# New Properties on Degenerate Bell Polynomials

Authors: Taekyun Kim; Dae San Kim; Hyunseok Lee; Seongho Park; Jongkyum Kwon
Journal: Complexity
 (2021)

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## Abstract

The aim of this paper is to study the degenerate Bell numbers and polynomials which are degenerate versions of the Bell numbers and polynomials. We derive some new identities and properties of those numbers and polynomials that are associated with the degenerate Stirling numbers of both kinds.


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## Body

## 1. Introduction

The Bell numberBeln counts the number of partitions of a set with n elements into disjoint nonempty subsets. The Bell polynomials Belnx, also called Touchard or exponential polynomials, are natural extensions of Bell numbers. The partial and complete Bell polynomials, which are multivariate generalizations of the Bell polynomials, have diverse applications not only in mathematics but also in physics and engineering as well (see [1]).For instance, the following formula, due to Faà di Bruno formula:(1)dndtnf°gt=∑k=0nfkgtBn,kg′t,g′′t,…,gn−k+1t,gives an explicit formula for higher derivatives of composite functions. Here, the partial Bell polynomials Bn,kx1,x2,…,xn−k+1 are defined by(2)Bn,kx1,x2,…,xn−k+1=∑n!∏l=1n−k+1il!∏l=1n−k+1xll!il,n≥k≥0,where the sum runs over all nonnegative integers i1,i2,…,in−k+1, satisfying i1+i2+⋯+in−k+1=k and i1+2i2+⋯n−k+1in−k+1=n (see [1], p. 133). Then, the complete Bell polynomials are given by Bnx|x1,…,xn=∑k=1nBn,kx1,x2,…,xn−k+1xk,n≥1, and Bnx|1,1,…,1=∑k=1nBn,k1,1,…,1xk=Belx,n≥1.As a degenerate version of those Bell polynomials and numbers, the degenerate Bell polynomialsBeln,λx and numbers Beln,λ (see (17)) are introduced and studied under the different names of the partially degenerate Bell polynomials and numbers in [2]. Some interesting identities for them were obtained in connection with Stirling numbers of the first and second kinds [2]. We hope that we will be able to find many interesting applications of these polynomials and numbers in near future.In [3], Carlitz initiated the exploration of degenerate Bernoulli and Euler polynomials, which are degenerate versions of the ordinary Bernoulli and Euler polynomials. Along the same line as Carlitz’s pioneering work, intensive studies have been done for degenerate versions of quite a few special polynomials and numbers (see [2–10] and the references therein). It is worthwhile to mention that these studies of degenerate versions have been done not only for some special numbers and polynomials but also for transcendental functions like gamma functions (see [8]). The studies have been carried out by various means like combinatorial methods, generating functions, differential equations, umbral calculus techniques, p-adic analysis, and probability theory.The aim of this paper is to further investigate the degenerate Bell polynomials and numbers by means of generating functions. In more detail, we derive several properties and identities of those numbers and polynomials which include recurrence relations for degenerate Bell polynomials (see Theorems1, 3, 4, and 8), and expressions for them that can be derived from repeated applications of certain operators to the exponential functions (see Theorem 2, Proposition 1), the derivatives of them (Corollary 1), the antiderivatives of them (see Theorem 6), and some identities involving them (see Theorems 5, 9). For the rest of this section, we recall some necessary facts that are needed throughout this paper.For anyλ∈ℝ, the degenerate exponential functions are defined by(3)eλxt=∑k=0∞xk,λk!tk,(see [9]), where(4)x0,λ=1,xn,λ=xx−λ⋯x−n−1λ,n≥1.Whenx=1, we see use the notation eλt=eλ1t.In [3], Carlitz introduced the degenerate Bernoulli numbers given by(5)teλt−1=∑n=0∞βn,λtnn!.Note thatlimλ⟶0βn,λ=Bn, where Bn are the ordinary Bernoulli numbers given by(6)tet−1=∑n=0∞Bntnn!,(see [1–14]).From (5), we deduce that(7)β0,λ=1,∑l=0n−1nl1n−l,λβl,λ=0,n≥2,from which we compute the first few values of βn,λ as follows:(8)β0,λ=1,β1,λ=12λ−12,β2,λ=−16λ2+16,β3,λ=14λ3−14λ,β4,λ=−1930λ4+23λ2−130,β5,λ=94λ5−52λ3+14λ,β6,λ=−86384λ6+12λ4−74λ2+142,β7,λ=137524λ7−70λ5+1058λ3−512λ,β8,λ=−3395390λ8+480λ6−162415λ4+509λ2−130.It is well known that the Stirling numbers of the first kind are defined by(9)xn=∑k=0nS1n,kxk,1k!log1+tk=∑n=k∞S1n,ktnn!,(see [14]), where x0=1,xn=xx−1⋯x−n+1,n≥1.As the inversion formula of (9), the Stirling numbers of the second kind are given by(10)xn=∑k=0nS2n,kxk,1k!et−1k=∑n=k∞S2n,ktnn!,(see [14]).The degenerate Stirling numbers of the first kind are defined by(11)xn=∑k=0nS1,λn,kxk,λ,1k!logλ1+tk=∑n=k∞S1,λn,ktnn!,(see [5]), and the degenerate Stirling numbers of the second kind are given by(12)xn,λ=∑k=0nS2,λn,kxk,1k!eλt−1k=∑n=k∞S2,λn,ktnn!,(see [5, 7]).Here,logλ1+t is the degenerate logarithm given by (18).We also recall the degenerate absolute Stirling numbers of the first kind that are defined by(13)xn=∑k=0nnkλxk,λ,(see [10]), where(14)x0=1,xn=xx+1⋯x+n−1,n≥1,x0,λ=1,xn,λ=xx+λx+2λ⋯x+n−1λ,n≥1.It is well known that the Bell polynomials are defined by(15)exet−1=∑n=0∞Belnxtnn!,(see [12, 13]).Whenx=1,Beln=Beln1 are called the Bell numbers.From (12), we note that(16)Belnx=∑k=0nS2n,kxk,(see [12, 13]).In [2], the degenerate Bell polynomials are defined by(17)exeλt−1=∑n=0∞Beln,λxtnn!.Note thatlimλ⟶0Beln,λx=Belnx. For x=1, Beln,λ=Beln,λ1 are called the degenerate Bell numbers.The compositional inverse ofeλt is given by logλt, namely, eλlogλt=t=logλeλt, where(18)logλ1+t=1λ1+tλ−1=∑n=1∞λn−11n,1λtnn!,(see [5]).Note thatlimλ⟶0logλ1+t=log1+t.From (17), we note that(19)Beln,λx=e−x∑k=0∞kn,λk!xk=∑k=0nS2,λn,kxk,(see [2]).From (12), we can deduce the recurrence relation given by(20)S2,λn+1,k=S2,λn,k−1+k−nλS2,λn,k,n≥k≥1,and the values(21)S2,λn,0=0,n≥1,S2,λn,n=1,n≥0,Now, we compute from (19), (3), and (21) the first few degenerate Bell polynomials as follows:(22)Bel0,λx=1,Bel1,λx=x,Bel2,λx=−xλ+x2+x,Bel3,λx=2xλ2+−3x2−3xλ+x3+3x2+x,Bel4,λx=−6xλ3+11x2+12xλ2+−6x3−18x2−7xλ+x4+6x3+7x2+x,Bel5,λx=24xλ4+−50x2−60xλ3+35x3+110x2+50xλ2+−10x4−60x3−75x2−15xλ+x5+10x4+25x3+15x2+x,Bel6,λx=−120xλ5+274x2+274xλ4+−225x3−675x2−225xλ3+85x4+510x3+595x2+85xλ2+−15x5−150x4−375x3−225x2−15xλ+x6+15x5+65x4+90x3+31x2+x,Bel7,λx=720xλ6+−1764x2−1764xλ5+1624x3+4872x2+1624xλ4+−735x4−4410x3−5145x2−735xλ3+175x5+1750x4+4375x3+2625x2+175xλ2+−21x6−315x5−1365x4−1890x3−651x2−21xλ+x7+21x6+140x5+350x4+301x3+63x2+x,Bel8,λx=−5040xλ7+13068x2+13068xλ6+−13132x3−39396x2−13132xλ5+6769x4+40614x3+47383x2+6769xλ4+−1960x5−19600x4−49000x3−29400x2−1960xλ3+332x6+4830x5+20930x4+28980x3+9982x2+322xλ2+−28x7−588x6−3920x5−9800x4−8428x3−1764x2−28xλ+x8+28x7+266x6+1050x5+1701x4+966x3+127x2+x.In Figure1, we plot the shapes of Bell polynomial Belk,λx. The upper-left graph looks different from the others. However, of course they all go to infinity as x tends to infinity.Figure 1
The shapes of Bell polynomialsBelk,λx.


## 2. Some New Properties on Degenerate Bell Polynomials

Leta be a nonzero constant. First, we observe that(23)dndtneaeλt=dndtn∑k=0∞akk!eλkt=∑k=0∞akk!kn,λeλk−nλt=∑k=0∞kn,λk!akeλkt11+λtn=∑k=0∞kn,λk!aeλtke−aeλteaeλt11+λtn=11+λtnBeln,λaeλteaeλt.Therefore, by (23), we obtain the following lemma.Lemma 1.
Forn≥0, the nth derivative of eaeλt is given by(24)dndtneaeλt=11+λtnBeln,λaeλteaeλt.Letx=eλt in (23). Then, we have(25)ddt=dxdtddx=11+λteλtddx=x1−λddx.By Lemma1 and (25), we get(26)x1−λddxneax=x−nλBeln,λaxeax,n≥0.Let(27)Sn,λ=∑k=0∞kn,λk!,n=0,1,2,….Then, we note from (19) that we have(28)eBeln,λ=Sn,λ.The generating function ofSn,λ is given by(29)eeλt=∑n=0∞Sn,λtnn!.Indeed, this can be seen from the following:(30)∑n=0∞Sn,λtnn!=e∑n=0∞Beln,λ1tnn!=eeeλt−1=eeλt.Taking the derivative with respect tot on both sides of (30), we have(31)∑n=0∞Sn+1,λtnn!=ddteeλt=eλ1−λteeλt=∑l=0∞1−λl,λtll!∑m=0∞Sm,λtmm!=∑n=0∞∑m=0nnmSm,λ1−λn−m,λtnn!=∑n=0∞∑m=0nnmSm,λ1n−m+1,λtnn!.Thus, by comparing the coefficients on both sides of (31) and from (28), we obtain the following theorem.Theorem 1.
Forn≥0, the following recurrence relation holds:(32)Beln+1,λ=∑m=0nnmBelm,λ1n−m+1,λ.Assume that the following identity holds:(33)x1−λddxnex=∑k=0∞kn,λk!xk−nλ.Then, we have(34)x1−λddxn+1ex=x1−λddx∑k=0∞kn,λk!xk−nλ=x1−λ∑k=0∞kn,λk!k−nλxk−nλ−1=∑k=0∞kn+1,λk!xk−n+1λ.This together with (26) gives the next result.Theorem 2.
Forn≥0, the following relations hold true:(35)x1−λddxnex=∑k=0∞kn,λk!xk−nλ=x−nλBeln,λxex.From the first equality in (35) and (27), we see that we have(36)Sn,λ=x1−λddxnex|x=1.Clearly,S0,λ=S1,λ=e. We can check that(37)x1−λddx2ex=1−λx1−2λex+x2−2λex,x1−λddx3ex=x1−3λex1−λ2,λ+1−λx+x2−3λex2−2λ+x.From (36) and (37), we have S2,λ=2−λe,S3,λ=2λ2−6λ+5e.By takingxd/dx in the second equality of (35), on the one hand, we have(38)xddxx−nλBeln,λxex=∑k=0∞kn+1,λk!xk−nλ.On the other hand, we also have(39)xddxx−nλBeln,λxex=xx−nλBeln,λ′x+Beln,λxex−nλx−nλBeln,λxex,where Beln,λ′x=d/dxBeln,λx.From (38) and (39) and Theorem 2, we note that(40)∑k=0∞kn+1,λk!xk=∑k=0∞kn+1,λk!xke−xex=Beln+1,λxex=xBeln,λ′x+Beln,λxex−nλBeln,λxex.Therefore, by (40) and Theorem 2, we obtain the following theorem.Theorem 3.
Forn≥0, the following identity holds:(41)Beln+1,λx=xBeln,λ′x+Beln,λx−nλBeln,λx,where Beln,λ′x=d/dxBeln,λx.From (17), we note that(42)∑n=0∞ddxBeln,λxtnn!=∂∂xexeλt−1=eλt−1exeλt−1=∑l=0∞1l,λtll!−1∑m=0∞Belm,λxtmm!=∑n=0∞∑m=0nnmBelm,λx1n−m,λ−Beln,λxtnn!=∑n=0∞∑m=0n−1nmBelm,λx1n−m,λtnn!.Thus, by comparing the coefficients on both sides of (42), we get(43)ddxBeln,λx=Beln,λ′x=∑m=0n−1nmBelm,λx1n−m,λ,n≥1.Taking the derivative with respect tot on both sides of (17), we have(44)ddtexeλt−1=∑n=0∞Beln+1,λxtnn!.On the other hand,(45)ddtexeλt−1=xeλ1−λtexeλt−1=x∑l=0∞1−λl,λtll!∑m=0∞Belm,λxtmm!=x∑n=0∞∑m=0nnmBeln,λx1−λn−m,λtnn!=∑n=0∞x∑m=0nnmBelm,λx1n−m+1,λtnn!.Therefore, by (44) and (45), we obtain the following theorem.Theorem 4.
. Forn≥0, the following recurrence relation is valid:(46)Beln+1,λx=x∑m=0nnmBelm,λx1n−m+1,λ.Remark 1.
Theorems3 and 4 and (43) give us the following:(47)Beln+1,λx=x∑m=0nnmBelm,λx1n−m,λ−nλBeln,λx=x∑m=0nnmBelm,λx1n−m,λ1−n−mλ.
This implies that the following identity must hold true:(48)nBeln,λx=x∑m=0nnmn−mBelm,λx1n−m,λ,the validity of which follows from Theorem 4.From Theorem3, we note that(49)xBeln,λ′x=xddxBeln,λx=Beln+1,λx−xBeln,λx+nλBeln,λx=Beln+1,λx−x−nλBeln,λx=x∑m=0n−1nmBelm,λx1n+1−m,λ+nλBeln,λx.Therefore, by (49), we obtain the following corollary.Corollary 1.
Forn≥1, we have the following identity:(50)xddxBeln,λx=x∑m=0n−1nmBelm,λx1n+1−m,λ+nλBeln,λx.We observe that(51)x1−λddxx−nλBeln,λxex=x1−λddxx−nλ∑k=0∞kn,λk!xk=∑k=0∞kn+1,λk!xk−n+1λ=x−n+1λ∑k=0∞kn+1,λk!xke−xex=x−n+1λBeln+1,λxex,n≥0.Thus, by (51), we get(52)x1−λddxx−nλBeln,λxex=x−n+1λBeln+1,λxex,n≥0.From (17), we have(53)∑n=0∞Beln,λx+ytnn!=ex+yeλt−1=exeλt−1⋅eyeλt−1=∑l=0∞Bell,λxtll!∑m=0∞Belm,λxtmm!=∑n=0∞∑l=0nnlBell,λxBeln−l,λytnn!.Therefore, by comparing the coefficients on both sides of (53), we obtain the following theorem.Theorem 5.
Forn≥0, the following binomial identity holds:(54)Beln,λx+y=∑l=0nnlBell,λxBeln−l,λy.From (17), we note that(55)∑n=0∞∫0xBeln,λxdxtnn!=∫0xexeλt−1dx.On the other hand, we also have(56)∫0xexeλt−1dx=1eλt−1exeλt−10x=1eλt−1exeλt−1−1=1eλt−1∑k=1∞Belk,λxtkk!=teλt−1∑k=0∞Belk+1,λxk+1tkk!=∑l=0∞βl,λtll!∑k=0∞Belk+1,λxk+1tkk!=∑n=0∞∑k=0nnkBelk+1,λxk+1βn−k,λtnn!=∑n=0∞1n+1∑k=0nn+1k+1Belk+1,λxβn−k,λtnn!=∑n=0∞1n+1∑k=1n+1n+1kBelk,λxβn+1−k,λtnn!.Therefore, by (55) and (56), we obtain the following theorem.Theorem 6.
Forn≥0, the antiderivative of Beln,λx is given by(57)∫0xBeln,λxdx=1n+1∑k=1n+1n+1kβn+1−k,λBelk,λx,where βn,λ are Carlitz’s degenerate Bernoulli numbers given by t/eλt−1=∑n=0∞βn,λtn/n!.Fork≥0, by (12), we get(58)∑n=k∞S2,λn,ktnn!=1k!eλt−1k=1k!∑j=0k−1k−jeλjtkj=∑n=0∞1k!∑j=0kkj−1k−jjn,λtnn!.By comparing the coefficients on both sides of (58), we have(59)1k!∑j=0kkj−1k−jjn,λ=S2,λn,k,ifn≥k,0,if $0≤n≤k−1$.LetD=d/dx, and let y=xp. As x1−λD=py1−λ/pd/dy, we have(60)x1−λDneaxp=py1−λ/pddyneay=pny1−λ/pddyneay=pny−nλ/pBeln,λ/payeay=pnx−nλBeln,λ/paxpeaxp.Thus, we have(61)x1−λDneaxp=pnx−nλBeln,λ/paxpeaxp,n≥0.Therefore, by (61), we obtain the following proposition.Proposition 1.
Forn≥0, we have the following operational formula:(62)xnλx1−λDneaxp=pnBeln,λ/paxpeaxp,where D=d/dx.From (12), we note that(63)∑k=0n+1S2,λn+1,kxk=xn+1,λ=x−nλxn,λ=x−nλ∑k=0nS2,λn,kxk=∑k=0nS2,λn,kx−k+k−nλxk=∑k=0nS2,λn,kxk+1+∑k=0nS2,λn,kk−nλxk=∑k=1n+1S2,λn,k−1xk+∑k=0nS2,λn,kk−nλxk=∑k=0n+1S2,λn,k−1+S2,λn,kk−nλxk.By (63), we get(64)S2,λn+1,k=S2,λn,k−1+k−nλS2,λn,k,where 0≤k≤n+1.We prove the next theorem by induction onn.Theorem 7.
Assume thatf is an infinitely differentiable function. Then, for n≥0, the following operational formula holds:(65)x1−λDnf=∑k=0nS2,λn,kxk−nλDkf,where D=d/dx.Proof.
The statement is obviously true forn=0. Assume that it is true for n,n≥0.(66)x1−λDn+1fx=x1−λD∑k=0nS2,λn,kxk−nλDkfx=x1−λ∑k=0nS2,λn,kk−nλxk−1−nλDkfx+xk−nλDk+1fx=∑k=0nS2,λn,kk−nλxk−n+1λDkfx+xk+1−n+1λDk+1fx=∑k=0n+1S2,λn,kk−nλxk−n+1λDkfx+∑k=0n+1S2,λn,k−1xk−n+1λDkfx=∑k=0n+1S2,λn,kk−nλ+S2,λn,k−1xk−n+1λDkfx=∑k=0n+1S2,λn+1,kxk−n+1λDkfx.
Letfx=ex. Then, we have(67)xnλx1−λDnex=∑k=0nS2,λn,kxkex=Beln,λxex.
Observe that, for anyα, we have(68)x1−λDnxα=αn,λxα−nλ.
By the Leibniz rule, we get(69)x1−λDnfg=∑l=0nnlx1−λDn−lfx1−λDlg.
From Theorem2, we note that(70)x−n+mλexBeln+m,λx=x1−λDn+mex=x1−λDnx1−λDmex=x1−λDnx−mλBelm,λxex.
By (69) and (70), we get(71)x−n+mλexBeln+m,λx=x1−λDnx−mλBelm,λex=∑k=0nnkx1−λDn−kx−mλBelm,λxx1−λDkex=∑k=0nnkx−kλBelk,λxexx1−λDn−kx−mλBelm,λx.
On the other hand,(72)x1−λDn−kx−mλBelm,λx=∑j=0mS2,λm,jx1−λDn−kxj−mλ=∑j=0mS2,λm,jj−mλn−k,λxj−mλ−n−kλ=∑j=0mS2,λm,jj−mλn−k,λxj−m+nλ+kλ=∑j=0mS2,λm,jjm+n−k,λjm,λxj−m+nλ+kλ.
By (71) and (72), we get(73)x−n+mλexBeln+m,λx=∑k=0nnkx−kλBelk,λxex∑j=0mS2,λm,jjm+n−k,λjm,λxj−m+nλ+kλ=x−m+nλex∑k=0n∑j=0mnkS2,λm,jBelk,λxjm+n−k,λjm,λxj.
Therefore, by comparing the coefficients on both sides of (73), we obtain the following theorem.Theorem 8.
Form,n≥0, we have the following expression:(74)Beln+m,λx=∑k=0n∑j=0mnkS2,λm,jBelk,λxjm+n−k,λjm,λxj.Takingx=1 in (74), we have(75)Beln+m,λ=∑k=0n∑j=0mnkS2,jm,jBelk,λjn+m−k,λjm,λ.From (19) and (68), we note that(76)x1−λDnBelm,λx=x1−λDn∑k=0mS2,λm,kxk=∑k=0mS2,λm,kkn,λxk−nλ.On the other hand, by Leibniz rule (69) and Theorem 2, we get(77)x1−λDnBelm,λx=x1−λDne−xxmλBelm,λxexx−mλ=∑k=0nnkx1−λDn−kxmλe−xx1−λDkBelm,λxexx−mλ=∑k=0nnkx1−λDn−kxmλe−xx1−λDm+kex=∑k=0nnkx1−λDn−kxmλe−xx−m+kλBelm+k,λxex.By (68) and (69) and Theorem 2, we easily get(78)x1−λDn−kxmλe−x=∑j=0n−kn−kjx1−λDjxmλx1−λDn−k−je−x=∑j=0n−kn−kjmλj,λxmλ−jλe−xBeln+k−j,λ−xx−n−k−jλ=∑j=0n−kn−kjmλj,λxmλ−nλ+kλBeln−k−j,λ−xe−x.From (77) and (78), we have(79)x1−λDnBelm,λx=∑k=0nnkx−mλ−kλBelm+k,λxex∑j=0n−kn−kjmλj,λxmλ−nλ+kλBeln−k−j,λ−xe−x=∑k=0n∑j=0n−knkn−kjBelm+k,λxBeln−k−j,λ−xmλj,λx−nλ.Therefore, by (76) and (79), we obtain the following theorem.Theorem 9.
Form,n≥0, the following identity holds true.(80)∑k=0mS2,λm,kkn,λxk=∑k=0n∑j=0n−knkn−kjBelm+k,λxBeln−k−j,λ−xmλj,λ.By (11) and (13), we easily get(81)−1n−kS1,λn,k=nkλ,0≤k≤n.Indeed,(82)∑n=0∞xntnn!=11−tx=eλ−xlogλ1−t=∑k=0∞−xk,λ1k!logλ1−tk=∑k=0∞−1kxk,λ∑n=k∞S1,λn,k−tnn!=∑n=0∞∑k=0n−1n−kS1,λn,kxk,λtnn!.Therefore, by (82), we get(83)xn=∑k=0n−1n−kS1,λn,kxk,λ.Replacingt by logλ1+t in (17), we get(84)ext=∑k=0∞Belk,λx1k!logλ1+tk=∑k=0∞Belk,λx∑n=k∞S1,λn,ktnn!=∑n=0∞∑k=0nBelk,λxS1,λn,ktnn!.Thus, from (81) and (84), we get(85)xn=∑k=0nBelk,λx−1n−knkλ.From (13), we note that(86)∑k=0n+1n+1kλxk,λ=xn+1=x+nxn=x+n∑k=0nnkλxk,λ=∑k=0nnkλx+kλ+n−kλxk,λ=∑k=0nnkλxk+1,λ+∑k=0nn−kλnkλxk,λ=∑k=0n+1nk−1λxk,λ+∑k=0n+1n−kλnkλxk,λ=∑k=0n+1nk−1λ+n−kλnkλxk,λ.Thus, by comparing the coefficients on both sides of (86), we get(87)n+1kλ=nk−1λ+n−kλnkλ,0≤k≤n+1.


## 3. Conclusion

Here, we studied by means of generating functions the degenerate Bell polynomials which are degenerate versions of the Bell polynomials. In more detail, we derived recurrence relations for degenerate Bell polynomials (see Theorems1, 3, 4, and 8), and expressions for them that can be derived from repeated applications of certain operators to the exponential functions (see Theorem 2 and Proposition 1), the derivatives of them (Corollary 1), the antiderivatives of them (see Theorem 6), and some identities involving them (see Theorems 5 and 9).As one of our future projects, we would like to continue to study degenerate versions of certain special polynomials and numbers and their applications to physics, science, and engineering as well as to mathematics. An earlier version of this paper has been presented as preprint in [15].



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