Research
Open Access
On Chebyshev polynomials and their applications
- Xingxing Lv1 and
- Shimeng Shen1Email authorView ORCID ID profile
Advances in Difference Equations20172017:343
https://doi.org/10.1186/s13662-017-1387-8
© The Author(s) 2017
Received: 7 September 2017
Accepted: 2 October 2017
Published: 24 October 2017
Abstract
The main purpose of this paper is, using some properties of the Chebyshev polynomials, to study the power sum problems for the sinx and cosx functions and to obtain some interesting computational formulas.
Keywords
Chebyshev polynomialstrigonometric power sumscomputational formulas
MSC
11B39
1 Introduction
As is well known, the Chebyshev polynomials of the first kind \(\{T_{n}(x)\}\) and the Chebyshev polynomials of the second kind \(\{U_{n}(x)\}\) are defined by \(T_{0}(x)=1\), \(T_{1}(x)=x\), \(U_{0}(x)=1\), \(U_{1}(x)=2x\) and \(T_{n+2}(x)=2xT_{n+1}(x)-T_{n}(x)\), \(U_{n+2}(x)=2xU _{n+1}(x)-U_{n}(x)\) for all integers \(n\geq0\). If we write \(\alpha(x)=x+\sqrt{x^{2}-1}\) and \(\beta(x)=x-\sqrt{x^{2}-1}\), then the explicit expressions of \(T_{n}(x)\) and \(U_{n}(x)\) are given by
and
$$\begin{aligned} T_{n}(x)=\frac{1}{2} \bigl( \alpha^{n}(x)+ \beta^{n}(x) \bigr) = \frac{n}{2}\sum _{k=0}^{ [ \frac{n}{2} ] }(-1)^{k} \frac{(n-k-1)!}{k!(n-2k)!}(2x)^{n-2k}, \quad \vert x\vert < 1 \end{aligned}$$
(1)
$$\begin{aligned} U_{n}(x)=\frac{1}{\alpha-\beta} \bigl( \alpha^{n+1}(x)-\beta ^{n+1}(x) \bigr) =\sum_{k=0}^{ [ \frac{n}{2} ] }(-1)^{k} \frac{(n-k)!}{k!(n-2k)!}(2x)^{n-2k}, \quad \vert x\vert < 1. \end{aligned}$$
(2)
If we take \(x=\cos\theta\), then
$$\begin{aligned} T_{n}(\cos\theta)=\cos(n\theta), \quad\quad U_{n}(\cos \theta)=\frac{\sin((n+1)\theta)}{\sin\theta}. \end{aligned}$$
(3)
These polynomials are very important in different areas of mathematics, and many scholars have studied various properties of them and obtained a series of interesting and important results. Some related papers can be found in [1–13]. Recently, Li Xiaoxue [6] established some identities involving the power sums of \(T_{n}(x)\) and \(U_{n}(x)\), by virtue of which she showed some divisible properties involving the Chebyshev polynomials of the first kind and the second kind, as follows:
$$\begin{aligned} U_{0}(x)U_{2}(x)U_{4}(x)\cdots U_{2n}(x)\cdot\sum_{m=1}^{h} T_{2m} ^{2n+1}(x) \equiv0\bmod \bigl( U_{2h}(x)-1 \bigr) . \end{aligned}$$
In a private communication with professor Wenpeng Zhang, he suggested us to give some explicit formulas for the following trigonometric power sums:
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n} \biggl( \frac{\pi a}{q} \biggr) \quad \textrm{and} \quad \sum _{a=0}^{q-1}\sin^{2n} \biggl( \frac{\pi a}{q} \biggr) . \end{aligned}$$
As far as we know, it seems that nobody has studied these problems yet. In this paper, we shall use the properties of the Chebyshev polynomials of the first kind to obtain some closed formulas for the above trigonometric power sums. These results are stated in the following theorems.
Theorem 1
Let
q
and
n
be positive integers with
\(q\geq2\). Then we have
where
\(\binom{m}{n}=\frac{m!}{n!(m-n)!}\), and
\([x]\)
denotes the greatest integer ≤x.
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n} \biggl( \frac{\pi a}{q} \biggr) = \frac{2q}{4^{n}}\cdot\sum _{k=0}^{ [ \frac{n}{q} ] } \binom{2n}{n-qk}-\frac{q}{4^{n}}\cdot \binom{2n}{n}, \end{aligned}$$
Theorem 2
Let
q
and
n
be positive integers with
\(q\geq2\). Then we have
$$\begin{aligned} \sum_{a=0}^{q-1}\sin^{2n} \biggl( \frac{\pi a}{q} \biggr) = \frac{2q}{4^{n}}\sum _{k=0}^{ [ \frac{n}{q} ] }(-1)^{kq} \binom{2n}{n-qk}- \frac{q}{4^{n}}\cdot\binom{2n}{n}. \end{aligned}$$
Theorem 3
Let
q
be a positive integer with
\(2\mid q\). Then, for any non-negative integer
n, we have the identity
$$\begin{aligned} \sum_{a=0}^{q-1}\sin^{2n+1} \biggl( \frac{\pi a}{q} \biggr) = \frac{1}{4^{n}}\sum_{k=0}^{n}(-1)^{k} \binom{2n+1}{n-k} \cot \biggl( \frac{ \pi(2k+1)}{2q} \biggr) . \end{aligned}$$
It is clear that from Theorem 1 and Theorem 2 we may immediately deduce the following corollary.
Corollary
For any positive integers
q
and
n
with
\(q > n\), we have the identity
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n} \biggl( \frac{\pi a}{q} \biggr) =\sum_{a=0}^{q-1} \sin^{2n} \biggl( \frac{\pi a}{q} \biggr) =\frac{q}{4^{n}}\cdot \binom{2n}{n}. \end{aligned}$$
Remark 1
In order to make our result described in Theorem 3 look simple and nice, we only consider the case \(2\mid q\). For the case \(2\nmid q\), we can also give a corresponding result, but the formula in this case is much more complicated. In fact, by substituting the cosine function in Theorem 3 for the sine function, we can obtain that, for positive integer q with \(2\mid q\),
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n+1} \biggl( \frac{\pi a}{q} \biggr) =1+\sum_{a=1} ^{\frac{q}{2}-1}\cos^{2n+1} \biggl( \frac{\pi a}{q} \biggr) + \sum _{a=1} ^{\frac{q}{2}-1}\cos^{2n+1} \biggl( \frac{\pi(q-a)}{q} \biggr) . \end{aligned}$$
2 Several simple lemmas
To complete the proofs of our theorems, we need some new properties of Chebyshev polynomials, which we summarize as the following lemmas.
Lemma 1
For any non-negative integer
k, we have the identities
and
$$ T_{2k} ( \sin x ) =(-1)^{k}\cos ( 2kx ) $$
$$ T_{2k+1} ( \sin x ) =(-1)^{k}\sin \bigl( (2k+1)x \bigr) . $$
Proof
Taking \(x=\sin(\theta)\) in (1) and noting that \(x^{2}-1=-\cos^{2}(\theta)=i^{2}\cos^{2}(\theta)\), from Euler’s formula, we have
This proves the first formula of Lemma 1.
$$\begin{aligned} T_{2k} \bigl( \sin(\theta) \bigr)& = \frac{1}{2} \bigl[ \bigl( \sin( \theta)+ \sqrt{\sin^{2}(\theta)-1} \bigr) ^{2k}+ \bigl( \sin( \theta)- \sqrt{\sin^{2}(\theta)-1} \bigr) ^{2k} \bigr] \\ & = \frac{1}{2} \bigl[ \bigl( \sin(\theta)+i\cos(\theta) \bigr) ^{2k}+ \bigl( \sin(\theta)-i\cos(\theta) \bigr) ^{2k} \bigr] \\ & = \frac{i^{2k}}{2} \bigl[ \bigl( \cos(\theta)-i\sin(\theta) \bigr) ^{2k}+ \bigl( \cos(\theta)+i\sin(\theta) \bigr) ^{2k} \bigr] \\ & = \frac{(-1)^{k}}{2} \bigl[ \cos(2k\theta)-i\sin(2k\theta)+\cos(2k \theta)+i \sin(2k\theta) \bigr] \\ & = (-1)^{k}\cdot\cos(2k\theta). \end{aligned}$$
Similarly, we also have the identity
This completes the proof of Lemma 1. □
$$\begin{aligned}& T_{2k+1} \bigl( \sin(\theta) \bigr) \\ & \quad = \frac{1}{2} \bigl[ \bigl( \sin(\theta)+ \sqrt{\sin^{2}( \theta)-1} \bigr) ^{2k+1}+ \bigl( \sin(\theta)- \sqrt{ \sin^{2}( \theta)-1} \bigr) ^{2k+1} \bigr] \\ & \quad = \frac{1}{2} \bigl[ \bigl( \sin(\theta)+i\cos(\theta) \bigr) ^{2k+1}+ \bigl( \sin(\theta)-i\cos(\theta) \bigr) ^{2k+1} \bigr] \\ & \quad = \frac{1}{2} \bigl[ i^{2k+1} \bigl( \cos(\theta)-i\sin(\theta) \bigr) ^{2k+1}+(-i)^{2k+1} \bigl( \cos(\theta)+i\sin(\theta) \bigr) ^{2k+1} \bigr] \\ & \quad = \frac{(-1)^{k}}{2} \bigl[ i\cos\bigl((2k+1)\theta\bigr)+\sin\bigl((2k+1) \theta\bigr)- i\cos\bigl((2k+1)\theta\bigr)+\sin\bigl((2k+1)\theta\bigr) \bigr] \\ & \quad = (-1)^{k}\cdot\sin\bigl((2k+1)\theta\bigr). \end{aligned}$$
Lemma 2
For any non-negative integer
n, we have the identities
and
$$\begin{aligned} x^{2n}= \frac{\binom{2n}{n}}{4^{n}}T_{0}(x)+ \frac{2}{4^{n}}\cdot \sum_{k=1}^{n}\binom{2n}{n-k}\cdot T_{2k}(x) \end{aligned}$$
$$\begin{aligned} x^{2n+1}= \frac{1}{4^{n}}\cdot\sum_{k=0}^{n} \binom{2n+1}{n-k}\cdot T _{2k+1}(x). \end{aligned}$$
Proof
This is Lemma 4 in Ma Rong and Zhang Wenpeng [7]. □
Lemma 3
For any non-negative integer
k
and positive integer
q
with
\(q\nmid(2k+1)\), we have the identities
$$ \sum_{a=0}^{q-1}\cos \biggl( \frac{\pi(2k+1)a}{q} \biggr) =1 \quad \textit{and} \quad \sum_{a=0}^{q-1} \sin \biggl( \frac{\pi(2k+1)a}{q} \biggr) = \cot \biggl( \frac{\pi(2k+1)}{2q} \biggr) . $$
Proof
Let \(e(x)=e^{2\pi ix}\), note that \(e ( \frac{(2k+1)}{2} ) =-1\), then, applying the summation formula for geometric series, we have
Applying Euler’s formula \(e^{ix}=\cos(x)+i\sin(x)\) and comparing the real and imaginary parts in (4), we may immediately deduce Lemma 3. □
$$\begin{aligned} \sum_{a=0}^{q-1}e \biggl( \frac{(2k+1)a}{2q} \biggr) =& \frac{-2}{e ( \frac{(2k+1)}{2q} ) -1}=\frac{-2e ( \frac{- (2k+1)}{4q} ) }{e ( \frac{(2k+1)}{4q} ) -e ( \frac {-(2k+1)}{4q} ) } \\ =&\frac{-2\cos ( \frac{\pi(2k+1)}{2q} ) +2i \sin ( \frac{ \pi(2k+1)}{2q} ) }{2i\sin ( \frac{\pi(2k+1)}{2q} ) }=1+i \cot \biggl( \frac{\pi(2k+1)}{2q} \biggr) . \end{aligned}$$
(4)
3 Proofs of the theorems
In this section, we shall complete the proofs of our theorems. First we prove Theorem 1.
Proof of Theorem 1
Replace x by \(\cos ( \frac{\pi a}{q} ) \) in the first formula of Lemma 2 and make the summation for a with \(0\leq a\leq q-1\), from (3) we have
Note the trigonometric identity
Combining (5) and (6) we have
This proves Theorem 1. □
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n} \biggl( \frac{\pi a}{q} \biggr) =&\frac{ \binom{2n}{n}}{4^{n}}\sum _{a=0}^{q-1}1+ \frac{2}{4^{n}}\cdot\sum _{k=1} ^{n}\binom{2n}{n-k}\cdot\sum _{a=0}^{q-1}T_{2k} \biggl( \cos \biggl( \frac{ \pi a}{q} \biggr) \biggr) \\ =&\frac{\binom{2n}{n}}{4^{n}}\cdot q + \frac{2}{4^{n}}\cdot\sum _{k=1} ^{n}\binom{2n}{n-k}\cdot\sum _{a=0}^{q-1}\cos \biggl( \frac{2\pi ka}{q} \biggr) . \end{aligned}$$
(5)
$$\begin{aligned} \sum_{a=0}^{q-1} \biggl( \cos \biggl( \frac{2\pi n a}{q} \biggr) +i \sin \biggl( \frac{2\pi n a}{q} \biggr) \biggr) = \sum_{a=0}^{q-1}e \biggl( \frac{na}{q} \biggr) = \textstyle\begin{cases} q, &\textrm{if }q\mid n, \\ 0, &\textrm{if }q\nmid n. \end{cases}\displaystyle \end{aligned}$$
(6)
$$\begin{aligned} \sum_{a=0}^{q-1}\cos^{2n} \biggl( \frac{\pi a}{q} \biggr) = &\frac{ \binom{2n}{n}}{4^{n}}\cdot q + \frac{2}{4^{n}}\cdot \mathop{\sum_{k=1}^{n}}_{q\mid k} \binom{2n}{n-k}\cdot q \\ =&\frac{2q}{4^{n}}\cdot\sum_{k=0}^{ [ \frac{n}{q} ] } \binom{2n}{n-qk}-q\cdot\frac{\binom{2n}{n}}{4^{n}}. \end{aligned}$$
Now we prove Theorem 2.
Proof of Theorem 2
Replace x by \(\sin ( \frac{\pi a}{q} ) \) in the first formula of Lemma 2 and summarize on \(0\leq a\leq q-1\), from (6) and Lemma 1 we have
This proves Theorem 2. □
$$\begin{aligned} \sum_{a=0}^{q-1}\sin^{2n} \biggl( \frac{\pi a}{q} \biggr) =&\frac{ \binom{2n}{n}}{4^{n}}\sum _{a=0}^{q-1}1+ \frac{2}{4^{n}}\cdot\sum _{k=1} ^{n}\binom{2n}{n-k}\cdot\sum _{a=0}^{q-1}T_{2k} \biggl( \sin \biggl( \frac{ \pi a}{q} \biggr) \biggr) \\ =&\frac{\binom{2n}{n}}{4^{n}}\cdot q + \frac{2}{4^{n}}\cdot\sum _{k=1} ^{n}\binom{2n}{n-k}\cdot\sum _{a=0}^{q-1}(-1)^{k} \cos \biggl( \frac{2 \pi ka}{q} \biggr) \\ =&\frac{\binom{2n}{n}}{4^{n}}\cdot q + \frac{2}{4^{n}}\cdot \mathop{\sum _{k=1}^{n}}_{q\mid k}(-1)^{k} \binom{2n}{n-k}\cdot\sum_{a=0}^{q-1} 1 \\ =&\frac{2q}{4^{n}}\sum_{k=0}^{ [ \frac{n}{q} ] }(-1)^{kq} \binom{2n}{n-qk}-\frac{\binom{2n}{n}}{4^{n}}\cdot q. \end{aligned}$$
We now use the similar methods of proving Theorem 1 and Theorem 2 to complete the proof of Theorem 3.
Proof of Theorem 3
Note that if \(2\mid q\), then \(q\nmid(2k+1)\) for any non-negative integer k. So, by substituting \(\sin(\frac{\pi a}{q})\) for x in the second formula of Lemma 2 and making the summation for a with \(0\leq a \leq q-1\), with the help of Lemma 1 and Lemma 3, we deduce Theorem 3 immediately. □
Declarations
Acknowledgements
The authors would like to thank the referees for their very helpful and detailed comments which have significantly improved the presentation of this paper. This work was supported by the N.S.F. (Grant No. 11771351) of P.R. China.
Authors’ contributions
All authors read and approved the final manuscript.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
(1)
School of Mathematics, Northwest University, Xi’an, P.R. China
References
- Bircan, N, Pommerenke, C: On Chebyshev polynomials and \(\mathbf{GL}(2,\mathrm{Z}/p\mathrm{Z})\). Bull. Math. Soc. Sci. Math. Roum. 103, 353-364 (2012) MATHGoogle Scholar
- Clemente, C: Identities and generating functions on Chebyshev polynomials. Georgian Math. J. 19, 427-440 (2012) MATHMathSciNetGoogle Scholar
- Chan-Lye, L, Wong, KB: On Chebyshev’s polynomials and certain combinatorial identities. Bull. Malays. Math. Soc. 34, 279-286 (2011) MATHMathSciNetGoogle Scholar
- Doha, EH, Bhrawy, AH, Ezz-Eldien, SS: Numerical approximations for fractional diffusion equations via a Chebyshev spectral-tau method. Cent. Eur. J. Phys. 11, 1494-1503 (2013) MATHGoogle Scholar
- Li, X: Some identities involving Chebyshev polynomials. Math. Probl. Eng. 2015, Article ID 950695 (2015) MathSciNetGoogle Scholar
- Ma, R, Zhang, W: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 45, 164-170 (2007) MATHMathSciNetGoogle Scholar
- Ma, Y, Lv, X: Several identities involving the reciprocal sums of Chebyshev polynomials. Math. Probl. Eng. 2017, Article ID 4194579 (2017) Google Scholar
- Wang, T, Zhang, H: Some identities involving the derivative of the first kind Chebyshev polynomials. Math. Probl. Eng. 2015, Article ID 146313 (2015) MathSciNetGoogle Scholar
- Zhang, W, Wang, T: Two identities involving the integral of the first kind Chebyshev polynomials. Bull. Math. Soc. Sci. Math. Roum. 108, 91-98 (2017) MathSciNetMATHGoogle Scholar
- Carlos, M, Da Fonseca, M, Glasser, L, Kowalenko, V: Basic trigonometric power sums with applications. Ramanujan J. 42, 401-428 (2017) View ArticleMATHMathSciNetGoogle Scholar
- Kim, D, Kim, T, Lee, S: Some identities for Bernoulli polynomials involving Chebyshev polynomials. J. Comput. Anal. Appl. 16, 172-180 (2014) MATHMathSciNetGoogle Scholar
- Kim, D, Dolgy, D, Kim, T, Rim, S: Identities involving Bernoulli and Euler polynomials arising from Chebyshev polynomials. Proc. Jangjeon Math. Soc. 15, 361-370 (2012) MATHMathSciNetGoogle Scholar
- Kim, T, Kim, D, Seo, J, Dolgy, D: Some identities of Chebyshev polynomials arising from non-linear differential equations. J. Comput. Anal. Appl. 23, 820-832 (2017) MathSciNetGoogle Scholar
Copyright
© The Author(s) 2017
