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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

# 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
• Email
• Other articles by this author:
/ 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.

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, $\begin{array}{}{b}_{m}^{\left(r\right)}={b}_{m}^{\left(r\right)}\left(0\right)\end{array}$ 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 $\begin{array}{}{b}_{m}\left(x\right)={b}_{m}^{\left(1\right)}\left(x\right)\text{\hspace{0.17em}and\hspace{0.17em}}{b}_{m}={b}_{m}^{\left(1\right)}.\end{array}$ 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 $bm=∑n=0mn!S2(m,n)=∑n=0∞nm2n+1,(m≥0).$(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 bm(x) has degree m and is a monic polynomial with integral coefficients, as we can see from $b0(x)=1,bm(x)=xm+∑l=0m−1mlbl(x),(m≥1)$(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 Bm(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 Bm(< x >):

1. for m ≥ 2, $Bm()=−m!∑n=−∞,n≠0∞e2πinx(2πin)m,$(7)

2. for m = 1, $−∑n=−∞,n≠0∞,e2πinx2πin=B1(),for x∈Zc,0,for x∈Z,$(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. $\begin{array}{}{\alpha }_{m}\left(\right)=\sum _{k=0}^{m}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k},\left(m\ge 1\right);\end{array}$

2. $\begin{array}{}{\beta }_{m}\left(\right)=\sum _{k=0}^{m}\frac{1}{k!\left(m-k\right)!}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k},\left(m\ge 1\right);\end{array}$

3. $\begin{array}{}\left(\right)=\sum _{k=1}^{m-1}\frac{1}{k\left(m-k\right)}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k},\left(m\ge 2\right).\end{array}$

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

where $\begin{array}{}{H}_{m}=\sum _{j=1}^{m}\frac{1}{j}\end{array}$ are the harmonic numbers and $\begin{array}{}{\mathit{\Lambda }}_{l}=\sum _{k=1}^{l-1}\frac{1}{k\left(l-k\right)}\left(2{b}_{k}^{\left(r\right)}-{b}_{k}^{\left(r-1\right)}\right).\end{array}$ 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 $\begin{array}{}\sum _{k=1}^{m-1}\frac{1}{k\left(m-k\right)}{B}_{k}\left(〈x〉\right){B}_{m-k}\left(〈x〉\right)\end{array}$ 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 $\begin{array}{}{\alpha }_{m}\left(x\right)=\sum _{k=0}^{m}{b}_{k}^{\left(r\right)}\left(x\right){x}^{m-k},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(m\ge 1\right).\end{array}$ Then we will consider the function $αm()=∑k=0mbk(r)()m−k,$(10)

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

where $An(m)=∫01αm()e−2πinxdx=∫01αm(x)e−2πinxdx.$(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 $\begin{array}{}{A}_{n}^{\left(m\right)}\end{array}$.

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)

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 $αm()=1m+2Δm+1+∑n=−∞,n≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx=1m+2Δm+1+1m+2∑j=1mm+2jΔm−j+1−j!∑n=−∞,n≠0∞e2πin(2πin)j=1m+2Δm+1+1m+2∑j=2mm+2jΔm−j+1Bj()+Δm×B1(), for x∈Zc,0,for x∈Z.$(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 Δm = 0. Then we have the following.

1. $\begin{array}{}\sum _{k=0}^{m}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k}\end{array}$ has the Fourier series expansion $∑k=0mbk(r)()m−k=1m+2Δm+1+∑n=−∞,n≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx,$(24)

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

2. $∑k=0mbk(r)()m−k=1m+2Δm+1+1m+2∑j=2mm+2jΔm−j+1Bj(),$(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 $Δl=∑k=0l(2bk(r)−bk(r−1))−bl(r).$(27)

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

1. $1m+2Δm+1+∑n=−∞,≠0∞−1m+2∑j=1m(m+2)j(2πin)jΔm−j+1e2πinx=∑k=0mbk(r)()m−k,forx∈Zc,bm(r)+12Δm,forx∈Z.$(28)

2. $1m+2∑j=0mm+2jΔm−j+1Bj()=∑k=0mbk(r)()m−k, for x∈Zc;1m+2∑j=0,j≠1mm+2jΔm−j+1Bj()=bm(r)+12Δm, for x∈Z.$(29)

## 3 Fourier series of functions of the second type

Let $\begin{array}{}{\beta }_{m}\left(x\right)=\sum _{k=0}^{m}\frac{1}{k!\left(m-k\right)!}{b}_{k}^{\left(r\right)}\left(x\right){x}^{m-k},\left(m\ge 1\right).\end{array}$ Then we will consider the function $βm()=∑k=0m1k!(m−k)!bk(r)()m−k,$(30)

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

where $Bn(m)=∫01βm()e−2πinxdx=∫01βm(x)e−2πinxdx.$(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 $\begin{array}{}{B}_{n}^{\left(m\right)}\end{array}$.

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)

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 $βm()=12Ωm+1+∑n=−∞,n≠0∞∑j=1m2j−1(2πin)jΩm−j+1e2πinx=12Ωm+1+∑j=1m2j−1j!Ωm−j+1−j!∑n=−∞,n≠0∞e2πinx(2πin)j=12Ωm+1+∑j=2m2j−1j!Ωm−j+1Bj()+Ωm×B1(),forx∈Zc,0,forx∈Z.$(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 Ωm = 0. Then we have the following.

1. $\sum _{k=0}^{m}\frac{1}{k!\left(m-k\right)!}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k}$ has the Fourier series expansion $∑k=0m1k!(m−k)!bk(r)()m−k=12Ωm+1+∑n=−∞n≠0∞,−∑j=1m2j−1(2πin)jΩm−j+1e2πinx,$(44)

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

2. $∑k=0m1k!(m−k)!bk(r)()m−k=∑j=0,j≠1m2j−1j!Ωm−j+1Bj(),$(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 $Ωl=∑k=0l1k!(l−k)!(2bk(r)−bk(r−1))−1l!bl(r).$(47)

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

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

2. $∑j=0m2j−1j!Ωm−j+1Bj()=∑k=0m1k!(m−k)!bk(r)()m−k,$(49)

for x ∈ ℤc; $∑j=0,j≠1m2j−1j!Ωm−j+1Bj()=1m!bm(r)+12Ωm,$(50)

for x ∈ ℤ.

## 4 Fourier series of functions of the third type

Let ${\gamma }_{m}\left(x\right)=\sum _{k=1}^{m-1}\frac{1}{k\left(m-k\right)}{b}_{k}^{\left(r\right)}\left(x\right){x}^{m-k},\phantom{\rule{thinmathspace}{0ex}}\left(m\ge 2\right).$ Then we will consider the function. $γm()=∑k=1m−11k(m−k)bk(r)()m−k,$(51)

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

where $Cn(m)=∫01γm()e−2πinxdx=∫01γm(x)e−2πinxdx.$(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 ${C}_{n}^{\left(m\right)}.$

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)

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 $γm()=1m(Λm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)))+∑n=−∞,n≠0∞{−1m∑s=1m(m)s(2πin)s(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))}e2πinx=1m(Λm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1)))+1m∑s=1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))−s!∑n=−∞,≠0∞e2πinx(2πin)s=1m∑s=0×s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs()+Λm×B1(),forx∈Zc,0,forx∈Z.$(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 Λm = 0, for an integer m ≥ 2. Then we have the following.

1. $\sum _{k=1}^{m-1}\frac{1}{k\left(m-k\right)}{b}_{k}^{\left(r\right)}\left(\right)}^{m-k}$ has the Fourier series expansion $∑k=1m−11k(m−k)bk(r)()m−k=1mΛm+1−1m(m+1)(1+bm+1(r)−bm+1(r−1))+∑n=−∞,n≠0∞−1m∑s=1m(m)s(2πin)s(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))e2πinx,$(71)

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

2. $∑k=1m−11k(m−k)bk(r)()m−k=1m∑s=0,s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs(),$(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 $Λl=∑k=1l−11k(l−k)(2bk(r)−bk(r−1)),$(74)

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

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

2. $1m∑s=0mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs()=∑k=1m−11k(m−k)bk(r)()m−k,$(76)

for x ∈ ℤc and $1m∑s=0,s≠1mms(Λm−s+1+Hm−1−Hm−sm−s+1(1+bm−s+1(r)−bm−s+1(r−1)))Bs()=12Λm,$(77)

for x ∈ ℤ.

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Accepted: 2017-12-04

Published Online: 2017-12-29

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1606–1617, ISSN (Online) 2391-5455,

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