/3,2/3 for j=0 and develop it here.[Exampleμ=1odd and λ=2/3] In this case,
(323)f1,2/3t=∑k=−∞∞−1kexp−qktqkk−1/2/3fort≥0.One sees thatμ+1/2=1+1/2=1/1=α/β and λ/2=1/3=γ/δ. Thus, α=β=γ=1, while δ=3, and ω=e2πi/δ=e2πi/3, consistent with Examples [M=δ=p,β=1] (with the prime p taken to be 3) in Section 6.1 and [M=δ=3, β=1] in Section 6.2 below. Also, we record that B=M/β=δ/β=3/1=3. Since δ=3, f1,2/3t has canonical extension W1,2/3 given by (307). That is,
(324)W1,2/3t=∑ℓ=0δ−1ωℓ+1B⋅αf~1,2/3ωℓ+1Dγt=∑ℓ=02e2πiℓ+1/33⋅1f~1,2/3e2πiℓ+1/31⋅1t=∑ℓ=02f~1,2/3e2πiℓ+1/3t=f~1,2/3e2πi/3t+f~1,2/3e4πi/3t+f~1,2/3t=0+0+f1,2/3t,fort≥0−1f1,2/3e2πi/3t+−1f1,2/3e4πi/3t+0,fort<0(325)=∑k=−∞∞−1kexp−qktqkk−1/2/3,fort≥0,−1∑k=−∞∞−1kexp−qke2πi/3tqkk−1/2/3+−1∑k=−∞∞−1kexp−qke4πi/3tqkk−1/2/3,fort<0,where (324) follows from Definition 15 and (325) follows from (323) and (9). From Theorem 40, one has that W1,2/3t satisfies properties (1) through (6), including that W1,2/3t is a Schwartz wavelet with all moments vanishing, satisfying the MADE W1,2/33t=q3W1,2/3qt, generating a frame for L2ℝ, and having Fourier transform −3μq33x2/2πθq3;−ix3.One computes(326)W1,2/30=f1,2/30=∑k=−∞∞−1k1qkk−1/2/3=∑k=−∞∞−1kq3kk−1/2=θq3;−1=0,while the first derivative at the origin is
(327)W1,2/3′0=f1,2/3′0=−f5/3,2/30=−∑k=−∞∞−1k1qkk−1−2/3/2/3(328)=−∑k=−∞∞−qkq3kk−1/2=−θq3;−q≠0,where the penultimate equality in each of (326) and (328) follows from (10), the second equality in (327) follows from the fact that fμ,λ′t=−fμ+λ,λt (as is proven in equation (11) of [2]), and the inequality giving nonvanishing in (328) follows from (13). While we are here, we also compute the second derivative at the origin by
(329)W1,2/3′′0=f1,2/3′′0=f7/3,2/30=∑k=−∞∞−1k1qkk−1−4/3/2/3=∑k=−∞∞−q2kq3kk−1/2=θq3;−q2≠0,NormalizeW1,2/3t by W1,2/3′0=−θq3;−q to obtain gq;t≡W1,2/3t/W1,2/3′0, where gq;t satisfies (1) through (6) of Theorem 40, while having Fourier transform 3μq33x2/2πθq3;−qθq3;−ix3 and satisfying the same MADE as W1,2/3t, namely,
(330)g3q;t=q3gq;qt,with initial conditions
(331)gq;0=0,g′q;0=−θq3;−q/−θq3;−q=1andg″q;0=θq3;−q2/−θq3;−q=−q.This second derivative reducing to−q in (331) follows from Lemma 41 below. For small q>1, as q⟶1+, (330) and (331) can be considered to be a perturbation of the ODE initial value problem (332) and (333), where
(332)f3t=ft,with initial conditions
(333)f0=0,f′0=1,f″0=−1,which is solved by ft=2exp−t/2sin3t/2/3. Figure 3 provides graphical evidence that W1,2/3t/W1,2/3′0 converges to ft=2exp−t/2sin3t/2/3 near t=0. We have scaled Figure 3(b) by expt/2 to better visualize the graph on the negative t axis.Lemma 41.
Forq>1 the Jacobi theta function (10) satisfies
(334)θq3;q2θq3;q=q=θq3;−q2θq3;−q.Proof.
We show both equalities in (334) together by requiring that each ± below be always simultaneously + or always simultaneously −. One has
(335)θq3;±q2=θq3;q3±1q=q3±1qθq3;±1q(336)=q±qθq3;±1q=qθq3;±q,where the second equality in (335) is given by the left hand equation in (12) and the last equality in (336) is given by the right hand equation in (12). Dividing (335) and (336) through by θq3;±q gives (334), and the lemma is shown.Figure 3
(a)y=2exp−t/2sin3t/2/3 (dashed) approximated by f1,2/3t/f1,2/3′0 for q=1.3 (solid red). (b) Scaled y=2sin3t/2/3 (dashed) approximated by the scaled canonical extension expt/2W1,2/3t/W1,2/3′0 (solid blue/red) for q=1.3 and t near 0. Each is scaled by expt/2.
(a)(b)In the next two propositions, we see forδ>3 that there are numerous cases for which Wμ,λt is not an extension of a given fμ,λt.Proposition 42.
Let the notation of Theorem23 hold. Let δ>3 with δ≠0mod4 and δ=0modβ be as in (337). Recall that for given fμ,λt one has an integer α and natural numbers β,γ,δ with
(337)μ+12=αβinreducedform,andλ2=γδinreducedform,while B=M/β=δ/β, D=M/δ=δ/δ=1, and ω=e2πi/δ. We observe that there are an infinite number of n∈ℕ such that
(338)αβ+nγδ∉ℤ.
Furthermore, recall that(339)+=ℓ∣Rωγℓ+1>0⊂0,1,2⋯δ−1.
If for any suchn as in (338), one has that
(340)∑+ωℓ+1Bα+nγ≠1,
then(341)Wμ,λt0,∞≠fμ,λt.
That is, the functionWμ,λt naturally generated by fμ,λt does not restrict to fμ,λt and is not a canonical extension of fμ,λt.Proof.
Note first that there are no two consecutive integersn and n+1 with
(342)αβ+nγδ∈ℤ,andαβ+n+1γδ∈ℤ,
for if there were two such consecutive integers, by subtracting one equation from the other in (342), one would conclude that γ/δ∈ℤ also. However, this is impossible as γ/δ is in a reduced form with δ>3, while γ/δ being both an integer and in a reduced form would require δ=1. We conclude that there is an infinite number of n∈ℕ with (338) holding.Fix anyn satisfying (338). For this n, from (138) and from the discussion between (138) through (144), one concludes
(343)Wμ,λn0=∑+ωℓ+1Bα+nDγ−1nfμ+nλ,λ0=∑+ωℓ+1Bα+nDγ−1nθQ;−Qμ+nλ−1/2=∑+ωℓ+1Bα+nDγ−1nθQ;−Qα/β+nγ/δ−1.Furthermore, for our same fixedn, from the fact that fμ,λ′t=−fμ+λ,λt, [2], one has
(344)fμ,λn0=−1nfμ+nλ,λ0=−1nθQ;−Qα/β+nγ/δ−1.Since (338) holds θQ;−Qα/β+nγ/δ−1≠0 by (10). If (343) were to equal (344), then dividing through by θQ;−Qα/β+nγ/δ−1 and recalling that D=1 in our setting would require that
(345)∑+ωℓ+1Bα+nDγ=∑+ωℓ+1Bα+nγ=1,which is disallowed by the hypothesis (340). Thus, if n satisfies (338) and (340), one cannot have equality of (343) and (344), whence Wμ,λt does not restrict to fμ,λt on 0,∞. The proposition is proven.We next harness Proposition42 to show that for λ with δ=5,6,7, none of the functions Wμ,λt restricts to fμ,λt on 0,∞.Proposition 43.
Under the same setting, notation, and assumptions as Proposition42, for fμ,λt with α,β,γ as in (337) but with the restriction that δ=5,6, or 7, one has Wμ,λt does not restrict to fμ,λt on 0,∞. Thus, fμ,λt on 0,∞ does not have a canonical extension for these values of δ.Proof.
Letδ=5,6 or 7. For given γ, define γ^ to be the unique value with γ=γ^modδ and 0<γ^<δ−1. Observe that in (339) for fixed δ one has ωγ=e2πiγ/δ=e2πiγ∧/δ=ωγ∧. Thus, the set + associated to γ,δ is identical to the set + associated to γ^,δ.In Table1, for the given value of δ and associated possible γ^ one has, by inspection, the unique values 1≤ℓ+1≤δ with Rωγℓ+1>0 that determine the set +. We denote the values ℓ+1 by σ,τ,δ. This last δ is the same value giving λ/2=γ/δ, where δ=5,6,7.Table 1
Notation and cases as used in Proposition44.
δγ^ℓ+1≡σ,τ,δ+−511,4,50,3,41,222,3,51,2,40,332,3,51,2,40,341,4,50,3,41,2611,5,60,4,51,2,351,5,60,4,51,2,3711,6,70,5,61,2,3,423,4,72,3,60,1,4,532,5,71,4,60,2,3,542,5,71,4,60,2,3,553,4,72,3,60,1,4,561,6,70,5,61,2,3,4For instance, to obtain the second row of the above Table1, when δ=5 and γ has associated value γ^=2, one has
(346)Rωγℓ+1=Re2πiγ/δℓ+1=Re2πiγ∧/5ℓ+1=Re2πi2/5ℓ+1>0precisely when theℓ+1 in (346) assume the values ℓ+1=2,3, or 5, that is, when
(347)Re2πi2/52=cos2π4/5>0,Re2πi2/53=cos2π6/5>0,andRe2πi2/55=1>0.In this case, one hasℓ+1=σ=2,τ=3,δ=5 and then +=ℓ=2−1,3−1,5−1=1,2,4; the remaining values of ℓ form −=0,3.Now note that, in all cases in Table1,
(348)σ+τ=δ,whence∀j∈ℤonehase2πiσj/δande2πiτj/δareconjugates,which follows from the fact that e2πiσj/δ⋅e2πiτj/δ=e2πiσ+τj/δ=e2πiδj/δ=1. We conclude that for all δ=5,6,7 and γ with γ/δ in a reduced form, and for all integers j one has
(349)∑+ωjℓ+1=e2πij/δσ+e2πij/δτ+e2πij/δδ=2cos2πσj/δ+1≠1.The inequality in (349) follows from the fact that no integer multiple of 2π/δ for δ=5,6,7 is an odd multiple of π/2.Now, we have seen in Proposition42 that for each of δ=5,6,7 there are infinitely many n with (338) holding. Fix such an n, and set j = Bα+nγ in (349) to obtain
(350)∑+ωBα+nγℓ+1=∑+ωℓ+1Bα+nγ≠1,from which we see that both (338) and now (340) hold. We conclude from Proposition 42 that Wμ,λt does not restrict to fμ,λt on 0,∞ for any λ with δ=5,6,7. Therfore, such fμ,λt do not have canonical extensions. This completes the proof of the proposition.We remark that forδ>8 and not divisible by 4, results similar to Proposition 43, with more tables analogous to Table 1 (but having many more cases), and with inequalities analogous to (349) (but with multiple cosine terms), should give that canonical extensions are at least rare, if not nonexistent.


## 6. Examples

The goals of this section are twofold:(1)
To classify thoseμ and λ for low values of M=δ such that fμ,λt and Wμ,λt meet the assumptions of Theorem 23includingδ=0modβ and satisfy (180)-(182);(2)
To fill out selections from the examples in goal 1 in detailThroughout the discussion, the notation given in (96) holds, and M=lcmβ,δ=δ.


### 6.1. Classifyingμ and λ in Theorem 23 with Low M=δ Values

Recall that the fractionsα/β and γ/δ in (96) are in reduced form.[M=δ=1]: since M=δ=lcmβ,δ=1, we conclude β=1=δ. This is equivalent to
(351)μ+12=αβ=α1λ2=γδ=γ1M=1.Now (351) holds if and only if μ=2α−1 is an odd integer and λ=2γ is an even positive integer. To match notation in previous work [2], we relabeled μ=2N+1 and λ=2n. Such f2N+1,2nt are precisely the fμ,λt that are flat at t=0 (as is shown in Proposition 2.2 of [2]). We further analyze this case in detail below and record now that
(352)M=δ=1⇔μ∈ℤwithμ=1mod2,andλ∈ℕwithλ=0mod2.M=δ=pisprime: since M=δ=lcmβ,δ=p, we conclude β∈1,p and δ=p. Thus, there are two subcases.[M=δ=p, β=p]:
(353)μ+12=αβ=αp⇔μ=2α−ppwithα≠0modp,λ2=γδ=γp⇔λ=2γp>0withγ≠0modp.[M=δ=p, β=1]:
(354)μ+12=αβ=α1⇔μ=2α−1⇔μ∈ℤwithμ=1mod2,λ2=γδ=γp⇔λ=2γp>0withγ≠0modp.Thus, we have(355)M=δ=pisprime⇔therightmostpairsofconditionsonμandλin353or354hold.[M=4]: this case is disallowed.[M=δ=6]: since M=lcmβ,δ=6, we conclude β∈1,2,3,6 and δ=6. There are now four subcases.[M=δ=6, β=6]:
(356)μ+12=αβ=α6⇔μ=α−33withα≠0mod2andα≠0mod3,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.Please see Section7 for a detailed convergence study involving the current example (356) with α=−1,β=6,γ=1, and δ=6.[M=δ=6, β=3]:
(357)μ+12=αβ=α3⇔μ=2α−33withα≠0mod3,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.[M=δ=6, β=2]:
(358)μ+12=αβ=α2⇔μ=α−1withα≠0mod2,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.[M=δ=6, β=1]:
(359)μ+12=αβ=α1⇔μ=2α−1⇔μ∈ℤwithμ=1mod2,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.Thus, we have(360)M=6=δ⇔anyoftherightmostpairsofconditionsonμandλin356through359hold.Otherμ and λ with fμ,λ meeting the criteria of Theorem 23 and M not among the above cases can be determined similarly via the following proposition.Proposition 44.
LetM=∏j=1Jpjnj be the prime factorization of M>1 and M≠0mod4. The μ and λ with fμ,λ meeting the criteria of Theorem 23 are
(361)μ=2α−ββ,(362)λ=2γδthat satisfy the conditions
(363)M=δ=∏j=1Jpjnj,(364)β=∏j=1Jpjkjwith0≤kj≤nj∀j,(365)α≠0modpjwhenkj>0,(366)γ≠0modpj∀pj.Proof.
From Proposition21, one sees that the hypothesis that δ=0modβ in Theorem 23 is required in order to have a non-identically vanishing of (180)–(182). From this, we conclude M=lcmβ,δ=δ, giving (363). Since β is a divisor of δ=M, one obtains (364). Equations (365) and (366) follow from the fact that α/β and γ/δ are in a reduced form. (361) and (362) follow from (96). The proposition is shown.


### 6.2. Selected Examples from Section6.1 in Further Depth: Canonical Extensions

[M=δ=1, μ=2N+1, λ=2n]: we expand on [Example M=1] in Section 6.1, where it was shown that μ=2N+1, λ=2n, and β=δ=1, which specifies (96) as
(367)μ+12=2N+1+12=N+11=αβλ2=2n2=n1=γδM=1Q=q2/λ=q1/nB=Mβ=11=1D=Mδ=1Bα=1α=N+1Dγ=1n=nω=e2πi/M=e2πi=1ω~=e2πi/n.Inserting these values in (131) gives (368), while insertion in (180)–(182) gives (369) and (370) below. That is,
(368)W2N+1,2nt=∑ℓ=01−11Nf~2N+1,2n1nt=f~2N+1,2nt.Furthermore,(369)11FW2N+1,2ntx=Ff~2N+1,2ntx(370)=μQ3eiπN+12π1−ix1n∑κ=0n−1ω~κz3N+1θQ;ω~κz3(371)=μQ3−1N+12π1−ix1n∑κ=0n−11ω~κz3−N−1θQ;ω~κz3(372)=μQ3−1N+12π1−ix⋅1n∑κ=0n−1QNN+1/2Q−N−1−N/2ω~κz3−N−1θQ;ω~κz3(373)=μQ3−1N+12πQNN+1/2−ix1n∑κ=0n−11θQ;ω~κz3Q−N−1,where (371) follows from moving ω~ℓz3N+1 from the numerator to the denominator; (372) is obtained by multiplying numerator and denominator by QNN+1/2; and (373) follows from (12). Also, from (184), z3 in (373) is any fixed nth root of z3n=−e−πinix. This example recovers Theorem 6.3 in [2]. Also, if one sets N=−1 and n=1, one has W2N+1,2nt=W−1,2t=f~−1,2t which is discussed earlier in [Example μ=−1odd and λ=2] in Section 5, and which was denoted Kt in [3]. This recovers the inaugural wavelet in the first paper that inspired the current direction of study.Finally,W2N+1,2nt=f~2N+1,2nt satisfies the same multiplied advanced differential equation on ℝ as does f2N+1,2nt on 0,∞. That is, (185) becomes
(374)W2N+1,2n1t=−1n+1qn+2N+1/2W2N+1,2nqnt.[M=δ=2]: we expand on [Example M=p] in the previous section for the case p=2. There are two subcases, when β=2 or β=1.[M=δ=2, β=2]:
(375)μ+12=αβ=α2λ2=γδ=γ2M=2Q=q2/λ=q2/γB=Mβ=22=1D=Mδ=22=1Bα=1α=αDγ=1γ=γω=e2πi/2=eπi=−1ω~=e2πi/Dγ=e2πi/γ.Note that in this case, sinceα and γ must be odd, μ=α−1 is even, and λ=γ is odd, so we set μ=2N and λ=2n+1=γ. Then, Bα=2N+1 and Dγ=2n+1. Inserting these values in (131) gives (376)–(378), namely,
(376)W2N,2n+1t=−12N+1f~2N,2n+1−12n+1t+−122N+1f~2N,2n+1−122n+1t(377)=−1f~2N,2n+1−1t+1f~2N,2n+11t(378)=−1−1f2N,2n+1−tχ−∞,0+f2N,2n+1tχ0,∞(379)=∑k=−∞∞−1ke−qktqkk−2N/2n+1.Note that (378) and (379) show that W2N,2n+1t is an even function. Also, from (180)-(182) along with (184), one has
(380)FW2N,2n+1tx=2μQ3eiπ2N+1/22π1−ix12n+1∑κ=02ne2πiκ/2n+1z32N+1θQ;e2πiκ/2n+1z32,with
(381)z32n+1=−e−πi2n+1/2ix,where Q=q2/2n+1. Finally, the MADE (185) becomes
(382)W2N,2n+12t=−12n+1+2q2n+12n+1+2N/2n+1W2N,2n+1q2n+1t=−q2n+1+2NW2N,2n+1q2n+1t.Whenn=0, the above results recover the even case of Theorem 6.5 of [2].Whenn=0 and N=0, the above results give
(383)W0,1t=∑k=−∞∞−1ke−qktqkk/1,with(384)W0,10=∑k=−∞∞−1k1qk2=∑k=−∞∞−1kq−kq2kk−1/2(385)=θq2;−1q=μq2∏n=0∞1−1q2n+12>0.In this setting (383)–(385) we recover [Example μ=0 even and λ=1] from Section 5, with normalization W0,1t/W0,10=qCost.[M=δ=2, β=1]:
(386)μ+12=αβ=α1λ2=γδ=γ2M=2Q=q2/λ=q2/γB=Mβ=21=2D=Mδ=22=1Bα=2αDγ=1γ=γω=e2πi/2=eπiω~=e2πi/Dγ=e2πi/γ.Note that in this caseμ=2α−1 is odd, and λ=γ is odd, so we set μ=2N+1 and λ=2n+1=γ. Then, Bα=2N+1 and Dγ=2n+1. Inserting these values in (131) gives (387)–(390), namely,
(387)W2N+1,2n+1t=−12N+1f~2N+1,2n+1−12n+1t(388)+−122N+1f~2N+1,2n+1−122n+1t(389)=1f~2N+1,2n+1−1t+1f~2N+1,2n+11t(390)=1−1f2N+1,2n+1−tχ−∞,0+f2N+1,2n+1tχ0,∞(391)=signt∑k=−∞∞−1ke−qktqkk−2N−1/2n+1.Note that (390) and (391) show that W2N+1,2n+1t is an odd function. Also, from (180)–(182) along with (184), one has
(392)FW2N+1,2n+1tx=2μQ3eiπN+12π1−ix12n+1∑κ=02ne2πiκ/2n+1z32N+1θQ;e2πiκ/2n+1z32,with
(393)z32n+1=−e−πi2n+1/2ix,where Q=q2/2n+1. Finally, the MADE (185) becomes
(394)W2N+1,2n+12t=−12n+1+2q2n+12n+1+2N+1/2n+1W2N+1,2n+1q2n+1t=−q2n+2+2NW2N+1,2n+1q2n+1t.Whenn=0, the above results recover the odd case of Theorem 6.5 of [2].Whenn=0 and N=0, the above results give
(395)W1,1t=signt∑k=−∞∞−1ke−qktqkk−1/1.In this setting, (395) recovers [Example μ=1 odd and λ=1] from Section 5, with normalization W1,1t/W0,10=qSint.[M=δ=3]: we expand on [Example M=p] in the previous section for the case p=3. There are two subcases.[M=δ=3, β=3]:
(396)μ+12=αβ=α3λ2=γδ=γ3M=3Q=q2/λ=q3/γB=Mβ=33=1D=Mδ=33=1Bα=1α=αDγ=1γ=γω=e2πi/3ω~=e2πi/Dγ=e2πi/γ.Note thatμ=2α/3−1 with α≠0mod3 and λ=2γ/3 with γ≠0mod3. Thus, we set α=3N+J and γ=3n+j where J,j∈1,2. Hence, μ=2N−1+2J/3 and λ=23n+j/3. Then, Bα=3N+J and Dγ=3n+j where J,j∈1,2. Inserting these values in (131) gives (397)–(400), namely,
(397)W2N−1+2J/3,23n+j/3t=e2πi/33N+Jf~2N−1+2J/3,23n+j/3e2πi/33n+jt+e4πi/33N+Jf~2N−1+2J/3,23n+j/3e4πi/33n+jt+13N+Jf~2N−1+2J/3,23n+j/313n+jt=e2πJi/3f~2N−1+2J/3,23n+j/3e2πji/3t(398)+e4πJi/3f~2N−1+2J/3,23n+j/3e4πji/3t+1f~2N−1+2J/3,23n+j/3(399)t=f2N−1+2J/3,23n+j/3t,fort≥0e2πJi/3−1f2N−1+2J/3,23n+j/3e2πji/3t+e4πJi/3−1f2N−1+2J/3,23n+j/3e4πji/3t,fort<0(400)=∑k=−∞∞−1kexp−qktqkk−2N+1−2J/3/23n+j/3,fort≥0,e2πJi/3∑k=−∞∞−1k+1exp−qke2πji/3tqkk−2N+1−2J/3/23n+j/3+e4πJi/3∑k=−∞∞−1k+1exp−qke4πji/3tqkk−2N+1−2J/3/23n+j/3,fort<0.Also, from (180)–(182) along with (184), one has:
(401)FW2N−1+2J/3,23n+j/3tx=3μQ3eiπN+J/32π1−ix13n+j∑κ=03n+j−1e2πiκ/3n+jz33N+JθQ;e2πiκ/3n+jz33,with
(402)z33n+j=−e−πi3n+j/3ix,where Q=q3/3n+j. Finally, the MADE (185) becomes
(403)W2N−1+2J/3,23n+j/33t=−13n+j+3q3n+j+2N−1+2J/3/2/3⋅W2N−1+2J/3,23n+j/3q3n+jt.[M=δ=3, β=1]:
(404)μ+12=αβ=α1λ2=γδ=γ3M=3Q=q2/λ=q3/γB=Mβ=31=3D=Mδ=33=1Bα=3αDγ=1γ=γω=e2πi/3ω~=e2πi/Dγ=e2πi/γ.Note thatμ=2α−1 is odd and λ=2γ/3 with γ≠0mod3. Thus, we set μ=2N+1 with γ=3n+j where j=1,2 and λ=23n+j/3. Then, Bα=3N+1 and Dγ=3n+j where j=1,2. Inserting these values in (131) gives (405)–(408), namely,
(405)W2N+1,23n+j/3t=e2πi/33N+1f~2N+1,23n+j/3e2πi/33n+jt+e4πi/33N+1f~2N+1,23n+j/3e4πi/33n+jt+13N+1f~2N+1,23n+j/313n+jt(406)=1f~2N+1,23n+j/3e2πji/3t+1f~2N+1,23n+j/3e4πji/3t+1f~2N+1,23n+j/3t(407)=f2N+1,23n+j/3t,fort≥0−1f2N+1,23n+j/3e2πji/3t+−1f2N+1,23n+j/3e4πji/3t,fort<0(408)=∑k=−∞∞−1kexp−qktqkk−2N−1/23n+j/3fort≥0,−1∑k=−∞∞−1kexp−qke2πi/3tqkk−2N−1/23n+j/3+−1∑k=−∞∞−1kexp−qke4πi/3tqkk−2N−1/23n+j/3,fort<0,where each of the three summations in (408) vanishes at t=0 as one can compute that
(409)∑k=−∞∞−1kqkk−2N−1/23n+j/3=θQ;−QN=0,with Q as in (404), where the vanishing follows from (13).Also, from (180)–(182) along with (184) one has
(410)FW2N+1,23n+j/3tx=3μQ3eiπN+12π1−ix13n+j∑κ=03n+j−1e2πiκ/3n+jz33N+1θQ;e2πiκ/3n+jz33,with
(411)z33n+j=−e−πi3n+j/3ix,where Q=q3/3n+j. Finally, the MADE (185) becomes
(412)W2N+1,23n+j/33t=−13n+j+3q3n+j+2N+1/2/3W2N+1,23n+j/3q3n+jt.WhenN=0,n=0, and j=1, equations (410) and (412) recover, respectively, the Fourier transform and the MADE satisfied by W1,2/3t as seen in [Example μ=1 odd and λ=2/3] in Section 5.


## 6.1. Classifyingμ and λ in Theorem 23 with Low M=δ Values

Recall that the fractionsα/β and γ/δ in (96) are in reduced form.[M=δ=1]: since M=δ=lcmβ,δ=1, we conclude β=1=δ. This is equivalent to
(351)μ+12=αβ=α1λ2=γδ=γ1M=1.Now (351) holds if and only if μ=2α−1 is an odd integer and λ=2γ is an even positive integer. To match notation in previous work [2], we relabeled μ=2N+1 and λ=2n. Such f2N+1,2nt are precisely the fμ,λt that are flat at t=0 (as is shown in Proposition 2.2 of [2]). We further analyze this case in detail below and record now that
(352)M=δ=1⇔μ∈ℤwithμ=1mod2,andλ∈ℕwithλ=0mod2.M=δ=pisprime: since M=δ=lcmβ,δ=p, we conclude β∈1,p and δ=p. Thus, there are two subcases.[M=δ=p, β=p]:
(353)μ+12=αβ=αp⇔μ=2α−ppwithα≠0modp,λ2=γδ=γp⇔λ=2γp>0withγ≠0modp.[M=δ=p, β=1]:
(354)μ+12=αβ=α1⇔μ=2α−1⇔μ∈ℤwithμ=1mod2,λ2=γδ=γp⇔λ=2γp>0withγ≠0modp.Thus, we have(355)M=δ=pisprime⇔therightmostpairsofconditionsonμandλin353or354hold.[M=4]: this case is disallowed.[M=δ=6]: since M=lcmβ,δ=6, we conclude β∈1,2,3,6 and δ=6. There are now four subcases.[M=δ=6, β=6]:
(356)μ+12=αβ=α6⇔μ=α−33withα≠0mod2andα≠0mod3,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.Please see Section7 for a detailed convergence study involving the current example (356) with α=−1,β=6,γ=1, and δ=6.[M=δ=6, β=3]:
(357)μ+12=αβ=α3⇔μ=2α−33withα≠0mod3,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.[M=δ=6, β=2]:
(358)μ+12=αβ=α2⇔μ=α−1withα≠0mod2,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.[M=δ=6, β=1]:
(359)μ+12=αβ=α1⇔μ=2α−1⇔μ∈ℤwithμ=1mod2,λ2=γδ=γ6⇔λ=γ3>0withγ≠0mod2andγ≠0mod3.Thus, we have(360)M=6=δ⇔anyoftherightmostpairsofconditionsonμandλin356through359hold.Otherμ and λ with fμ,λ meeting the criteria of Theorem 23 and M not among the above cases can be determined similarly via the following proposition.Proposition 44.
LetM=∏j=1Jpjnj be the prime factorization of M>1 and M≠0mod4. The μ and λ with fμ,λ meeting the criteria of Theorem 23 are
(361)μ=2α−ββ,(362)λ=2γδthat satisfy the conditions
(363)M=δ=∏j=1Jpjnj,(364)β=∏j=1Jpjkjwith0≤kj≤nj∀j,(365)α≠0modpjwhenkj>0,(366)γ≠0modpj∀pj.Proof.
From Proposition21, one sees that the hypothesis that δ=0modβ in Theorem 23 is required in order to have a non-identically vanishing of (180)–(182). From this, we conclude M=lcmβ,δ=δ, giving (363). Since β is a divisor of δ=M, one obtains (364). Equations (365) and (366) follow from the fact that α/β and γ/δ are in a reduced form. (361) and (362) follow from (96). The proposition is shown.


## 6.2. Selected Examples from Section6.1 in Further Depth: Canonical Extensions

[M=δ=1, μ=2N+1, λ=2n]: we expand on [Example M=1] in Section 6.1, where it was shown that μ=2N+1, λ=2n, and β=δ=1, which specifies (96) as
(367)μ+12=2N+1+12=N+11=αβλ2=2n2=n1=γδM=1Q=q2/λ=q1/nB=Mβ=11=1D=Mδ=1Bα=1α=N+1Dγ=1n=nω=e2πi/M=e2πi=1ω~=e2πi/n.Inserting these values in (131) gives (368), while insertion in (180)–(182) gives (369) and (370) below. That is,
(368)W2N+1,2nt=∑ℓ=01−11Nf~2N+1,2n1nt=f~2N+1,2nt.Furthermore,(369)11FW2N+1,2ntx=Ff~2N+1,2ntx(370)=μQ3eiπN+12π1−ix1n∑κ=0n−1ω~κz3N+1θQ;ω~κz3(371)=μQ3−1N+12π1−ix1n∑κ=0n−11ω~κz3−N−1θQ;ω~κz3(372)=μQ3−1N+12π1−ix⋅1n∑κ=0n−1QNN+1/2Q−N−1−N/2ω~κz3−N−1θQ;ω~κz3(373)=μQ3−1N+12πQNN+1/2−ix1n∑κ=0n−11θQ;ω~κz3Q−N−1,where (371) follows from moving ω~ℓz3N+1 from the numerator to the denominator; (372) is obtained by multiplying numerator and denominator by QNN+1/2; and (373) follows from (12). Also, from (184), z3 in (373) is any fixed nth root of z3n=−e−πinix. This example recovers Theorem 6.3 in [2]. Also, if one sets N=−1 and n=1, one has W2N+1,2nt=W−1,2t=f~−1,2t which is discussed earlier in [Example μ=−1odd and λ=2] in Section 5, and which was denoted Kt in [3]. This recovers the inaugural wavelet in the first paper that inspired the current direction of study.Finally,W2N+1,2nt=f~2N+1,2nt satisfies the same multiplied advanced differential equation on ℝ as does f2N+1,2nt on 0,∞. That is, (185) becomes
(374)W2N+1,2n1t=−1n+1qn+2N+1/2W2N+1,2nqnt.[M=δ=2]: we expand on [Example M=p] in the previous section for the case p=2. There are two subcases, when β=2 or β=1.[M=δ=2, β=2]:
(375)μ+12=αβ=α2λ2=γδ=γ2M=2Q=q2/λ=q2/γB=Mβ=22=1D=Mδ=22=1Bα=1α=αDγ=1γ=γω=e2πi/2=eπi=−1ω~=e2πi/Dγ=e2πi/γ.Note that in this case, sinceα and γ must be odd, μ=α−1 is even, and λ=γ is odd, so we set μ=2N and λ=2n+1=γ. Then, Bα=2N+1 and Dγ=2n+1. Inserting these values in (131) gives (376)–(378), namely,
(376)W2N,2n+1t=−12N+1f~2N,2n+1−12n+1t+−122N+1f~2N,2n+1−122n+1t(377)=−1f~2N,2n+1−1t+1f~2N,2n+11t(378)=−1−1f2N,2n+1−tχ−∞,0+f2N,2n+1tχ0,∞(379)=∑k=−∞∞−1ke−qktqkk−2N/2n+1.Note that (378) and (379) show that W2N,2n+1t is an even function. Also, from (180)-(182) along with (184), one has
(380)FW2N,2n+1tx=2μQ3eiπ2N+1/22π1−ix12n+1∑κ=02ne2πiκ/2n+1z32N+1θQ;e2πiκ/2n+1z32,with
(381)z32n+1=−e−πi2n+1/2ix,where Q=q2/2n+1. Finally, the MADE (185) becomes
(382)W2N,2n+12t=−12n+1+2q2n+12n+1+2N/2n+1W2N,2n+1q2n+1t=−q2n+1+2NW2N,2n+1q2n+1t.Whenn=0, the above results recover the even case of Theorem 6.5 of [2].Whenn=0 and N=0, the above results give
(383)W0,1t=∑k=−∞∞−1ke−qktqkk/1,with(384)W0,10=∑k=−∞∞−1k1qk2=∑k=−∞∞−1kq−kq2kk−1/2(385)=θq2;−1q=μq2∏n=0∞1−1q2n+12>0.In this setting (383)–(385) we recover [Example μ=0 even and λ=1] from Section 5, with normalization W0,1t/W0,10=qCost.[M=δ=2, β=1]:
(386)μ+12=αβ=α1λ2=γδ=γ2M=2Q=q2/λ=q2/γB=Mβ=21=2D=Mδ=22=1Bα=2αDγ=1γ=γω=e2πi/2=eπiω~=e2πi/Dγ=e2πi/γ.Note that in this caseμ=2α−1 is odd, and λ=γ is odd, so we set μ=2N+1 and λ=2n+1=γ. Then, Bα=2N+1 and Dγ=2n+1. Inserting these values in (131) gives (387)–(390), namely,
(387)W2N+1,2n+1t=−12N+1f~2N+1,2n+1−12n+1t(388)+−122N+1f~2N+1,2n+1−122n+1t(389)=1f~2N+1,2n+1−1t+1f~2N+1,2n+11t(390)=1−1f2N+1,2n+1−tχ−∞,0+f2N+1,2n+1tχ0,∞(391)=signt∑k=−∞∞−1ke−qktqkk−2N−1/2n+1.Note that (390) and (391) show that W2N+1,2n+1t is an odd function. Also, from (180)–(182) along with (184), one has
(392)FW2N+1,2n+1tx=2μQ3eiπN+12π1−ix12n+1∑κ=02ne2πiκ/2n+1z32N+1θQ;e2πiκ/2n+1z32,with
(393)z32n+1=−e−πi2n+1/2ix,where Q=q2/2n+1. Finally, the MADE (185) becomes
(394)W2N+1,2n+12t=−12n+1+2q2n+12n+1+2N+1/2n+1W2N+1,2n+1q2n+1t=−q2n+2+2NW2N+1,2n+1q2n+1t.Whenn=0, the above results recover the odd case of Theorem 6.5 of [2].Whenn=0 and N=0, the above results give
(395)W1,1t=signt∑k=−∞∞−1ke−qktqkk−1/1.In this setting, (395) recovers [Example μ=1 odd and λ=1] from Section 5, with normalization W1,1t/W0,10=qSint.[M=δ=3]: we expand on [Example M=p] in the previous section for the case p=3. There are two subcases.[M=δ=3, β=3]:
(396)μ+12=αβ=α3λ2=γδ=γ3M=3Q=q2/λ=q3/γB=Mβ=33=1D=Mδ=33=1Bα=1α=αDγ=1γ=γω=e2πi/3ω~=e2πi/Dγ=e2πi/γ.Note thatμ=2α/3−1 with α≠0mod3 and λ=2γ/3 with γ≠0mod3. Thus, we set α=3N+J and γ=3n+j where J,j∈1,2. Hence, μ=2N−1+2J/3 and λ=23n+j/3. Then, Bα=3N+J and Dγ=3n+j where J,j∈1,2. Inserting these values in (131) gives (397)–(400), namely,
(397)W2N−1+2J/3,23n+j/3t=e2πi/33N+Jf~2N−1+2J/3,23n+j/3e2πi/33n+jt+e4πi/33N+Jf~2N−1+2J/3,23n+j/3e4πi/33n+jt+13N+Jf~2N−1+2J/3,23n+j/313n+jt=e2πJi/3f~2N−1+2J/3,23n+j/3e2πji/3t(398)+e4πJi/3f~2N−1+2J/3,23n+j/3e4πji/3t+1f~2N−1+2J/3,23n+j/3(399)t=f2N−1+2J/3,23n+j/3t,fort≥0e2πJi/3−1f2N−1+2J/3,23n+j/3e2πji/3t+e4πJi/3−1f2N−1+2J/3,23n+j/3e4πji/3t,fort<0(400)=∑k=−∞∞−1kexp−qktqkk−2N+1−2J/3/23n+j/3,fort≥0,e2πJi/3∑k=−∞∞−1k+1exp−qke2πji/3tqkk−2N+1−2J/3/23n+j/3+e4πJi/3∑k=−∞∞−1k+1exp−qke4πji/3tqkk−2N+1−2J/3/23n+j/3,fort<0.Also, from (180)–(182) along with (184), one has:
(401)FW2N−1+2J/3,23n+j/3tx=3μQ3eiπN+J/32π1−ix13n+j∑κ=03n+j−1e2πiκ/3n+jz33N+JθQ;e2πiκ/3n+jz33,with
(402)z33n+j=−e−πi3n+j/3ix,where Q=q3/3n+j. Finally, the MADE (185) becomes
(403)W2N−1+2J/3,23n+j/33t=−13n+j+3q3n+j+2N−1+2J/3/2/3⋅W2N−1+2J/3,23n+j/3q3n+jt.[M=δ=3, β=1]:
(404)μ+12=αβ=α1λ2=γδ=γ3M=3Q=q2/λ=q3/γB=Mβ=31=3D=Mδ=33=1Bα=3αDγ=1γ=γω=e2πi/3ω~=e2πi/Dγ=e2πi/γ.Note thatμ=2α−1 is odd and λ=2γ/3 with γ≠0mod3. Thus, we set μ=2N+1 with γ=3n+j where j=1,2 and λ=23n+j/3. Then, Bα=3N+1 and Dγ=3n+j where j=1,2. Inserting these values in (131) gives (405)–(408), namely,
(405)W2N+1,23n+j/3t=e2πi/33N+1f~2N+1,23n+j/3e2πi/33n+jt+e4πi/33N+1f~2N+1,23n+j/3e4πi/33n+jt+13N+1f~2N+1,23n+j/313n+jt(406)=1f~2N+1,23n+j/3e2πji/3t+1f~2N+1,23n+j/3e4πji/3t+1f~2N+1,23n+j/3t(407)=f2N+1,23n+j/3t,fort≥0−1f2N+1,23n+j/3e2πji/3t+−1f2N+1,23n+j/3e4πji/3t,fort<0(408)=∑k=−∞∞−1kexp−qktqkk−2N−1/23n+j/3fort≥0,−1∑k=−∞∞−1kexp−qke2πi/3tqkk−2N−1/23n+j/3+−1∑k=−∞∞−1kexp−qke4πi/3tqkk−2N−1/23n+j/3,fort<0,where each of the three summations in (408) vanishes at t=0 as one can compute that
(409)∑k=−∞∞−1kqkk−2N−1/23n+j/3=θQ;−QN=0,with Q as in (404), where the vanishing follows from (13).Also, from (180)–(182) along with (184) one has
(410)FW2N+1,23n+j/3tx=3μQ3eiπN+12π1−ix13n+j∑κ=03n+j−1e2πiκ/3n+jz33N+1θQ;e2πiκ/3n+jz33,with
(411)z33n+j=−e−πi3n+j/3ix,where Q=q3/3n+j. Finally, the MADE (185) becomes
(412)W2N+1,23n+j/33t=−13n+j+3q3n+j+2N+1/2/3W2N+1,23n+j/3q3n+jt.WhenN=0,n=0, and j=1, equations (410) and (412) recover, respectively, the Fourier transform and the MADE satisfied by W1,2/3t as seen in [Example μ=1 odd and λ=2/3] in Section 5.


## 7. Convergence of MADEs to Classical Solutions, an Example

The purpose of this section is to provide an overview of an example of aWμ,λt satisfying a MADE where the normalization Wμ,λt/Wμ,λ0 converges to the corresponding solution of its classical analogue (the ODE obtained when the parameter q in the original MADE is set to 1). Our example is W−4/3,1/3t, which is not a canonical extensions as will be seen below. As such, it represents a new phenomenon with corresponding new challenges. First, W−4/3,1/3t can be obtained as the output of a reproducing kernel computation for an input wavelet W−1,2/3t, in the same manner of computation as in [4]. Second, since it satisfies a MADE of order 6, there are 6 derivatives to determine and use to compute initial conditions at the origin. Third, determining the initial conditions for the analogous ODE will be accomplished with the aide of a significant new result, namely the generalized q-Wallis formulas in Theorem 46. This theorem is also crucial in our proof of convergence of the normalized solution of the MADE to the solution of the analogous ODE. The normalization W−4/3,1/3t/W−4/3,1/30 will be our main object of study as the parameter q⟶1+. From (425), the parameters μ=−4/3 and λ=1/3 show our example W−4/3,1/3t to be from [M=δ=6, β=6] in Section 6.


### 7.1. Preliminaries

Forb≥0, the reproducing kernel computation relevant to our setting is
(413)∫−∞∞W−1,2/3tW−1,2/3t−bdt(414)=−3iμq342μq62⋅∑ℓ=02exp2πiℓ+πi6−1f−4/3,1/3−iexp2πiℓ+iπ6b(415)=−3iμq342μq62−i⋅e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b(416)=−3iμq342μq62−iW−4/3,1/3b.Moving from (413) to (414) is accomplished via Plancherel’s theorem where the expressions for the Fourier transform of W−1,2/3 are given by Theorem 23. The resulting integral of Fourier transforms is evaluated with a residue computation in the upper half-plane (for b≥0) similar in nature to the computation of reproducing kernels for b≥0 in [4]. This computation is lengthy, and the details are left as part of a more general set of reproducing kernel computations in an upcoming work. A direct computation moves one from line (414) to (415), while equation (132) of Definition 16 gives for b≥0 that
(417)W−4/3,1/3b=e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b,which justifies movement from (415) to (416).Settingb=0 in (413)–(416) yields
(418)∫−∞∞W−1,2/3tW−1,2/3tdt=W−1,2/3t22=−3iμq342μq62−iW−4/3,1/30,which in turn gives the functional identity
(419)1W−4/3,1/30=−3μq342μq62W−1,2/3t22.Normalizing (413)–(416) by the squared L2 norm and applying the functional identity in (419), one observes that
(420)1W−1,2/3t22∫−∞∞W−1,2/3tW−1,2/3t−bdt=−3μq342μq62W−1,2/3t22W−4/3,1/3b=W−4/3,1/3bW−4/3,1/30.Recalling thatW−4/3,1/3t is real valued, and applying Cauchy-Schwartz to (420) yields
(421)1=W−1,2/3tW−1,2/3t22W−1,2/3t−bW−1,2/3t22≥1W−1,2/3t22∫−∞∞W−1,2/3tW−1,2/3t−bdt=W−4/3,1/3bW−4/3,1/30,which gives a unit global bound on the normalization W−4/3,1/3b/W−4/3,1/30 independent of q for all b≥0. Now, equation (132) of Definition 16 also gives for b<0 that
(422)W−4/3,1/3b=e−4πi/6−1f−4/3,1/3e4πi/6b+e−6πi/6−1f−4/3,1/3e6πi/6b+e−8πi/6−1f−4/3,1/3e8πi/6b=e2πi/6f−4/3,1/3e−2πi/6−1b+f−4/3,1/3−1b+e−2πi/6f−4/3,1/3e2πi/6−1b=e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b(423)=W−4/3,1/3b,where comparison of (422) with (417) allows movement to (423). So W−4/3,1/3b is an even Schwartz function and consequently the bound in (421) extends to a global bound: for all b∈ℝ and for each q>1, one has
(424)1≥W−4/3,1/3bW−4/3,1/30.From (415) and (416), one sees that W−4/3,1/3t satisfies the same MADE as does f−4/3,1/3t, where μ=−4/3 and λ=1/3. From these values, we deduce that
(425)μ+12=−4/3+12=−1/32=−16=αβ,λ2=1/32=16=γδ,by which one sees thatα=−1,β=6,γ=1,δ=6. From these parameter values, one sees that W−4/3,1/3t falls in the example class [M=δ=6,β=6] with (356) holding. Since δ=6, we conclude from Proposition 43 that W−4/3,1/3t is not a canonical extension of f−4/3,1/3t. See also Table 1 where W−4/3,1/3t fall in the category of row δ=6,γ^=1,ℓ+1=1,5,6,+=0,4,5, and −=1,2,3. From (2), one determines the MADE for f−4/3,1/3t to be
(426)f−4/3,1/36t=−11+6q11+−4/3/1/3f−4/3,1/3q1t=−q−1f−4/3,1/3qt.And hence,(427)W−4/3,1/36t=−q−1W−4/3,1/3qt,while(428)W−4/3,1/3tW−4/3,1/306=−q−1W−4/3,1/3qtW−4/3,1/30.Each of (426)–(428) we take to be our MADE under study for this example. We take the classical analogue of our MADE to be the ODE obtained by setting q=1 in the original MADE. In our case, the classical analogue of (426)–(428) in this example is
(429)g6t=−gt.Next, we turn to the computation of our derivatives, which will in turn lead to the initial conditions for our MADE and (by taking the limit asq⟶1+) the initial conditions of our analogous ODE (429). Relying on (417) and the fact that
(430)fμ,λ′t=−fμ+λ,λt,(which can be directly computed from (1)), one sees that
(431)W−4/3,1/3mt=e2πi/6f−4/3,1/3me−2πi/6te−2πi/6m+f−4/3,1/3mt+e−2πi/6f−4/3,1/3me2πi/6te2πi/6m=−1mf−4/3+m/3,1/3e−2πi/6te−m−12πi/6+f−4/3+m/3,1/3t+f−4/3+m/3,1/3e2πi/6tem−12πi/6.Whenm=6, we have a direct check that the MADE given in (427) holds, after noting that f−4/3+2,1/3t=−q−1f−4/3,1/3qt which follows from (441) below with ℓ=1. Evaluating (431) at t=0 yields
(432)W−4/3,1/3m0=−1me−m−12πi/6+1+em−12πi/6f−4/3+m/3,1/30=−1m1+2cosm−12π/6f−4/3+m/3,1/30=−1m1+2cosm−12π/6θq6;−1q3−7/3+m/3,where we have used the fact that
(433)fμ,λ0=∑k=−∞∞−1kqkk−μ/λ=∑k=−∞∞−1kqkμ/λq−k/λqkk/λq−k/λ=∑k=−∞∞−qμ−1/λkq2/λkk−1/2=θq2/λ;−qμ−1/λ,to move to (432). Note that the last equality in (433) follows from (10).Settingm=6ℓ+j with j∈0,1,2,⋯,5 in (432) and relying on (12), it follows that
(434)W−4/3,1/36ℓ+j0=−16ℓ+j1+2cos6ℓ+j−12π/6θq6;−1q3−7/3+6ℓ+j/3=−1j1+2cosj−12π/6θq6;−1q6ℓ−1+j−1=−1j1+2cosj−12π/6q6ℓ−1ℓ/2−qj−1ℓ−1θq6;−qj−1.From (434), we see that for j=1 the theta factor θq6;−qj−1=θq6;−1=0 by (13). Also from (414), for j=3,5 one has 1+2cosj−12π/6=0. We conclude that W−4/3,1/36ℓ+j0=0 for j=1,3,5 odd. Setting ℓ=0,j=0 in (434) gives that W−4/3,1/30=1+2cos−12π/6−q−1−1θq6;−q−1=−2qθq6;−q−1. Thus, the derivatives of all orders of the normalized function at t=0 are expressed via
(435)W−4/3,1/36ℓ+j0W−4/3,1/30=0,j=1,3,5−1ℓ1+2cosj−12π/62q3ℓ−1ℓqj−1ℓ−1−1θq6;−qj−1θq6;−q−1,j=0,2,4(436)=0,j=1,3,5−1ℓq3ℓ−1ℓq−1ℓ−1−1θq6;−q−1θq6;−q−1,j=0−1ℓq3ℓ−1ℓq1ℓ−1−1θq6;−q1θq6;−q−1,j=2−1ℓ/2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4(437)=0,j=1,3,5−1ℓq3ℓ−1ℓq−1ℓ−1−1,j=0−1ℓq3ℓ−1ℓq1ℓ−1−1−q,j=2−1ℓ/2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4where moving to the j=2 case in (437) is facilitated by the fact that −qθq6;−q−1=θq6;−q1 which is obtained from the right hand equation in (12) when q is replaced by q6 and u is replaced by −q. We remark that, from (430) and (433), the derivatives of all orders of all normalized fμ,λt (and functions constructed from the fμ,λt such as Wμ,λt) at t=0 are expressible in terms of ratios of theta functions, as is seen overtly in (436) above for the normalized Wμ,λt/Wμ,λ0.We next determine the derivatives of all orders for the ODE in (429) analogous to our MADE in (428) by taking the limits as q⟶1+ of the derivatives obtained in (437). That is
(438)g6ℓ+j0≡limq⟶1+W−4/3,1/36ℓ+j0W−4/3,1/30=0,j=1,3,5limq⟶1+−1ℓq3ℓ−1ℓq−1ℓ−1−1,j=0limq⟶1+−1ℓq3ℓ−1ℓq1ℓ−1−1−q,j=2limq⟶1+−1ℓ2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4(439)=0,j=1,3,5−1ℓ,j=0−1ℓ−1,j=2−1ℓ2−1limq⟶1+θq6;−q3θq6;−q−1=−1ℓ2−1−2=−1ℓ,j=4where we have relied on the power of the generalized q-Wallis formulas for ratios of theta functions obtained in Section 7.3 and given by (470), (482), and (485) below to evaluate the limit in the case j=4 of (438) above. More precisely, by (482) in the case at hand, we have
(440)limq⟶1+θq6;−q3θq6;−q−1=limq⟶1+θq6;−q60+3θq6;−q6−1+5=−10−1sin3π/6sin5π/6=−11/2=−2.One final observation regarding higher order derivatives of order0mod6 follows. From (430), one has
(441)f−4/3,1/36ℓt=−16ℓf−4/3+6ℓ/3,1/3t=f−4/3+2ℓ,1/3t=∑k=−∞∞−1ke−qktqkk+4/3−2ℓ/1/3=∑k=−∞∞−1k−ℓ+ℓe−qk−ℓqℓtqk−ℓ+ℓk−ℓ−ℓ+4/3/1/3=∑m=−∞∞−1m+ℓe−qmqℓtqm+ℓm−ℓ+4/3/1/3=−1ℓ∑m=−∞∞−1me−qmqℓtqm2−ℓ2+4/3m+4/3ℓ/1/3=−1ℓqℓ2−4ℓ/3/1/3∑m=−∞∞−1me−qmqℓtqmm+4/3/1/3=−1ℓqℓℓ−4/3/1/3f−4/3,1/3qℓt.


### 7.2. Convergence

We now are prepared to prove thatft=W−4/3,1/3t/W−4/3,1/30 satisfying the MADE f6t=−q−1fqt converges as q⟶1+ to the solution gt of the analogous ODE g6t=−gt having initial conditions given by (439) with ℓ=0. That is, the initial conditions are given by
(442)g0=1,g10=0,g20=−1,g30=0,g40=1,g50=0.The unique ODE solution satisfying (442) is given by gt=cost, by inspection. Convergence of ft to gt=cost will be in the sense of uniform convergence on all compact subsets S of ℝ. The proof will hinge on three factors (for K sufficiently large): (1) proximity of ft to the Taylor polynomial PKft on S; (2) proximity of gt to the Taylor polynomial PKgt on S; and (3) proximity of PKft to PKgt on S. These in turn then force proximity of ft to gt.Proposition 45.
For any compact setS in ℝ, ft=W−4/3,1/3t/W−4/3,1/30 converges uniformly to gt=cost on S as q⟶1+.Proof.
First, the6N+5-degree Taylor polynomials P6N+5gt,P6N+5ft of g and f, respectively, expanded about t=0 are given by
(443)P6N+5gt=∑n=06N+5gn0n!tn=∑k=0Ng6k06k!t6k+∑k=0Ng6k+206k+2!t6k+2+∑k=0Ng6k+406k+4!t6k+4=∑k=0N−1k6k!t6k+∑k=0N−1k+16k+2!t6k+2+∑k=0N−1k6k+4!t6k+4(444)P6N+5ft=∑n=06N+5fn0n!tn=∑k=0Nf6k06k!t6k+∑k=0Nf6k+206k+2!t6k+2+∑k=0Nf6k+406k+4!t6k+4=∑k=0N−1kq3kk−1q−1k−1−16k!t6k(445)+∑k=0N−1kq3kk−1qk−1−1−q6k+2!t6k+2(446)+∑k=0N−1kq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−16k+4!t6k+4,where (443) follows from (439), and (444)–(446) follow from (437). These have respective remainder terms
(447)R6N+5gt=g6N+6ξ6N+6!t6N+6=−1N+1gξ6N+6!t6N+6,
for someξ between 0 and t, and
(448)R6N+5ft=f6N+6ζ6N+6!t6N+6=−1N+1qN+1N+1−4/3/1/3fqN+1ζ6N+6!t6N+6,
for someζ between 0 and t, where (448) follows from (441). Given a compact set S in ℝ there is a ρ>0 such that S⊆−ρ,ρ, and thus it is sufficient to prove uniform convergence on −ρ,ρ. For t∈−ρ,ρ, from the triangle inequality one has
(449)ft−gt≤ft−P6N+5ft+P6N+5ft−P6N+5gt+P6N+5gt−gt=R6N+5ft+P6N+5ft−P6N+5gt+R6N+5gt.
Now fort∈−ρ,ρ and relying on (448), one sees
(450)R6N+5ft=−1N+1qN+1N+1−4/3/1/3fqN+1ζ6N+6!t6N+6≤qN+1N+1−4/3/1/3ρ6N+66N+6!fqN+1ζ(451)≤qN+1N+1−4/3/1/3ρ6N+66N+6!1,where moving from (450) to (451) is obtained from the global bound (424). Similarly, from (447), one has
(452)R6N+5gt=−1N+1gξ6N+6!t6N+6≤ρ6N+66N+6!,as gξ=cosξ≤1. Also, from (443) and (444)–(446) and for N≥2, we have
(453)P6N+5ft−P6N+5gt=∑k=0N−1kq3kk−1q−1k−1−1−16k!t6k+∑k=0N−1k+1q3kk−1qk−1−16k+2!t6k+2+∑k=0N−1kq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−16k+4!t6k+4≤∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−16k+4!ρ6k+4≤∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−q3kk−1q3k−1−16k+4!ρ6k+4+∑k=0Nq3kk−1q3k−1−1−16k+4!ρ6k+4,which, after rearranging and factoring, equals
(454)∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1−16k+4!ρ6k+4+θq6;−q3−2θq6;−q−1−1∑k=0Nq3kk−1q3k−1−16k+4!ρ6k+4≤q3NN−1q3N−1−1−1∑k=0Nρ6k6k!+∑k=0Nρ6k+26k+2!+∑k=0Nρ6k+46k+4!+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1∑k=0Nρ6k+46k+4!≤q3NN−1q3N−1−1−1eρ+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1eρ.
Applying (451), (454), and (452), respectively, to the terms in (449), one has that for each N≥2(455)ft−gt≤qN+1N+1−4/3/1/3ρ6N+66N+6!1+q3NN−1q3N−1−1−1eρ(456)+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1eρ+ρ6N+66N+6!.
Givenε>0, choose N0≥2 sufficiently large such that one has ρ6N0+6/6N0+6!<ε/4 to begin bounding the second term in (456). To handle the terms in (455), note that
(457)1<ε/46N0+6!/ρ6N0+6and,automatically,1<1+ε/4eρ,and pick q0>1 so that
(458)q0N0+1N0+1−4/3/1/3<ε/46N0+6!ρ6N0+6,(459)q03N0N0−1q03N0−1−1<1+ε4eρ
Then, for1<q<q0, one has
(460)qN0+1N0+1−4/3/1/3<ε/46N0+6!ρ6N0+6,(461)q3N0N0−1q3N0−1−1<1+ε4eρ.
Hence, for1<q<q0(462)ρ6N0+66N0+6!<qN0+1N0+1−4/3/1/3ρ6N0+66N0+6!<ε/4,and
(463)q3N0N0−1q3N0−1−1−1eρ<ε/4,where (462)–(463) control the terms in (455) as well as the last term in (456). To handle the remaining term in (456), pick q1 with 1<q1<q0 such that for all 1<q<q1 one has
(464)θq6;−q3−2θq6;−q−1−1<ε/4q03N0N0−1q03N0−1−1eρ,which follows from the fact that
(465)limq⟶1+θq6;−q3−2θq6;−q−1=1,via (440) above and/or the generalized q-Wallis formula (482) below (in the case that m=6,K=−1, and J=3). Then for 1<q<q1, it follows that
(466)θq6;−q3−2θq6;−q−1−1q3N0N0−1q3N0−1−1eρ<θq6;−q3−2θq6;−q−1−1q03N0N0−1q03N0−1−1eρ<ε4.
Applying (462) and (463) along with (466) to (455) and (456) with N taken to be N0 one has that for 1<q<q1(467)ft−gt<ε4+ε4+ε4+ε4=ε,
fort∈−ρ,ρ. So ft=W−4/3,1/3t/W−4/3,1/30 approaches gt=cost uniformly on −ρ,ρ as q⟶1+, and the proposition is proven.


### 7.3. Generalizedq-Wallis Formulas

In [6], we have proven a q-Wallis formula given by the first equality in (468)
(468)limq⟶1+lnqμq23θq2,−1/q=π2=∏n=1∞2n22n−12n+1=2123⋅4345⋅6567⋯,where the last two equalities of (468) are Wallis’ formula for π/2. We now finish the paper by generalizing the above result in order to provide a number of related generalized q-Wallis formulas.Theorem 46.
Letm∈ℕ with m≥2, and let j∈0,1,⋯,m−1 with k∈1,2,⋯,m−1. Then the following families of generalized q-Wallis formulas hold:
(469)limq⟶1+lnqμqm3θqm,−qk=−π/msinkπ/m,(470)limq⟶1+θqm,−qjθqm,−qk=sinjπ/msinkπ/m.Proof.
The proof relies on the following factorization ofsinx/x:
(471)sinxx=∏n=1∞1−x2π2n2.
From (10) and (11), one has
(472)lnqμqm3θqm,−qk=lnqμqm∏n≥11−1/qmn1−1/qmnμqm∏n≥01−qk/qmn1−1/qkqmn+1=lnq∏n≥11−1/qmn1−1/qmn1−qk∏n≥01−qk/qmn+11−1/qkqmn+1=lnq∏n≥11−1/qmn1−1/qmn−1q−1∑ℓ=0k−1qℓ∏n≥11−1/qmn−k1−1/qmn+k=−lnqq−1∑ℓ=0k−1qℓ∏n≥11/q2mnqmn−1qmn−11/q2mnqmn−k−1qmn+k−1=−lnqq−1∑ℓ=0k−1qℓ∏n≥1q−12∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓq−12∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ=−lnqq−1∑ℓ=0k−1qℓ∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ,where we have reindexed replacing n+1 by n in the denominator to obtain (472). Relying on the facts that limq⟶1+lnq/q−1=1 and limq⟶1+∑ℓ=0pqℓ=∑ℓ=0p1=p+1, one sees (since k∈1,2,⋯,m−1) that
(473)limq⟶1+lnqμqm3θqm,−qk=limq⟶1+−lnqq−1∑ℓ=0k−1qℓ∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ(474)=−1k∏n≥1mn2mn−kmn+k=−1k∏n≥1mn2mn2−k2(475)=−1k∏n≥111−k/m2/n2=−1k1∏n≥11−kπ/m2/π2n2(476)=−1kkπ/msinkπ/m=−π/msinkπ/m,
giving (469), where (471) was used to obtain the first equality in (476). Now when j=0 (470) holds since θqm;−1=0 (which in turn follows from (13)). For j∈1,2,⋯,m−1, one now has from (469) that
(477)limq⟶1+θqm,−qjθqm,−qk=limq⟶1+θqm,−qjlnqμqm3lnqμqm3θqm,−qk=sinjπ/m−π/m−π/msinkπ/m=sinjπ/msinkπ/m,giving (470). This finishes the proof of the theorem.Remark 47.
We call the above results generalizedq-Wallis formulas via the following reasoning. First, note that the left-most infinite product
(478)∏n≥1mn2mn−kmn+k
in (474) generalizes the infinite product
(479)∏n=1∞2n22n−12n+1,
in the Wallis formula forπ/2 in (468). In particular note that when m=2 and k=1 the infinite product in (478) becomes the infinite product in (479), and the Wallis product for π/2 is then duplicated in (474)–(476). Second, note that the product
(480)∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ
in (473) is the q-analogue of the product (478). Hence, we are introducing generalized q-Wallis formulas.Corollary 48.
Letm∈ℕ satisfy m≥2. Let J=mL+j∈ℤ with j∈0,1,⋯,m−1 and K=mℓ+k∈ℤ with k∈1,⋯,m−1 then
(481)limq⟶1+lnqμqm3θqm,−qK=−1ℓ+1π/msinkπ/m,(482)limq⟶1+θqm,−qJθqm,−qK=−1L+ℓsinjπ/msinkπ/m.Proof.
Relying on (12), one has
(483)limq⟶1+lnqμqm3θqm,−qK=limq⟶1+lnqμqm3θqm,−qmℓ+k=limq⟶1+lnqμqm3qmℓℓ+1/2−qkℓθqm,−qk=−1ℓ−π/msinkπ/m,
giving (481), where the last equality follows from (469). If j=0 then (482) holds since then θqm;−qmL=0 by (13). If on the other hand j∈1,2,⋯,m−1 then from (481)
(484)limq⟶1+θqm,−qJθqm,−qK=limq⟶1+θqm,−qJlnqμqm3lnqμqm3θqm,−qK=sinjπ/m−1L+1π/m−1ℓ+1π/msinkπ/m=−1L+ℓsinjπ/msinkπ/m,
giving (482). Observe that when m=2 and K=−1=2−1+1=mℓ+j, then setting m=2, K=−1, ℓ=−1, and k=1 in (481) recovers the q-Wallis limit in (468). This proves the corollary.We point out that in the case of rational exponentslimq⟶1+θqm/n;−qa/b/θqm/n;−qc/d can be evaluated by obtaining a common denominator N for the fractions in the exponents and reexpressing them as m/n=M/N,a/b=A/N, and c/d=C/N to obtain:Corollary 49.
In the case of rational exponents(485)limq⟶1+θqm/n;−qa/bθqm/n;−qc/d=limq⟶1+θqM/N;−qA/NθqM/N;−qC/N=limQ⟶1+θQM;−QAθQM;−QC=−1L+ℓsinjπ/Msinkπ/M,where Q=q1/N and A=ML+j and C=Mℓ+k with j,k∈0,1,⋯,M−1 and k≠0.Proof.
One applies Corollary48 with A=ML+j and C=Mℓ+k with j,k∈0,1,⋯,M−1 and k≠0 to obtain the last equation in (485). This proves the corollary.We point out that, in our setting, important and useful applications of these generalizedq-Wallis formulas (especially those of type (470), (482), and (485)) occur in proving
(1)
initial conditions for a classical ODE analogous to a given MADE as seen earlier in (439) and (440)(2)
convergence of MADE solutions to their classical analogous ODE as seen in (464) and (465) in Proposition 45


## 7.1. Preliminaries

Forb≥0, the reproducing kernel computation relevant to our setting is
(413)∫−∞∞W−1,2/3tW−1,2/3t−bdt(414)=−3iμq342μq62⋅∑ℓ=02exp2πiℓ+πi6−1f−4/3,1/3−iexp2πiℓ+iπ6b(415)=−3iμq342μq62−i⋅e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b(416)=−3iμq342μq62−iW−4/3,1/3b.Moving from (413) to (414) is accomplished via Plancherel’s theorem where the expressions for the Fourier transform of W−1,2/3 are given by Theorem 23. The resulting integral of Fourier transforms is evaluated with a residue computation in the upper half-plane (for b≥0) similar in nature to the computation of reproducing kernels for b≥0 in [4]. This computation is lengthy, and the details are left as part of a more general set of reproducing kernel computations in an upcoming work. A direct computation moves one from line (414) to (415), while equation (132) of Definition 16 gives for b≥0 that
(417)W−4/3,1/3b=e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b,which justifies movement from (415) to (416).Settingb=0 in (413)–(416) yields
(418)∫−∞∞W−1,2/3tW−1,2/3tdt=W−1,2/3t22=−3iμq342μq62−iW−4/3,1/30,which in turn gives the functional identity
(419)1W−4/3,1/30=−3μq342μq62W−1,2/3t22.Normalizing (413)–(416) by the squared L2 norm and applying the functional identity in (419), one observes that
(420)1W−1,2/3t22∫−∞∞W−1,2/3tW−1,2/3t−bdt=−3μq342μq62W−1,2/3t22W−4/3,1/3b=W−4/3,1/3bW−4/3,1/30.Recalling thatW−4/3,1/3t is real valued, and applying Cauchy-Schwartz to (420) yields
(421)1=W−1,2/3tW−1,2/3t22W−1,2/3t−bW−1,2/3t22≥1W−1,2/3t22∫−∞∞W−1,2/3tW−1,2/3t−bdt=W−4/3,1/3bW−4/3,1/30,which gives a unit global bound on the normalization W−4/3,1/3b/W−4/3,1/30 independent of q for all b≥0. Now, equation (132) of Definition 16 also gives for b<0 that
(422)W−4/3,1/3b=e−4πi/6−1f−4/3,1/3e4πi/6b+e−6πi/6−1f−4/3,1/3e6πi/6b+e−8πi/6−1f−4/3,1/3e8πi/6b=e2πi/6f−4/3,1/3e−2πi/6−1b+f−4/3,1/3−1b+e−2πi/6f−4/3,1/3e2πi/6−1b=e2πi/6f−4/3,1/3e−2πi/6b+f−4/3,1/3b+e−2πi/6f−4/3,1/3e2πi/6b(423)=W−4/3,1/3b,where comparison of (422) with (417) allows movement to (423). So W−4/3,1/3b is an even Schwartz function and consequently the bound in (421) extends to a global bound: for all b∈ℝ and for each q>1, one has
(424)1≥W−4/3,1/3bW−4/3,1/30.From (415) and (416), one sees that W−4/3,1/3t satisfies the same MADE as does f−4/3,1/3t, where μ=−4/3 and λ=1/3. From these values, we deduce that
(425)μ+12=−4/3+12=−1/32=−16=αβ,λ2=1/32=16=γδ,by which one sees thatα=−1,β=6,γ=1,δ=6. From these parameter values, one sees that W−4/3,1/3t falls in the example class [M=δ=6,β=6] with (356) holding. Since δ=6, we conclude from Proposition 43 that W−4/3,1/3t is not a canonical extension of f−4/3,1/3t. See also Table 1 where W−4/3,1/3t fall in the category of row δ=6,γ^=1,ℓ+1=1,5,6,+=0,4,5, and −=1,2,3. From (2), one determines the MADE for f−4/3,1/3t to be
(426)f−4/3,1/36t=−11+6q11+−4/3/1/3f−4/3,1/3q1t=−q−1f−4/3,1/3qt.And hence,(427)W−4/3,1/36t=−q−1W−4/3,1/3qt,while(428)W−4/3,1/3tW−4/3,1/306=−q−1W−4/3,1/3qtW−4/3,1/30.Each of (426)–(428) we take to be our MADE under study for this example. We take the classical analogue of our MADE to be the ODE obtained by setting q=1 in the original MADE. In our case, the classical analogue of (426)–(428) in this example is
(429)g6t=−gt.Next, we turn to the computation of our derivatives, which will in turn lead to the initial conditions for our MADE and (by taking the limit asq⟶1+) the initial conditions of our analogous ODE (429). Relying on (417) and the fact that
(430)fμ,λ′t=−fμ+λ,λt,(which can be directly computed from (1)), one sees that
(431)W−4/3,1/3mt=e2πi/6f−4/3,1/3me−2πi/6te−2πi/6m+f−4/3,1/3mt+e−2πi/6f−4/3,1/3me2πi/6te2πi/6m=−1mf−4/3+m/3,1/3e−2πi/6te−m−12πi/6+f−4/3+m/3,1/3t+f−4/3+m/3,1/3e2πi/6tem−12πi/6.Whenm=6, we have a direct check that the MADE given in (427) holds, after noting that f−4/3+2,1/3t=−q−1f−4/3,1/3qt which follows from (441) below with ℓ=1. Evaluating (431) at t=0 yields
(432)W−4/3,1/3m0=−1me−m−12πi/6+1+em−12πi/6f−4/3+m/3,1/30=−1m1+2cosm−12π/6f−4/3+m/3,1/30=−1m1+2cosm−12π/6θq6;−1q3−7/3+m/3,where we have used the fact that
(433)fμ,λ0=∑k=−∞∞−1kqkk−μ/λ=∑k=−∞∞−1kqkμ/λq−k/λqkk/λq−k/λ=∑k=−∞∞−qμ−1/λkq2/λkk−1/2=θq2/λ;−qμ−1/λ,to move to (432). Note that the last equality in (433) follows from (10).Settingm=6ℓ+j with j∈0,1,2,⋯,5 in (432) and relying on (12), it follows that
(434)W−4/3,1/36ℓ+j0=−16ℓ+j1+2cos6ℓ+j−12π/6θq6;−1q3−7/3+6ℓ+j/3=−1j1+2cosj−12π/6θq6;−1q6ℓ−1+j−1=−1j1+2cosj−12π/6q6ℓ−1ℓ/2−qj−1ℓ−1θq6;−qj−1.From (434), we see that for j=1 the theta factor θq6;−qj−1=θq6;−1=0 by (13). Also from (414), for j=3,5 one has 1+2cosj−12π/6=0. We conclude that W−4/3,1/36ℓ+j0=0 for j=1,3,5 odd. Setting ℓ=0,j=0 in (434) gives that W−4/3,1/30=1+2cos−12π/6−q−1−1θq6;−q−1=−2qθq6;−q−1. Thus, the derivatives of all orders of the normalized function at t=0 are expressed via
(435)W−4/3,1/36ℓ+j0W−4/3,1/30=0,j=1,3,5−1ℓ1+2cosj−12π/62q3ℓ−1ℓqj−1ℓ−1−1θq6;−qj−1θq6;−q−1,j=0,2,4(436)=0,j=1,3,5−1ℓq3ℓ−1ℓq−1ℓ−1−1θq6;−q−1θq6;−q−1,j=0−1ℓq3ℓ−1ℓq1ℓ−1−1θq6;−q1θq6;−q−1,j=2−1ℓ/2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4(437)=0,j=1,3,5−1ℓq3ℓ−1ℓq−1ℓ−1−1,j=0−1ℓq3ℓ−1ℓq1ℓ−1−1−q,j=2−1ℓ/2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4where moving to the j=2 case in (437) is facilitated by the fact that −qθq6;−q−1=θq6;−q1 which is obtained from the right hand equation in (12) when q is replaced by q6 and u is replaced by −q. We remark that, from (430) and (433), the derivatives of all orders of all normalized fμ,λt (and functions constructed from the fμ,λt such as Wμ,λt) at t=0 are expressible in terms of ratios of theta functions, as is seen overtly in (436) above for the normalized Wμ,λt/Wμ,λ0.We next determine the derivatives of all orders for the ODE in (429) analogous to our MADE in (428) by taking the limits as q⟶1+ of the derivatives obtained in (437). That is
(438)g6ℓ+j0≡limq⟶1+W−4/3,1/36ℓ+j0W−4/3,1/30=0,j=1,3,5limq⟶1+−1ℓq3ℓ−1ℓq−1ℓ−1−1,j=0limq⟶1+−1ℓq3ℓ−1ℓq1ℓ−1−1−q,j=2limq⟶1+−1ℓ2−1q3ℓ−1ℓq3ℓ−1−1θq6;−q3θq6;−q−1,j=4(439)=0,j=1,3,5−1ℓ,j=0−1ℓ−1,j=2−1ℓ2−1limq⟶1+θq6;−q3θq6;−q−1=−1ℓ2−1−2=−1ℓ,j=4where we have relied on the power of the generalized q-Wallis formulas for ratios of theta functions obtained in Section 7.3 and given by (470), (482), and (485) below to evaluate the limit in the case j=4 of (438) above. More precisely, by (482) in the case at hand, we have
(440)limq⟶1+θq6;−q3θq6;−q−1=limq⟶1+θq6;−q60+3θq6;−q6−1+5=−10−1sin3π/6sin5π/6=−11/2=−2.One final observation regarding higher order derivatives of order0mod6 follows. From (430), one has
(441)f−4/3,1/36ℓt=−16ℓf−4/3+6ℓ/3,1/3t=f−4/3+2ℓ,1/3t=∑k=−∞∞−1ke−qktqkk+4/3−2ℓ/1/3=∑k=−∞∞−1k−ℓ+ℓe−qk−ℓqℓtqk−ℓ+ℓk−ℓ−ℓ+4/3/1/3=∑m=−∞∞−1m+ℓe−qmqℓtqm+ℓm−ℓ+4/3/1/3=−1ℓ∑m=−∞∞−1me−qmqℓtqm2−ℓ2+4/3m+4/3ℓ/1/3=−1ℓqℓ2−4ℓ/3/1/3∑m=−∞∞−1me−qmqℓtqmm+4/3/1/3=−1ℓqℓℓ−4/3/1/3f−4/3,1/3qℓt.


## 7.2. Convergence

We now are prepared to prove thatft=W−4/3,1/3t/W−4/3,1/30 satisfying the MADE f6t=−q−1fqt converges as q⟶1+ to the solution gt of the analogous ODE g6t=−gt having initial conditions given by (439) with ℓ=0. That is, the initial conditions are given by
(442)g0=1,g10=0,g20=−1,g30=0,g40=1,g50=0.The unique ODE solution satisfying (442) is given by gt=cost, by inspection. Convergence of ft to gt=cost will be in the sense of uniform convergence on all compact subsets S of ℝ. The proof will hinge on three factors (for K sufficiently large): (1) proximity of ft to the Taylor polynomial PKft on S; (2) proximity of gt to the Taylor polynomial PKgt on S; and (3) proximity of PKft to PKgt on S. These in turn then force proximity of ft to gt.Proposition 45.
For any compact setS in ℝ, ft=W−4/3,1/3t/W−4/3,1/30 converges uniformly to gt=cost on S as q⟶1+.Proof.
First, the6N+5-degree Taylor polynomials P6N+5gt,P6N+5ft of g and f, respectively, expanded about t=0 are given by
(443)P6N+5gt=∑n=06N+5gn0n!tn=∑k=0Ng6k06k!t6k+∑k=0Ng6k+206k+2!t6k+2+∑k=0Ng6k+406k+4!t6k+4=∑k=0N−1k6k!t6k+∑k=0N−1k+16k+2!t6k+2+∑k=0N−1k6k+4!t6k+4(444)P6N+5ft=∑n=06N+5fn0n!tn=∑k=0Nf6k06k!t6k+∑k=0Nf6k+206k+2!t6k+2+∑k=0Nf6k+406k+4!t6k+4=∑k=0N−1kq3kk−1q−1k−1−16k!t6k(445)+∑k=0N−1kq3kk−1qk−1−1−q6k+2!t6k+2(446)+∑k=0N−1kq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−16k+4!t6k+4,where (443) follows from (439), and (444)–(446) follow from (437). These have respective remainder terms
(447)R6N+5gt=g6N+6ξ6N+6!t6N+6=−1N+1gξ6N+6!t6N+6,
for someξ between 0 and t, and
(448)R6N+5ft=f6N+6ζ6N+6!t6N+6=−1N+1qN+1N+1−4/3/1/3fqN+1ζ6N+6!t6N+6,
for someζ between 0 and t, where (448) follows from (441). Given a compact set S in ℝ there is a ρ>0 such that S⊆−ρ,ρ, and thus it is sufficient to prove uniform convergence on −ρ,ρ. For t∈−ρ,ρ, from the triangle inequality one has
(449)ft−gt≤ft−P6N+5ft+P6N+5ft−P6N+5gt+P6N+5gt−gt=R6N+5ft+P6N+5ft−P6N+5gt+R6N+5gt.
Now fort∈−ρ,ρ and relying on (448), one sees
(450)R6N+5ft=−1N+1qN+1N+1−4/3/1/3fqN+1ζ6N+6!t6N+6≤qN+1N+1−4/3/1/3ρ6N+66N+6!fqN+1ζ(451)≤qN+1N+1−4/3/1/3ρ6N+66N+6!1,where moving from (450) to (451) is obtained from the global bound (424). Similarly, from (447), one has
(452)R6N+5gt=−1N+1gξ6N+6!t6N+6≤ρ6N+66N+6!,as gξ=cosξ≤1. Also, from (443) and (444)–(446) and for N≥2, we have
(453)P6N+5ft−P6N+5gt=∑k=0N−1kq3kk−1q−1k−1−1−16k!t6k+∑k=0N−1k+1q3kk−1qk−1−16k+2!t6k+2+∑k=0N−1kq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−16k+4!t6k+4≤∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−16k+4!ρ6k+4≤∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1θq6;−q3/−2θq6;−q−1−q3kk−1q3k−1−16k+4!ρ6k+4+∑k=0Nq3kk−1q3k−1−1−16k+4!ρ6k+4,which, after rearranging and factoring, equals
(454)∑k=0Nq3kk−1q−1k−1−1−16k!ρ6k+∑k=0Nq3kk−1qk−1−16k+2!ρ6k+2+∑k=0Nq3kk−1q3k−1−1−16k+4!ρ6k+4+θq6;−q3−2θq6;−q−1−1∑k=0Nq3kk−1q3k−1−16k+4!ρ6k+4≤q3NN−1q3N−1−1−1∑k=0Nρ6k6k!+∑k=0Nρ6k+26k+2!+∑k=0Nρ6k+46k+4!+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1∑k=0Nρ6k+46k+4!≤q3NN−1q3N−1−1−1eρ+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1eρ.
Applying (451), (454), and (452), respectively, to the terms in (449), one has that for each N≥2(455)ft−gt≤qN+1N+1−4/3/1/3ρ6N+66N+6!1+q3NN−1q3N−1−1−1eρ(456)+θq6;−q3−2θq6;−q−1−1q3NN−1q3N−1−1eρ+ρ6N+66N+6!.
Givenε>0, choose N0≥2 sufficiently large such that one has ρ6N0+6/6N0+6!<ε/4 to begin bounding the second term in (456). To handle the terms in (455), note that
(457)1<ε/46N0+6!/ρ6N0+6and,automatically,1<1+ε/4eρ,and pick q0>1 so that
(458)q0N0+1N0+1−4/3/1/3<ε/46N0+6!ρ6N0+6,(459)q03N0N0−1q03N0−1−1<1+ε4eρ
Then, for1<q<q0, one has
(460)qN0+1N0+1−4/3/1/3<ε/46N0+6!ρ6N0+6,(461)q3N0N0−1q3N0−1−1<1+ε4eρ.
Hence, for1<q<q0(462)ρ6N0+66N0+6!<qN0+1N0+1−4/3/1/3ρ6N0+66N0+6!<ε/4,and
(463)q3N0N0−1q3N0−1−1−1eρ<ε/4,where (462)–(463) control the terms in (455) as well as the last term in (456). To handle the remaining term in (456), pick q1 with 1<q1<q0 such that for all 1<q<q1 one has
(464)θq6;−q3−2θq6;−q−1−1<ε/4q03N0N0−1q03N0−1−1eρ,which follows from the fact that
(465)limq⟶1+θq6;−q3−2θq6;−q−1=1,via (440) above and/or the generalized q-Wallis formula (482) below (in the case that m=6,K=−1, and J=3). Then for 1<q<q1, it follows that
(466)θq6;−q3−2θq6;−q−1−1q3N0N0−1q3N0−1−1eρ<θq6;−q3−2θq6;−q−1−1q03N0N0−1q03N0−1−1eρ<ε4.
Applying (462) and (463) along with (466) to (455) and (456) with N taken to be N0 one has that for 1<q<q1(467)ft−gt<ε4+ε4+ε4+ε4=ε,
fort∈−ρ,ρ. So ft=W−4/3,1/3t/W−4/3,1/30 approaches gt=cost uniformly on −ρ,ρ as q⟶1+, and the proposition is proven.


## 7.3. Generalizedq-Wallis Formulas

In [6], we have proven a q-Wallis formula given by the first equality in (468)
(468)limq⟶1+lnqμq23θq2,−1/q=π2=∏n=1∞2n22n−12n+1=2123⋅4345⋅6567⋯,where the last two equalities of (468) are Wallis’ formula for π/2. We now finish the paper by generalizing the above result in order to provide a number of related generalized q-Wallis formulas.Theorem 46.
Letm∈ℕ with m≥2, and let j∈0,1,⋯,m−1 with k∈1,2,⋯,m−1. Then the following families of generalized q-Wallis formulas hold:
(469)limq⟶1+lnqμqm3θqm,−qk=−π/msinkπ/m,(470)limq⟶1+θqm,−qjθqm,−qk=sinjπ/msinkπ/m.Proof.
The proof relies on the following factorization ofsinx/x:
(471)sinxx=∏n=1∞1−x2π2n2.
From (10) and (11), one has
(472)lnqμqm3θqm,−qk=lnqμqm∏n≥11−1/qmn1−1/qmnμqm∏n≥01−qk/qmn1−1/qkqmn+1=lnq∏n≥11−1/qmn1−1/qmn1−qk∏n≥01−qk/qmn+11−1/qkqmn+1=lnq∏n≥11−1/qmn1−1/qmn−1q−1∑ℓ=0k−1qℓ∏n≥11−1/qmn−k1−1/qmn+k=−lnqq−1∑ℓ=0k−1qℓ∏n≥11/q2mnqmn−1qmn−11/q2mnqmn−k−1qmn+k−1=−lnqq−1∑ℓ=0k−1qℓ∏n≥1q−12∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓq−12∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ=−lnqq−1∑ℓ=0k−1qℓ∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ,where we have reindexed replacing n+1 by n in the denominator to obtain (472). Relying on the facts that limq⟶1+lnq/q−1=1 and limq⟶1+∑ℓ=0pqℓ=∑ℓ=0p1=p+1, one sees (since k∈1,2,⋯,m−1) that
(473)limq⟶1+lnqμqm3θqm,−qk=limq⟶1+−lnqq−1∑ℓ=0k−1qℓ∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ(474)=−1k∏n≥1mn2mn−kmn+k=−1k∏n≥1mn2mn2−k2(475)=−1k∏n≥111−k/m2/n2=−1k1∏n≥11−kπ/m2/π2n2(476)=−1kkπ/msinkπ/m=−π/msinkπ/m,
giving (469), where (471) was used to obtain the first equality in (476). Now when j=0 (470) holds since θqm;−1=0 (which in turn follows from (13)). For j∈1,2,⋯,m−1, one now has from (469) that
(477)limq⟶1+θqm,−qjθqm,−qk=limq⟶1+θqm,−qjlnqμqm3lnqμqm3θqm,−qk=sinjπ/m−π/m−π/msinkπ/m=sinjπ/msinkπ/m,giving (470). This finishes the proof of the theorem.Remark 47.
We call the above results generalizedq-Wallis formulas via the following reasoning. First, note that the left-most infinite product
(478)∏n≥1mn2mn−kmn+k
in (474) generalizes the infinite product
(479)∏n=1∞2n22n−12n+1,
in the Wallis formula forπ/2 in (468). In particular note that when m=2 and k=1 the infinite product in (478) becomes the infinite product in (479), and the Wallis product for π/2 is then duplicated in (474)–(476). Second, note that the product
(480)∏n≥1∑ℓ=0mn−1qℓ∑ℓ=0mn−1qℓ∑ℓ=0mn−k−1qℓ∑ℓ=0mn+k−1qℓ
in (473) is the q-analogue of the product (478). Hence, we are introducing generalized q-Wallis formulas.Corollary 48.
Letm∈ℕ satisfy m≥2. Let J=mL+j∈ℤ with j∈0,1,⋯,m−1 and K=mℓ+k∈ℤ with k∈1,⋯,m−1 then
(481)limq⟶1+lnqμqm3θqm,−qK=−1ℓ+1π/msinkπ/m,(482)limq⟶1+θqm,−qJθqm,−qK=−1L+ℓsinjπ/msinkπ/m.Proof.
Relying on (12), one has
(483)limq⟶1+lnqμqm3θqm,−qK=limq⟶1+lnqμqm3θqm,−qmℓ+k=limq⟶1+lnqμqm3qmℓℓ+1/2−qkℓθqm,−qk=−1ℓ−π/msinkπ/m,
giving (481), where the last equality follows from (469). If j=0 then (482) holds since then θqm;−qmL=0 by (13). If on the other hand j∈1,2,⋯,m−1 then from (481)
(484)limq⟶1+θqm,−qJθqm,−qK=limq⟶1+θqm,−qJlnqμqm3lnqμqm3θqm,−qK=sinjπ/m−1L+1π/m−1ℓ+1π/msinkπ/m=−1L+ℓsinjπ/msinkπ/m,
giving (482). Observe that when m=2 and K=−1=2−1+1=mℓ+j, then setting m=2, K=−1, ℓ=−1, and k=1 in (481) recovers the q-Wallis limit in (468). This proves the corollary.We point out that in the case of rational exponentslimq⟶1+θqm/n;−qa/b/θqm/n;−qc/d can be evaluated by obtaining a common denominator N for the fractions in the exponents and reexpressing them as m/n=M/N,a/b=A/N, and c/d=C/N to obtain:Corollary 49.
In the case of rational exponents(485)limq⟶1+θqm/n;−qa/bθqm/n;−qc/d=limq⟶1+θqM/N;−qA/NθqM/N;−qC/N=limQ⟶1+θQM;−QAθQM;−QC=−1L+ℓsinjπ/Msinkπ/M,where Q=q1/N and A=ML+j and C=Mℓ+k with j,k∈0,1,⋯,M−1 and k≠0.Proof.
One applies Corollary48 with A=ML+j and C=Mℓ+k with j,k∈0,1,⋯,M−1 and k≠0 to obtain the last equation in (485). This proves the corollary.We point out that, in our setting, important and useful applications of these generalizedq-Wallis formulas (especially those of type (470), (482), and (485)) occur in proving
(1)
initial conditions for a classical ODE analogous to a given MADE as seen earlier in (439) and (440)(2)
convergence of MADE solutions to their classical analogous ODE as seen in (464) and (465) in Proposition 45



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*Source: 6721360-2022-07-07.xml*