Fourier series of functions involving higher-order ordered Bell polynomials

Taekyun Kim; Dae San Kim; Gwan-Woo Jang; Lee Chae Jang

doi:10.1515/math-2017-0134

Article Open Access

Fourier series of functions involving higher-order ordered Bell polynomials

Taekyun Kim , Dae San Kim , Gwan-Woo Jang and Lee Chae Jang

Published/Copyright: December 29, 2017

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$Open Mathematics$

From the journal Open Mathematics Volume 15 Issue 1

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

12−errext=∑m=0∞bm(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 b_m(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=0∞nm2n+1,(m≥0). (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 b_m(x) has degree m and is a monic polynomial with integral coefficients, as we can see from

b0(x)=1,bm(x)=xm+∑l=0m−1mlbl(x),(m≥1) (3)

(see [11]). From (1), we can derive the following.

ddxbm(r)=mbm−1(r),(m≥1),bm(r)(x+1)−bm(r)(x)=bm(r)(x)−bm(r−1)(x),(m≥0). (4)

Also, from these we immediately get

bm(r)(1)−bm(r)=bm(r)−bm(r−1),(m≥0),∫01bm(r)(x)dx=1m+1(bm+1(r)(1)−bm+1(r))=1m+1(bm+1(r)−bm+1(r−1)). (5)

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

ter−1ext=∑m=0∞Bm(x)tmm!. (6)

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

for m ≥ 2,
Bm(<x>)=−m!∑n=−∞,n≠0∞e2πinx(2πin)m, (7)
for m = 1,
−∑n=−∞,n≠0∞,e2πinx2πin=B1(<x>),for x∈Zc,0,for x∈Z, (8)

where ℤ^c = ℝ − ℤ. Here we will consider the following three types of functions α_m(< x >), β_m(< x >), and γ_m(< 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:

αm(<X>)=∑k=0mbk(r)(<x>)<x>m−k,(m≥1);
βm(<x>)=∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k,(m≥1);
(<x>)=∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k,(m≥2).

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

As to γ_m(< 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 γ_m(< x >) .

∑k=1m−11k(m−k)bk(r)(x)xm−k=1m∑s=0mmsΛm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1))Bs(x), (9)

where Hm=∑j=1m1j are the harmonic numbers and Λl=∑k=1l−11k(l−k)(2bk(r)−bk(r−1)). The obvious polynomial identities can be derived also for α_m(< x >) and β_m(< 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=1m−11k(m−k)Bk(〈x〉)Bm−k(〈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)xm−k,(m≥1). Then we will consider the function

αm(<x>)=∑k=0mbk(r)(<x>)<x>m−k, (10)

defined on ℝ which is periodic of period 1. The Fourier series of α_m(< x >) is

∑n=−∞∞An(m)e2πinx, (11)

where

An(m)=∫01αm(<x>)e−2πinxdx=∫01αm(x)e−2πinxdx. (12)

To proceed further, we need to observe the following.

αmr(x)=∑k=0m{kbk−1(r)(x)xm−k+(m−k)bk)(r)(x)xm−k−1}=∑k=1mkbk−1(r)(x)xm−k+∑k=0m−1(m−k)bk(r)(x)xm−k−1=∑(=0m−1(k+1)bk(r)(x)xm−1−k+∑k=0m−1(m−k)bk(r)(x)xm−1−k=(m+1)αm−1(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(r−1)−bk(r)δm,k=∑l⇒=0m2bk(r)−bk(r−1)−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)e−2πinxdx=−12πinαm(x)e−2πinx01+12πin∫01αm′(x)e−2πinxdx=m+12πinAm(m−1)−12πinΔm, (19)

from which by induction on m we can deduce that

An(m)=−∑j=1m(m+1)j−1(2πin)jΔm−j+1=−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1. (20)

Case 2: n = 0.

A0(m)=∫01αm(x)dx=1m+2Δm+1. (21)

α_m(< x >), (m ≥ 1) is piecewise C^∞. In addition, α_m(< x >) is continuous for those positive integers with Δ_m = 0 and discontinuous with jump discontinuities at integers with Δ_m ≠ 0. Assume first that m is a positive integer with Δ_m = 0. Then α_m(0) = α_m(1). Thus the Fourier series of α_m(< x >) converges uniformly to α_m(< x >), and

αm(<x>)=1m+2Δm+1+∑n=−∞,n≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx=1m+2Δm+1+1m+2∑j=1mm+2jΔm−j+1−j!∑n=−∞,n≠0∞e2πin(2πin)j=1m+2Δm+1+1m+2∑j=2mm+2jΔm−j+1Bj(<x>)+Δm×B1(<x>), for x∈Zc,0,for x∈Z. (22)

Now, we can state our first result.

Theorem 2.1

For each positive integer l, we let

Δl=∑k=0l(2bk(r)−bk(r−1))−bl(r). (23)

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

∑k=0mbk(r)(<x>)<x>m−k has the Fourier series expansion
∑k=0mbk(r)(<x>)<x>m−k=1m+2Δm+1+∑n=−∞,n≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx, (24)

for all x ∈ ℝ, where the convergence is uniform.
∑k=0mbk(r)(<x>)<x>m−k=1m+2Δm+1+1m+2∑j=2mm+2jΔm−j+1Bj(<x>), (25)

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

Assume next that Δ_m ≠ 0, for a positive integer m. Then α_m(0) ≠α_m(1). Hence α_m(< x >) is piecewise C^∞, and discontinuous with jump discontinuities at integers. The Fourier series of α_m(< x >) converges pointwise to α_m(< x >), for x ∈ ℤ^c, 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(r−1))−bl(r). (27)

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

1m+2Δm+1+∑n=−∞,≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx=∑k=0mbk(r)(<x>)<x>m−k,forx∈Zc,bm(r)+12Δm,forx∈Z. (28)
1m+2∑j=0mm+2jΔm−j+1Bj(<x>)=∑k=0mbk(r)(<x>)<x>m−k, for x∈Zc;1m+2∑j=0,j≠1mm+2jΔm−j+1Bj(<x>)=bm(r)+12Δm, for x∈Z. (29)

3 Fourier series of functions of the second type

Let βm(x)=∑k=0m1k!(m−k)!bk(r)(x)xm−k,(m≥1). Then we will consider the function

βm(<x>)=∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k, (30)

defined on ℝ which is periodic with period 1. The Fourier series of β_m(< x >) is

∑n=−∞∞Bn(m)e2πinx, (31)

where

Bn(m)=∫01βm(<x>)e−2πinxdx=∫01βm(x)e−2πinxdx. (32)

To proceed further, we need to observe the following.

βm′(x)=∑k=0mkk!(m−k)!bk−1(r)(x)xm−k+m−kk!(m−k)!bk(r)(x)xm−k−1=∑k=1m1(k−1)!(m−k)!bk−1(r)(x)xm−k+∑k=0m−11k!(m−k−1)!bk(r)(x)xm−k−1=∑k=0m−11k!(m−1−k)!bk(r)(x)xm−1−k+∑k=0m−11k!(m−k−1)!bk(r)(x)xm−1−k=2βm−1(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!(m−k)!bk(r)(1)−bk(r)δm,k=∑k=0m1k!(m−k)!2bk(r)−bk(r−1)−bk(r)δm,k=∑k=0m1k!(m−k)!2bk(r)−bk(r−1)−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)e−2πinxdx=−12πinβm(x)e−2πinx01+12πin∫01βm′(x)e−2πinxdx=−12πin(βm(1)−βm(0))+22πin∫01βm−1(x)e−2πinxdx=22πinBn(m−1)−12πinΩm, (39)

from which by induction on m we can easily get

Bn(m)=−∑j=1m2j−1(2πin)jΩm−j+1. (40)

Case 2: n = 0.

B0(m)=∫01βm(x)dx=12Ωm+1. (41)

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

Assume first that Ω_m = 0, for a positive integer m. Then β_m(0) = β_m(1). Hence β_m(< x >) is piecewise C^∞, and continuous. Thus the Fourier series of β_m(< x >) converges uniformly to β_m(< x >), and

βm(<x>)=12Ωm+1+∑n=−∞,n≠0∞∑j=1m2j−1(2πin)jΩm−j+1e2πinx=12Ωm+1+∑j=1m2j−1j!Ωm−j+1−j!∑n=−∞,n≠0∞e2πinx(2πin)j=12Ωm+1+∑j=2m2j−1j!Ωm−j+1Bj(<x>)+Ωm×B1(<x>),forx∈Zc,0,forx∈Z. (42)

Now, we state our first result.

Theorem 3.1

For each positive integerl, we let

Ωl=∑k=0l1k!(l−k)!(2bk(r)−bk(r−1))−1l!bl(r). (43)

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

∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k has the Fourier series expansion
∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k=12Ωm+1+∑n=−∞n≠0∞,−∑j=1m2j−1(2πin)jΩm−j+1e2πinx, (44)

for all x ∈ ℝ, where the convergence is uniform.
∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k=∑j=0,j≠1m2j−1j!Ωm−j+1Bj(<x>), (45)

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

Assume next that m is a positive integer with Ω_m ≠ 0. Then β_m(0) ≠β_m(1). Hence β_m(< x >) is piecewise C^∞, and discontinuous with jump discontinuities at integers. The Fourier series of β_m(< x >) converges pointwise to β_m(< x >), for x ∈ ℤ^c, 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!(l−k)!(2bk(r)−bk(r−1))−1l!bl(r). (47)

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

12Ωm+1+∑n=−∞,n≠0∞−∑j=1m2j−1(2πin)jΩm−j+1e2πinx=∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k,forx∈Zc,1m!bm(r)+12Ωm, forx∈Z. (48)
∑j=0m2j−1j!Ωm−j+1Bj(<x>)=∑k=0m1k!(m−k)!bk(r)(<x>)<x>m−k, (49)

for x ∈ ℤ^c;
∑j=0,j≠1m2j−1j!Ωm−j+1Bj(<x>)=1m!bm(r)+12Ωm, (50)

for x ∈ ℤ.

4 Fourier series of functions of the third type

Let γm(x)=∑k=1m−11k(m−k)bk(r)(x)xm−k,(m≥2). Then we will consider the function.

γm(<x>)=∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k, (51)

defined on ℝ, which is periodic with period 1. The Fourier series of γ_m(< x >) is

∑n=−∞∞Cm(r)e2πinx, (52)

where

Cn(m)=∫01γm(<x>)e−2πinxdx=∫01γm(x)e−2πinxdx. (53)

To proceed further, we need to observe the following.

γmr(x)=∑k=1m−11m−kbk−1(r)(x)xm−k+∑k=1m−11kbk(r)(x)xm−k−1=∑k=0m−21m−1−kbk(r)(x)xm−1−k+∑k=1m−11kbk(r)(x)xm−1−k=∑k=1m−21m−1−k+1kbk(r)(x)xm−1−k+1m−1xm−1+1m−1bm−1(r)(x)=(m−1)γm−1(x)+1m−1xm−1+1m−1bm−1(r)(x). (54)

From this, we see that

1m(γm+1(x)−1m(m+1)xm+1−1m(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=1m−11k(m−k)(bk(r)(1)−bk(r)δm,k)=∑k=1m−11k(m−k)(2bk(r)−bk(r−1)). (57)

Notice here that,

γm(0)=γm(1)⟺Λm=0, (58)

and

∫01γm(x)dx=1mΛm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)). (59)

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

Case 1: n ≠ 0.

Cn(m)=∫01γm(x)e−2πinxdx=−12πin[γm(x)e−2πinx]01+12πin∫01γm′(x)e−2πinxdx=−12πin(γm(1)−γm(0))+12πin∫01(m−1)γm−1(x)+1m−1Xm−1+1m−1bm−1(r)(x)e−2πinxdx=m−12πinCn(m−1)−12πinΛm+12πin(m−1)∫01xm−1e−2πinxdx+12πin(m−1)∫01bm−1(r)(x)e−2πinxdx. (60)

We can show that

∫01xle−2πinxdx=−∑k=1l(l)k−1(2πin)k,forn≠0,1l+1,forn=0. (61)

Also, from the paper [12], we have

∫01bl(r)(x)e−2πinxdx=−∑k=1l(l)k−1(2πin)+(bl−k+1(r)−bl−k+1(r−1)),forn≠0,1l+1(bl+1(r)−bl+1(r−1)),forn=0. (62)

From (60), (61), and (62), we have

Cn(m)=m−12πinCn(m−1)−12πinΛm−12πin(m−1)Φm, (63)

where

Φm=∑k=1m−1(m−1)k−1(2πin)k(1+bm−k(r)−bm−k(r−1)). (64)

From (63) by induction on m we can deduce that

Cn(m)=−∑j=1m−1(m−1)j−1(2πin)jΛm−j+1−∑j=1m−1(m−1)j−1(2πin)j(m−j)Φm−j+1. (65)

We note here that

∑j=1m−1(m−1)j−1(2πin)j(m−j)Φm−j+1=∑j=1m−1(m−1)j−1(2πin)j(m−j)∑k=1m−j(m−j)l)−1(2πin)k(1+bm−j−k+1(r)−bm−j−k+1(r−1))=∑j=1m−11m−j∑k=1m−j(m−1)j+k−2(2πin)j+k(1+bm−j−k+1(r)−bm−j−k+1(r−1))=∑j=1m−11m−j∑s=j+1m(m−1)s−2(2πin)s(1+bm−s+1(r)−bm−s+1(r−1))=∑s=2m(m−1)s−2(2πin)s(1+bm−s+1(r)−bm−s+1(r−1))∑j=1s−11m−j=1m∑s=1m(m)s(2πin)sHm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)). (66)

Putting everything altogether, from (65), we finally obtain

Cn(m)=−1m∑s=1m(m)s(2πin)sΛm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)). (67)

Case 2: n = 0.

C0(m)=∫01γm(x)dx=1mΛm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)). (68)

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

Assume first that Λ_m = 0, for an integer m ≥ 2. Then γ_m(0) = γ_m(1). Hence γ_m(< x >) is piecewise C^∞, and continuous. Thus the Fourier series of γ_m(< x >) converges uniformly to γ_m(< x >), and

γm(<x>)=1m(Λm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)))+∑n=−∞,n≠0∞{−1m∑s=1m(m)s(2πin)s(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))}e2πinx=1m(Λm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)))+1m∑s=1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))−s!∑n=−∞,≠0∞e2πinx(2πin)s=1m∑s=0×s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs(<x>)+Λm×B1(<x>),forx∈Zc,0,forx∈Z. (69)

Now, we are going to state our first result.

Theorem 4.1

For each integer l ≥ 2, we let

Λl=∑k=1l−11k(l−k)(2bk(r)−bk(r−1)), (70)

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

∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k has the Fourier series expansion
∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k=1mΛm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1))+∑n=−∞,n≠0∞−1m∑s=1m(m)s(2πin)s(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))e2πinx, (71)

for all x ∈ ℝ, where the convergence is uniform.
∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k=1m∑s=0,s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs(<x>), (72)

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

Assume next that Λ_m ≠ 0, for an integer m ≥ 2. Then γ_m (0) ≠ γ_m(1). Hence γ_m(< x >) is piecewise C^∞, and discontinuous with jump discontinuities at integers.Thus the Fourier series of γ_m(< x >) converges pointwise to γ_m(< x >), for x ∈ ℤ^c, 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=1l−11k(l−k)(2bk(r)−bk(r−1)), (74)

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

1m(Λm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)))+∑n=−∞,n≠0∞{−1m∑s=1m(m)s(2πin)s(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))}e2πinx=∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k,forx∈Zc,12Λm,forx∈Z. (75)
1m∑s=0mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs(<x>)=∑k=1m−11k(m−k)bk(r)(<x>)<x>m−k, (76)

for x ∈ ℤ^c and
1m∑s=0,s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs(<x>)=12Λm, (77)

for x ∈ ℤ.

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Received: 2017-4-7

Accepted: 2017-12-4

Published Online: 2017-12-29

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