Some properties of Fibonacci and Chebyshev polynomials

Abstract

In this paper, we study the properties of Chebyshev polynomials of the first and second kind and those of Fibonacci polynomials and use an elementary method to give Chebyshev polynomials of the first and second kind in terms of Fibonacci polynomials and vice versa. Finally, we get some identities involving the Fibonacci numbers and the Lucas numbers.

Li Advances in Dierence Equations (2015) 2015:118 DOI 10.1186/s13662-015-0420-z

*Correspondence: mailto:[e-mail]

Web End =[e-mail] Department of Mathematics, Northwest University, Xian, Shaanxi, P.R. China

R E S E A R C H Open Access

Some properties of Fibonacci and Chebyshev polynomials

Yang Li*

Abstract

In this paper, we study the properties of Chebyshev polynomials of the rst and second kind and those of Fibonacci polynomials and use an elementary method to give Chebyshev polynomials of the rst and second kind in terms of Fibonacci polynomials and vice versa. Finally, we get some identities involving the Fibonacci numbers and the Lucas numbers.

Keywords: elementary method; Chebyshev polynomials; Fibonacci polynomials; Fibonacci numbers; Lucas numbers

1 Introduction

As we know, the Chebyshev polynomials and Fibonacci polynomials are usually dened as follows: Chebyshev polynomials of the rst kind are Tn+(x) = xTn+(x) Tn(x), n , with the initial values T(x) = , T(x) = x; Chebyshev polynomials of the second kind are

Un+(x) = xUn+(x) Un(x), n , with the initial values U(x) = , U(x) = x; Fibonacci polynomials are Fn+(x) = xFn+(x) + Fn(x), n , with the initial values F(x) = , F(x) = .

From the second-order linear recurrence sequences, we have

Tn(x) =

x +

Un(x) =

x

Fn(x) =

nx +

x n + x x n ,

x + x n+ x x n+ ,

x + x + n x x + n .

These polynomials play a very important role in the study of the theory and application of mathematics, and they are closely related to the famous Fibonacci numbers {Fn} and

Lucas numbers {Ln} which are dened by the second-order linear recurrence sequences

Fn+ = Fn+ + Fn,

Ln+ = Ln+ + Ln,

where n , F = , F = , L = and L = . Therefore, many authors have investigated these polynomials and got many properties and corollaries. For example, Zhang [] uses

2015 Li; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Li Advances in Dierence Equations (2015) 2015:118 Page 2 of 12

the Chebyshev polynomials and has obtained the general formulas involving Fn and Ln,

a+a++ak+=n

Fm(a+) Fm(a+) Fm(ak++)

= (i)mn Fk+m

k k!

U(k)n+k

imLm

,

a+a++ak+=n+k+

Lm(a+) Lm(a+) Lm(ak++)

= (i)m(n+k+) k!

k+

h=

im+Lm

h (k + )!h!(k + h)!U(k)n+k+h

imLm

,

where k, m are any positive integers, a, a, . . . , ak+ are nonnegative integers and i is the square root of . Falcn and Plaza [, ] presented many formulas about Fibonacci polynomials. This fact allowed them to present a family of integer sequences in a new and direct way. Zhang [] also used the Chebyshev polynomials to solve some calculating problems of the general summations. Wu and Yang [] studied Chebyshev polynomials and got a lot of properties.

In this paper, we combine Sergio Falcn and Wenpeng Zhangs ideas. Then we obtain the following theorems and corollaries. These results strengthen the connections of two kinds of polynomials. They are also helpful in dealing with some calculating problems of the general summations or studying some integer sequences.

Theorem For any positive integer n, we have the identities

Tn(x) =

n+

s=

n

k=

k (sn n)(n + k )!Fs(x)

()s+n(n k)!(k + s)!(k s + )!,

Tn(x) =

k(ns s)(n + k )!Fs(x)

()s+n(n k)!(k + s)!(k s)! .

Theorem For any positive integer n, we have

Un(x) =

n+

s=

n

s=

n

k=

n

k=

k( s)(n + k)!Fs(x) ()s+n(n k)!(k + s)!(k s + )!,

Un(x) =

k+s(n + k )!()s+nFs(x) (n k + )!(k + s + )!(k s + )!.

Theorem For any positive integer n, we have the following forms:

Fn(x) =

n

k=

n

s=

n

k=

n

j=

j+n(k + )(n j )!j!(n k j )!(n + k j)!Uk(x),

Fn(x) =

n

k=

n

j=

jn+k(n j )!

j!(n k j)!(n + k j)!Uk(x).

Li Advances in Dierence Equations (2015) 2015:118 Page 3 of 12

Theorem For any positive integer n, we have the following forms:

Fn(x) =

n

k=

n

j=

j+n(n j )!Tk(x) j!(n + k j )!(n k j)!,

Fn(x) =

j+n(n j )! j!(n j )!(n j )!.

Corollary For any positive integer n, we have the following identities:

Fn imLm

=

n

k=

n

j=

j+n(n j )!Tk(x) j!(n + k j )!(n k j )! +

n

j=

n

k=

n

j=

j+n(k + )(n j )!Fm(k+) j!(n k j )!(n + k j)!()kmFm ,

Fn imLm

=

n

k=

n

j=

jn+k(n j )!(i)Fkm ()kmj!(n k j)!(n + k j)!Fm .

Corollary For any positive integer n, we have the following identities:

Fn imLm

=

n

k=

n

j=

j+n(n j )!()km(i)Lkmm j!(n + k j )!(n k j)! ,

Fn imLm

=

n

k=

n

j=

j+n(n j )!()kmLkm j!(n + k j )!(n k j )!

+

n

j=

j+n(n j )! j!(n j )!(n j )!.

2 Some lemmas

Lemma For any nonnegative integers m and n, we have these identities

Tm(x)Tn(x)

x dx =

, m = n,

, m = n > , , m = n = ;

()

Um(x)Un(x)

x dx =

, m = n,

, m = n;

()

Tn(cos ) = cos n; ()

Un(cos ) =

sin (n + )sin . ()

Proof See reference [].

Lemma For any positive integers m and n, we have these identities

Un i

= inFn+,

Tn i

= in

Ln+,

Li Advances in Dierence Equations (2015) 2015:118 Page 4 of 12

Tn Tm i

= imnLmn,

Un Tm i

= imn Fm(n+)

Fm .

Proof See reference [].

Lemma For any positive integer n, we have

Tn Tm(x) = Tnm(x),

Un Tm(x) = Um(n

+)(x)

Um(x) .

Proof See reference [].

Lemma For any positive integer n, let

Fn(x) =

+

k=

ankUk(x) ()

and

Fn(x) =

bnT(x) +

+

k=

bnkTk(x), ()

then we can get

ank =

n

j=

(k+)(nj)! j!(nkj)!!(n+k+j)!! , k + n is odd, , otherwise,

()

()

Proof To begin with, we multiply xUm(x) to both sides of (), then integrate it from to , we can get the following identity by applying property ():

xFn(x)Um(x) dx =

bnk =

n

j=

(nj)! j!(n+kj)!!(nkj)!! , k + n is odd, , otherwise.

k=

ank

xUm(x)Uk(x) dx =

anm,

and then we have

anm =

xFn(x)Um(x) dx.

From reference [] we know

Fn(x) =

n

j=

n j j

xnj, ()

Li Advances in Dierence Equations (2015) 2015:118 Page 5 of 12

where n . We dene

wnk =

cosn sin(k + ) sin d.

From reference [] we know

wnk =

(k+)n!(n+k+)!!(nk)!! , k + n is even and n k,, otherwise, ()

where n and k are any nonnegative integers. Let x = cos , then we can get the following identity by applying property () and property ():

anm =

cos ) sin(m + ) sin d

=

n

j=

n j j

cosnj sin(m + ) sin d

n

j=

n j j

wnj,m,

and then we have anm = if n + m is even. If n + m is odd, we have

anm =

=

n

j=

n j j

(m + )(n j )!(n m j )!!(n + m + j)!!

=

n

j=

(n j )! j!(n j )!

(m + )(n j )!(n m j )!!(n + m + j)!!

n

j=

(m + )(n j )!j!(n m j )!!(n + m + j)!!.

We nish proving property ().

In order to prove property (), we must multiply Tm(x)

x to both sides of (), then integrate it from to , we can get the following identity by applying property ():

Fn(x)Tm(x)

x dx =

=

bnT(x)Tm(x)

x dx +

k=

bnkTm(x)Tn(x)

x dx =

bnm,

and then we have

bnm =

Fn(x)Tm(x)

x dx.

We dene

qnk =

cosn cos k d.

Li Advances in Dierence Equations (2015) 2015:118 Page 6 of 12

From reference [] we know

qnk =

n!

(n+k)!!(nk)!! , k + n is even and n k,, otherwise, ()

where n and k are any nonnegative integers. Let x = cos , then we can get the following identity by applying property () and property ():

bnm =

Tm(cos )Fn(cos )sin sin d

=

n

j=

n j j

cosnj cos m d

=

n

j=

n j j

qnj,m,

and then we have bnm = if n + m is even. If n + m is odd, we have

bnm =

n

j=

n j j

(n j )!(n + m j )!!(n m j )!!

=

n

j=

(n j )! j!(n j )!

(n j )!(n + m j )!!(n m j )!!

=

n

j=

(n j )!j!(n + m j )!!(n m j )!!.

This proves Lemma .

Lemma For any positive integers m and n, we have the following identities:

Fn(i cos ) = in+

sin nsin , ()

i

i

x + Fm(x)Fn(x) dx =

im, m = n > ,

, otherwise. ()

Proof As we know,

Fn(x) =

nx +

x + x + n x x + n .

Let x = i cos , then we have

Fn(i cos ) =

sin (i

cos + sin )n (i cos sin )n

=

sin inein inein

Li Advances in Dierence Equations (2015) 2015:118 Page 7 of 12

= in

sin (

cos n i sin n cos n i sin n)

= in+

sin n sin .

This proves property (). Let x = i cos in the following identity:

A =

i

i

x + Fm(x)Fn(x) dx,

then we can get

A =

sin Fn(i cos )Fm(i cos )i sin d

=

sin in+ sin n

sin

im+ sin m sin d

= in+m

sin n sin m d

= in+m

cos (n m) cos (m + n) d.

Then we can get property (). This proves Lemma .

Lemma For any positive integer n, we have

Tn(x) =

n

k=

()nk k n n + k

n + k k

xk,

Tn+(x) =

n

k=

()nk k (n + ) n + k +

n + k + k +

xk+,

Un(x) =

n

k=

()nk k (k + ) n + k +

n + k +

k +

xk,

Un+(x) =

n

k=

()nk k+ (k + ) n + k +

n + k +

k +

xk+.

Proof From Theorem of reference [], we can get the following result easily:

Tn(x) =

n

k=

()nk k n + k n k

k n + k

k

xk

n

k=

= ()nk k

(n + k)!

(k)!(n k)!

(n + k )! (k )!(n k)!

xk

n

k=

= ()nk k

n(n + k )! (k)!(n k)! xk

=

n

k=

()nk k n n + k

n + k

k

xk.

Li Advances in Dierence Equations (2015) 2015:118 Page 8 of 12

From Theorem of reference [], we know

Tn+(x) =

n

k=

()nk k+ n + k + n k

k n + k k

xk+.

In a similar way, we can get

Tn+(x) =

n

k=

()nk k (n + ) n + k +

n + k + k +

xk+.

We can get T n(x) = nUn(x) easily from the denition of the Chebyshev polynomials. If we derive both sides of the above properties, we will get

Un(x) =

n

k=

()nk k (k + ) n + k +

n + k + k +

xk,

Un+(x) =

n

k=

()nk k+ (k + ) n + k +

n + k + k +

xk+.

This proves Lemma .

Lemma For any positive integer n, let

Tn(x) =

+

s=

cn,sFs(x) ()

and

Tn(x) =

+

s=

cn,sFs(x), ()

then we can get

cn,s =

n

k=

k+sni

s+n+(n+k)!(nk)!(k+s+)!!(ks+)!! , s is odd,, otherwise, ()

cn,s =

n

k=

(nss)is+n(n+k)! k(nk)!(k+s)!!(ks)!! , s is even,

, otherwise. ()

Proof At rst, we multiply x + Fm(x) to both sides of (), then integrate it from i to i. We can get the following identity by applying Lemma , where m is any positive integer:

i

i

x + Fm(x)Tn(x) dx =

s=

i

i cn,s

x + Fs(x)Fm(x) dx = imcn,m,

and then we have

cn,m = (i)m

i

i

x + Fm(x)Tn(x) dx.

Li Advances in Dierence Equations (2015) 2015:118 Page 9 of 12

Let x = i cos , then we can get the following identity by applying Lemma and property ():

cn,m = (i)m

Tn(i

cos )Fm(i cos )i sin d

= im

Tn(i

cos )

im+ sin msin ( sin ) d

= im+

Tn(i

cos ) sin m sin d

k n im+n+ n + k

n + k k

wk,m,

so when m is even, we have cn,m = . When m is odd, we have

cn,m =

n

k=

=

n

k=

k n im+n+ n + k

n + k k

m (k)!

(k + m + )!!(k m + )!!

=

n

k=

k n im+n+ n + k

(n + k)! (k)!(n k)!

m (k)!

(k + m + )!!(k m + )!!

k+ mn im+n+(n + k )!

= (n k)!(k + m + )!!(k m + )!!.

In a similar way, we can get the following result easily:

cn,m = (i)m

n

k=

Tn(i

cos )Fm(i cos )i sin d

=

n

k=

k (n ) im+n

n + k

n + k

k

wk,m,

and we have cn,m = if m is odd. If m is even, we have

cn,m =

n

k=

k (n ) im+n

n + k

n + k

k

m(k )!

(k + m)!!(k m)!!

(nm m)im+n(n + k )!

= k(n k)!(k + m)!!(k m)!!.

This proves Lemma .

Lemma For any positive integer n, let

Un(x) =

+

s=

n

k=

dn,sFs(x) ()

and

Un(x) =

+

s=

dn,sFs(x), ()

Li Advances in Dierence Equations (2015) 2015:118 Page 10 of 12

then we can get

dn,s =

n

k=

k+i

s+n+s(n+k)!(nk)!(k+s+)!!(ks+)!! , s is odd,, otherwise, ()

dn,s =

n

k=

k+i

s+ns(n+k)!(nk)!(k+s+)!!(k+s)!! , s is even,, otherwise. ()

Proof At rst, we multiply x + Fm(x) to both sides of (), then integrate it from i to i, we can get the following identities by applying Lemma , where m is any positive integer:

i

i

x + Fm(x)Un(x) dx =

k=

i

i dns

x + Fs(x)Fm(x) dx = imdn,m,

then we have

dn,m = (i)m

x + Fm(x)Un(x) dx.

Let x = i cos , then we can get the following identity by applying Lemma and property ():

dn,m = im+

i

i

Un(i

cos )Fm(i cos )i sin d

= im+

cos )im+

sin msin ( sin ) d

= im+

Un(i

Un(i

cos ) sin m sin d

=

n

k=

k (k + ) im++n

n + k +

n + k +

k +

wk,m,

so we have dn,m = if m is even. If m is odd, we have

dn,m =

n

k=

k (k + ) im++n

n + k +

n + k +

k +

m (k)!

(k + m + )!!(k m + )!!

=

n

k=

k (k + ) im++n

n + k +

(n + k + )! (k + )!(n k)!

m (k)!

(k + m + )!!(k m + )!!

=

k+ im+n+m(n + k)!(n k)!(k + m + )!!(k m + )!!.

In a similar way, we have

dn,m = (i)m

n

k=

Un(i

cos )Fm(i cos )i sin d

=

n

k=

k+ (k + ) im+n

n + k +

n + k +

k +

wk+,m.

Li Advances in Dierence Equations (2015) 2015:118 Page 11 of 12

When m is odd, we have dn,m = . When m is even, we have

dn,m =

n

k=

k+ (k + ) im+n

n + k +

n + k + k +

m(k + )!(k + m + )!!(k + m)!!

k+ im+nm(n + k )!(n k )!(k + m + )!!(k + m)!!.

This proves Lemma .

3 Proof of the theorems and corollaries

In this section, we will prove our theorems and corollaries. First of all, we can prove all the theorems from Lemma , Lemma and Lemma easily.

Proof of Corollary We can get the following properties from Lemma and Lemma by letting x = Tm(x) in Theorem :

Fn Tm(x) =

n

k=

=

n

k=

n

j=

j+n(k + )(n j )!j!(n k j )!(n + k j)!Uk T

m(x) ,

Fn Tm(x) =

n

k=

n

j=

jn+k(n j )!

j!(n k j)!(n + k j)!Uk T

m(x) .

Then, taking x = i in the above identities, according to Lemma , we can get Corollary .

Proof of Corollary We can get the following properties from Lemma and Lemma by letting x = Tm(x) in Theorem :

Fn Tm(x) =

n

k=

n

j=

j+n(n j )!Tk(Tm(x)) j!(n + k j )!(n k j)!,

Fn Tm(x) =

n

k=

n

j=

j+n(n j )!Tk(Tm(x)) j!(n + k j )!(n k j )!

+

n

j=

j+n(n j )!j!(n j )!(n j )!T T

m(x) .

Then, taking x = i in the above identities, according to Lemma , we can get Corollary .

Competing interests

The author declares that there is no conict of interests regarding the publication of the article.

Acknowledgements

The author would like to thank the referees for their very helpful and detailed comments which have signicantly improved the presentation of this paper.

Received: 28 November 2014 Accepted: 18 February 2015

Li Advances in Dierence Equations (2015) 2015:118 Page 12 of 12

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3. Falcn, S, Plaza, : The k-Fibonacci sequences and the Pascal 2-triangle. Chaos Solitons Fractals 38, 38-49 (2007)4. Wu, X, Yang, G: The general formula of Chebyshev polynomials. J. Wuhan Transp. Univ. 24, 573-576 (2000)5. Abramowitz, M, Stegun, IA: Handbook of Mathematical Functions: With Formulas, Graphs and Mathematical Tables, chap. 22, pp. 771-802. Dover New York (1965)

6. Ma, R, Zhang, W: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 5, 164-171 (2007)