1 Introduction and preliminaries
The Chebyshev polynomials \(T_{n}(x)\) of the first kind, the Chebyshev polynomials \(U_{n}(x)\) of the second kind, and the Fibonacci polynomials \(F_{n}(x)\) are respectively defined by the recurrence relations as follows (see [13‐15]):
When \(x=1\), \(F_{n}=F_{n}(1)\) (\(n\geq0\)) is the Fibonacci sequence.
$$\begin{aligned}& T_{n+2}(x)=2xT_{n+1}(x)-T_{n}(x)\quad (n\geq0), \qquad T_{0}(x)=1,\qquad T_{1}(x)=x, \end{aligned}$$
(1.1)
$$\begin{aligned}& U_{n+2}(x)=2xU_{n+1}(x)-U_{n}(x)\quad (n\geq0), \qquad U_{0}(x)=1,\qquad U_{1}(x)=2x, \end{aligned}$$
(1.2)
$$\begin{aligned}& F_{n+2}(x)=xF_{n+1}(x)+F_{n}(x)\quad (n\geq0),\qquad F_{0}(x)=0,\qquad F_{1}(x)=1. \end{aligned}$$
(1.3)
From (1.1), (1.2), and (1.3), it can be easily shown that the generating functions for \(T_{n}(x)\), \(U_{n}(x)\), and \(F_{n}(x)\) are respectively given by (see [13‐15]):
$$\begin{aligned}& \frac{1-xt}{1-2xt+t^{2}}=\sum_{n=0} ^{\infty} T_{n}(x)t^{n}, \end{aligned}$$
(1.4)
$$\begin{aligned}& F(t,x)=\frac{1}{1-2xt+t^{2}}=\sum_{n=0} ^{\infty}U_{n} (x)t^{n}, \end{aligned}$$
(1.5)
$$\begin{aligned}& G(t,x)=\frac{1}{1-xt-t^{2}}=\sum_{n=0} ^{\infty}F_{n+1}(x)t^{n}. \end{aligned}$$
(1.6)
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As is well known, the Bernoulli polynomials
\(B_{n}(x)\) are defined by the generating function
$$ \frac{t}{e^{t}-1}e^{xt}=\sum _{n=0} ^{\infty}B_{n}(x)\frac{t^{n}}{n!}. $$
(1.7)
For any real number x, we let
denote the fractional part of x, where \([x]\) indicates the greatest integer ≤x.
$$ \langle x\rangle =x- [x ]\in [0, 1 ) $$
(1.8)
For any integers m, r with \(m,r\geq1\), we put
where the sum runs over all nonnegative integers \(i_{1},i_{2},\ldots ,i_{r+1}\) with \(i_{1}+i_{2}+\cdots+i_{r+1}=m\).
$$ \alpha_{m,r}(x)=\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m} U_{i_{1}}(x)U_{i_{2}}(x)\cdots U_{i_{r+1}}(x), $$
(1.9)
Then we will consider the function \(\alpha_{m,r}(\langle x\rangle)\) and derive their Fourier series expansions. As a corollary to these Fourier series expansions, we will be able to express \(\alpha_{m,r}(x)\) in terms of Bernoulli polynomials \(B_{n}(x)\). Indeed, our result here is as follows.
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Theorem A
For any integers
m, r
with
\(m,r\geq1\), we let
Then we have the identity
Here
\((x)_{r}=x(x-1)\cdots(x-r+1)\)
for
\(r\geq1\), and
\((x)_{0}=1\).
$$ \Delta_{m,r}=\frac{1}{2^{r}r!}\sum_{k=0} ^{ [\frac{m-1}{2} ]}(-1)^{k}\binom{m+r-k}{k}(m+r-2k)_{r}2^{m+r-2k}. $$
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}(x)U_{i_{2}}(x) \cdots U_{i_{r+1}}(x) \\ &\quad =\frac{1}{2r}\sum_{j=0} ^{m} 2^{j}\binom{r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j}(x). \end{aligned}$$
(1.10)
Also, for any integers
m, r
with
\(m\geq1\), \(r\geq2\), we let
where the sum runs over all nonnegative integers
\(i_{1},i_{2},\ldots,i_{r}\)
with
\(i_{1}+i_{2}+\cdots+i_{r}=m\).
$$ \beta_{m,r}(x)=\sum_{i_{1}+i_{2}+\cdots +i_{r}=m}F_{i_{1}+1}(x)F_{i_{2}+1}(x) \cdots F_{i_{r}+1}(x), $$
(1.11)
Then we will consider the function \(\beta_{m,r}(\langle x\rangle)\) and derive their Fourier series expansions. Again, as an immediate corollary to these, we can express \(\beta_{m,r}(x)\) as a linear combination of Bernoulli polynomials. In detail, our result is as follows.
Theorem B
For any integers
m, r
with
\(m\geq1\), \(r\geq2\), we let
Then we have the identity
$$ \Omega_{m,r}=\sum_{l=0} ^{ [\frac{m-1}{2} ]} \binom {m+r-1-l}{l}\binom{m+r-1-2l}{r-1}. $$
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r}=m}F_{i_{1}+1}(x)F_{i_{2}+1}(x) \cdots F_{i_{r}+1}(x) \\ &\quad =\frac{1}{r-1}\sum_{j=0} ^{m} \binom{r-2+j}{j}\Omega_{m-j+1,r+j-1}B_{j}(x). \end{aligned}$$
(1.12)
One particular thing we have to note here is that neither \(U_{n}(x)\) nor \(F_{n}(x)\) is Appell polynomials, while all our related results so far have been only about Appell polynomials (see [1, 5‐8]).
Moreover, we will get some interesting identities that follow from Theorems A and B together with Lemmas 1 and 2 in [9].
As was mentioned in [7], studying these kinds of sums of finite products of special polynomials can be well justified by the following. Let us put
$$ \gamma_{m}(x)=\sum_{k=1} ^{m-1}\frac{1}{k(m-k)}B_{k}(x)B_{m-k}(x)\quad (m \geq2). $$
(1.13)
Then from the Fourier series expansion of \(\gamma_{m}(\langle x\rangle)\) we can express \(\gamma_{m}(x)\) in terms of Bernoulli polynomials just as in (1.10) and (1.12). Then, after some simple modification of this expression, we are able to obtain the famous Faber–Pandharipande–Zagier identity (see [3]) and some slightly different variant of Miki’s identity (see [2, 4, 10, 12]). For the details on this, the reader is referred to Introduction of the paper [7]. For some related results, we let the reader refer to the papers [1, 5‐8].
2 Fourier series expansions for functions associated with Chebyshev polynomials of the second kind
By differentiating equation (1.5) it was shown in [15] and mentioned in [13] that the sum of products in (1.9) can be neatly expressed as in the following. This will play a crucial role in this paper.
Lemma 2.1
Let
n, r
be nonnegative integers. Then we have the identity
where the sum runs over all nonnegative integers
\(i_{1},i_{2},\ldots ,i_{r+1}\)
with
\(i_{1}+i_{2}+\cdots+i_{r+1}=n\).
$$ \sum_{i_{1}+i_{2}+\cdots+i_{r+1}=n}U_{i_{1}}(x)U_{i_{2}}(x) \cdots U_{i_{r+1}}(x)=\frac{1}{2^{r}r!}U_{n+r} ^{(r)}(x), $$
(2.1)
It is well known that the Chebyshev polynomials of the second kind \(U_{n}(x)\) are explicitly given by (see [11, 13])
The rth derivative of (2.1) is given by
Then, combining (2.1) and (2.3), we obtain
$$ U_{n}(x)=\sum_{k=0} ^{ [\frac{n}{2} ]}(-1)^{k}\binom {n-k}{k}(2x)^{n-2k}. $$
(2.2)
$$ U_{n} ^{(r)}(x)=\sum _{k=0} ^{ [\frac{n-r}{2} ]}(-1)^{k}\binom {n-k}{k}(n-2k)_{r}2^{n-2k}x^{n-2k-r}. $$
(2.3)
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=n}U_{i_{1}}(x)U_{i_{2}}(x) \cdots U_{i_{r+1}}(x) \\ &\quad =\frac{1}{2^{r}r!}\sum_{k=0} ^{ [\frac{n}{2} ]}(-1)^{k}\binom{n+r-k}{k}(n+r-2k)_{r}2^{n+r-2k}x^{n-2k}. \end{aligned}$$
(2.4)
As in (1.9), we let
where \(m,r\geq1\), and the sum runs over all nonnegative integers \(i_{1},i_{2},\ldots,i_{r+1}\) with \(i_{1}+i_{2}+\cdots+i_{r+1}=m\).
$$ \alpha_{m,r}(x)=\sum_{i_{1}+i_{2}+\cdots +i_{r+1}=m}U_{i_{1}}(x)U_{i_{2}}(x) \cdots U_{i_{r+1}}(x), $$
Then we will consider the function
defined on \({\mathbb{R}}\), which is periodic with period 1.
$$ \alpha_{m,r}\bigl(\langle x\rangle\bigr)=\sum _{i_{1}+i_{2}+\cdots +i_{r+1}=m}U_{i_{1}}\bigl(\langle x\rangle \bigr)U_{i_{2}}\bigl(\langle x\rangle\bigr)\cdots U_{i_{r+1}}\bigl( \langle x\rangle\bigr), $$
(2.5)
The Fourier series of \(\alpha_{m,r}(\langle x\rangle)\) is
where
$$ \sum_{n=-\infty} ^{\infty}A_{n} ^{(m,r)}e^{2\pi inx}, $$
(2.6)
$$\begin{aligned} A_{n} ^{(m,r)}&= \int_{0} ^{1} \alpha_{m,r}\bigl(\langle x \rangle\bigr)e^{-2\pi inx}\,dx \\ &= \int_{0} ^{1} \alpha_{m,r}(x)e^{-2\pi inx} \,dx. \end{aligned}$$
(2.7)
For \(m,r\geq1\), we put
Then, for (2.4) and (2.8), we get
where we note that
$$\begin{aligned} \Delta_{m,r}&=\alpha_{m,r}(1)-\alpha_{m,r}(0) \\ &=\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m} \bigl(U_{i_{1}}(1)\cdots U_{i_{r+1}}(1)-U_{i_{1}}(0)\cdots U_{i_{r+1}}(0) \bigr). \end{aligned}$$
(2.8)
$$ \Delta_{m,r}=\frac{1}{2^{r}r!}\sum _{k=0} ^{ [\frac{m-1}{2} ]}(-1)^{k}\binom{m+r-k}{k}(m+r-2k)_{r}2^{m+r-2k}, $$
(2.9)
$$ \alpha_{m,r}(0)= \textstyle\begin{cases} 0,&\text{if }m\text{ is odd}, \\ (-1)^{\frac{m}{2}}\binom{\frac{m}{2}+r}{\frac{m}{2}},&\text{if } m\text{ is even}. \end{cases} $$
(2.10)
Now, using (2.1), we note the following:
Thus we have shown that
Replacing m by \(m+1\) and r by \(r-1\), from (2.11) we have
and
$$\begin{aligned} \frac{d}{dx}\alpha_{m,r}(x)&=\frac{d}{dx} \biggl( \frac {1}{2^{r}r!}U_{m+r} ^{(r)}(x) \biggr) \\ &=\frac{1}{2^{r}r!}U_{m+r} ^{(r+1)}(x) \\ &=2(r+1)\alpha_{m-1,r+1}(x). \end{aligned}$$
$$ \frac{d}{dx}\alpha_{m,r}(x)=2(r+1) \alpha_{m-1,r+1}(x). $$
(2.11)
$$\begin{aligned}& \frac{d}{dx} \biggl(\frac{\alpha_{m+1,r-1}(x)}{2r} \biggr)= \alpha_{m,r}(x), \end{aligned}$$
(2.12)
$$\begin{aligned}& \int_{0} ^{1} \alpha_{m,r}(x)\,dx= \frac{1}{2r}\Delta_{m+1,r-1}, \end{aligned}$$
(2.13)
$$ \alpha_{m,r}(0)=\alpha_{m,r}(1)\quad \Longleftrightarrow\quad \Delta_{m,r}=0. $$
(2.14)
We are now ready to determine the Fourier coefficients \(A_{n} ^{(m)}\).
Case 1: \(n\neq0\).
$$\begin{aligned} A_{n} ^{(m,r)} =& \int_{0} ^{1} \alpha_{m,r}(x)e^{-2\pi inx} \,dx \\ =&-\frac{1}{2\pi in} \bigl[\alpha_{m,r}(x)e^{-2\pi inx} \bigr]_{0} ^{1} +\frac{1}{2\pi in} \int_{0} ^{1} \biggl(\frac{d}{dx}\alpha _{m,r}(x) \biggr)e^{-2\pi inx}\,dx \\ =&\frac{2(r+1)}{2\pi in} \int_{0} ^{1}\alpha_{m-1,r+1}(x)e^{-2\pi inx} \,dx-\frac{1}{2\pi in} \bigl(\alpha_{m,r}(1)-\alpha_{m,r}(0) \bigr) \\ =&\frac{2(r+1)}{2\pi in}A_{n} ^{(m-1,r+1)}-\frac{1}{2\pi in}\Delta _{m,r} \\ =&\frac{2(r+1)}{2\pi in} \biggl(\frac{2(r+2)}{2\pi in}A_{n} ^{(m-2,r+2)}- \frac{1}{2\pi in}\Delta_{m-1,r+1} \biggr)-\frac{1}{2\pi in} \Delta_{m,r} \\ =&\frac{2^{2}(r+2)_{2}}{(2\pi in)^{2}}A_{n} ^{(m-2,r+2)}-\sum _{j=1} ^{2}\frac {2^{j-1}(r+j-1)_{j-1}}{(2\pi in)^{j}}\Delta_{m-j+1,r+j-1} \\ =&\cdots \\ =&\frac{2^{m}(r+m)_{m}}{(2\pi in)^{m}}A_{n} ^{(0,r+m)}-\sum _{j=1} ^{m}\frac {2^{j-1}(r+j-1)_{j-1}}{(2\pi in)^{j}}\Delta_{m-j+1,r+j-1} \\ =&-\sum_{j=1} ^{m}\frac{2^{j-1}(r+j-1)_{j-1}}{(2\pi in)^{j}}\Delta _{m-j+1,r+j-1} \\ =&-\frac{1}{2r}\sum_{j=1} ^{m} \frac{2^{j}(r+j-1)_{j}}{(2\pi in)^{j}}\Delta _{m-j+1,r+j-1}. \end{aligned}$$
(2.15)
Case 2: \(n=0\).
Before proceeding further, we recall here that
$$ A_{0} ^{(m,r)}= \int_{0} ^{1}\alpha_{m,r} (x)\,dx= \frac{1}{2r}\Delta_{m+1,r-1}. $$
(2.16)
(a)
for \(m\geq2\),
$$ B_{m}\bigl(\langle x\rangle\bigr)=-m!\sum _{\substack{n=-\infty\\n\neq0}} ^{\infty}\frac {e^{2\pi inx}}{(2\pi in)^{m}}, $$
(2.17)
(b)
for \(m=1\),
$$ -\sum_{\substack{n=-\infty\\n\neq0}} ^{\infty} \frac{e^{2\pi inx}}{2\pi in}= \textstyle\begin{cases} B_{1}(\langle x\rangle)&\text{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ 0&\text{for }x\in{\mathbb{Z}}. \end{cases} $$
(2.18)
From (2.15)–(2.18), we now obtain the Fourier series of \(\alpha_{m,r}(\langle x\rangle)\) given by
$$\begin{aligned} &\frac{1}{2r}\Delta_{m+1,r-1}-\sum_{\substack{n=-\infty\\n\neq0}} ^{\infty} \Biggl(\frac{1}{2r}\sum_{j=1} ^{m} \frac{2^{j}(r+j-1)_{j}}{(2\pi in)^{j}}\Delta_{m-j+1,r+j-1} \Biggr)e^{2\pi inx} \\ &\quad =\frac{1}{2r}\Delta_{m+1,r-1}+\frac{1}{2r}\sum _{j=1} ^{m} 2^{j}\binom {r+j-1}{r-1} \Delta_{m-j+1,r+j-1}\times \Biggl(-j!\sum_{\substack {n=-\infty\\n\neq0}} ^{\infty}\frac{e^{2\pi inx}}{(2\pi in)^{j}} \Biggr) \\ &\quad =\frac{1}{2r}\Delta_{m+1,r-1}+\frac{1}{2r}\sum _{j=2} ^{m} 2^{j} \binom {r+j-1}{r-1} \Delta_{m-j+1,r+j-1} \\ &\qquad {}+\Delta_{m,r}\times \textstyle\begin{cases} B_{1}(\langle x\rangle)&\text{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ 0&\text{for } x\in{\mathbb{Z}}. \end{cases}\displaystyle \\ &\quad =\frac{1}{2r}\sum_{\substack{j=0\\j\neq1}} ^{m} 2^{j} \binom {r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j}\bigl( \langle x\rangle\bigr) \\ &\qquad {}+\Delta_{m,r}\times \textstyle\begin{cases} B_{1}(\langle x\rangle)&\text{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ 0&\text{for }x\in{\mathbb{Z}}. \end{cases}\displaystyle \end{aligned}$$
(2.19)
\(\alpha_{m,r}(\langle x\rangle)\) (\(m,r\geq1\)) is piecewise \(C^{\infty}\). Moreover, \(\alpha_{m,r}(\langle x\rangle)\) is continuous for those positive integers m, r with \(\Delta_{m,r}=0\), and discontinuous with jump discontinuities at integers for those positive integers m, r with \(\Delta_{m,r}\neq0\). Thus, for \(\Delta_{m,r}=0\), the Fourier series of \(\alpha_{m,r}(\langle x\rangle)\) converges uniformly to \(\alpha_{m,r}(\langle x\rangle)\). On the other hand, for \(\Delta_{m,r}\neq0\), the Fourier series of \(\alpha_{m,r}(\langle x\rangle)\) converges pointwise to \(\alpha_{m,r}(\langle x\rangle)\) for \(x\in{\mathbb{R}}-{\mathbb{Z}}\), and converges to
for \(x\in{\mathbb{Z}}\).
$$ \frac{1}{2} \bigl(\alpha_{m,r}(0)+ \alpha_{m,r}(1) \bigr)=\alpha _{m,r}(0)+\frac{1}{2} \Delta_{m,r} $$
(2.20)
Theorem 2.2
For any integers
m, r
with
\(m,r\geq1\), we let
Assume that
\(\Delta_{m,r}=0\)
for some positive integers
\(m,r\). Then we have the following:
$$ \Delta_{m,r}=\frac{1}{2^{r}r!}\sum_{k=0} ^{ [\frac{m-1}{2} ]}(-1)^{k}\binom{m+r-k}{k}(m+r-2k)_{r}2^{m+r-2k}. $$
(a)
has the Fourier series expansion
for all
\(x\in{\mathbb{R}}\), where the convergence is uniform.
$$ \sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}\bigl(\langle x\rangle \bigr)U_{i_{2}}\bigl(\langle x\rangle\bigr)\cdots U_{i_{r+1}}\bigl( \langle x\rangle\bigr) $$
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}\bigl(\langle x\rangle \bigr)U_{i_{2}}\bigl(\langle x\rangle\bigr)\cdots U_{i_{r+1}}\bigl( \langle x\rangle\bigr) \\ &\quad =\frac{1}{2r}\Delta_{m+1,r-1}-\sum _{\substack{n=-\infty\\n\neq 0}} ^{\infty} \Biggl(\frac{1}{2r}\sum _{j=1} ^{m} \frac {2^{j}(r+j-1)_{j}}{(2\pi in)^{j}}\Delta_{m-j+1,r+j-1} \Biggr)e^{2\pi inx} \end{aligned}$$
(b)
for all
\(x\in{\mathbb{R}}\).
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}\bigl(\langle x\rangle \bigr)U_{i_{2}}\bigl(\langle x\rangle\bigr)\cdots U_{i_{r+1}}\bigl( \langle x\rangle\bigr) \\ &\quad =\frac{1}{2r}\sum_{\substack{j=0\\j\neq1}} ^{m} 2^{j}\binom {r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j}\bigl( \langle x\rangle\bigr) \end{aligned}$$
Theorem 2.3
For any integers
m, r
with
\(m,r\geq1\), we let
Assume that
\(\Delta_{m,r}\neq0\)
for some positive integers
m, r. Then we have the following:
$$ \Delta_{m,r}=\frac{1}{2^{r}r!}\sum_{k=0} ^{ [\frac{m-1}{2} ]}(-1)^{k}\binom{m+r-k}{k}(m+r-2k)_{r}2^{m+r-2k}. $$
(a)
$$\begin{aligned} &\frac{1}{2r}\Delta_{m+1,r-1}-\sum_{\substack{n=-\infty\\n \neq 0}} ^{\infty} \Biggl(\frac{1}{2r}\sum_{j=1} ^{m} \frac {2^{j}(r+j-1)_{j}}{(2\pi in)^{j}}\Delta_{m-j+1,r+j-1} \Biggr)e^{2\pi inx} \\ &\quad = \textstyle\begin{cases} \sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}(\langle x\rangle)U_{i_{2}}(\langle x\rangle)\cdots U_{i_{r+1}}(\langle x\rangle)&\textit{for }x\in\mathbb{R}-\mathbb{Z}, \\ \frac{1}{2}\Delta_{m,r}& \textit{for }x\in{\mathbb{Z}} \textit{ and }m \textit{ odd}, \\ (-1)^{\frac{m}{2}}\binom{\frac{m}{2}+r}{\frac{m}{2}}+\frac {1}{2}\Delta_{m,r} &\textit{for }x\in{\mathbb{Z}}\textit{ and }m\textit{ even}. \end{cases}\displaystyle \end{aligned}$$
(b)
$$\begin{aligned} &\frac{1}{2r}\sum_{j=0} ^{m} 2^{j} \binom{r+j-1}{r-1}\Delta _{m-j+1,r+j-1}B_{j}\bigl( \langle x\rangle\bigr) \\ &\quad =\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}U_{i_{1}}\bigl(\langle x \rangle\bigr)U_{i_{2}}\bigl(\langle x\rangle\bigr)\cdots U_{i_{r+1}} \bigl(\langle x\rangle\bigr)\quad \textit{for }x\in{\mathbb{R}}-{\mathbb{Z}}; \\ &\frac{1}{2r}\sum_{\substack{j=0\\j\neq1}} ^{m} 2^{j} \binom {r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j}\bigl( \langle x\rangle\bigr) \\ &\quad = \textstyle\begin{cases} \frac{1}{2}\Delta_{m,r}&\textit{for }x\in\mathbb{Z}\textit{ and }m \textit{ odd}, \\ (-1)^{\frac{m}{2}}\binom{\frac{m}{2}+r}{\frac{m}{2}}+\frac {1}{2}\Delta_{m,r}&\textit{for }x\in{\mathbb{Z}}\textit{ and }m\textit{ even}. \end{cases}\displaystyle \end{aligned}$$
3 Fourier series expansions for functions associated with Fibonacci polynomials
The following lemma is stated as equation (7) in [14] which is important for our purpose.
Lemma 3.1
Let
n, r
be integers with
\(n\geq0\), \(r\geq1\). Then we have the identity
where the sum runs over all nonnegative integers
\(i_{1},i_{2},\ldots,i_{r}\)
with
\(i_{1}+i_{2}+\cdots+i_{r}=n\).
$$ \sum_{i_{1}+i_{2}+\cdots+i_{r}=n}F_{i_{1}+1}(x)F_{i_{2}+1}(x) \cdots F_{i_{r}+1}(x)=\frac{1}{(r-1)!}F_{n+r} ^{(r-1)}(x), $$
(3.1)
An explicit expression for \(F_{n+1}(x)\) (\(n\geq0\)) is stated in equation (9) of [14].
$$ F_{n+1}(x)=\sum_{l=0} ^{ [\frac{n}{2} ]}\binom{n-l}{l}x^{n-2l}. $$
(3.2)
As was noted in (10) of [14], the \((r-1)\)th derivative of \(F_{n+r}(x)\) is
In addition, it was also noted in [14] that, combining (3.1) and (3.3), we have
$$ F_{n+r} ^{(r-1)}(x)=\sum _{l=0} ^{ [\frac{n}{2} ]}\frac {(n+r-1-l)!}{l!(n-2l)!}x^{n-2l}. $$
(3.3)
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r}=n}F_{i_{1}+1}(x)F_{i_{2}+1}(x) \cdots F_{i_{r}+1}(x) \\ &\quad =\sum_{l=0} ^{ [\frac{n}{2} ]}\binom{n+r-1-l}{l} \binom {n+r-1-2l}{r-1}x^{n-2l}. \end{aligned}$$
(3.4)
As in (1.11), we let
where \(m\geq1\), \(r\geq2\), and the sum runs over all nonnegative integers \(i_{1},i_{2},\ldots,i_{r}\) with \(i_{1}+i_{2}+\cdots+i_{r}=m\).
$$ \beta_{m,r}(x)=\sum_{i_{1}+i_{2}+\cdots +i_{r}=m}F_{i_{1}+1}(x)F_{i_{2}+1}(x) \cdots F_{i_{r}+1}(x), $$
Then we will consider the function
defined on \({\mathbb{R}}\), which is periodic with period 1.
$$ \beta_{m,r}\bigl(\langle x\rangle\bigr)=\sum _{i_{1}+i_{2}+\cdots +i_{r}=m}F_{i_{1}+1}\bigl(\langle x\rangle \bigr)F_{i_{2}+1}\bigl(\langle x\rangle\bigr)\cdots F_{i_{r}+1}\bigl( \langle x\rangle\bigr), $$
(3.5)
The Fourier series of \(\beta_{m,r}(\langle x\rangle)\) is
where
$$ \sum_{n=-\infty} ^{\infty} B_{n} ^{(m,r)}e^{2\pi inx}, $$
(3.6)
$$\begin{aligned} B_{n} ^{(m,r)}&= \int_{0} ^{1}\beta_{m,r}\bigl(\langle x \rangle\bigr)e^{-2\pi inx}\,dx \\ &= \int_{0} ^{1} \beta_{m,r}(x)e^{-2\pi inx} \,dx. \end{aligned}$$
(3.7)
For \(m\geq1\), \(r\geq2\), we set
Then, from (3.4) and (3.8), we have
In particular, we note that \(\Omega_{m,r}>0\) for any \(m\geq1\), \(r\geq 2\). Also, we note that
$$\begin{aligned} \Omega_{m,r}&=\beta_{m,r}(1)-\beta_{m,r}(0) \\ &=\sum_{i_{1}+i_{2}+\cdots+i_{r}=m} \bigl(F_{i_{1}+1}(1)\cdots F_{i_{r}+1}(1)-F_{i_{1}+1}(0)\cdots F_{i_{r}+1}(0) \bigr). \end{aligned}$$
(3.8)
$$ \Omega_{m,r}=\sum_{l=0} ^{ [\frac{m-1}{2} ]}\binom {m+r-1-l}{l}\binom{m+r-1-2l}{r-1}. $$
(3.9)
$$ \beta_{m,r}(0)= \textstyle\begin{cases} 0,&\text{if }m\mbox{ is odd}, \\ \binom{\frac{m}{2}+r-1}{\frac{m}{2}},&\text{if }m\mbox{ is even}. \end{cases} $$
(3.10)
Now, using (3.1), we see the following:
Thus we have shown that
Replacing m by \(m+1\) and r by \(r-1\), from (3.11) we get
and
We are now going to determine the Fourier coefficients \(B_{n} ^{(m)}\).
$$\begin{aligned} \frac{d}{dx}\beta_{m,r}(x)&=\frac{d}{dx} \biggl( \frac {1}{(r-1)!}F_{m+r} ^{(r-1)}(x) \biggr) \\ &=\frac{1}{(r-1)!}F_{m+r} ^{(r)}(x) \\ &=r\beta_{m-1,r+1}(x). \end{aligned}$$
$$ \frac{d}{dx}\beta_{m,r}(x)=r\beta_{m-1,r+1}(x). $$
(3.11)
$$\begin{aligned}& \frac{d}{dx} \biggl(\frac{\beta_{m+1,r-1}(x)}{r-1} \biggr)= \beta_{m,r}(x), \end{aligned}$$
(3.12)
$$\begin{aligned}& \int_{0} ^{1} \beta_{m,r}(x)\,dx= \frac{1}{r-1}\Omega_{m+1,r-1}, \end{aligned}$$
(3.13)
$$ \beta_{m,r}(0)=\beta_{m,r}(1) \quad \Longleftrightarrow\quad \Omega_{m,r}=0. $$
(3.14)
Case 1: \(n\neq0\).
$$\begin{aligned} B_{n} ^{(m,r)}&= \int_{0} ^{1} \beta_{m,r}(x)e^{-2\pi inx} \,dx \\ &=-\frac{1}{2\pi in} \bigl[\beta_{m,r}(x)e^{-2\pi inx} \bigr]_{0} ^{1} +\frac{1}{2\pi in} \int_{0} ^{1} \biggl(\frac{d}{dx} \beta_{m,r}(x) \biggr)e^{-2\pi inx}\,dx \\ &=\frac{r}{2\pi in} \int_{0} ^{1}\beta_{m-1,r+1}(x)e^{-2\pi inx} \,dx-\frac {1}{2\pi in} \bigl(\beta_{m,r}(1)-\beta_{m,r}(0) \bigr) \\ &=\frac{r}{2\pi in}B_{n} ^{(m-1,r+1)}-\frac{1}{2\pi in} \Omega_{m,r} \\ &=\frac{r}{2\pi in} \biggl(\frac{r+1}{2\pi in}B_{n} ^{(m-2,r+2)}- \frac {1}{2\pi in}\Omega_{m-1,r+1} \biggr)-\frac{1}{2\pi in} \Omega_{m,r} \\ &=\frac{(r+1)_{2}}{(2\pi in)^{2}}B_{n} ^{(m-2,r+2)}-\sum _{j=1} ^{2}\frac {(r+j-2)_{j-1}}{(2\pi in)^{j}}\Omega_{m-j+1,r+j-1} \\ &=\cdots \\ &=\frac{(r+2)_{m}}{(2\pi in)^{m}}B_{n} ^{(0,r+m)}-\sum _{j=1} ^{m}\frac {(r+j-2)_{j-1}}{(2\pi in)^{j}}\Omega_{m-j+1,r+j-1} \\ &=-\sum_{j=1} ^{m}\frac{(r+j-2)_{j-1}}{(2\pi in)^{j}}\Omega _{m-j+1,r+j-1} =-\frac{1}{r-1}\sum_{j=1} ^{m} \frac{(r+j-2)_{j}}{(2\pi in)^{j}}\Omega _{m-j+1,r+j-1}. \end{aligned}$$
(3.15)
Case 2: \(n=0\).
$$ B_{0} ^{(r,m)}= \int_{0} ^{1}\beta_{m,r}(x)\,dx= \frac{1}{r-1}\Omega_{m+1,r-1}. $$
(3.16)
From (2.17), (2.18), (3.15), and (3.16), we now have the following Fourier series expansion of \(\beta_{m,r}(\langle x\rangle)\) given by
$$\begin{aligned} &\frac{1}{r-1}\Omega_{m+1,r-1}-\sum_{\substack{n=-\infty\\n\neq 0}} ^{\infty} \Biggl(\frac{1}{r-1}\sum_{j=1} ^{m}\frac {(r+j-2)_{j}}{(2\pi in)^{j}}\Omega_{m-j+1,r+j-1} \Biggr)e^{2\pi inx} \\ &\quad =\frac{1}{r-1}\Omega_{m+1,r-1} \\ &\qquad {}+\frac{1}{r-1}\sum_{j=1} ^{m} \binom{r-2+j}{j}\Omega _{m-j+1,r+j-1}\times \Biggl(-j!\sum _{\substack{n=-\infty\\n\neq0}} ^{\infty}\frac{e^{2\pi inx}}{(2\pi in)^{j}} \Biggr) \\ &\quad =\frac{1}{r-1}\Omega_{m+1,r-1}+\frac{1}{r-1}\sum _{j=2} ^{m} \binom {r-2+j}{j}\Omega_{m-j+1,r+j-1}B_{j} \bigl(\langle x\rangle\bigr) \\ &\qquad {}+\Omega_{m,r}\times \textstyle\begin{cases} B_{1} (\langle x\rangle)&\text{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ 0&\text{for }x\in{\mathbb{Z}} \end{cases}\displaystyle \\ &\quad =\frac{1}{r-1}\sum_{\substack{j=0\\j\neq1}} ^{m} \binom {r-2+j}{j}\Omega_{m-j+1,r+j-1}B_{j}\bigl(\langle x\rangle\bigr) \\ &\qquad {}+\Omega_{m,r}\times \textstyle\begin{cases} B_{1}(\langle x\rangle)&\text{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ 0&\text{for }x\in{\mathbb{Z}}. \end{cases}\displaystyle \end{aligned}$$
(3.17)
\(\beta_{m,r}(\langle x\rangle)\) (\(m\geq1\), \(r\geq2\)) is piecewise \(C^{\infty}\) and discontinuous with jump discontinuities at integers, as \(\Omega _{m,r}>0\) for any \(m\geq1\), \(r\geq2\). Thus the Fourier series of \(\beta_{m,r}(\langle x\rangle)\) converges pointwise to \(\beta_{m,r}(\langle x\rangle)\) for \(x\in{\mathbb{R}}-{\mathbb{Z}}\), and converges to
for \(x\in{\mathbb{Z}}\).
$$ \frac{1}{2} \bigl(\beta_{m,r}(0)+ \beta_{m,r}(1) \bigr)=\beta _{m,r}(0)+\frac{1}{2} \Omega_{m,r} $$
(3.18)
Theorem 3.2
For any integers
m, r
with
\(m\geq1\), \(r\geq2\), we let
Then we have the following:
$$ \Omega_{m,r}=\sum_{l=0} ^{ [\frac{m-1}{2} ]} \binom {m+r-1-l}{l}\binom{m+r-1-2l}{r-1}. $$
(a)
$$\begin{aligned} &\frac{1}{r-1}\Omega_{m+1,r-1}-\sum_{\substack{n=-\infty\\n\neq 0}} ^{\infty} \Biggl(\frac{1}{r-1}\sum_{j=1} ^{m} \frac {(r-2+j)_{j}}{(2\pi in)^{j}}\Omega_{m-j+1,r+j-1} \Biggr)e^{2\pi inx} \\ &\quad = \textstyle\begin{cases} \sum_{i_{1}+i_{2}+\cdots+i_{r}=m}F_{i_{1}+1}(\langle x\rangle)F_{i_{2}+1}(\langle x\rangle)\cdots F_{i_{r}+1}(\langle x\rangle)&\textit{for }x\in{\mathbb{R}}-{\mathbb{Z}}, \\ \frac{1}{2}\Omega_{m,r}& \textit{for }x\in{\mathbb{Z}}\textit{ and }m\textit{ odd}, \\ \binom{\frac{m}{2}+r-1}{\frac{m}{2}}+\frac{1}{2}\Omega_{m,r}& \textit{for }x\in{\mathbb{Z}}\textit{ and }m\textit{ even}. \end{cases}\displaystyle \end{aligned}$$
(b)
$$\begin{aligned} &\frac{1}{r-1}\sum_{j=0} ^{m} \binom{r-2+j}{j}\Omega _{m-j+1,r+j-1}B_{j}\bigl(\langle x\rangle \bigr) \\ &\quad =\sum_{i_{1}+i_{2}+\cdots+i_{r}=m}F_{i_{1}+1}\bigl(\langle x \rangle\bigr)F_{i_{2}+1}\bigl(\langle x\rangle\bigr)\cdots F_{i_{r}+1} \bigl(\langle x\rangle\bigr)\quad \textit{for }x\in{\mathbb{R}}-{\mathbb{Z}}; \\ &\frac{1}{r-1}\sum_{\substack{j=0\\j\neq1}} ^{m} \binom {r-2+j}{j}\Omega_{m-j+1,r+j-1}B_{j}\bigl(\langle x\rangle\bigr) \\ &\quad = \textstyle\begin{cases} \frac{1}{2}\Omega_{m,r}&\textit{for }x\in{\mathbb{Z}}\textit{ and }m \textit{ odd}, \\ \binom{\frac{m}{2}+r-1}{\frac{m}{2}}+\frac{1}{2}\Omega _{m,r}&\textit{for }x\in{\mathbb{Z}}\textit{ and }m\textit{ even}. \end{cases}\displaystyle \end{aligned}$$
4 Applications
Let \(T_{n}(x)\) (\(n\geq0\)) be the Chebyshev polynomials of the first kind given by (1.1) or (1.4). We need the following lemma from [9].
Lemma 4.1
([9, Lemmas 1, 2])
Let
\(n\geq0\), \(m\geq1\)
be integers. Then we have the following:
$$\begin{aligned}& U_{n} \biggl(\frac{\sqrt{-1}}{2} \biggr)= (\sqrt{-1} )^{n}F_{n+1}, \end{aligned}$$
(4.1)
$$\begin{aligned}& U_{n} \biggl(T_{m} \biggl(\frac{\sqrt{-1}}{2} \biggr) \biggr)= (\sqrt {-1} )^{mn}\frac{F_{m(n+1)}}{F_{m}}, \end{aligned}$$
(4.2)
$$\begin{aligned}& U_{n} \bigl(T_{m}(x) \bigr)=\frac{U_{m(n+1)-1}(x)}{U_{m-1}(x)}. \end{aligned}$$
(4.3)
Substituting \(x=\frac{\sqrt{-1}}{2}\) into (1.10) and using (4.1), we have
On the other hand, with \(x=1\) and replacing r by \(r+1\), from (1.12) we obtain
Combining (4.4) and (4.5), we get
Substituting \(T_{a} (\frac{\sqrt{-1}}{2} )\) for x in (1.10) and using (4.2), we get: for any positive integer a,
$$\begin{aligned} &(\sqrt{-1})^{m}\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m} F_{i_{1}+1}F_{i_{2}+1}\cdots F_{i_{r+1}+1} \\ &\quad =\frac{1}{2r}\sum_{j=0} ^{m} 2^{j}\binom{r+j-1}{r-1}\Delta _{m-j+1,r+j-1}B_{j} \biggl( \frac{\sqrt{-1}}{2} \biggr). \end{aligned}$$
(4.4)
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m} F_{i_{1}+1}F_{i_{2}+1}\cdots F_{i_{r+1}+1} \\ &\quad =\frac{1}{r}\sum_{j=0} ^{m} \binom{r-1+j}{j}\Omega_{m-j+1,r+j}B_{j}(1). \end{aligned}$$
(4.5)
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m} F_{i_{1}+1}F_{i_{2}+1}\cdots F_{i_{r+1}+1} \\ &\quad =\frac{1}{2r}(-\sqrt{-1})^{m}\sum _{j=0} ^{m} 2^{j} \binom {r+j-1}{r-1} \Delta_{m-j+1,r+j-1}B_{j} \biggl(\frac{\sqrt{-1}}{2} \biggr) \\ &\quad =\frac{1}{r}\sum_{j=0} ^{m} \binom{r-1+j}{j}\Omega _{m-j+1,r+j}B_{j}+\Omega_{m,r+1}. \end{aligned}$$
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots+i_{r+1}=m}F_{a(i_{1}+1)}F_{a(i_{2}+1)}\cdots F_{a(i_{r+1}+1)} \\ &\quad =\frac{1}{2r}(-\sqrt{-1})^{am}F_{a} ^{r+1} \sum_{j=0} ^{m} 2^{j} \binom {r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j} \biggl(T_{a} \biggl(\frac{\sqrt {-1}}{2} \biggr) \biggr). \end{aligned}$$
Finally, replacing x by \(T_{a}(x)\) in (1.10) and using (4.3), we have: for any positive integer a,
$$\begin{aligned} &\sum_{i_{1}+i_{2}+\cdots +i_{r+1}=m}U_{a(i_{1}+1)-1}(x)U_{a(i_{2}+1)-1}(x) \cdots U_{a(i_{r+1}+1)-1}(x) \\ &\quad =\frac{1}{2r} \bigl(U_{a-1} (x) \bigr)^{r+1}\sum _{j=1} ^{m} 2^{j} \binom{r+j-1}{r-1}\Delta_{m-j+1,r+j-1}B_{j} \bigl(T_{a}(x) \bigr). \end{aligned}$$
5 Results and discussion
In this paper, we study sums of finite products of Chebyshev polynomials of the second kind and of Fibonacci polynomials and derive Fourier series expansions of functions associated with them. From these Fourier series expansions, we can express those sums of finite products in terms of Bernoulli polynomials and obtain some identities by using those expressions. The Fourier series expansion of the Chebyshev polynomials and Fibonacci polynomials are useful in computing the special values of zeta function or some special functions (see [5, 7, 9, 11, 13‐15]). It is expected that the Fourier series of those polynomials will find some applications in relationship to the generalizations of the special zeta functions.
6 Conclusion
In this paper, we considered the Fourier series expansions of functions associated with Chebyshev polynomials of the second kind and of Fibonacci polynomials. The Fourier series are determined completely.
Competing interests
The authors declare that they have no competing interests.
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