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Research article

New double-sum expansions for certain Mock theta functions

  • Received: 06 June 2022 Revised: 18 July 2022 Accepted: 19 July 2022 Published: 22 July 2022
  • MSC : Primary 05A30, 30C45; Secondary 11B65, 47B38

  • The study of expansions of certain mock theta functions in special functions theory has a long and quite significant history. Motivated by recent correlations between q-series and mock theta functions, we establish a new q-series transformation formula and derive the double-sum expansions for mock theta functions. As an application, we state new double-sum representations for certain mock theta functions.

    Citation: Qiuxia Hu, Bilal Khan, Serkan Araci, Mehmet Acikgoz. New double-sum expansions for certain Mock theta functions[J]. AIMS Mathematics, 2022, 7(9): 17225-17235. doi: 10.3934/math.2022948

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  • The study of expansions of certain mock theta functions in special functions theory has a long and quite significant history. Motivated by recent correlations between q-series and mock theta functions, we establish a new q-series transformation formula and derive the double-sum expansions for mock theta functions. As an application, we state new double-sum representations for certain mock theta functions.



    Let A denote the class of functions f which are analytic in the open unit disk Δ={zC:|z|<1}, normalized by the conditions f(0)=f(0)1=0. So each fA has series representation of the form

    f(z)=z+n=2anzn. (1.1)

    For two analytic functions f and g, f is said to be subordinated to g (written as fg) if there exists an analytic function ω with ω(0)=0 and |ω(z)|<1 for zΔ such that f(z)=(gω)(z).

    A function fA is said to be in the class S if f is univalent in Δ. A function fS is in class C of normalized convex functions if f(Δ) is a convex domain. For 0α1, Mocanu [23] introduced the class Mα of functions fA such that f(z)f(z)z0 for all zΔ and

    ((1α)zf(z)f(z)+α(zf(z))f(z))>0(zΔ). (1.2)

    Geometrically, fMα maps the circle centred at origin onto α-convex arcs which leads to the condition (1.2). The class Mα was studied extensively by several researchers, see [1,10,11,12,24,25,26,27] and the references cited therein.

    A function fS is uniformly starlike if f maps every circular arc Γ contained in Δ with center at ζ Δ onto a starlike arc with respect to f(ζ). A function fC is uniformly convex if f maps every circular arc Γ contained in Δ with center ζ Δ onto a convex arc. We denote the classes of uniformly starlike and uniformly convex functions by UST and  UCV, respectively. For recent study on these function classes, one can refer to [7,9,13,19,20,31].

    In 1999, Kanas and Wisniowska [15] introduced the class k-UCV (k0) of k-uniformly convex functions. A function fA is said to be in the class k-UCV if it satisfies the condition

    (1+zf(z)f(z))>k|zf(z)f(z)|(zΔ). (1.3)

    In recent years, many researchers investigated interesting properties of this class and its generalizations. For more details, see [2,3,4,14,15,16,17,18,30,32,35] and references cited therein.

    In 2015, Sokół and Nunokawa [33] introduced the class MN, a function fMN if it satisfies the condition

    (1+zf(z)f(z))>|zf(z)f(z)1|(zΔ).

    In [28], it is proved that if (f)>0 in Δ, then f is univalent in Δ. In 1972, MacGregor [21] studied the class B of functions with bounded turning, a function fB if it satisfies the condition (f)>0 for zΔ. A natural generalization of the class B is B(δ1) (0δ1<1), a function fB(δ1) if it satisfies the condition

    (f(z))>δ1(zΔ;0δ1<1), (1.4)

    for details associated with the class B(δ1) (see [5,6,34]).

    Motivated essentially by the above work, we now introduce the following class k-Q(α) of analytic functions.

    Definition 1. Let k0 and 0α1. A function fA is said to be in the class k-Q(α) if it satisfies the condition

    ((zf(z))f(z))>k|(1α)f(z)+α(zf(z))f(z)1|(zΔ). (1.5)

    It is worth mentioning that, for special values of parameters, one can obtain a number of well-known function classes, some of them are listed below:

    1. k-Q(1)=k-UCV;

    2. 0-Q(α)=C.

    In what follows, we give an example for the class k-Q(α).

    Example 1. The function f(z)=z1Az(A0) is in the class k-Q(α) with

    k1b2bb(1+α)[b(1+α)+2]+4(b=|A|). (1.6)

    The main purpose of this paper is to establish several interesting relationships between k-Q(α) and the class B(δ) of functions with bounded turning.

    To prove our main results, we need the following lemmas.

    Lemma 1. ([8]) Let h be analytic in Δ with h(0)=1, β>0 and 0γ1<1. If

    h(z)+βzh(z)h(z)1+(12γ1)z1z,

    then

    h(z)1+(12δ)z1z,

    where

    δ=(2γ1β)+(2γ1β)2+8β4. (2.1)

    Lemma 2. Let h be analytic in Δ and of the form

    h(z)=1+n=mbnzn(bm0)

    with h(z)0 in Δ. If there exists a point z0(|z0|<1) such that |argh(z)|<πρ2(|z|<|z0|) and |argh(z0)|=πρ2 for some ρ>0, then z0h(z0)h(z0)=iρ, where

    :{n2(c+1c)(argh(z0)=πρ2),n2(c+1c)(argh(z0)=πρ2),

    and (h(z0))1/ρ=±ic(c>0).

    This result is a generalization of the Nunokawa's lemma [29].

    Lemma 3. ([37]) Let ε be a positive measure on [0,1]. Let ϝ be a complex-valued function defined on Δ×[0,1] such that ϝ(.,t) is analytic in Δ for each t[0,1] and ϝ(z,.) is ε-integrable on [0,1] for all zΔ. In addition, suppose that (ϝ(z,t))>0, ϝ(r,t) is real and (1/ϝ(z,t))1/ϝ(r,t) for |z|r<1 and t[0,1]. If ϝ(z)=10ϝ(z,t)dε(t), then (1/ϝ(z))1/ϝ(r).

    Lemma 4. ([22]) If 1D<C1, λ1>0 and (γ2)λ1(1C)/(1D), then the differential equation

    s(z)+zs(z)λ1s(z)+γ2=1+Cz1+Dz(zΔ)

    has a univalent solution in Δ given by

    s(z)={zλ1+γ2(1+Dz)λ1(CD)/Dλ1z0tλ1+γ21(1+Dt)λ1(CD)/Ddtγ2λ1(D0),zλ1+γ2eλ1Czλ1z0tλ1+γ21eλ1Ctdtγ2λ1(D=0).

    If r(z)=1+c1z+c2z2+ satisfies the condition

    r(z)+zr(z)λ1r(z)+γ21+Cz1+Dz(zΔ),

    then

    r(z)s(z)1+Cz1+Dz,

    and s(z) is the best dominant.

    Lemma 5. ([36,Chapter 14]) Let w, x and\ y0,1,2, be complex numbers. Then, for (y)>(x)>0, one has

    1. 2G1(w,x,y;z)=Γ(y)Γ(yx)Γ(x)10sx1(1s)yx1(1sz)wds;

    2. 2G1(w,x,y;z)= 2G1(x,w,y;z);

    3. 2G1(w,x,y;z)=(1z)w2G1(w,yx,y;zz1).

    Firstly, we derive the following result.

    Theorem 1. Let 0α<1 and k11α. If fk-Q(α), then fB(δ), where

    δ=(2μλ)+(2μλ)2+8λ4(λ=1+αkk(1α);μ=kαk1k(1α)). (3.1)

    Proof. Let f=, where is analytic in Δ with (0)=1. From inequality (1.5) which takes the form

    (1+z(z)(z))>k|(1α)(z)+α(1+z(z)(z))1|=k|1α(z)+α(z)αz(z)(z)|,

    we find that

    ((z)+1+αkk(1α)z(z)(z))>kαk1k(1α),

    which can be rewritten as

    ((z)+λz(z)(z))>μ(λ=1+αkk(1α);μ=kαk1k(1α)).

    The above relationship can be written as the following Briot-Bouquet differential subordination

    (z)+λz(z)(z)1+(12μ)z1z.

    Thus, by Lemma 1, we obtain

    1+(12δ)z1z, (3.2)

    where δ is given by (3.1). The relationship (3.2) implies that fB(δ). We thus complete the proof of Theorem 3.1.

    Theorem 2. Let 0<α1, 0<β<1, c>0, k1, nm+1(m N ), ||n2(c+1c) and

    |αβ±(1α)cβsinβπ2|1. (3.3)

    If

    f(z)=z+n=m+1anzn(am+10)

    and fk-Q(α), then fB(β0), where

    β0=min{β:β(0,1)}

    such that (3.3) holds.

    Proof. By the assumption, we have

    f(z)=(z)=1+n=mcnzn(cm0). (3.4)

    In view of (1.5) and (3.4), we get

    (1+z(z)(z))>k|(1α)(z)+α(1+z(z)(z))1|.

    If there exists a point z0Δ such that

    |arg(z)|<βπ2(|z|<|z0|;0<β<1)

    and

    |arg(z0)|=βπ2(0<β<1),

    then from Lemma 2, we know that

    z0(z0)(z0)=iβ,

    where

    ((z0))1/β=±ic(c>0)

    and

    :{n2(c+1c)(arg(z0)=βπ2),n2(c+1c)(arg(z0)=βπ2).

    For the case

    arg(z0)=βπ2,

    we get

    (1+z0(z0)(z0))=(1+iβ)=1. (3.5)

    Moreover, we find from (3.3) that

    k|(1α)(z0)+α(1+z0(z0)(z0))1|=k|(1α)((z0)1)+αz0(z0)(z0)|=k|(1α)[(±ic)β1]+iαβ|=k(1α)2(cβcosβπ21)2+[αβ±(1α)cβsinβπ2]21. (3.6)

    By virtue of (3.5) and (3.6), we have

    (1+z(z0)(z0))k|(1α)(z0)+α(1+z0(z0)(z0))1|,

    which is a contradiction to the definition of k-Q(α). Since β0=min{β:β(0,1)} such that (3.3) holds, we can deduce that fB(β0).

    By using the similar method as given above, we can prove the case

    arg(z0)=βπ2

    is true. The proof of Theorem 2 is thus completed.

    Theorem 3. If 0<β<1 and 0ν<1. If fk-Q(α), then

    (f)>[2G1(2β(1ν),1;1β+1;12)]1,

    or equivalently, k-Q(α)B(ν0), where

    ν0=[2G1(2β(1μ),1;1β+1;12)]1.

    Proof. For

    w=2β(1ν), x=1β, y=1β+1,

    we define

    ϝ(z)=(1+Dz)w10tx1(1+Dtz)wdt=Γ(x)Γ(y) 2G1(1,w,y;zz1). (3.7)

    To prove k-Q(α)B(ν0), it suffices to prove that

    inf|z|<1{(q(z))}=q(1),

    which need to show that

    (1/ϝ(z))1/ϝ(1).

    By Lemma 3 and (3.7), it follows that

    ϝ(z)=10ϝ(z,t)dε(t),

    where

    ϝ(z,t)=1z1(1t)z(0t1),

    and

    dε(t)=Γ(x)Γ(w)Γ(yw)tw1(1t)yw1dt,

    which is a positive measure on [0,1].

    It is clear that (ϝ(z,t))>0 and ϝ(r,t) is real for |z|r<1 and t[0,1]. Also

    (1ϝ(z,t))=(1(1t)z1z)1+(1t)r1+r=1ϝ(r,t)

    for |z|r<1. Therefore, by Lemma 3, we get

    (1/ϝ(z))1/ϝ(r).

    If we let r1, it follows that

    (1/ϝ(z))1/ϝ(1).

    Thus, we deduce that k-Q(α)B(ν0).

    Theorem 4. Let 0α<1 and k11α. If fk-Q(α), then

    f(z)s(z)=1g(z),

    where

    g(z)=2G1(2λ,1,1λ+1;zz1)(λ=1+αkk(1α)).

    Proof. Suppose that f=. From the proof of Theorem 1, we see that

    (z)+z(z)1λ(z)1+(12μ)z1z1+z1z(λ=1+αkk(1α);μ=kαk1k(1α)).

    If we set λ1=1λ, γ2=0, C=1 and D=1 in Lemma 4, then

    (z)s(z)=1g(z)=z1λ(1z)2λ1/λz0t(1/λ)1(1t)2/λdt.

    By putting t=uz, and using Lemma 5, we obtain

    (z)s(z)=1g(z)=11λ(1z)2λ10u(1/λ)1(1uz)2/λdu=[2G1(2λ,1,1λ+1;zz1)]1,

    which is the desired result of Theorem 4.

    The present investigation was supported by the Key Project of Education Department of Hunan Province under Grant no. 19A097 of the P. R. China. The authors would like to thank the referees for their valuable comments and suggestions, which was essential to improve the quality of this paper.

    The authors declare no conflict of interest.



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