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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo


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Volume 15, Issue 1

Issues

Volume 13 (2015)

Fourier series of functions involving higher-order ordered Bell polynomials

Taekyun Kim
  • Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin 300160, China
  • Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
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/ Dae San Kim / Gwan-Woo Jang / Lee Chae Jang
Published Online: 2017-12-29 | DOI: https://doi.org/10.1515/math-2017-0134

Abstract

In 1859, Cayley introduced the ordered Bell numbers which have been used in many problems in number theory and enumerative combinatorics. The ordered Bell polynomials were defined as a natural companion to the ordered Bell numbers (also known as the preferred arrangement numbers). In this paper, we study Fourier series of functions related to higher-order ordered Bell polynomials and derive their Fourier series expansions. In addition, we express each of them in terms of Bernoulli functions.

Keywords: Fourier series; Bernoulli functions; Higher-order ordered Bell polynomials

MSC 2010: 11B73; 11B83; 05A19

1 Introduction

The ordered Bell polynomials of order r(r ∈ ℤ > 0) are defined by the generating function 12errext=m=0bm(r)(x)tmm!,(see [2,3,7,8,15,18,20]).(1)

When x = 0, bm(r)=bm(r)(0) are called the ordered Bell numbers of order r. These higher-order ordered Bell polynomials and numbers are further generalizations of the ordered Bell polynomials and numbers which are respectively given by bm(x)=bm(1)(x) and bm=bm(1). The ordered Bell polynomials bm(x) were defined in [11] as a natural companion to the ordered Bell numbers which were introduced in 1859 by Cayley to count certain plane trees with m+1 totally ordered leaves. The ordered Bell numbers have been used in many counting problems in number theory and enumerative combinatorics since its first appearance. They are all positive integers, as we can see from bm=n=0mn!S2(m,n)=n=0nm2n+1,(m0).(2)

Here we would like to point out that the ordered Bell numbers are also known (or mostly known) as the preferred arrangement numbers (see [8]). The ordered Bell polynomial bm(x) has degree m and is a monic polynomial with integral coefficients, as we can see from b0(x)=1,bm(x)=xm+l=0m1mlbl(x),(m1)(3)

(see [11]). From (1), we can derive the following. ddxbm(r)=mbm1(r),(m1),bm(r)(x+1)bm(r)(x)=bm(r)(x)bm(r1)(x),(m0).(4)

Also, from these we immediately get bm(r)(1)bm(r)=bm(r)bm(r1),(m0),01bm(r)(x)dx=1m+1(bm+1(r)(1)bm+1(r))=1m+1(bm+1(r)bm+1(r1)).(5)

For any real number x, we let < x > = x−[x] ∈ [0, 1) denote the fractional part of x. Let Bm(x) be the Bernoulli polynomials given by the generating function ter1ext=m=0Bm(x)tmm!.(6)

For later use, we will state the following facts about Bernoulli functions Bm(< x >):

  1. for m ≥ 2, Bm(<x>)=m!n=,n0e2πinx(2πin)m,(7)

  2. for m = 1, n=,n0,e2πinx2πin=B1(<x>),for xZc,0,for xZ,(8)

    where ℤc = ℝ − ℤ. Here we will consider the following three types of functions αm(< x >), βm(< x >), and γm(< x >) involving higher order ordered Bell polynomials.

In this paper, we will derive their Fourier series expansions and in addition express each of them in terms of Bernoulli functions:

  1. αm(<X>)=k=0mbk(r)(<x>)<x>mk,(m1);

  2. βm(<x>)=k=0m1k!(mk)!bk(r)(<x>)<x>mk,(m1);

  3. (<x>)=k=1m11k(mk)bk(r)(<x>)<x>mk,(m2).

The reader may refer to any book for elementary facts about Fourier analysis (for example, see [1,16,21]).

As to γm(< x >), we note that the polynomial identity (9) follows immediately from Theorems 4.1 and 4.2, which is in turn derived from the Fourier series expansion of γm(< x >) . k=1m11k(mk)bk(r)(x)xmk=1ms=0mmsΛms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1))Bs(x),(9)

where Hm=j=1m1j are the harmonic numbers and Λl=k=1l11k(lk)(2bk(r)bk(r1)). The obvious polynomial identities can be derived also for αm(< x >) and βm(< x >) from Theorems 2.1 and 2.2, and Theorems 3.1 and 3.2, respectively. It is noteworthy that from the Fourier series expansion of the function k=1m11k(mk)Bk(x)Bmk(x) we can derive the Faber-Pandharipande-Zagier identity (see [6]) and the Miki’s identity (see [5,17,19]). Some related works on Fourier series expansions for analogous functions can be found in the recent papers [9,10,13,14]. From now on, we will assume that r ≥ 2. The case of r = 1 has been treated as a special case of the result in [4].

2 Fourier series of functions of the first type

Let αm(x)=k=0mbk(r)(x)xmk,(m1). Then we will consider the function αm(<x>)=k=0mbk(r)(<x>)<x>mk,(10)

defined on ℝ which is periodic of period 1. The Fourier series of αm(< x >) is n=An(m)e2πinx,(11)

where An(m)=01αm(<x>)e2πinxdx=01αm(x)e2πinxdx.(12)

To proceed further, we need to observe the following. αmr(x)=k=0m{kbk1(r)(x)xmk+(mk)bk)(r)(x)xmk1}=k=1mkbk1(r)(x)xmk+k=0m1(mk)bk(r)(x)xmk1=(=0m1(k+1)bk(r)(x)xm1k+k=0m1(mk)bk(r)(x)xm1k=(m+1)αm1(x).(13)

From this, we get αm+1(x)m+2=αm(x),(14)

and 01αm(x)dx=1m+2(αm+1(1)αm+1(0)).(15)

For m ≥ 1, we put Δm=αm(1)αm(0)=k=0mbk(r)(1)bk(r)δm,k=k=0m2bk(r)bk(r1)bk(r)δm,k=l⇒=0m2bk(r)bk(r1)bm(r).(16)

We note from this that αm(0)=αm(1)Δm=0,(17)

and 01αm(x)dx=1m+2Δm+1.(18)

We are now going to determine the Fourier coefficients An(m).

Case 1: n ≠ 0. An(m)=01αm(x)e2πinxdx=12πinαm(x)e2πinx01+12πin01αm(x)e2πinxdx=m+12πinAm(m1)12πinΔm,(19)

from which by induction on m we can deduce that An(m)=j=1m(m+1)j1(2πin)jΔmj+1=1m+2j=1m(m+2)j(2πin)jΔmj+1.(20)

Case 2: n = 0. A0(m)=01αm(x)dx=1m+2Δm+1.(21)

αm(< x >), (m ≥ 1) is piecewise C. In addition, αm(< x >) is continuous for those positive integers with Δm = 0 and discontinuous with jump discontinuities at integers with Δm ≠ 0. Assume first that m is a positive integer with Δm = 0. Then αm(0) = αm(1). Thus the Fourier series of αm(< x >) converges uniformly to αm(< x >), and αm(<x>)=1m+2Δm+1+n=,n01m+2j=1m(m+2)j(2πin)jΔmj+1e2πinx=1m+2Δm+1+1m+2j=1mm+2jΔmj+1j!n=,n0e2πin(2πin)j=1m+2Δm+1+1m+2j=2mm+2jΔmj+1Bj(<x>)+Δm×B1(<x>), for xZc,0,for xZ.(22)

Now, we can state our first result.

Theorem 2.1

For each positive integer l, we let Δl=k=0l(2bk(r)bk(r1))bl(r).(23)

Assume that m is a positive integer with Δm = 0. Then we have the following.

  1. k=0mbk(r)(<x>)<x>mk has the Fourier series expansion k=0mbk(r)(<x>)<x>mk=1m+2Δm+1+n=,n01m+2j=1m(m+2)j(2πin)jΔmj+1e2πinx,(24)

    for all x ∈ ℝ, where the convergence is uniform.

  2. k=0mbk(r)(<x>)<x>mk=1m+2Δm+1+1m+2j=2mm+2jΔmj+1Bj(<x>),(25)

    for all x ∈ ℝ, where Bj(< x >) is the Bernoulli function.

Assume next that Δm ≠ 0, for a positive integer m. Then αm(0) ≠αm(1). Hence αm(< x >) is piecewise C, and discontinuous with jump discontinuities at integers. The Fourier series of αm(< x >) converges pointwise to αm(< x >), for x ∈ ℤc, and converges to 12(αm(0)+αm(1))=αm(0)+12Δm,(26)

for x ∈ č. We can now state our second result.

Theorem 2.2

For each positive integer l, we let Δl=k=0l(2bk(r)bk(r1))bl(r).(27)

Assume that m is a positive integer with Δm ≠ 0. Then we have the following.

  1. 1m+2Δm+1+n=,01m+2j=1m(m+2)j(2πin)jΔmj+1e2πinx=k=0mbk(r)(<x>)<x>mk,forxZc,bm(r)+12Δm,forxZ.(28)

  2. 1m+2j=0mm+2jΔmj+1Bj(<x>)=k=0mbk(r)(<x>)<x>mk, for xZc;1m+2j=0,j1mm+2jΔmj+1Bj(<x>)=bm(r)+12Δm, for xZ.(29)

3 Fourier series of functions of the second type

Let βm(x)=k=0m1k!(mk)!bk(r)(x)xmk,(m1). Then we will consider the function βm(<x>)=k=0m1k!(mk)!bk(r)(<x>)<x>mk,(30)

defined on ℝ which is periodic with period 1. The Fourier series of βm(< x >) is n=Bn(m)e2πinx,(31)

where Bn(m)=01βm(<x>)e2πinxdx=01βm(x)e2πinxdx.(32)

To proceed further, we need to observe the following. βm(x)=k=0mkk!(mk)!bk1(r)(x)xmk+mkk!(mk)!bk(r)(x)xmk1=k=1m1(k1)!(mk)!bk1(r)(x)xmk+k=0m11k!(mk1)!bk(r)(x)xmk1=k=0m11k!(m1k)!bk(r)(x)xm1k+k=0m11k!(mk1)!bk(r)(x)xm1k=2βm1(x).(33)

From this, we have βm+1(x)2=βm(x),(34)

and 01βm(x)dx=12(βm+1(1)βm+1(0)).(35)

For m ≥ 1, we have Ωm=βm(1)βm(0)=k=0m1k!(mk)!bk(r)(1)bk(r)δm,k=k=0m1k!(mk)!2bk(r)bk(r1)bk(r)δm,k=k=0m1k!(mk)!2bk(r)bk(r1)1m!bm(r).(36)

From this, we note that βm(0)=βm(1)Ωm=0,(37)

and 01βm(x)dx=12Ωm+1.(38)

We are now going to determine the Fourier coefficients Bn(m).

Case 1: n ≠ 0. Bn(m)=01βm(x)e2πinxdx=12πinβm(x)e2πinx01+12πin01βm(x)e2πinxdx=12πin(βm(1)βm(0))+22πin01βm1(x)e2πinxdx=22πinBn(m1)12πinΩm,(39)

from which by induction on m we can easily get Bn(m)=j=1m2j1(2πin)jΩmj+1.(40)

Case 2: n = 0. B0(m)=01βm(x)dx=12Ωm+1.(41)

βm(< x >), (m ≥ 1) is piecewise C. Moreover, βm(< x >) is continuous for those positive integers m with Ωm = 0, and discontinuous with jump discontinuities at integers for those positive integers m with Ωm ≠ 0.

Assume first that Ωm = 0, for a positive integer m. Then βm(0) = βm(1). Hence βm(< x >) is piecewise C, and continuous. Thus the Fourier series of βm(< x >) converges uniformly to βm(< x >), and βm(<x>)=12Ωm+1+n=,n0j=1m2j1(2πin)jΩmj+1e2πinx=12Ωm+1+j=1m2j1j!Ωmj+1j!n=,n0e2πinx(2πin)j=12Ωm+1+j=2m2j1j!Ωmj+1Bj(<x>)+Ωm×B1(<x>),forxZc,0,forxZ.(42)

Now, we state our first result.

Theorem 3.1

For each positive integerl, we let Ωl=k=0l1k!(lk)!(2bk(r)bk(r1))1l!bl(r).(43)

Assume that m is a positive integer with Ωm = 0. Then we have the following.

  1. k=0m1k!(mk)!bk(r)(<x>)<x>mk has the Fourier series expansion k=0m1k!(mk)!bk(r)(<x>)<x>mk=12Ωm+1+n=n0,j=1m2j1(2πin)jΩmj+1e2πinx,(44)

    for all x ∈ ℝ, where the convergence is uniform.

  2. k=0m1k!(mk)!bk(r)(<x>)<x>mk=j=0,j1m2j1j!Ωmj+1Bj(<x>),(45)

    for all x ∈ ℝ, where Bj(< x >) is the Bernoulli function.

Assume next that m is a positive integer with Ωm ≠ 0. Then βm(0) ≠βm(1). Hence βm(< x >) is piecewise C, and discontinuous with jump discontinuities at integers. The Fourier series of βm(< x >) converges pointwise to βm(< x >), for x ∈ ℤc, and converges to 12(βm(0)+βm(1))=βm(0)+12Ωm.(46)

for x ∈ ℤ. Now we state our second result.

Theorem 3.2

For each positive integer l, we let Ωl=k=0l1k!(lk)!(2bk(r)bk(r1))1l!bl(r).(47)

Assume that Ωm ≠ 0, for a positive integer m. Then we have the following.

  1. 12Ωm+1+n=,n0j=1m2j1(2πin)jΩmj+1e2πinx=k=0m1k!(mk)!bk(r)(<x>)<x>mk,forxZc,1m!bm(r)+12Ωm, forxZ.(48)

  2. j=0m2j1j!Ωmj+1Bj(<x>)=k=0m1k!(mk)!bk(r)(<x>)<x>mk,(49)

    for x ∈ ℤc; j=0,j1m2j1j!Ωmj+1Bj(<x>)=1m!bm(r)+12Ωm,(50)

    for x ∈ ℤ.

4 Fourier series of functions of the third type

Let γm(x)=k=1m11k(mk)bk(r)(x)xmk,(m2). Then we will consider the function. γm(<x>)=k=1m11k(mk)bk(r)(<x>)<x>mk,(51)

defined on ℝ, which is periodic with period 1. The Fourier series of γm(< x >) is n=Cm(r)e2πinx,(52)

where Cn(m)=01γm(<x>)e2πinxdx=01γm(x)e2πinxdx.(53)

To proceed further, we need to observe the following. γmr(x)=k=1m11mkbk1(r)(x)xmk+k=1m11kbk(r)(x)xmk1=k=0m21m1kbk(r)(x)xm1k+k=1m11kbk(r)(x)xm1k=k=1m21m1k+1kbk(r)(x)xm1k+1m1xm1+1m1bm1(r)(x)=(m1)γm1(x)+1m1xm1+1m1bm1(r)(x).(54)

From this, we see that 1m(γm+1(x)1m(m+1)xm+11m(m+1)bm+1(r)(x))=γm(x)(55)

and 01γm(x)dx=1m(γm+1(1)γm+1(0)1m(m+1)1m(m+1)(bm+1(r)(1)bm+1(r))).(56)

For m ≥ 2, we put Λm=γm(1)γm(0)=k=1m11k(mk)(bk(r)(1)bk(r)δm,k)=k=1m11k(mk)(2bk(r)bk(r1)).(57)

Notice here that, γm(0)=γm(1)Λm=0,(58)

and 01γm(x)dx=1mΛm+11m(m+1)(1+bm+1(r)bm+1(r1)).(59)

We are now going to determine the Fourier coefficients Cn(m).

Case 1: n ≠ 0. Cn(m)=01γm(x)e2πinxdx=12πin[γm(x)e2πinx]01+12πin01γm(x)e2πinxdx=12πin(γm(1)γm(0))+12πin01(m1)γm1(x)+1m1Xm1+1m1bm1(r)(x)e2πinxdx=m12πinCn(m1)12πinΛm+12πin(m1)01xm1e2πinxdx+12πin(m1)01bm1(r)(x)e2πinxdx.(60)

We can show that 01xle2πinxdx=k=1l(l)k1(2πin)k,forn0,1l+1,forn=0.(61)

Also, from the paper [12], we have 01bl(r)(x)e2πinxdx=k=1l(l)k1(2πin)+(blk+1(r)blk+1(r1)),forn0,1l+1(bl+1(r)bl+1(r1)),forn=0.(62)

From (60), (61), and (62), we have Cn(m)=m12πinCn(m1)12πinΛm12πin(m1)Φm,(63)

where Φm=k=1m1(m1)k1(2πin)k(1+bmk(r)bmk(r1)).(64)

From (63) by induction on m we can deduce that Cn(m)=j=1m1(m1)j1(2πin)jΛmj+1j=1m1(m1)j1(2πin)j(mj)Φmj+1.(65)

We note here that j=1m1(m1)j1(2πin)j(mj)Φmj+1=j=1m1(m1)j1(2πin)j(mj)k=1mj(mj)l)1(2πin)k(1+bmjk+1(r)bmjk+1(r1))=j=1m11mjk=1mj(m1)j+k2(2πin)j+k(1+bmjk+1(r)bmjk+1(r1))=j=1m11mjs=j+1m(m1)s2(2πin)s(1+bms+1(r)bms+1(r1))=s=2m(m1)s2(2πin)s(1+bms+1(r)bms+1(r1))j=1s11mj=1ms=1m(m)s(2πin)sHm1Hmsms+1(1+bms+1(r)bms+1(r1)).(66)

Putting everything altogether, from (65), we finally obtain Cn(m)=1ms=1m(m)s(2πin)sΛms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)).(67)

Case 2: n = 0. C0(m)=01γm(x)dx=1mΛm+11m(m+1)(1+bm+1(r)bm+1(r1)).(68)

γm(< x >), (m ≥ 2) is piecewise C. Moreover, γm(< x >) is continuous for those integers m ≥ 2 with Λm = 0, and discontinuous with jump discontinuities at integers for those integers m ≥ 2 with Λm ≠ 0.

Assume first that Λm = 0, for an integer m ≥ 2. Then γm(0) = γm(1). Hence γm(< x >) is piecewise C, and continuous. Thus the Fourier series of γm(< x >) converges uniformly to γm(< x >), and γm(<x>)=1m(Λm+11m(m+1)(1+bm+1(r)bm+1(r1)))+n=,n0{1ms=1m(m)s(2πin)s(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))}e2πinx=1m(Λm+11m(m+1)(1+bm+1(r)bm+1(r1)))+1ms=1mms(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))s!n=,0e2πinx(2πin)s=1ms=0×s1mms(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))Bs(<x>)+Λm×B1(<x>),forxZc,0,forxZ.(69)

Now, we are going to state our first result.

Theorem 4.1

For each integer l ≥ 2, we let Λl=k=1l11k(lk)(2bk(r)bk(r1)),(70)

with Λ1 = 0. Assume that Λm = 0, for an integer m ≥ 2. Then we have the following.

  1. k=1m11k(mk)bk(r)(<x>)<x>mk has the Fourier series expansion k=1m11k(mk)bk(r)(<x>)<x>mk=1mΛm+11m(m+1)(1+bm+1(r)bm+1(r1))+n=,n01ms=1m(m)s(2πin)s(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))e2πinx,(71)

    for all x ∈ ℝ, where the convergence is uniform.

  2. k=1m11k(mk)bk(r)(<x>)<x>mk=1ms=0,s1mms(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))Bs(<x>),(72)

    for all x ∈ ℝ, where Bs(< x >) is the Bernoulli function.

Assume next that Λm ≠ 0, for an integer m ≥ 2. Then γm (0) ≠ γm(1). Hence γm(< x >) is piecewise C, and discontinuous with jump discontinuities at integers.Thus the Fourier series of γm(< x >) converges pointwise to γm(< x >), for x ∈ ℤc, and converges to 12(γm(0)+γm(1))=γm(0)+12Λm,(73)

for x ∈ ℤ. Next, we are going to state our second result.

Theorem 4.2

For each integer l ≥ 2, we let Λl=k=1l11k(lk)(2bk(r)bk(r1)),(74)

with Λ1 = 0. Assume that Λm ≠ 0, for an integer m ≥ 2. Then we have the following.

  1. 1m(Λm+11m(m+1)(1+bm+1(r)bm+1(r1)))+n=,n0{1ms=1m(m)s(2πin)s(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))}e2πinx=k=1m11k(mk)bk(r)(<x>)<x>mk,forxZc,12Λm,forxZ.(75)

  2. 1ms=0mms(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))Bs(<x>)=k=1m11k(mk)bk(r)(<x>)<x>mk,(76)

    for x ∈ ℤc and 1ms=0,s1mms(Λms+1+Hm1Hmsms+1(1+bms+1(r)bms+1(r1)))Bs(<x>)=12Λm,(77)

    for x ∈ ℤ.

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About the article

Received: 2017-04-07

Accepted: 2017-12-04

Published Online: 2017-12-29


Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1606–1617, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0134.

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© 2017 Kim et al.. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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