# Q-Analogues of Symbolic Operators

Authors: Michael J. Dancs; Tian-Xiao He
Journal: Journal of Discrete Mathematics
 (2013)

---

## Abstract

Here presented areq-extensions of several linear operators including a novel q-analogue of the derivative operator D. Some q-analogues of the symbolic substitution rules given by He et al., 2007, are obtained. As sample applications, we show how these q-substitution rules may be used to construct symbolic summation and series transformation formulas, including q-analogues of the classical Euler transformations for accelerating the convergence of alternating series.


---

## Body

## 1. Definitions and Basic Identities

Unless otherwise stated, we consider all operators to act on formal power series in the single variablet, with coefficients possibly depending on q. We assume that 0<|q|<1. Issues of convergence will be addressed in a later paper.We will use1 to denote the identity operator and define the following operators:  (1)
Eqf(t)=f(tq) (forward multiplicative shift),(2)
Δqf(t)=f(tq)-f(t) (forward q-difference),(3)
Lqf(t)=t(logq)f′(t) (forward logarithmic shift).The first two of these can be regarded asq-analogues of the ordinary (additive) shift and forward difference operators, respectively. Lq will play a role similar to that of the derivative D.The operator inverse ofEq (which we denote as Eq-1) clearly exists and is equal to Eq-1. We define the central q-difference operator δq by
(1)δq=f(tq1/2)-f(tq-1/2),
and note that δq=ΔqEq-1/2=Δq/Eq1/2,δq2k=Δq2kEq-k.The previousq-operators are linear and satisfy some familiar identities, for example, Eq=1+Δq. The binomial identity
(2)Δqn=∑k=0n‍(-1)n-k(nk)Eqk
can be established by induction or by considering the operator expansion of (Eq-1)n.Treating these operators formally, we need only to consider their effect on nonnegative integer powers oft.  Eq,  Δq, and Lq are “diagonal” in the sense that each maps tk↦M(q,k)tk, with the function M depending on the particular operator. For example, Δq[tk]=(qk-1)tk for k>0, and Δq[1]=0. Similarly, Lq[tk]=tklog(qk).With this observation, it is easy to verify many additional identities. For example, consider the alternating geometric series∑n=0∞(-1)nΔqn applied to tk. We have
(3)∑n=0∞‍(-1)nΔqn[tk]=tk∑n=0∞‍(-1)n(qk-1)n=tk11-(1-qk)=tkq-k.
In other words, this formal power series gives the operator Eq-1. Stated differently,
(4)(1+Δq)-1=(Eq)-1=Eq-1=∑n=0∞‍(-1)nΔqn,
which is exactly the result we should expect. We may establish the following identities in similar fashion:
(5)(1-Δq)-1=∑n=0∞‍Δqn,(6)log(1+Δq)=∑n=1∞‍(-1)n+1nΔqn=Lq,(7)eLq=∑n=0∞‍1n!Lqn=Eq.
In addition to these last two identities, Lq obeys the product rule
(8)Lq[f(t)g(t)]=Lq[f(t)]g(t)+f(t)Lq[g(t)],
so that Lq is a q-analogue of the ordinary derivative operator D.


## 2. Main Results

We begin with someq-analogues of the symbolic substitution rules in [1] (specifically, (2.4) and (2.5)).Proposition 1.
LetF(t) have the formal power series expansion F(t)=∑k≥0fktk, with coefficients possibly dependent on q. One may obtain operational formulas according to the following rules.(1)
The substitutiont↦Eqleads to the symbolic formula(9)F(Eq)=∑k=0∞‍fkEqk.(2)
IfF(t)=G(t,et), the substitutiont↦Lqleads to(10)G(Lq,Eq)=∑k=0∞‍fkLqk.(3)
IfF(t)=G(t,log(1+t)),
the substitutiont↦Δqleads to(11)G(Δq,Lq)=∑k=0∞‍fkΔqk.Note that each of the identities in (5)–(7) can be obtained from elementary Maclaurin series by applying one of these substitution rules. We now present a less trivial example.Fork a positive integer, let αk(x) denote the Eulerian fraction (cf. [2], p. 245). It is well known that
(12)∑j=0∞‍jkxj=Ak(x)(1-x)k+1=αk(x),(|x|<1),
where Ak(x) is the kth Eulerian polynomial. Additionally, ([3], p. 24) gives the formula
(13)(1-xet)-1=∑k=0∞‍αk(x)tkk!.
Substituting t↦Lq leads to the formal identity
(14)(1-xEq)-1=∑k=0∞‍αk(x)k!Lqk.We can obtain additional identities in this fashion from other expansions of(1-xet)-1. For example, if x≠0 and x≠1, we have the following analogues of (3.1)–(3.4) in [4]:
(15)(1-xEq)-1=∑k=0∞‍xk(1-x)k+1Δqk,(16)(1-xEq)-1=∑k=0∞‍(x(1-x)2)k+1(x-1Δq2kEqk-Δq2kEqk+1),(17)(1-xEq)-1=1+∑k=0∞‍(x(1-x)2)k+1(Δq2kEqk-1-xΔq2kEqk).
Direct proofs of (14)–(17) are given in Section 5.Proposition 2.
For a given functionf(t), define Fq(x)=∑k≥0f(qk)xk. If x≠0 and x≠1,
(18)Fq(x)=∑k=0∞‍αk(x)k!Lqkf(1),(19)Fq(x)=∑k=0∞‍xk(1-x)k+1Δqkf(1),(20)Fq(x)=∑k=0∞‍(x(1-x)2)k+1(x-1δq2kf(1)-δq2kf(q-1)),(21)Fq(x)=1+∑k=0∞‍(x(1-x)2)k+1(δq2kf(q)-xδq2kf(1)).Proof.
Clearly, these follow by applying the operators in (14)–(17) to the function f(t) and then evaluating at t=1.


## 3. Some Applications

As an application, takingf(t)=1/(logq(t)+1),  x=-1 in (19) leads to
(22)∑k≥0‍(-1)k1k+1=∑k≥0‍(-1)k2k+1∑j=0k‍(kj)(-1)k-j1j+1=∑k=0∞‍1(k+1)2k+1∑j=0k‍(-1)j(k+1j+1)=∑k=0∞‍1(k+1)2k+1=∑k=1∞‍1k2k,
which gives
(23)ln2=12+12·22+13·23+14·24+⋯.
The rate of convergence of this series is O(1/2n), much faster than
(24)ln2=∑k≥0‍(-1)k1k+1,
whose convergence rate is O(1/n).As for a second application, we may substitutex=-1 in Proposition 2, obtaining the following series transformation formulas:
(25)∑k≥0‍(-1)kf(qk)=∑k=0∞‍αk(-1)k!Lqkf(1),∑k≥0‍(-1)kf(qk)=∑k=0∞‍(-1)k2k+1Δqkf(1),∑k≥0‍(-1)kf(qk)=∑k=0∞‍(-1)k4k+1(δq2kf(1)+δq2kf(q-1)),∑k≥1‍(-1)kf(qk)=1+∑k=0∞‍(-14)k+1(δq2kf(q)+δq2kf(1)).
These four identities appear to be novel and could be used to accelerate slowly convergent alternating series ∑k=1∞‍(-1)kf(qk). We consider them as q-analogues of the ordinary Euler transformations.


## 4. Extensions of the Main Results

All operational formulas presented in Proposition1 can be extended and the corresponding symbolic substitution formulas established accordingly with an analogous form of (10). For example, we may consider a generating function of the form
(26)∑k≥0‍fktk=F(t,et,eαt).
Letting t↦Lq gives
(27)∑k≥0‍fkLqk=F(Lq,Eq,Eqα).
Applying this to the well-known identity
(28)∑k≥0‍4k(2k)!B2kt2k=tcotht=tet+e-tet-e-t,
with Bn being the nth Bernoulli number, we obtain
(29)∑k≥0‍4k(2k)!B2kLq2k=LqEq+Eq-1Eq-Eq-1.
Hence, we obtain a symbolic formula
(30)∑k≥0‍4k(2k)!B2kLq2k-1(Eq-Eq-1)=Eq+Eq-1.
Applying this to an infinitely differentiable function f(t) at t=1 yields
(31)∑k≥0‍4k(2k)!B2kLq2k-1(Eq-Eq-1)f(1)=(Eq+Eq-1)f(1).
Similarly, using the symbolic relation
(32)LqEq+Eq-1Eq-Eq-1=Lq(1+Δq-1-(Eq+1)-1),
we obtain another operational formula
(33)-1+∑k≥0‍4k(2k)!B2kLq2k-1+(Eq+1)-1=Δq-1,
from which one may construct a series transformation formula.Another extension is aq-analogue of the symbolic formulas presented in [5], which is actually a Newton series type extension of the symbolic expansions given in [1]. Consider
(34)(1+Eq)xf(1)=∑k≥0(xk)Eqkf(1)=∑k≥0f(qk)(x)kk!,
where (x)k=x(x-1)⋯(x-k+1). We have
(35)(1+Eq)x=2x∑k=0∞‍(x)k2kk!Δqk,(36)(1+Eq)x=∑k=0∞‍(xk)[F12(k-x,2k+1;k+1;-1)+x-kk+1x-kk+1×F21(k+1-x,2k+2;k+2;-1)x-kk+1×Eq-1Δq]δq2k,(37)(1+Eq)x=1+∑k=0∞‍(xk+1)×[F12(k+1-x,2k+2;k+2;-1)Eqx-k-1k+2-x-k-1k+2x-k-1k+2×F12(k+2-x,2k+2;k+3;-1)]δq2k.Finally, we present an extension of (14) using Bell polynomials (see, e.g., p. 134 in [2]) as follows:
(38)(1+Eq)x=2x∑k=0∞‍Pk(x)(12,12,…)Lqkk!,
where the values of potential Bell polynomials at (1/2,1/2,…) are defined by
(39)2xPk(x)(12,12,…)=∑ℓ=0∞‍ℓk(x)ℓℓ!.For a given functionf(t), define Fq(x)=∑k≥0f(qk)(x)k/k!. From (35)–(38), we obtain series transformation formulas by simply applying (35)–(37) to f:
(40)Fq(x)=2x∑k=0∞‍(x)k2kk!Δqkf(1),(41)Fq(x)=∑k=0∞‍(xk)[F12(k-x,2k+1;k+1;-1)+x-kk+1×F12(k+1-x,2k+2;k+2;-1)x-kk+1×Eq-1Δq]δq2kf(1),(42)Fq(x)=1+∑k=0∞‍(xk+1)×[F12(k+1-x,2k+2;k+2;-1)Eqx-k-1k+2-x-k-1k+22F1(k+2-x,2k+2;k+3;-1)]×δq2kf(1),(43)Fq(x)=2x∑k=0∞‍Pk(x)(12,12,…,)Lqkk!f(1).As an example, substitutingf(t)=tn into (43) and noting
(44)Lqf(1)=∑j=1∞‍(-1)j+1jΔqjf(1)=∑j=1∞‍(-1)j+1j(qn-1)j=log(1+(qn-1))=nlogq,
yields the series transformation formula from
(45)∑k=0∞‍qkn(x)kk!=2x∑j=0∞‍Pj(x)(12,12,…)(nlogq)jj!.


## 5. Selected Proofs

Here we present the proofs of (14)–(17) in the sense of symbolic calculus, namely, every series expansion is considered as a formal series.For proving (14), it suffices to make use of Eq=eLq and (12). Indeed we have
(46)(1-xEq)-1=(1-xeLq)-1=∑k=0∞‍xkekLq=∑k=0∞‍xk∑j=0∞‍(kLq)jj!=∑j=0∞‍(∑k=0∞‍xkkj)Lqjj!=∑j=0∞‍αj(x)Lqjj!.Equation (15) may be derived as follows:
(47)(1-xEq)-1=(1-x(1+Δq))-1=(1-x)-1(1-xΔq1-x)-1=∑k=0∞‍xkΔqk(1-x)k+1.To prove (16) and (17), we first establish the following lemma.Lemma 3.
Letβ=1+α with 0<α<1, and let x be any real number. One has symbolic identities involving the first Gauss series as follows:
(48)βx=∑k=0∞‍[(x+k2k)α2kβk+(x+k2k+1)α2k+1βk+1],
and a modified q-form of Gauss’s first symbolic expression (cf. Section  127 of [6]):
(49)Eqx=∑k=0∞‍[(x+k2k)Δq2kEqk+(x+k2k+1)Δq2k+1Eqk+1]=∑k=0∞‍[(x+k2k)δq2k+(x+k2k+1)Δqδq2kEq-1].Proof.
Starting from the following Newton’s formula:(50)βx=∑k=0∞‍(xk)αk,
we multiply (α+1)/β=1 to the summation from the term α up and obtain
(51)βx=1+(x1)αβ+(x1)α2β+∑k=2∞‍(xk)αk(1+α)β=1+(x1)αβ+(x+12)α2β+∑k=2∞‍αk+1β[(xk)+(xk+1)]=1+(x1)αβ+(x+12)α2β+∑k=2∞‍(x+1k+1)αk+1β=1+(x1)αβ+(x+12)α2β+(x+13)α3β+∑k=3∞‍(x+1k+1)αk+1β.
Repeating the operation on the series from the term α3 up yields
(52)βx=1+(x1)αβ+(x+12)α2β+(x+13)α3β2+(x+13)α4β2+∑k=3∞‍(x+1k+1)αk+1(1+α)β2=1+(x1)αβ+(x+12)α2β+(x+13)α3β2+(x+24)α4β2+∑k=3∞‍(x+2k+2)αk+2β2.
The above operation is repeated from α5 up, and so on. We obtain
(53)βx=∑k=0∞‍[(x+k2k)α2kβk+(x+k2k+1)α2k+1βk+1].
Substituting β=Eq and α=Δq into the above identity, we obtain the desired result.Equations (16) and (17) can be proved using the first Gauss symbolic expression (49) and the following q-form of the Everett’s symbolic expression (cf. [6], Section  129), respectively. 						(54)Eqx=∑k=0∞‍[(x+k2k+1)Δq2kEqk-1-(x+k-12k+1)Δq2kEqk]=∑k=0∞‍[(x+k2k+1)Eqδq2k-(x+k-12k+1)δq2k].
Indeed, using (49) and noting the identity
(55)∑m=k∞‍(mk)xm=xk(1-x)k+1(|x|<1),
one may derive (16) as follows:
(56)(1-xEq)-1=∑j=0∞‍(xEq)j=∑k=0∞‍{(∑j=0∞‍(j+k2k)xj)Δq2kEqk+(∑j=0∞‍(j+k2k+1)xj)Δq2k+1Eqk+1}=∑k=0∞‍{xk(1-x)2k+1Δq2kEqk+xk+1(1-x)2k+2Δq2k+1Eqk+1}=∑k=0∞‍(x(1-x)2)k+1(1-xxΔq2kEqk+Δq2k+1Eqk+1)=∑k=0∞‍(x(1-x)2)k+1(x-1Δq2kEqk-Δq2kEqk+1).
Equation (17) can be proved similarly using (54). However, it can also be verified by a direct symbolic computations. In fact, we have
(57)RHSof(17)=1+x(1-x)2∑k=0∞‍(x(1-x)2)k(Δq2Eq)k(Eq-x)=1+x(1-x)2Eq-x1-(x/(1-x)2)(Δq2/Eq)=1+x(Eq-x)(1-x)2-x(Δq2/Eq)=1+Eqx(Eq-x)(1-x)2Eq-x(Eq-1)2=1+Eqx1-xEq=(1-xEq)-1=LHSof(17).
This completes the proofs of (14)–(17).The proof of (35) is straightforward:
(58)(1+Eq)x=(2+Δq)x=2x(1+Δq2)x=2x∑k≥0‍(x)kk!2kΔqk.To prove (36), we use (49) as follows:
(59)(1+Eq)x=1+∑j≥1‍(xj)Eqj=1+∑j≥1‍(xj)∑k≥0‍[(k+j2k)Δq2kEqk+(k+j2k+1)Δq2k+1Eqk+1]=∑k≥0‍[Δq2kEqk∑j≥k‍(xj)(k+j2k)+Δq2k+1Eqk+1∑j≥k+1‍(xj)(k+j2k+1)]=∑k≥0‍[(xk)F12(k-x,2k+1;k+1;-1)Δq2kEqk+(xk+1)F12(k+1-x,2k+2;k+2;-1)Δq2kEqk+1],
which implies (36). Equation (37) can be proved similarly using Everett’s symbolic expression (54).For (38), we first have
(60)(1+Eq)x=(1+eLq)x=∑j≥0‍(xj)ejLq(61)=∑j≥0‍(xj)∑k≥0‍(jLq)kk!=∑k≥0‍(∑j≥0‍(x)jjkj!)Lqkk!.
Using (39), we may write the part in the parenthesis of the rightmost term as 2xPk(x)(1/2,1/2,…) to finish.



---

*Source: 487546-2013-07-14.xml*