Special classes of polynomials

Special classes of polynomials Gospava B. Djordjević Gradimir V. Milovanović

University of Niˇ s, Faculty of Technology Leskovac, 2014.

ii

Preface In this book we collect several recent results on special classes of polynomials. We mostly focus to classes of polynomials related to classical orthogonal polynomials. These classes are named as polynomials of Legendre, Gegenbauer, Chebyshev, Hermite, Laguerre, Jacobsthal, Jacobsthal – Lucas, Fibonacci, Pell, Pell – Lucas, Morgan – Voyce. Corresponding numbers are frequently investigated. We present new relations, explicit representations and generating functions. We are not able to collect all results in this topic, so we reduce material to subjects of our own interest. Authors Leskovac and Belgrade, 2014.

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Contents 1 Standard classes of polynomials 1.1 Bernoulli and Euler polynomials . . . . . . . . . . . 1.1.1 Introductory remarks . . . . . . . . . . . . . 1.1.2 Bernoulli polynomials . . . . . . . . . . . . . 1.1.3 Euler polynomials . . . . . . . . . . . . . . . 1.1.4 Zeros of Bernoulli and Euler polynomials . . 1.1.5 Real zeros of Bernoulli polynomials . . . . . . 1.2 Orthogonal polynomials . . . . . . . . . . . . . . . . 1.2.1 Moment–functional and orthogonality . . . . 1.2.2 Properties of orthogonal polynomials . . . . . 1.2.3 The Stieltjes procedure . . . . . . . . . . . . 1.2.4 Classical orthogonal polynomials . . . . . . . 1.2.5 Generating function . . . . . . . . . . . . . . 1.3 Gegenbauer polynomials and generalizations . . . . . 1.3.1 Properties of Gegenbauer polynomials . . . . 1.3.2 Generalizations of Gegenbauer polynomials . 1.3.3 The differential equation . . . . . . . . . . . 1.3.4 Polynomials in a parameter λ . . . . . . . . 1.3.5 Polynomials induced by polynomials pλn,m (x) 1.3.6 Special cases . . . . . . . . . . . . . . . . . . 1.3.7 Distribution of zeros . . . . . . . . . . . . . . 1.3.8 Generalizations of Dilcher polynomials . . . . 1.3.9 Special cases . . . . . . . . . . . . . . . . . .

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1 1 1 2 7 9 12 14 14 16 18 19 23 25 25 27 34 37 38 41 42 43 47

2 Horadam polynomials 2.1 Horadam polynomials . . . . . . . . . . . . . . . . . . . 2.1.1 Introductory remarks . . . . . . . . . . . . . . . 2.1.2 Pell and Pell–Lucas polynomials . . . . . . . . . 2.1.3 Convolutions of Pell and Pell–Lucas polynomials

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49 49 49 51 53

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vi

CONTENTS 2.1.4 2.1.5

Generalizations of the Fibonacci and Lucas polynomials . . . . . . . . . . . . . . . . . . . . . . . . (k) Polynomials Un,4 (x) . . . . . . . . . . . . . . . . . . . (k)

2.2

2.1.6 Polynomials Un,m (x) . . . . . . . . . 2.1.7 Some interesting identities . . . . . . 2.1.8 The sequence {Cn,3 (r)} . . . . . . . . 2.1.9 The sequence {Cn,4 (r)} . . . . . . . . Pell, Pell–Lucas and Fermat polynomials . . . k (x) and Qk (x) . . 2.2.1 Polynomials Pn,m n,m 2.2.2 Mixed convolutions . . . . . . . . . . . 2.2.3 Generalizations of Fermat polynomials 2.2.4 Numerical sequences . . . . . . . . . .

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3 Morgan–Voyce and Jacobsthal 3.1 Morgan–Voyce polynomials . . . . . . . . . . . . . . . . 3.1.1 Introductory remarks . . . . . . . . . . . . . . . 3.1.2 Polynomials Un,2 (p, q; x) . . . . . . . . . . . . . . 3.1.3 Polynomials Un,m (p, q; x) . . . . . . . . . . . . . 3.1.4 Particular cases . . . . . . . . . . . . . . . . . . . 3.1.5 Diagonal polynomials . . . . . . . . . . . . . . . 3.1.6 Generalizations of Morgan–Voyce polynomials . . (r) (r) 3.1.7 Polynomials Pn,m (x) and Qn,m (x) . . . . . . . . 3.2 Jacobsthal polynomials . . . . . . . . . . . . . . . . . . . 3.2.1 Introductory remarks . . . . . . . . . . . . . . . 3.2.2 Polynomials Jn,m (x) and jn,m (x) . . . . . . . . . 3.2.3 Polynomials Fn,m (x) and fn,m (x) . . . . . . . . 3.2.4 Polynomials related to generalized Chebyshev polynomials . . . . . . . . . . . . . . 3.2.5 Polynomials Pn,3 (x) and Chebyshev polynomials 3.2.6 General polynomials . . . . . . . . . . . . . . . . 3.2.7 Chebyshev and Jacobsthal polynomials . . . . . 3.2.8 Mixed convolutions of the Chebyshev type . . . 3.2.9 Incomplete generalized Jacobsthal and Jacobsthal–Lucas numbers . . . . . . . . . .

56 57

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61 63 69 75 80 80 83 86 91

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95 95 95 96 99 102 104 107 107 113 113 114 116

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123 125 127 129 132

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4 Hermite and Laguerre polynomials 143 4.1 Generalized Hermite polynomials . . . . . . . . . . . . . . . . 143 4.1.1 Introductory remarks . . . . . . . . . . . . . . . . . . 143 4.1.2 Properties of polynomials hλn,m (x) . . . . . . . . . . . 144

CONTENTS

4.2

4.3

4.1.3 Polynomials with two parameters . . . . . . . . . . . 4.1.4 Generalized polynomials with the parameter λ . . . . 4.1.5 Special cases and distribution of zeros . . . . . . . . . 4.1.6 The Rodrigues type formula . . . . . . . . . . . . . . . 4.1.7 Special cases . . . . . . . . . . . . . . . . . . . . . . . 4.1.8 The operator formula . . . . . . . . . . . . . . . . . . 4.1.9 Implications related to generalized polynomials . . . . Polynomials induced by generalized Hermite . . . . . . . . . . 4.2.1 Polynomials with two parameters . . . . . . . . . . . . 4.2.2 Distribution of zeros . . . . . . . . . . . . . . . . . . . 4.2.3 Polynomials related to the generalized Hermite polynomials . . . . . . . . . . . . . . . . . . . 4.2.4 Explicit formulas for the polynomials heνn,m (z, x; α, β) 4.2.5 Polynomials {Hnm (λ)} . . . . . . . . . . . . . . . . . . 4.2.6 Connection of the polynomials {Hnm (λ)} and the hyperbolic functions . . . . . . . . . . . . . . . . . m (λ)} . . . . . . . . . . . . . . . . . 4.2.7 Polynomials {Hr,n 4.2.8 A natural generating function . . . . . . . . . . . . . 4.2.9 A conditional generating function . . . . . . . . . . . Generalizations of Laguerre polynomials . . . . . . . . . . . . 4.3.1 Introductory remarks . . . . . . . . . . . . . . . . . . 4.3.2 Polynomials `sn,m (x) . . . . . . . . . . . . . . . . . . . 4.3.3 Generalization of Panda polynomials . . . . . . . . . . a (x) and ha (x) . . . . . . . . . . . . 4.3.4 Polynomials gn,m n,m 4.3.5 Convolution type equalities . . . . . . . . . . . . . . . 4.3.6 Polynomials of the Laguerre type . . . . . . . . . . . . c,r 4.3.7 Polynomials fn,m (x) . . . . . . . . . . . . . . . . . . . c,r (x) . . . . . . . . . . . . . . 4.3.8 Some special cases of fn,m 4.3.9 Some identities of the convolution type . . . . . . . .

vii 146 148 151 153 154 156 157 160 160 163 165 169 170 173 175 177 178 181 181 183 185 188 190 193 194 196 200

viii

CONTENTS

Chapter 1

Standard classes of polynomials 1.1 1.1.1

Bernoulli and Euler polynomials Introductory remarks

In 1713 Jacob Bernoulli introduced one infinite sequence of numbers in an elementary way. Bernoulli’s results appeared in his work “Ars conjectandi” for the first time. These numbers are known as Bernoulli numbers. Bernoulli investigated sums of the form Sp (n) = 1p + 2p + 3p + · · · + np . He obtained the result that these numbers can be written in the form of the following polynomials 1 1 p 1 p 1 p p+1 p−1 Sp (n) = n + n + An + Bnp−3 + . . . , p+1 2 2 1 4 3 whose coefficients contain the sequence of rational numbers 1 A= , 6

B=−

1 , 30

C=

1 , 42

D=−

1 ,.... 30

Later, Euler [23] investigated the same problem independently of Bernoulli. Euler also introduced the sequence of rational numbers A, B, C, D, . . . . In the work “Introductio in analusin infinitorum”, 1748, Euler noticed the connection between infinite sums s(2n) =

1 1 1 + 2n + 2n + · · · = αn π 2n 2n 1 2 3 1

2

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS

and rational coefficients αn which contain the same sequence of numbers A, B, C, D, . . . . Euler [23] obtained interesting results, showing that these numbers are contained in coefficients of the series expansion of functions x 7→ cot x, x 7→ tan x, x 7→ 1/ sin x. Euler admitted Bernoulli’s priority in this subject and named these rational numbers as Bernoulli numbers. It was shown that these numbers have lots of applications, so they became the subject of study of many mathematicians (Jacobi, Carlitz [9], Delange [13], Dilcher [17], [18] , [19], Rakoˇcević [97], etc.). In the same time properties and applications of Bernoulli polynomials Sp (n) are investigated. We can say that Bernoulli polynomials form a special class of polynomials because of their great applicability. The most important applications of these polynomials are in theory of finite differences, analytic number theory and lots of applications in classical analysis.

1.1.2

Bernoulli polynomials

The coefficient Bn of the Teylor expansion of the function t → 7 g(t) = t/(et − 1), i.e., ∞ X t Bn n g(t) = t = t . (1.2.1) e −1 n! n=0

Numbers Bn are rational. Namely, we have 1 1 1 1 B0 = 1, B1 = − , B2 = , B4 = − , B6 = , . . . , 2 6 30 42 and B2k+1 = 0, for all k ≥ 1. Remark 1.1.1. Bernoulli numbers Bn can be expressed by the following Euler’s formula 2(2k)! B2k = (−1)k+1 ζ(2k) (1.2.2) (2π)2k where

∞ X 1 ζ(z) = nz

( 1)

n=1

is the Riemann zeta function. Thus, according to (1.2.2), we can conclude that the following holds: (−1)k+1 B2k > 0

for all k ≥ 1.

1.1. BERNOULLI AND EULER POLYNOMIALS

3

Bernoulli polynomials Bn (x) are defined as +∞

X Bn (x) tetx = tn . t e −1 n!

(1.2.3)

n=0

By (1.2.1) and (1.2.3) we have t tetx = t · etx = t e −1 e −1

+∞ X Bn n=0

n!

! tn

+∞ X (tx)n n=0

!

n!

,

hence +∞ X Bn (x)

n!

n=0

+∞ X n X

Bk xn−k tn k!(n − k)! n=0 k=0 +∞ X n X n tn = Bk xn−k . n! k

tn =

n=0 k=0

According to the last equalities we obtain n X n Bn (x) = Bk xn−k . k

(1.2.4)

k=0

Using (1.2.4) we find 1 1 B0 (x) = 1, B1 (x) = x − , B2 (x) = x2 − x + , 2 6 3 2 1 1 3 4 3 B3 (x) = x − x + x, B4 (x) = x − 2x + x2 − , 2 2 30 5 1 5 B5 (x) = x5 − x4 + x3 − x, etc. 2 3 6 Differentiating both sides of (1.2.3) with respect to x, we get +∞

X tn t2 tx 0 e = B (x) , n et − 1 n! n=0

and obtain the equality Bn0 (x) = nBn−1 (x). We put x = 0 and x = 1, respectively, in (1.2.3) to obtain +∞

X t 1 −1+ t= Bn (0)tn t e −1 2 n=2

(1.2.5)

4


and

+∞

X tet 1 −1+ t= Bn (1)tn . t e −1 2 n=2

Since

1 tet 1 t − 1 + t = − 1 − t, t t e −1 2 e −1 2

we have Bn (1) − Bn (0) = 0, i.e., Bn (1) = Bn (0). Using (1.2.3) again, we get +∞ X

(Bn (x + 1) − Bn (x))

n=0

t (x+1)t tn = t e − etx n! e −1 = te

tx

+∞ X xn−1 n = t . (n − 1)! n=1

Comparing the coefficients with respect to tn , we establish the relation Bn (x + 1) − Bn (x) = nxn−1 ,

n = 0, 1, . . . .

(1.2.6)

From (1.2.6) it follows that Bn (1 − x) = (−1)n Bn (x).

(1.2.7)

Notice that B2n (1 − x) = B2n (x) is an immediate consequence of the equality (1.2.7). We put x = 1/2 + z and obtain 1 1 B2n + z = B2n −z . 2 2

(1.2.8)

Using (1.2.8) we can determine the values of Bernoulli polynomials of the even degree in the interval (1/2, 1), if these values are known in the interval (0, 1/2). The geometric interpretation of the equation (1.2.8) is that the curve y = B2n (x) is symmetric with respect to the line x = 1/2 in the interval (0, 1). The differentiation of (1.2.8) implies 1 1 + z = −B2n−1 −z , B2n−1 2 2

1.1. BERNOULLI AND EULER POLYNOMIALS

5

and we conclude that Bernoulli polynomials of the odd degree are symmetric with respect to the point (1/2, 0). We put z = 1/2 in (1.2.8) and obtain B2n−1 (1) = −B2n−1 (0). Hence, Bernoulli numbers of the odd index are equal to zero. If m ≥ 1, then the following equality can be proved Bn (x + m) − Bn (x) = n

m−1 X

(x + j)n−1 .

j=0

Since +∞ X n=0

Bn (x + 1)

te(t(x+1) tetx t tn = t = t e n! e −1 e −1 ! +∞ ! +∞ X X tn tn , = Bn (x) n! n! n=0

it follows that Bn (x + 1) =

n=0

n X n k=0

k

Bk (x).

(1.2.9)

Some particular values of Bernoulli polynomials are given below: Bn (0) = Bn (1) = Bn ; 1 Bn = (21−n − 1)Bn (n ≥ 2); 2 1 1 B2k = 1 − 2−2k+1 1 − 3−2k+1 B2k . 6 2

(1.2.10) (1.2.11) (1.2.12)

Also, for k ≥ 1 the following holds 1 . (1.2.13) 2 We also consider the Fourier expansion of Bernoulli polynomials Bn (x). Namely, for 0 ≤ x < 1 and k ≥ 1 we have (−1)k B2k−1 (x) > 0

k−1

B2k (x) = (−1)

B2k−1 (x) = (−1)k

when

0 1, y > 0, we get n X X n n n−j Bn (x + iy) = Bj (x)(iy) = B2j (x)(iy)n−j j 2j j=0 2j≤n X n + B2j+1 (x)(iy)n−2j−1 . 2j + 1 2j+1≤n

For the polynomial Bn (x + iy) the following statement holds (see [19]). Theorem 1.1.4. Let n ≥ 200 and x ≥ 1 such that (a) y ≥ 1 and 2 1/(n−1) πd (n − 1) n − 1 2 4πy/(n−1) 2 2 (x − 1) + y ≤ e , 32x2 2πe

12


where d = 1/4 − 1/2e−4π = 0.249998256, or (b) 0 < y ≤ 1 and x≤

7 (2π)3/2 n−1/2 5

1/n

n . 2πe

Then Bn (x + iy) 6= 0. Corollary 1.1.1. The Bernoulli polynomial Bn (z) does not have any nonreal zero in the set √ √ 1 − 0.0709 n − 1 ≤ Re(z) ≤ 0.0709 n − 1. Corollary 1.1.2. Bn (z + 1/2) 6= 0 holds for all n if z = x + iy belongs to a parabolic set 1 2 x≥0 i x+ ≤ D|y|, 2 where D = 0.0315843. We state some other Dilcher’s [19] results. Theorem 1.1.5. For n ≥ 200, the Bernoulli polynomial Bn (z + 1/2) does not have any zero in the parabolic set x2
1 the polynomial B2n+1 has three zeros: x = 0, 1/2, 1. We shall prove that these points are the only zeros of the Bernoulli polynomial of the odd degree in the interval 0 ≤ x ≤ 1. We use D to denote the standard differentiation operator (D = d/dx, m D = dm /dxm ). If the polynomial B2n+1 (x) has at least four zeros in the interval 0 ≤ x ≤ 1, then the polynomial D B2n+1 (x) has at least three zeros in the interior if this interval, and the polynomial D2 B2n+1 (x) has at least two zeros in the same interval. Since D2 B2n+1 (x) = (2n + 1)(2n)B2n−1 (x), we conclude that the polynomial B2n−1 (x) has at least four zeros in the interval 0 ≤ x ≤ 1. In the same way, according to the equality D2 B2n−1 (x) = (2n − 1)(2n − 2)B2n−3 (x), it follows that B2n−3 (x) must have at least four zeros in the interval 0 ≤ x ≤ 1. Thus, we get the conclusion that the polynomial B3 (x) must have at least four zeros in the interval 0 ≤ x ≤ 1, which is not possible, since the degree of the polynomial B3 (x) is equal to three. Hence, the polynomial B2n+1 (x) can have only three zeros in the interval 0 ≤ x ≤ 1, and these zeros are x = 0, 1/2 and 1. Since D B2n (x) = 2nB2n−1 (x), and knowing that the polynomial B2n−1 (x) does not have any zero in the interval 0 < x < 1/2, it follows that B2n (x) can not have more than one zero in the interval 0 ≤ x ≤ 1/2. However, since D B2n+1 (x) = (2n + 1)B2n (x), holds, according to the equality B2n+1 (0) = B2n+1 (1/2) = 0, it follows that the polynomial B2n (x) has at least one zero in the interval 0 < x < 1/2. Since it can not have more then one zero, we conclude that the polynomial B2n (x) has only one zero in the interval 0 ≤ x ≤ 1/2. Since the polynomial B2n (x) is symmetric with respect to the line x = 1/2 in the interval (0, 1), it follows that this polynomial has one more zero in the interval 1/2 ≤ x ≤ 1. Hence, the Bernoulli polynomial of the even degree has two zeros x1 and x2 in the interval 0 ≤ x ≤ 1 and these zeros belong to intervals 0 < x1 < 1/2 and 1/2 < x2 < 1.

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1.2 1.2.1

Orthogonal polynomials Moment–functional and orthogonality

Orthogonal polynomials represent an important class of polynomials. In this book we shall consider various classes of generalizations of orthogonal polynomials. Hence, this section is devoted to basic properties of orthogonal polynomials. Classical orthogonal polynomials are also considered. Definition 1.2.1. The function x 7→ p(x), which is defined on a bounded interval (a, b), is a weighted function if it satisfies the following conditions on (a, b): Z b p(x) ≥ 0 (x ∈ (a, b)), 0 < p(x)dx < +∞. (2.1.1) a

If the interval (a, b) is unbounded, for example one of the following types: (−∞, b),

(a, +∞),

(−∞, +∞),

then it is required that integrals Z ck =

b

xk p(x)dx

(k = 1, 2, . . . )

(2.1.2)

a

are absolutely convergent. Integrals ck , which appear in (2.1.2), are called moments of the weighted function p. Let L2 (a, b) denote the set of all real functions, which are integrable on the set (a, b) with respect to the measure dµ(x) = p(x)dx. Here x 7→ p(x) is a weighted function on (a, b). We use (·, ·) to denote the scalar product in L2 (a, b), defined as Z (f, g) =

b

p(x)f (x)g(x)dx

(f, g ∈ L2 (a, b)).

(2.1.3)

a

Two functions are orthogonal if (f, g) = 0. Definition 1.2.2. The sequence {Qk }k of polynomials is orthogonal in the interval (a, b) with respect to the weighted function x 7→ p(x), if it is orthogonal with respect to the scalar product defined with (2.1.3).

1.2. ORTHOGONAL POLYNOMIALS

15

Orthogonal polynomials can be defined in a more general way. Let P denote the set of all algebraic polynomials, let {Ck }k be a sequence of complex numbers and let L : P → R denote the linear functional defined by L[xk ] = Ck ,

(k = 0, 1, 2, . . . ),

L(αP (x) + βQ(x)] = αL[P (x)] + βL[Q(x)]

(α, β ∈ C, P, Q ∈ P).

Then we say that L is the moment–functional determined by the sequence {Ck }k . We say that Ck is the moment of the order equal to k. Notice that it is easy to verify that L is linear. Definition 1.2.3. The sequence of polynomials {Qk }k is orthogonal with respect to the moment–functional L, if the following is satisfied: (1) dg(Qk ) = k; (2) L[Qk Qn ] = 0 for k 6= n; (3) L[Q2k ] 6= 0. for all k, n ∈ N0 . Conditions (2) and (3) in the previous definition can be replaced by the condition L[Qk Qn ] = Kn δkn , Kn 6= 0, and δkn is the Kronecker delta. Notice that orthogonal polynomials with respect to the moment-functional L, which is defined as Z L[f ] = (f, 1) =

b

p(f )f (x)dx

(f ∈ P),

a

are exactly orthogonal polynomials described in Definition 2.1.2. The following statement can be proved by induction on n. Theorem 1.2.1. Let {Qk }k denote the set of orthogonal polynomials with respect to the moment–functional L. Then every polynomial P of degree n can be represented as n X P (x) = αk Qk (x), k=0

where coefficients αk are give as αk =

L[P (x)Qk (x)] L[Qk (x)2 ]

(k = 0, 1, . . . , n).

16


1.2.2

Properties of orthogonal polynomials

Let {Qn }n∈N0 denote the sequence of polynomials, which are orthogonal on the interval (a, b) with respect to the weighted function x 7→ p(x). Theorem 1.2.2. All zeros of the polynomial Qn , n ≥ 1, are real, different and contained in the interval (a, b). Proof. From the orthogonality Z b 0= p(x)Qk (x)dx = a

1 (Qk , Q0 ) Q0 (x)

(k = 1, 2, . . . )

we conclude that the polynomial Qk (x) change the sign in at least one point form the interval (a, b). Suppose that x1 , . . . , xm (m ≤ k) are real zeros of the odd order of the polynomial Qk (x), which are contained in (a, b). Define the polynomial P (x) as P (x) = Qk (x)ω(x), where ω(x) = (x − x1 ) · · · (x − xm ). Since the polynomial ω(x) can be represented by m X ω(x) = αi Qi (x), i=0

we have Z

b

p(x)P (x)dx = (Qk , ω) = a

m X

αi (Qk , Qi ),

i=0

i.e., ( 0, (m < k), (Qk , ω) = 2 αk kQk k , (m = k). On the other hand, notice that the polynomial P (x) does not change the sign in the set (a, b). Thus, it follows that (Qk , ω) 6= 0. We conclude that we must have m = k and the proof is completed. Orthogonal polynomials satisfy the three term recurrence relation. Theorem 1.2.3. The following three term recurrence relation follows for the sequence {Qn }n : Qn+1 (x) − (αn x + βn )Qn (x) + γn Qn−1 (x) = 0, where αn , βn , γn are constants.

(2.2.1)


17

Proof. We can take αn such that the degree of the polynomial Rn (x) = Qn+1 (x) − αn xQn (x) is equal to n. From Theorem 2.1.4 this polynomial can be represented as Rn (x) = Qn+1 (x) − αn xQn (x) = βn Qn (x) −

n X

γj Qj−1 (x).

j=1

Now we have (Qn+1 , Qi ) − αn (Qn , xQi ) = βn (Qn , Qi ) −

n X

γj (Qj−1 , Qi ).

j=1

If 0 ≤ i ≤ n − 2, using the orthogonality of the sequence {Qn }n , we conclude that γj = 0 (j = 1, . . . , n − 1). Hence, the last formula reduce to (2.2.1). Since dg Qn = n, we can write Qn (x) = an xn + bn xn−1 + . . . . Coefficients αn and βn from (2.2.1) can be determined by formulae an+1 bn bn+1 αn = , βn = an − (n ∈ N). (2.2.2) an an+1 an In order to determine γn , we start with the recurrence relation Qn (x) − (αn−1 x + βn−1 )Qn−1 (x) + γn Qn−2 (x) = 0.

(2.2.3)

Multiplying (2.2.3) by p(x)Qn (x), integrating from a to b, then using the orthogonality of the sequence {Qn }n∈N0 , we get Z b 2 ||Qn || = αn−1 p(x)xQn−1 (x)Qn (x)dx. a

Similarly, multiplying (2.2.1) with p(x)Qn−1 (x) and then integrating from a to b, we get the equality Z b αn p(x)xQn−1 (x)Qn (x)dx − γn ||Qn−1 ||2 = 0. a

From the last two equalities we obtain ||Qn || 2 an+1 an−1 αn ||Qn || 2 = . γn = αn−1 ||Qn−1 || a2n ||Qn−1 ||

(2.2.4)

If the leading coefficient of the polynomial Qn (x) is equal to 1, then the polynomial Qn (x) is called monic. For these polynomials the following statement holds.

18


Theorem 1.2.4. For the monic sequence of orthogonal polynomials {Qn }n the following recurrence relation holds: Qn+1 (x) = (x − βn )Qn (x) − γn Qn−1 (x),

(2.2.5)

where βn and γn are real constants and γn > 0. Proof. The last relation (2.2.5) is an immediate consequence of the relation (2.2.2). Namely, since an = 1, from (2.2.2) and (2.2.4) we get αn = 1,

βn = bn − bn+1 ,

γn =

||Qn || ||Qn−1 ||

2 > 0.

The three term recurrence relation (2.2.3) is fundamental in constructive theory of orthogonal polynomials. Three important reasons are mentioned below. 1◦ The sequence of orthogonal polynomial {Qn }n∈N0 can be easily generated provided that sequences {βn } and {γn } are known. 2◦ Coefficients γn given in (2.2.4) determine the norm of polynomials Qn : ||Qn ||2 =

Z

b

p(x)Qn (x)2 dx = γ0 γ1 · · · · · γn ,

a

where we take γ0 = c0 =

Rb a

p(x)dx.

3◦

Coefficients βn and γn form a symmetric three-diagonal matrix, so called Jacobi matrix, which is important in construction of Gauss quadrature formulae for numerical integration.

1.2.3

The Stieltjes procedure

Let {Qn }n∈N0 be the monic set of orthogonal polynomials in the interval (a, b) with respect to the weighted function x 7→ p(x). Coefficients βn and γn from the recurrence relation (2.2.4) can be easily expressed in the form βn =

(xQn , Qn ) (Qn , Qn )

γn =

(Qn , Qn ) (Qn−1 , Qn−1 )

(n = 0, 1, 2, . . . ),

(n = 1, 2, . . . ),


19

i.e., Rb

p(x)xQn (x)2 dx βn = Ra b 2 a p(x)Qn (x) dx Rb p(x)Qn (x)2 dx γn = R ba 2 a p(x)Qn−1 (x) dx where γ0 =

Rb a

(n = 0, 1, 2, . . . ),

(2.3.1)

(n = 1, 2, . . . ),

(2.3.2)

p(x)dx.

Namely, Stieltjes proved that the recurrence relation (2.2.3) and formulae (2.3.1) and (2.3.2) can be successively applied to generate polynomials Q1 (x), Q2 (x),... . The procedure of alternating applications of formulae ( 2.3.1) and (2.3.2) and the recurrence relation (2.2.3) is known as the Stieltjes procedure.

1.2.4

Classical orthogonal polynomials

Classical orthogonal polynomials form one special class of orthogonal polynomials. They are solutions of some problems in mathematical physics, quantum mechanics, and especially in approximation theory and numerical integration. We present the definition and the most important properties of classical orthogonal polynomials. Definition 1.2.4. Let {Qn }n∈N0 be the sequence of orthogonal polynomials in (a, b) with respect to the weighted function x 7→ p(x). Polynomials {Qn }n∈N0 are called classical, if the weighted function satisfies the differential equation (A(x)p(x))0 = B(x)p(x), (2.4.1) where x 7→ B(x) is the polynomial of the degree 1, and the function x 7→ A(x), depending of a and b, has the form   (x − a)(b − x), a and b are finite ,    x − a, a is finite and b = +∞, A(x) =  b − x, a = −∞ and b is finite,    1, a = −∞, b = +∞. The weighted function of classical orthogonal polynomials satisfies the conditions lim xm A(x)p(x) = 0

x→a+

i

lim xm A(x)p(x) = 0.

x→b−

(2.4.2)

20

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS Solving the differential equation (2.4.1) we obtain Z C B(x) p(x) = exp dx , A(x) A(x)

where C is an arbitrary constant. Namely, with respect to the choice of a and b we have   (b − x)α (x − a)β , a and b are finite,    (x − a)s erx , a is finite, b = +∞, p(x) =  (b − x)t e−rx , a = −∞, b is finite,    exp R B(x)dx , a = −∞, b = +∞, where B(a) B(b) − 1, β = − − 1, b−a b−a s = B(a) − 1, t = −B(b) − 1, B(x) = rx + q. α=

(2.4.3)

If a is finite, the boundary conditions (2.4.2) require B(a) > 0; if b is finite, then (2.4.2) require B(b) < 0; also r = B 0 (x) < 0 must hold. However, if a and b are both finite, then a condition r = B 0 (x) < 0 is a corollary of the first two conditions. The interval (a, b) can be mapped into one of intervals (−1, 1), (0, +∞), (−∞, +∞) by a linear function. Hence, we can take the weighted function p(x) to be equal, respectively to one of the following: x 7→ (1 − x)α (1 + x)β ,

x 7→ xs e−x ,

2

x 7→ e−x .

Parameters α, β and s must obey conditions (according to (2.4.2) and (2.4.3)) α > −1, β > −1, s > −1. We have three essentially different cases. 1◦

Let p(x) = (1 − x)α (1 + x)β and (a, b) = (−1, 1). Then A(x) = 1 − x2 and B(x) =

1 d (A(x)p(x)) = β − α − (α + β + 2)x. p(x) dx

Corresponding orthogonal polynomials are called Jacobi polynomials and (α,β) they are denoted by Pn (x). Special cases of Jacobi polynomials follow: 1.1. Legendre polynomials Pn (x) ( α = β = 0),


21

1.2. Chebyshev polynomials of the first kind Tn (x) ( α = β = −1/2), 1.3. Chebyshev polynomials of the second kind Sn (x) (α = β = 1/2 ), 1.4. Gegenbauer polynomials Gλn (x) ( α = β = λ − 1/2 ). 2◦ Let p(x) = xs e−x and (a, b) = (0, +∞) . Then A(x) = x and B(x) = 1 + s − x . Corresponding orthogonal polynomials are called generalized Laguerre polynomials and they are denoted by Lsn (x). For s = 0 these polynomials are known as Laguerre polynomials Ln (x). 2

3◦ Let p(x) = e−x and (a, b) = (−∞, +∞). Then A(x) = 1 and B(x) = −2x. Corresponding orthogonal polynomials are called Hermite polynomials and they are denoted by Hn (x). (α,β) Characteristics of Jacobi Pn (x), Laguerre Lsn (x) and Hermite Hn (x) polynomials are given in the following table.

Classification of classical orthogonal polynomials

(a, b)

p(x)

A(x)

B(x)

(−1, 1)

(1 − x)α (1 + x)β

1 − x2 —

β − α − (α + β + 2)x

(0, +∞)

xs e−x

(−∞, +∞)

e−x

2

Qn (x) (α,β)

Pn

x

1+s−x

1

−2x

Hn (x)

Theorem 1.2.5. Classical orthogonal polynomials {Qn }n satisfy the formula Qn (x) =

(x)

Lsn (x)

Cn dn (A(x)p(x)) p(x) dxn

(n = 0, 1, . . . ),

where Cn are non-zero constants. The formula (2.4.4) is called the Rodrigues formula. One way to choose constants Cn is the following  (−1)n  (α,β)  for Pn (x),   2n n! Cn = 1 for Lsn (x),    (−1)n for Hn (x).

(2.4.4)

22


Constants Cn for Gegenbauer polynomials Gλn (x), for Chebyshev polynomials of the first kind Tn (x) and for Chebyshev polynomials of the second kind Sn (x), respectively, are taken as follows: (2λ)n P (α,α) (x) (α = λ − 1/2), (λ + 1/2)n n n! Tn (x) = P (−1/2,−1/2) (x), (1/2)n n (n + 1)! (1/2,1/2) Sn (x) = P (x), (3/2)n n Gλn (x) =

where (s)n = s(s + 1) · · · · · (s + n − 1) =

Γ(s + n) Γ(s)

(Γ is the gamma function).

Let {Qn }n be classical orthogonal polynomials in the interval (a, b). Theorem 1.2.6. The polynomial x 7→ Qn (x) is one particular solution of the linear homogenous differential equation of the second order L(y) = A(x)y 00 (x) + B(x)y 0 (x) + λn y = 0,

(2.4.5)

where λn = −n

1 (n − 1)A00 (0) + B 0 (0) 2

(n ∈ N0 ).

(2.4.6)

Using (2.4.5) and (2.4.6) we find, respectively, the following differential equations (1 − x2 )y 00 − (2λ + 1)xy 0 + n(n + 2λ)y = 0, (1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0, (1 − x2 )y 00 − xy 0 + n2 y = 0, (1 − x2 )y 00 − 3xy 0 + n(n + 2)y = 0, xy 00 + (1 + s − x)y 0 + ny = 0, y 00 − 2xy 0 + 2ny = 0, corresponding to the polynomials Gλn (x), Pn (x), Tn (x), Sn (x), Lsn (x), Hn (x).


1.2.5

23

Generating function

We begin with a general definition of the generating function. Definition 1.2.5. The function (x, t) 7→ F (x, t) is called the generating function for the class of polynomials {Qn }n∈N0 if, for some small t, the following holds: ∞ X Qn (x) tn F (x, t) = , Cn n! n=0

where Cn are normalization constants. In the case of classical orthogonal polynomials, Cn is given in the Rodrigues formula (2.4.4). Classical orthogonal polynomials can be defined as coefficients of Taylor expansion of certain generating functions. Generating functions of Gegenbauer, Laguerre and Hermite polynomials, respectively, are given as follows: 2 −λ

(1 − 2xt + t )

=

(1 − x)−(s+1) e−tx/(1−x) = 2

e2xt−t =

∞ X n=0 ∞ X n=0 ∞ X

Gλn (x)tn , Lsn (x)

tn , n!

Hn (x)tn .

n=0

Using previous relations, we easily find the three term recurrence relation as well as the explicit representation of Gegenbauer polynomials, i.e., nGλn (x) = 2x(λ + n − 1)Gλn−1 (x) − (n + 2λ − 2)Gλn−2 (x), with starting values Gλ0 (x) = 1 and Gλ1 (x) = 2λx: [n/2]

Gλn (x)

=

X k=0

(−1)k

(λ)n−k (2x)n−2k . k!(n − 2k)!

Notice that Legendre polynomials Pn (x) are a special case of Gegenbauer 1/2 polynomials, i.e, Pn (x) = Gn (x). Hence, the corresponding recurrence relation is nPn (x) = x(2n − 1)Pn−1 (x) − (n − 1)Pn−2 (x), P0 (x) = 1, P1 (x) = x.

24


The explicit representation of Legendre polynomials is [n/2]

Pn (x) =

X

(−1)k

k=0

(2n − 2k − 1)!! (2x)n−2k . 2n−k k!(n − 2k)!

Classical Chebyshev polynomials of the second kind Sn (x) and of the first kind Tn (x) are also special cases of Gegenbauer polynomials. Precisely, we have Sn (x) = G1n (x) and G0n (x) = 2/nTn (x). Thus, the following representations hold: [n/2]

Sn (x) =

X

k

(−1)

k=0

n−k (2x)n−2k k

and [n/2] n−k 1X k n (2x)n−2k , T0 (x) = 1. (−1) Tn (x) = k 2 n−k k=0

Similarly, we find the three term recurrence relation for Laguerre polynomials Lsn+1 (x) = (2n + s + 1 − x)Lsn (x) − n(n + s)Lsn−1 (x), Ls0 (x) = 1,

Ls1 (x) = s + 1 − x,

and Hermite polynomials nHn (x) = 2xHn−1 (x) − 2Hn−2 (x), H0 (x) = 1, H1 (x) = 2x. From corresponding generating functions we have representations n X s k n Ln (x) = (−1) (s + k + 1)n−k xk , k k=0

[n/2]

Hn (x) =

X

(−1)k

k=0

(2x)n−2k . k!(n − 2k)!

Remark 1.2.1. Chebyshev polynomials Tn (x) and Sn (x) are special cases of Gegenbauer polynomials, i.e., Gλn (x) 2 = Tn (x) λ→0 λ n lim

(n = 1, 2, . . . ),

Sn (x) = G1n (x),

so we easily find three term recurrence relations and explicit representations.

1.3. GEGENBAUER POLYNOMIALS AND GENERALIZATIONS

1.3 1.3.1

25

Gegenbauer polynomials and generalizations Properties of Gegenbauer polynomials

Gegenbauer polynomials Gλn (x) are classical polynomials orthogonal on the interval (−1, 1) with respect to the weight function x 7→ (1 − x2 )λ−1/2 (λ > −1/2). We presented general properties of the orthogonal polynomials in the previous section. Here we shall mention the most important properties of Gegenbauer polynomials. Polynomials Gλn (x) are defined by the following expansion: Gλ (x, t) = (1 − 2xt + t2 )−λ =

+∞ X

Gλn (x)tn .

(3.1.1)

n=0

The function Gλ (x, t) is the generating function of polynomials Gλn (x) . Corresponding three term recurrence relation is nGλn (x) = 2x(λ + n − 1)Gλn−1 (x) − (n + 2λ − 2)Gλn−2 (x),

(3.1.2)

for n ≥ 2, where Gλ0 (x) = 1 , Gλ1 (x) = 2λx . Hence, starting from (3.1.2) with the initial values Gλ0 (x) = 1 and λ G1 (x) = 2λx, we easily generate the sequence of polynomials {Gλn (x)}: Gλ0 (x) = 1, Gλ1 (x) = 2λx, (λ)2 Gλ2 (x) = (2x)2 − λ, 2! (λ)2 (λ)3 (2x)3 − (2x), Gλ3 (x) = 3! 1! (λ)4 (λ)3 (λ)2 Gλ4 (x) = (2x)4 − (2x)2 + , 4! 2! 2!

etc.

By the series expansion of the generating function Gλn (x) in powers of t, and then comparing the coefficients with respect to tn , we get the representation [n/2] X (λ)n−k λ Gn (x) = (−1)k (2x)n−2k . (3.1.3) k!(n − 2k)! k=0

Gegenbauer polynomials Gλn (x) can be represented in several ways. For example, we can start from the equality −λ t2 (x2 − 1) λ −2λ G (x, t) = (1 − xt) 1− , (1 − xt)2

26


and obtain the representation (Rainville [96]) [n/2]

Gλn (x) =

X (2λ)n xn−2k (x2 − 1)k . k!(n − 2k)!22k (λ/2)k k=0

Using the well-known result for Jacobi polynomials (see [96]) and the equality (2λ)n (λ− 12 ,λ− 12 ) Gλn (x) = P (x), n λ + 21 n we get the following representations: 1 1 − x (2x)n 2 F1 −n, 2λ + n; λ + ; n! 2 2 n X k x−1 (2λ)n+k , = 1 2 k!(n − k)! λ + 2 k k=0 x + 1 n (2λ) 1 1 x − 1 n Gλn (x) = − λ − n; λ + ; 2 F1 −n, n! 2 2 2 x+1 n x − 1 k x + 1 n−k X (2λ)n λ + 12 n , = 1 1 2 2 k!(n − k)! λ + λ + 2 k 2 n−k k=0 n n 1 (2λ)n n 1 x2 − 1 Gλn (x) = x 2 F1 − , − + ; λ + ; , n! 2 2 2 2 x2 n n 1 (2λ)n Gλn (x) = (2x)n 2 F1 − , − + ; 1 − n − λ; x−2 , n! 2 2 2 Gλn (x) =

where 2 F1 is a hypergeometric function, defined as 2 F1 (a, b; c; x)

=

+∞ X (a)k (b)k k=0

k!(c)k

xk ,

where a, b, c are real parameters and c 6= 0, −1, −2, . . . . The Rodrigues formula for Gegenbauer polynomials is Gλn (x) =

(−1)n (2λ)n (1 − x2 )−λ+1/2 Dn (1 − x2 )n+λ−1/2 , 2n n! λ + 21 n

where D denotes the standard differentiation operator (D = d/dx). Remark 1.3.1. Some other representations of Gegenbauer polynomials can be found, for example, in [6], [14], [75], [94], [96], [100], [107].


27

Let Gλ (x, t) be the function defined in (3.1.1). The next theorem gives some differential-difference relations for polynomials Gλn (x). Theorem 1.3.1. Gegenbauer polynomials Gλn (x) satisfy the following equalities Dm Gλn (x) = 2m (λ)m Gλ+m n−m (x),

dm ; dxm

(3.1.4)

(n ≥ 1);

(3.1.5)

Dm ≡

nGλn (x) = x D Gλn (x) − D Gλ+m n−1 (x)

D Gλn+1 (x) = (n + 2λ)Gλn (x) + x D Gλn (x); 2λGλn (x) = D Gλn+1 (x) − 2x D Gλn (x) + D Gλn−1 (x)

(3.1.6) (n ≥ 1);

(3.1.7)

[(n−2)/2]

D Gλn (x)

=2

X

(λ + n − 1 − 2k)Gλn−1−2k (x);

(3.1.8)

k=0

λ (x) − xG (x) ; (n + 2λ)Gλn (x) = 2λ Gλ+1 n n−1 1 Dk G1/2 n (x); (2k − 1)!! X Dk Pn+k (x) = (2k − 1)!! Pi1 (x)Pi2 (x) · · · Pi2k+1 (x), Gk+1/2 (x) = n

(3.1.9) (3.1.10) (3.1.11)

i1 +···+i2k+1 =n

where Pn (x) is the Legendre polynomial. We omit the proof of this theorem. In the next section we shall give the proof of the corresponding theorem for generalized Gegenbauer polynomials pλn,m (x), which reduce to polynomials Gλn (x) for m = 2. Notice that Popov [95] proved the equality (3.1.11).

1.3.2

Generalizations of Gegenbauer polynomials

In 1921 Humbert [95] defined the class of polynomials {Πλn,m }n∈N0 using the generating function m −λ

(1 − mxt + t )

=

∞ X

Πλn,m (x) tn .

(3.2.1)

n=0

Differentiating (3.2.1) with respect to t, and then comparing coefficients with respect to tn , we obtain the recurrence relation (n + 1)Πλn+1,m (x) − mx(n + λ)Πλn,m (x) − (n − mλ − m)Πλn−m+1,m (x) = 0.

28


For m = 2 in (3.2.1), we obtain the generating function of the Gegenbauer polynomials. Also, we can see that Pincherle’s polynomials Pn (x) are a special case of Humbert’s polynomials. Namely, the following holds (see [81],[25]) −1/2

Gλn (x) = Πλn,2 (x) and Pn (x) = Πn,3 (x). Later, Gould [54] investigated the class of generalized Humbert polynomials Pn (m, x, y, p, C) , which are defined as (C − mxt + ytm )p =

∞ X

Pn (m, x, y, p, C) tn

(m ≥ 1).

(3.2.2)

n=0

From (3.2.2), we get the recurrence relation CnPn − m(n − 1 − p)xPn−1 + (n − m − mp)yPn−m = 0

(n ≥ m ≥ 1),

where the notation Pn = Pn (m, x, y, p, C) is introduced. Horadam and Pethe [67] investigated polynomials pλn (x), which are associated to Gegenbauer polynomials Gλn (x). Namely, writing polynomials Gλn (x) horizontally with respect to the powers of x, and then taking sums along the growing diagonals, Horadam and Pethe obtained polynomials pλn (x), whose generating function is (1 − 2xt + t3 )−λ =

+∞ X

pλn (x) tn−1 .

(3.2.3)

n=1

Remark 1.3.2. Some special cases of polynomials pλn (x) are investigated in papers of Horadam ([57], [58]) and Jaiswal [70]. Comparing (3.2.1) and (3.2.3), it can be seen that polynomials {pλn (x)} are a special case of Humbert polynomials {Πλn,m (x)} , i.e., the following equality holds: 2x pλn+1 (x) = Πλn,3 . 3 We can be seen that polynomials Gλn (x) are a special case of generalized Humbert polynomials Pn , i.e., Gλn (x) = Pn (2, x, 1, −λ, 1). Polynomials {pλn,m (x)} are investigated in the paper [82]. These polynomials are defined as 2x pλn,m (x) = Πλn,m . m


29

Further, we point out the most important properties of these polynomials. The generating function is given as Gλm (x, t) = (1 − 2xt + tm )−λ =

+∞ X

pλn,m (x) tn .

(3.2.4)

n=0

It can easily be seen that polynomials Gλn (x), Pnλ (x), Pn (x) are closely connected with polynomials pλn,m (x). Namely, the following equalities hold: pλn,2 (x) = Gλn (x), pλn,3 (x) = pλn+1 (x), pλn,m (x) = Pn (m, 2x/m, 1, −λ, 1), 2 λ Pn (m, x, m/2, −λ, m/2). pλn,m (x) = m By the series expansion of the function Gλm (x, t) = (1 − 2xt + tm )−λ in powers of t, and then comparing coefficients with respect to tn , we get the representation [n/m]

pλn,m (x)

=

X

(−1)k

k=0

(λ)n−(m−1)k (2x)n−mk . k!(n − mk)!

(3.2.5)

Differentiating both sides of (3.2.4) with respect to t and comparing the corresponding coefficients, for n ≥ m ≥ 1 we get npλn,m (x) = 2x(λ + n − 1)pλn−1,m (x) − (n + mλ − m)pλn−m,m (x),

(3.2.6)

with starting values pλn,m (x) =

(λ)n (2x)n , n!

n = 0, 1, . . . , m − 1.

The recurrence relation (3.2.6) for corresponding monic polynomials pˆλn,m (x) is pˆλn,m (x) = xˆ pλn−1,m (x) − bn pˆλn−m,m (x), n ≥ m ≥ 1, with starting values pˆλn,m (x) = xn , n = 0, 1, . . . , m − 1, where bn =

(n − 1)! n + m(λ − 1) · m . (m − 1)! 2 (λ + n − m)m

(3.2.7)

30


It is interesting to consider the relation between two terms of the sequence of coefficients {bn }. Namely, by (3.2.7), we find that the following holds ([25]) bn+1 n!(n + 1 + m(λ − 1)) (m − 1)!2m (λ + n − m)m = m · bn 2 (m − 1)!(λ + n + 1 − m)m (n − 1)1(n + m(λ − 1)) (n + m(λ − 1) + 1)n(λ + n − m) = ∼ n. (n + m(λ − 1))(λ + n) Remark 1.3.3. We can obtain polynomials pλn,m (x) in the following constructive way, so called the “repeated diagonal process”. We describe this process. Polynomials [n/m]

pλn,m (x)

=

X

aλn,m (k)(2x)n−mk ,

k=0

where

(λ)n−(m−1)k , k!(n − mk)! are written horizontally, one below another, likewise it is shown in Table 3.2.1. aλn,m (k) = (−1)k

Table 3.2.1 n

pλn,m (x)

0

1

1

2λx

2 .. . m m+1 m+2 .. .

(λ)2 2!

(λ)m m!

(2x)2 .. . 1 (2x)m - (λ) 0!

(λ)m+1 m+1 - (λ)2 (2x) 1! (m+1)! (2x) (λ)m+2 (λ)3 m+2 - 2! (2x)2 (m+2)! (2x)

.. .

Since aλn,m (k) = (−1)k

(λ)n−k−(m−1)k = aλn,m+1 (k), k!(n − k − mk)!

m ≥ 1,


31

summing along growing diagonals yields polynomials pλ0,m+1 (x) = 1, pλ1,m+1 (x) = 2λx, .. . (λ)m pλm,m+1 (x) = (2x)m , m! (λ)m+1 pλm+1,m+1 (x) = (2x)m+1 − (m + 1)! (λ)m+2 (2x)m+2 − pλm+2,m+1 (x) = (m + 2)!

(λ)1 , 0! (λ)2 (2x), etc. 1!

Let pλk,m (x) = 0 for k ≤ 0. In the following theorem we present several important properties of polynomials pλn,m (x). As a particular case, these properties hold for classical Gegenbauer polynomials Gλn (x). Theorem 1.3.2. Polynomials pλn,m satisfy the following equalities λ+k (x); Dk pλn+k,m (x) = 2k (λ)k pn,m

(3.2.8)

2npλn,m (x) = 2x D pλn,m (x) − m D pλn−m+1,m (x); m D pλn+1,m (x) = 2(n + mλ)pλn,m (x) + 2x(m − 1) D pλn,m (x); 2λpλn,m (x) = D pλn+1,m (x) − 2x D pλn,m (x) + D pλn−m+1,m (x);

(3.2.9) (3.2.10) (3.2.11)

[(n−m)/m]

D pλn,m (x)

X

=2

(λ + n − 1 − mk)pλn−1−mk,m (x)

k=0 [(n−m)/m]

+ (m − 2)

X

D pλn−m(k+1),m (x);

(3.2.12)

k=0 λ 2λxpλn−1,m (x) − mλpλ+1 n−m,m (x) = npn,m (x);

mλ)pλn,m (x)

mλpλ+1 n,m (x)

1)λxpλ+1 n−1,m (x);

= − 2(m − 1 k+1/2 1/2 pn−k,m (x) = Dk pn,m (x); (2k − 1)!! X 1/2 1/2 1/2 Dk pn+k,m (x) = (2k − 1)!! pi1 ,m (x) · · · pi2k+1 (x),

(n +

(3.2.13) (3.2.14) (3.2.15) (3.2.16)

i1 +···+i2k+1 =n 1/2

where pn,m (x) is a polynomial associated with the Legendre polynomial.

32


Proof. Differentiating the polynomial pλn+k,m (x) one by one k-times, we obtain [n/m] k

D

pλn+k,m (x)

=

X

(−1)j

(λ)n+k−(m−1)j k 2 (2x)n−mj j!(n + k − mj)!

(−1)j

(λ)k (λ + k)n−(m−1)j k 2 (2x)n−mj j!(n − mj)!

j=0 [n/m]

=

X j=0

= 2k (λ)k pλ+k n (x). Notice that (3.2.8) is an immediate corollary of the last equalities. Differentiating the function Gλm (x, t) with respect to t and x, we get 2t

∂Gλm (x, t) ∂Gλm (x, t) − 2x − mtm−1 = 0. ∂t ∂x

Now, according to (3.2.4), we obtain the equality (3.2.9). Now, differentiating (3.2.6) and changing n by n + 1, we get the equality (3.2.10). The equality (3.2.11) can be proved differentiating the equality (3.2.4) with respect to x, and comparing the corresponding coefficients. Multiplying (3.2.9) by λ + n, we get (2x)(λ + n) D pλn,m (x) = 2n(λ + n)pλn,m (x) + m(λ + n) D pλn+1−m,m (x).

(3.2.17)

Using (3.2.6), and changing n by n + 1, we get the equality (n + 1)pλn+1,m (x) = 2(λ + n)pλn,m (x) + 2x(λ + n) D pλn,m (x) − (m(λ − 1) + n + 1) D pλn+1−m,m (x).

(3.2.18)

Next, from (3.2.17) and (3.2.18) we obtain 2(λ + n)pλn,m (x) = D pλn+1,m (x) − (m − 1) D pλn−m,m (x). We change n, respectively, by n + 1, n − m, n − 2m, . . ., n − mk, where k ≤ [(n − m)/m]. Thus, the last equation generate the following system of equations: 2(λ + n − 1 − mk)pλn−1−mk,m (x) = D pλn−mk,m (x) − (m − 1) D pλn−m(k+1),m (x).

(3.2.19)


33

Summing the obtained equalities (3.2.19) we get (3.2.12). Using the representation (3.2.5) we easily obtain the equality (3.2.13). For λ = 1/2, from (3.2.5) we get the polynomial [n/m]

p1/2 n,m (x)

=

X

(−1)j

j=0

(2n − 1 − 2(m − 1)j)!! (2x)n−mj . 2n−(m−1)j j!(n − mj)!

1/2

Hence, differentiating pn,m (x) one by one k-times, we get Dk p1/2 n,m (x) = [(n−k)/m]

X

(−1)j

j=0

(2n − 1 − 2(m − 1)j)!! (2x)n−k−mj . j!(n − k − mj)!2n−k−(m−1)j

(3.2.20)

Since [(n−k)/m] k+1/2 pn−k,m (x)

=

X j=0

=

(−1)j

(k + 1/2)n−(m−1)j−k (2x)n−k−mj n−k−(m−1)j 2 j!(n − k − mj)!

1 1/2 Dk pn,m (x), (2k − 1)!!

form (3.2.20) we conclude that (3.2.15) holds. Now, differentiating (3.2.4) with respect to x, one by one k-times, we get dk Gλm (x, t) = (2k − 1)!!(Gλm (x, t))2k+1 . dxk Since (Gλm (x, t))2k+1 =

∞ X

1/2

Dk pn+k,m (x)tn+k ,

(3.2.21)

(3.2.22)

n=0

from (3.2.21) and (3.2.22), we conclude that (3.2.16) holds. Thus, Theorem 3.2.1 is proved. Corollary 1.3.1. Form m = 2 equalities (3.2.8)–(3.2.16) correspond to Gegenbauer polynomials Gλn (x). Corollary 1.3.2. For m = 1 equalities (3.2.10)–(3.2.16) correspond to polynomials pλn,1 (x). Namely, these equalities, respectively, reduce to the follo-

34


wing: 2npλn,1 (x) = (2x − 1) D pλn,1 (x), D pλn+1,1 (x) = 2(n + λ)pλn,1 (x), 2λpλn,1 (x) = D pλn+1,1 (x), 2λpλn,1 (x) = D pλn+1,1 (x) + (1 − 2x) D pλn,1 (x), D pλn,1 (x) = 2 −

n−1 X

(λ + n − 1 − k)pλn−1−k,1 (x)

k=0 n−1 X

D pλn−1−k,1 (x), n ≥ 1,

k=0

npλn,1 (x)

= λ(2x − 1)pλ+1 n−1,1 (x),

(n + λ)pλn,1 (x) = λpλ+1 n,1 (x), 1 1/2 Dk pn,1 (x), (2k − 1)!! X 1/2 Dk pn+k,1 (x) = (2k − 1)!! k+1/2

pn−k,1 (x) =

1/2

1/2

pi1 ,1 (x) · · · pi2k+1 ,1 (x).

i1 +···+ik+1 =n

1.3.3

The differential equation

We prove one more important property for polynomials pλn,m (x). Namely, we find a differential equation which has the polynomial pλn,m (x) as one of its solutions. In order to prove this result we define the sequence {fr }nr=0 , and operators ∆ and E. Let {fr }nr=0 be the sequence defined as fr = f (r), where n − t + m(λ + t) f (t) = (n − t) . m m−1 Let 4 and E, respectively, denote the finite difference operator and the translation operator (the shift operator), defined as (see also Milovanović, Djordjević [78]) 4fr = fr+1 − fr , Efr = fr+1 , 40 fr = fr , 4k fr = 4 4k−1 fr , E k fr = fr+k . The following result holds.


35

Theorem 1.3.3. The polynomial pλn,m (x) is one particular solution of the homogenous differential equation of the m-th order y (m) (x) +

m X

as xs y (s) (x) = 0,

(3.3.1)

s=0

with coefficients as =

2m s 4 f0 s!m

(s = 0, 1, . . . , m).

(3.3.2)

Proof. From (3.2.5) we get [(n−s)/m]

x

s

Ds pλn,m (x)

=

X

(−1)k

k=0

(λ)n−(m−1)k (2x)n−mk k!(n − s − mk)!

(3.3.3)

and Dm pλn,m (x)

p−1 X = (−1)k k=0

(λ)n−(m−1)k 2m (2x)n−m(k+1) , k!(n − m(k + 1))!

(3.3.4)

where p = [(n − s)/m], s ≤ q; p − 1 = [(n − s)/m], s > q; q = 0, 1, . . . , m − 1. Taking (3.3.3) and (3.3.4) in the differential equation (3.3.1), and comparing the corresponding coefficients we get equalities: m X n − mk s! as = 2m k(λ + n − (m − 1)k)m−1 , (3.3.5) s s=0

k = 0, 1, . . . , p − 1, and q X n − mp s! as = 2m p(λ + n − (m − 1)p)m−1 . s

(3.3.6)

s=0

The equality (3.3.6) can be presented as q m X q 2 s=0

s

m

s

mn

4 f0 = 2

−q m

n−q λ+q+ m

. m−1

Since (1 + 4)q f0 = E q f0 = fq = f (q), it follows that the equality (3.3.6) holds.

36

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS The equality (3.3.5) can also be presented as m X n − mk s=0

s

4s f0 = fn−mk

(k = 0, 1, . . . , p − 1).

(3.3.7)

The formula (3.3.7) is an interpolation formula for the function f (t) in the point t = n − mk. The degree of the polynomial f (t) is equal to m, and the interpolation formula is constructed in m + 1 points. Hence, the interpolation formula coincides with the polynomial. It follows that the equality (3.3.5) holds. Taking m = 1, 2, 3, respectively, from (3.3.1) we get differential equations: (1 − 2x)y 0 (x) + 2ny(x) = 0, (1 − x2 )y 00 (x) − (2λ + 1)xy 0 (x) + n(n + 2λ)y(x) = 0, 32 3 000 16 1 − x y (x) − (2λ + 3)x2 y 00 (x) 27 9 8 − (3n(n + 2λ + 1) − (3λ + 2)(3λ + 5))xy 0 (x) 27 8 + n(n + 3λ)(n + 3λ + 3)y(x) = 0. 27 These equations, respectively, correspond to polynomials pλn,1 (x), Gλn (x), pλn,3 (x). For λ = 1/2 the second equation becomes (1 − x2 )y 00 − 2xy 0 + n(n + 1)y = 0, and corresponds to Legendre polynomials; for λ = 1 this equation becomes (1 − x2 )y 00 − 3xy 0 + n(n + 2)y = 0, and one solution of this equation is Sn (x), the Chebyshev polynomial of the second kind; for a λ = 0 we get the differential equation (1 − x2 )y 00 − xy 0 + n2 y = 0, which corresponds to Tn (x), the Chebyshev polynomial of the first kind.


1.3.4

37

Polynomials in a parameter λ

We can consider the Gegenbauer polynomials Gλn (x) as the function of a parameter λ, and this polynomial can be represented in the form (see [85]) Gλn (x) =

n X

gj,n (x)λj ,

(3.4.1)

j=1

where gj,n (x) (j = 1, 2, . . . , n) are polynomials of the degree equal to n. Polynomials gj,n (x) (j = 1, 2, . . . , n) can be represented in the form Mn (j) n−j

gj,n (x) = (−1)

X k=0

(j)

Sn−k (2x)n−2k . k!(n − 2k)! (j)

Here Mn (j) = min([n/2], n − j) and Sn kind, defined as x

(n)

(3.4.2)

are Stirling numbers of the first

= x(x − 1) · · · (x − n + 1) =

n X

Sn(j) xj .

j=1

For j = 1, from (3.4.2) we get the equality g1,n (x) =

2 Tn (x). n

(3.4.3)

Using the generating function Gλ (x, t) of Gegenbauer polynomials Gλn (x), the following equality can be proved: g2,n (x) = 2

n−1 X j=1

1 1 Tj (x) Tn−j (x). j n−j

(3.4.4)

The generating function of polynomials gj,n (x) is (see [120] ) j log

(−1)

j

∞

(1 − 2xt + t2 ) X = gj,n (x)tn . j! n=0

Starting from (3.4.5) and from the generating function −1/2 log(1 − 2xt + t2 ) =

∞ X Tn (x) n=1

n

tn

(3.4.5)

38


of Chebyshev polynomials Tn (x) (see Wrigge [120]) we obtain ∞ X X Tij (x) n Ti1 (x) j 2 j j ··· t , log (1 − 2xt + t ) = (−1) 2 i1 ij n=j i1 +···+ij =n

i.e., 2j gj,n (x) = j!

X i1 +···+ij =n

Ti (x) Ti1 (x) ··· j i1 ij

.

(3.4.6)

Polynomials gj,n (x) can also be expressed in terms of symmetric functions (see [85]). Similarly, we can consider polynomials pλn,m (x). Wrigge ([120]) investigated polynomials pλn,m (x) as functions of the parameter λ. The polynomial pλn,m (x) has the representation pλn,m (x) =

n X

hj,n (x) · λj .

(3.4.7)

j=1

Thus, from (3.4.7) and (3.2.5) we get the explicit representation of polynomials hj,n (x), j = 1, 2, · · · , n, i.e., (j)

Mn (j) n−j

hj,n (x) = (−1)

X k=0

mk

(−1)

Sn−(m−1)k k!(n − mk)!

(2x)n−mk ,

(3.4.8)

where Mn (j) = min([n/m], [(n − j)/(m − 1)]). Expanding the function Gλm (x, t) (given in (3.2.4)) in powers of λ, and then using (3.4.8), we obtain ∞

X (−1)j logj (1 − 2xt + tm ) = hj,n (x)tn . j! n=j

Remark 1.3.4. More details about polynomials hj,n (x), j = 1, 2, . . . , n, can be found in the paper of Milovanović and Marinković [85].

1.3.5

Polynomials induced by polynomials pλn,m (x) (m,q,λ)

In this section we consider polynomials QN (t), which are induced by λ generalized Gegenbauer polynomials pn,m (x) (see [84]). In order to de(m,q,λ)

fine polynomials QN q ∈ {0, 1, . . . , m − 1}.

(t), let n = mN + q, where N = [n/m] and


39

The explicit representation of polynomials pλn,m (x) now has the form pλn,m (x) =

N X (λ)mN +q−(m−1)k (2x)mN +q−mk , k!(mN + q − mk)! k=0

wherefrom we obtain (m,q,λ)

pλn,m (x) = (2x)q QN

(t),

(3.5.1)

(λ)mN +q−(m−1)k n−k t . k!(mN + q − mk)!

(3.5.2)

where (2x)m = t, and (m,q,λ) QN (t)

=

N X

(−1)k

k=0

If x = 0, then (m,q,λ)

QN

(0) = (−1)N

Γ(λ + N + q) . Γ(λ)Γ(N + 1)Γ(q + 1)

From (3.5.1), (3.5.2) and the recurrence relation npλn,m (x) = 2x(λ + n − 1)pλn−1,m (x) − (n + m(λ − 1))pλn−m,m (x), we can prove the next statement. (m,q,λ)

Theorem 1.3.4. Polynomials QN relations: For q ∈ {1, 2, . . . , m − 1}, (m,q,λ)

(mN + q)QN

(t) satisfy the following recurrence

(m,q−1,λ)

(t) = (λ + mN + q − 1)QN − (mN + q + m(λ −

(t)

(m,q,λ) 1))QN −1 (t).

(3.5.3)

For q = 0, (m,0,λ)

mN QN

(m,m−1,λ)

(t) = (λ + mN − 1)tQN −1 − m(N + λ −

(m,0,λ) 1)QN −1 (t).

Using the well-known equality (see [25]) Dk pλn+k,m (x) = 2k (λ)k pλ+k n,m (x), we can prove the next statement.

(t) (3.5.4)

40

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS (m,q,λ)

Theorem 1.3.5. Polynomials QN relations: For 0 ≤ q ≤ m − 2, (m,q+1,λ)

(q + 1)QN

(t) (λ > −1/2) satisfy the recurrence

(m,q+1,λ)

(t) + mt D QN

(m,q,λ+1)

(t) = λQN

(t).

For q = m − 1, (m,0,λ)

(m,m−1,λ+1)

m D QN +1 (t) = λQN

(t).

Also, the following statement holds. (m,q,λ)

Theorem 1.3.6. Polynomials QN recurrence relations: For 0 ≤ q ≤ m − 3 (m,q+2,λ)

(q + 1)(q + 2)QN

(m,q+2,λ)

+ m2 t2 D2 QN

(t) (λ > −1/2 , m ≥ 3 ) satisfy the

(m,q+2,λ)

(t) + m(m + q + 1)t D QN (m,q,λ+2)

(t) = (λ)2 QN

(t)

(t).

For q = m − 2 (m,0,λ)

(m,0,λ)

(m,m−2,λ+2)

m(m − 1) D QN +1 (t) + m2 t D2 QN +1 (t) = (λ)2 QN

(t).

For q = m − 1 (m,1,λ)

(m,1,λ)

(m,m−1,λ+2)

m2 D QN +1 (t) + m2 tQN +1 (t) = (λ)2 QN

(t).

(m,q,λ)

The recurrence relation for polynomials QN (t) is proved in [84], where parameters m, q and λ are fixed. Precisely, the next statement is valid. (m,q,λ)

Theorem 1.3.7. Polynomials QN (t) satisfy the (m+1)-term recurrence relation m X (m,q,λ) (m,q,λ) Ai,N,q QN +1−i (t) = BN,q tQN (t), i=0

where coefficients BN,q and Ai,N,q (i = 0, 1, . . . , m) depend on parameters m, q and λ.


1.3.6

41

Special cases (m,q,λ)

We consider polynomials QN (t) for λ = 1 and λ = 0, as well as polynomials induced by Gegenbauer polynomials Gλn (x) and Chebyshev polynomials Tn (x). 1◦ λ = 1 . Recurrence relations (3.5.3) and (3.5.4), respectively, become (m,q,1)

QN

(m,q−1,1)

(t) = QN

(m,q,1)

(t) − QN −1 (t)

(1 ≤ q ≤ m − 1)

and (m,0,1)

QN

(m,m−1,1)

(t) = tQN −1

(m,q,1)

Polynomials QN

(m,0,1)

(t) − QN −1 (t) (q = 0).

(t) satisfy the recurrence relation

m X m

i

i=0

(m,q,1)

(m,q,1)

qN +1−i (t) = tQN

(m,q,0)

2◦ λ = 0. We introduce polynomials QN

(t) in the following way

(m,q,0)

(m,q,0)

QN (m,q,0)

These polynomials QN m X i=0

(t) = lim

λ→0

QN

λ

(t).

(t)

.

(t) (0 ≤ q ≤ m − 1) satisfy the recurrence relation

m (m,q,0) (m,q,0) (m(N + 1 − i) + q) QN +1−i (t) = (mN + q)tQN (t). i (2,q,0)

For m = 2 we obtain polynomials QN (t) , q = 0, 1 , which are directly connected with Chebyshev polynomials of the first kind Tn (x). These polynomials satisfy the following relations (see [24]) (2,0,0)

T2N (x) = N QN

(t)

(2,1,0)

and T2N +1 (x) = (2N + 1)tQN

(t),

where t = (2x)2 . (m,q,λ) Gegenbauer polynomials Gλn (x) and polynomials QN (t) satisfies the relations (2,0,λ)

Gλ2N (x) = QN for t = (2x)2 .

(t)

(m,1,λ)

and Gλ2N +1 (x) = (2x)QN

(t),

42


1.3.7

Distribution of zeros

Some numerical examinations related to the distribution of zeros of induced (m,q,λ) polynomials QN (t) are established. We investigated fixed values of parameters m, N, q, λ. These values imply the degree of the polynomial n = mN + q. The following cases are considered: 1◦ The case λ = 1/2, for m ∈ {3, 4, 5, 6, 7, 8}, N ∈ {1, 2, 3, . . . , 15}, q ∈ {0, 1, . . . , 7}, n ∈ {3, 4, . . . , 127}. 2◦ The case λ = 0; other parameters are the same as in the case λ = 1/2. 3◦ The case λ = −0, 49990; other parameters are given by: m = 3, N ∈ {1, 2, . . . , 10}, q ∈ {0, 1, 2}, n ∈ {3, 4, . . . , 32}. 4◦ The case λ = 10; other parameters are the same as in the case 3◦ . From results obtained by numerical investigation, we notice some cha(m,q,λ) racteristic properties of polynomials QN (t) (see [25]). Thus, we get the following conclusions. (m,q,λ)

1. All zeros of polynomials QN in the interval (0, 2m ).

(t) (λ > −1/2) are real and contained (m,q,λ)

2. Two near-by terms of the sequence of polynomials {QN (t)} have the property that zeros of one polynomial are separated by zeros of the other polynomial, i.e., t01 < t1 < t02 < t2 < · · · < t0N < tN < t0N +1 (m,q,λ)

is satisfied, where t1 , t2 , . . . , tN are zeros of the polynomial QN (m,q,λ) t01 , t02 , . . . , t0N +1 are zeros of the polynomial QN +1 (t). (m,q,λ)

3. Zeros of the polynomial QN (m,q+1,λ) polynomial QN (t).

(t), and

(t) are separated by zeros of the

Remark 1.3.5. Some numerical results which illustrate previous conclusions, can be found in [25] and [83]. (m,q,λ)

Theorem 1.3.8. The polynomial QN real zeros.

(t) (λ > −1/2) has no negative

Proof. From (3.5.2) we get (m,q,λ)

QN

(t) = (−1)N

N X (λ)mN +q−(m−1)k N −k t . k!(q + m(N − k))! k=0


43

Notice that all coefficients on the right side have the same signa. Hence, the (m,q,λ) polynomial QN (t) has no negative zeros. According to a great number of numerical results, obtained for concrete values of parameters m, N and q, we formulate the following conjecture. (m,q,λ) Conjecture (a) All zeros of the polynomial QN (t) are real, mutually m different and contained in the interval (0, 2 ). (b) All zeros of the polynomial pλn,m (x) are contained in the unit disc. If (a) is true, then we prove that (b) holds. Let (a) be true. Then 0 < t < 2m ,

i.e.,

0 < (2x)m < 2m ,

and |x| < 1. Remark 1.3.6. Since t = (2x)m , from any zero t1 , t2 , . . . , tN of the poly(m,q,λ) nomial QN (t), we get the m-th zero of the polynomial pλn,m (x). Hence, zeros of the polynomial pλn,m (x) are contained in concentric circles of the ti radius ri = m , i = 1, . . . , [n/m], i.e., m-zeros are contained in N concentric 2 circles. Previous results, which concern the distribution of zeros of polynomials 1/2 1/2 are illustrated for polynomials: p14,3 (x), p055,8 (x) and p41,6 (x).

pλn,m (x),

1.3.8

Generalizations of Dilcher polynomials

In this section we consider Dilcher polynomials, which are connected to classical Gegenbauer polynomials. Dilcher (see [17]) considered polynomials (λ,ν) {fn (z)}, defined by G(λ,ν) (z, t) = (1 − (1 + z + z 2 )t + λz 2 t2 )−ν =

X

fn(λ,ν) (z)tn ,

(3.8.1)

n≥0

where ν > 1/2 and λ ≥ 0. Obviously, the degree of the polynomial fnλ,ν (z) is equal to 2n. Comparing (3.8.1) with the generating function Gνn of Gegenbauer polynomials (see [25], [96]), we get the following equality fn(λ,ν) (z)

n n/2

=z λ

Gνn

1 + z + z2 √ 2 λz

.

44

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS Using the recurrence relation for Gegenbauer polynomials nGνn (x) = 2x(ν + n − 1)Gνn−1 (x) − (n + 2(ν − 1))Gνn−2 (x), n ≥ m,

with starting values Gν0 (x) = 1, Gν1 (x) = 2νx, we get f0λ,ν (z) = 1, f1λ,ν (z) = ν(1 + z + z 2 ) and ν−1 ν−1 λ,ν 2 λ,ν λ,ν (1 + z + z )fn−1 (z) − 1 + 2 fn−2 (z). fn (z) = 1 + n n (λ,ν)

Polynomials fn

(z) are self-inversive (see [17], [18], [19]), i.e., (λ,ν) 2n (λ,ν) 1 fn (z) = z fn . z

(λ,ν)

Hence, polynomials fn

(z) can be represented in the form

λ,ν n+1 λ,ν n λ,ν λ,ν 2n λ,ν z + · · · + Cn,n z , z + Cn,1 z + · · · + Cn,0 fn(λ,ν) (z) = Cn,n + Cn,n−1 λ,ν λ,ν . = Cn,−k where Cn,k The most important results form [17] are related to determining coeffiλ,ν . Thus, the following formulae are obtained cients Cn,k

1 = Γ(ν)

λ,ν Cn,k

[(n−k)/2]

X

(−λ)s

s=0

Γ(ν + n − s) s!(n − 2s)!

[(n−k−2s)/2]

×

X j=0

2j + k j

n − 2s , 2j + k

and λ,ν Cn,k =

1 Γ(ν)

[(n−k)/2]

X

(−λ)s

s=0

n − k − s Γ(ν + n − s) (n−k−2s) B , s k!(n − k − s)! k

where (m) Bk

[m/2]

=

X j=0

If

i2

2j j

m k + j −1 . 2j j

= −1, then we can prove that the following formula holds: (m)

Bk

√ m!(2k)! k+1/2 √ = (−i 3)m G (1/ 3), (m + 2k)! m


45

λ,ν wherefrom we conclude that coefficients Cn,k obey the representation λ,ν Cn,k =

√ (2k)! (−i 3)n−k × S, Γ(ν)k!

where [(n−k)/2]

S=

X s=0

λ 3

s

√ Γ(ν + n − s) k+1/2 Gn−k−2s (i/ 3). s!(n + k − 2s)! (λ,ν)

Dilcher’s idea is used in [28], where polynomials {fn,m (z)} are defined and investigated. These polynomials are a generalization of polynomials λ,ν) fn (z), i.e., the following equality holds: (λ,ν)

fn,2 (z) = fn(λ,ν) (z). (λ,ν)

Polynomials {fn,m (z)} are determined by the expansion F (z, t) = 1 − (1 + z + z 2 )t + λz m tm

−ν

=

∞ X

(λ,ν) fn,m (z)tn .

(3.8.2)

n=0

Comparing (3.8.2) with (3.2.4), we obtain 1 + z + z2 (λ,ν) n n/m ν √ fn,m (z) = z λ pn,m . 2 m λz

(3.8.3)

Using the recurrence relation (see [25], [81]) npνn,m (z) = 2z(ν + n − 1)pνn−1,m (z) − ((n + m)ν − 1))pνn−m,m (z), n ≥ m, with starting values pνn,m (z) =

(ν)n (2z)n , n!

n = 0, 1, . . . , m − 1,

and also using (3.8.3), we get the following recurrence relation n−1 (λ,ν) (λ,ν) fn,m (z) = 1 + (1 + z + z 2 )fn−1,m (z) n m(ν − 1) (λ,ν) λz m fn−1,m (z), n ≥ m, − 1+ n with starting values (λ,ν) fn,m (z) =

(ν)n (1 + z + z 2 )n , n!

n = 0, 1, . . . , m − 1.

(3.8.4)

46

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS Changing z by 1/z in (3.8.3), we get (λ,ν) fn,m (z)

=

(λ,ν) z 2n fn,m

1 , z

(λ,ν)

and we conclude that the polynomial fn,m (z) is self-inversive. Hence, the (λ,ν) polynomial fn,m (z) (whose degree is equal to 2n) can be represented in the form (λ,ν)

(λ,ν)

(λ,ν) n fn,m (z) = p(λ,ν) n,n + pn,n−1 z + · · · + pn,0 z (λ,ν)

2n + pn,1 z n+1 + · · · + p(λ,ν) n,n z .

(3.8.5)

(λ,ν)

Using (3.8.4) we easily prove that coefficients pn,k , appearing in the formula (3.8.5), satisfy the recurrence relation ν − 1 (λ,ν) (λ,ν) (λ,ν) (λ,ν) pn,k = 1 + pn−1,k−1 + pn−1,k + pn−1,k+1 n m(ν − 1) (λ,ν) λpn−m,k , n ≥ m, (3.8.6) − 1− n (λ,ν)

(λ,ν)

where pn,k = pn,−k . The most important results from [28] are related to determining coeffi(λ,ν) cients pn,k . One such result is contained in the following theorem. (λ,ν)

Theorem 1.3.9. Coefficients pn,k (λ,ν)

pn,k =

1 Γ(ν)

are given by the formula

[(n−k)/m]

X

(−λ)s

s=0

Γ(ν + n − (m − 1)s) s!(n − ms)!

[(n−k−ms)/2]

X

×

j=0

n − ms 2j + k

2j + k . j

Proof. Using the explicit representation (see [22]) [n/m]

pνn,m (x) =

X

(−1)k

k=0

(ν)n−(m−1)k (2x)n−mk , k!(n − mk)!

and (3.8.3), we obtain (λ,ν) fn,m (z)

=

z n λn/m pνn,m [n/m]

=

X s=0

(−1)s

1 + z + z2 2λ1/m z

(ν)n−(m−1)s (1 + z + z 2 )n−ms z ms λs . s!(n − ms)!

(3.8.7)


47

The formula (3.8.7) follows from the last equalities and from the formula 2 r

(1 + z + z ) =

2r X

z

m

[m/2]

m=0

r m − 2j

X j=0

m−j m − 2j

(r ∈ N).

(λ,ν)

The next two results are related to coefficients pn,k . (λ,ν)

Theorem 1.3.10. Coefficients pn,k mula [(n−k)/m] (λ,ν) pn,k

=

X

s

(−λ)

s=0

can be expressed by the following for-

(ν)n−(m−1)s n − k − (m − 1)s (n−k−ms) B , k!(n − k − (m − 1)s)! k s

where (r) Bk

[r/2]

X 2j r k + j −1 = . j 2j j j=0

(λ,ν)

Theorem 1.3.11. Coefficients pn,k (λ,ν)

pn,k

1 = k+1

[(n−k)/m]

X j=0

are given by the formula

[r/2] X − 2r j 1−r 22j (−λ)s 2 j , s!(k!)2 (n − k − ms)! j! Γ(k + j) j=0

where r = n − k − ms.

1.3.9

Special cases

λ,ν We consider some special cases of polynomials fn,m (z). (λ,ν) ◦ 1 For m = 2 the formula (3.8.7) corresponds to the polynomial fn (z) (see [16]) and reduces to

(λ,ν) pn,k

1 = Γ(ν)

[(n−k)/2]

X

(−λ)s

s=0

Γ(ν + n − s) s!(n − 2s)!

(n−k−2s)/2]

×

X j=0

n − 2s 2j + k

2j + k . j

48

CHAPTER 1. STANDARD CLASSES OF POLYNOMIALS If z = 1, then we get the formula fn(λ,ν) (1) =

n X

(λ,ν)

pn,k = λn/2 pνn

k=−n

3 √

2 λ

.

(3.9.1)

2◦ From (3.9.1), for m = 2 and ν = 1 we obtain 3 (λ,1) n/2 √ fn (1) = λ Sn , 2 λ where Sn (x) is the Chebyshev polynomial of the second kind. From (3.8.3), for x = 1 we obtain n X 3 (λ,ν) (λ,ν) n/m ν √ fn,m (1) = pn,k = λ pn,m . 2mλ k=−n λ,ν Several interesting properties of polynomials fn,m (z) are given in the following theorem. (λ,ν)

Theorem 1.3.12. Polynomials fn,m (z) satisfy the following equalities: n (λ,ν) f (z); z n,m n! (λ,ν) Dk fn,m (z) = z −k f (λ,ν) (z); (n − k)! n,m (λ,ν) D fn,m (z) =

1 + z + z 2 (λ,ν) √ fn−1,m (z); 2mλ k (−1)k n! −k/m X (n)k−i (λ,1/2) 2k (λ,k+1/2) i k z fn−k,m (z) = λ (−1) f (z); (2k − 1)!! i (n − i)! n,m

(λ,ν) (λ,ν) (n + mν)fn,m (z) = mνfn,m (z) − ν(m − 1)

2

2

z D

(λ,ν) fn,m (z)

=

(λ,ν) nz D fn,m (z)

+

i=0 (λ,ν) nfn,m (z).

Corollary 1.3.3. For m = 2 previous equalities correspond to Dilcher poly(λ,ν) nomials fn (z). (λ,ν)

We get one more interesting result for polynomials fn,m (z). Let z 7→ g(z) be a differentiable function which is different from zero. Then the following equality holds: g −1 nz + 2z 2 g −1 D{g} − z 2 D2 −nz D{g 2 }g −1 n (λ,ν) +z 2 D2 {g}g −1 + 2z 2 D{g} D{g −1 } {gfn,m (z)}.

(λ,ν) fn,m (z) =

Chapter 2

Horadam polynomials and generalizations 2.1 2.1.1

Horadam polynomials Introductory remarks

In the paper [58] Horadam investigated polynomials An (x) and Bn (x), which are defined by the recurrence relations An (x) = pxAn−1 (x) + qAn−2 (x), A0 (x) = 0, A1 (x) = 1,

(1.1.1)

Bn (x) = pxBn−1 (x) + qBn−2 (x), B0 (x) = 2, B1 (x) = x.

(1.1.2)

and

Obviously, polynomials An (x) and Bn (x) include large families of polynomials obeying interesting properties, such as three term recurrence relation and homogenous differential equation of the second order. A. F. Horadam investigated several representatives of these polynomials. Sometimes Horadam’s collaborators took part in these investigations. This is the reason for the title of this chapter. For some fixed values of parameters p and q we have the following classes of polynomials: 1. for p = 1 and q = −2, An (x) are Fermat polynomials of the first kind; 2. for p = 1 and q = −2, Bn (x) are Fermat polynomials of the second kind; 49

50 3. kind; 4. 5. 6.

CHAPTER 2. HORADAM POLYNOMIALS for p = 2 and q = −1, An (x) are Chebyshev polynomials of the second for p = 2 and q = 1, An (x) are Pell polynomials; for p = 2 and q = 1, Bn (x) are Pell–Lucas polynomials; for p = 1 and q = 1, An (x) are Fibonacci polynomials.

Using standard methods, from relations (1.1.1) and (1.1.2) we find generating functions of the polynomials An (x) and Bn (x), respectively: F (x, t) = (1 − pxt − qt2 )−1 =

∞ X

An (x)tn

(1.1.3)

n=0

and

∞

X 1 + qt2 G(x, t) = = Bn (x)tn . 1 − pxt − qt2

(1.1.4)

n=0

Expanding the function F (x, t) from (1.1.3) in powers of t, then comparing coefficients with tn , we obtain the following explicit representation [n/2]

An (x) =

X k=0

qk

(n − k)! (px)n−2k . k!(n − 2k)!

We can prove that the polynomial x 7→ An (x) is one particular solution of the linear homogenous differential equation of the second order p2 2 00 3p2 0 p2 1+ x y + xy − n(n + 2)y = 0. (1.1.5) 4q 4q 4q The differential equation (1.1.5) reduces to several particular forms, depending on the choice of p and q. Thus we find: 1 2 00 3 0 1 1 − x y − xy + n(n + 2)y = 0, p = 1, q = −2; 8 8 8 (1 − x2 )y 00 − 3xy 0 + n(n + 2)y = 0, p = 2, q = −1; 1 3 1 1 + x2 y 00 + xy 0 − n(n + 2)y = 0, p = 1, q = 1; 4 4 4 (1 + x2 )y 00 + 3xy 0 − n(n + 2)y = 0, p = 2, q = 1. These differential equations, respectively, correspond to the Fermat polynomial of the first kind, the Chebyshev polynomial of the second kind, the Fibonacci polynomial and the Pell polynomial.

2.1. HORADAM POLYNOMIALS

2.1.2

51

Pell and Pell–Lucas polynomials

Now, we are interested in Pell and Pell–Lucas polynomials. We mentioned before that Pell polynomials Pn (x) and Pell–Lucas polynomials Qn (x) , respectively, a special cases of polynomials An (x) and Bn (x), taking p = 2 and q = 1. Thus, we have the following recurrence relations for polynomials Pn (x) and Qn (x) (see [62]): Pn (x) = 2xPn−1 (x) + Pn−2 (x), n ≥ 2, P0 (x) = 0, P1 (x) = 1,

(1.2.1)

Qn (x) = 2xQn−1 (x) + Qn−2 (x), n ≥ 2, Q0 (x) = 2, Q1 (x) = 2x.

(1.2.2)

and

Hence, we find: n

Pn (x)

Qn (x)

0 1 2 3 4 5 .. .

0 1 2x (2x)2 + 1 (2x)3 + 2(2x) (2x)4 + 3(2x)2 + 1 .. .

2 2x (2x)2 + 2 (2x)3 + 3(2x) (2x)4 + 4(2x)2 + 2 (2x)5 + 5(2x)3 + 5(2x) .. .

From (1.2.1) and (1.2.2) we notice that the following formula holds Qn (x) = Pn+1 (x) + Pn−1 (x).

(1.2.3)

If x = 1, then we get: Pn (1) = Pn the nth -Pell number; Qn (1) = Qn the nth -Pell–Lucas number; Pn (1/2) = Fn the nth -Fibonacci number; Qn (1/2) = Ln the nth -Lucas number. We also notice that Pn (x/2) = Fn (x) is the Fibonacci polynomial, and Qn (x/2) = Ln (x) is the Lucas polynomial. However, taking p = 2 and q = 1 in (1.1.3) and (1.1.4), and using (1.1.1) and (1.1.2), we find generating functions for Pell and Pell–Lucas polynomials: ∞ X

Pn+1 (x)tn = (1 − 2xt − t2 )−1 ,

n=0

and

∞ X n=0

Qn+1 (x)tn = (2x + 2t)(1 − 2xt − t2 )−1 .

52

CHAPTER 2. HORADAM POLYNOMIALS

Starting from generating functions and using standard methods we obtain representations of polynomials Pn (x) and Qn (x): [(n−1)/2]

Pn (x) =

n−k−1 (2x)n−2k−1 k

X k=0

and [n/2]

X k=0

n n−k (2x)n−2k . n−k k

We mention some very interesting properties of Pell and Pell–Lucas polynomials. These properties concern the relationship between these polynomials and Chebyshev polynomials of the first kind Tn (x) and of the second kind Sn (x), as well as Gegenbauer polynomials Gλn (x). Let i2 = −1 and let polynomials Pn (x) and Qn (x), respectively, be given by (1.2.1) and (1.2.2). Then the following equalities are satisfied: Pn (x) = (−i)n−1 Sn−1 (ix)

and

Qn (x) = 2(−1)n Tn (ix).

Hence, polynomials Pn (x) and Qn (x) are modified Chebyshev polynomials with the complex variable. From (1.1.3) we obtain the relation Pn+1 (ix) + Pn−1 (ix) = Qn (ix), wherefrom we conclude: Sn (ix) − Sn−2 (ix) = 2Tn (ix). Since the equalities Sn (x) = G1n (x)

and

Tn (x) = 2/nG0n (x)

are satisfied, we get that equalities : Pn (x) = (−i)n−1 G1n−1 (ix),

Qn (x) = n(−i)n G0n (ix) (n ≥ 1)

are also valid. We can easily prove that the following equalities are also satisfied: F1 = G10 (i/2) = 1,

Fn = (−i)n−1 G1n−1 (i/2),

L0 = 2G00 (i/2) = 2,

Ln = n(−i)n G0n (i/2).


2.1.3

53

Convolutions of Pell and Pell–Lucas polynomials

Further investigations of Pell and Pell–Lucas polynomials a presented according to papers of Horadam and Mahon (see [64], [65]). Horadam and (k) Mahon defined the k-th convolutions of Pell polynomials Pn (x), Pell-Lucas (k) (a,b) polynomials Qn (x) and mixed Pell polynomials πn (x). We present properties which are related to further generalizations of these polynomials. (k) Polynomials Pn (x) are defined by (see [26], [64]): n P (k−1)   Pi (x)Pn+1−i (x), k ≥ 1,     i=1 n P (1) (k−2) (k) Pi (x)Pn+1−i (x), k ≥ 2, Pn (x) =  i=1   n P (m) (k−1−m)    Pi (x)Pn+1−i (x), 0 ≤ m ≤ k − 1, i=1

(0)

(k)

where P0 (x) = Pn (x), P0 (x) = 0. The corresponding generating function is given by (1 − 2xt − t2 )−(k+1) =

∞ X

(k)

Pn+1 (x)tn .

(1.3.1)

n=0 (k)

The second class of polynomials Qn (x) is defined as Q(k) n (x) =

n X

(k−1)

Qi (x)Qn+1−i (x), k ≥ 1,

Q(0) n (x) = Qn (x),

i=1

wherefrom we find the generating function

∞

2x + 2t k+1 X (k) = Qn+1 (x)tn . 1 − 2xt − t2

(1.3.2)

n=0

Starting from (1.3.1) and (1.3.2), respectively, we obtain the following (k) (k) explicit formulas of the polynomials Pn (x) and Qn (x), respectively: Pn(k) (x)

[(n−1)/2]

=

k+n−1−r k

X r=0

and Q(k) n (x)

k+1

=2

n−1 X r=0

n−1−r (2x)n−2r−1 r

k + 1 k+1−r (k) x Pn−r (x). r

54

CHAPTER 2. HORADAM POLYNOMIALS Notice that the following equalities hold: (k−1)

Gkn (ix) = in Pn+1 (x)

(i2 = −1),

where Gkn (x) is the Gegenbauer polynomial (λ = k); (k)

Pn+1 (x) = Pn (2, x, −1, −(k + 1), 1), where Pn (m, x, y, p, C) is the Humbert polynomial defined by ([66]) ∞ X

(C − mxt + ytm )p =

Pn (m, x, y, p, C)tn

(m ≥ 1).

n=0 (k)

It is also interesting to consider the combination of polynomials Pn (x) (k) and Qn (x) ([65]). Thus, the notion of the mixed Pell convolution, as well as the convolution of the convolution are introduced. The mixed Pell con(a,b) volution πn (x) is defined by the expansion ∞ X

(a,b)

πn+1 (x)tn =

n=0

(2x + 2t)b , (1 − 2xt − t2 )a+b

a + b ≥ 1. (a,b)

We obtain the representation of the polynomials πn πn(a,b) (x)

=2

b−j

(1.3.3)

(x):

b−j X b − j b−j−i (a,b) x πn−i (x). i

(1.3.4)

i=0

(k)

(k)

Notice that polynomials Pn (x) and Qn (x) are special cases of the (a,b) polynomials πn (x). Namely, for b = 0 and a = k, from (1.3.3) we get πn(k,0) (x) = Pn(k−1) (x),

(1.3.5)

i.e., πn(1,0) (x) = Pn (x),

πn(2,0) (x) = Pn(1) (x).

If a = 0 and b = k, from (1.3.3) we also get πn(0,k) (x) = Qn(k−1) (x), i.e., πn(0,1) (x) = Qn (x),

πn(0,2) (x) = Q(1) n (x).

(1.3.6)


55

For j = 0 from (1.3.4) and using (1.3.5) we get the following representa(a,b) tion of polynomials πn (x): πn(a,b) (x)

b X b b−i (a+b−1) =2 x Pn−i (x). i b

(1.3.6’)

i=0

(k)

Using (1.3.1), (1.3.2) and the definition of polynomials Pn (x), we find that hold the following relation πn(a,b) (x)

=

n X

(a−1)

Pi

(b−1)

(a ≥ 1, b ≥ 1).

Qn+1−i (x)

i=1

Now, differentiating (1.3.3) with respect to t and then comparing coefficients with tn , we get the recurrence relation (a,b)

nπn+1 (x) = 2bπn(a+1,b+1) (x) + (a + b)πn(a,b+1) (x).

(1.3.7)

Using the equality (2x + 2t)b (2x + 2t)a (2x + 2t)a+b = · , (1 − 2xt − t2 )2a+2b (1 − 2xt − t2 )a+b (1 − 2xt − t2 )b+a we get the convolution of the convolution, i.e., πn(a+b,a+b) (x)

=

n X

(a,b)

πi

(b,a)

(x)πn+1−i (x).

(1.3.8)

i=1

Thus, for b = a in (1.3.8), follows formula πn(2a,2a) (x) =

n X

(a,a)

πi

(a,a)

(x)πn+1−i (x).

i=1

Taking b = 0 in (1.3.8), from (1.3.5) and (1.3.6) we obtain two represen(a,b) tations of polynomials πn (x):

πn(a,a) (x)

=

n X i=1

(a,0) (0,a) πi (x)πn+1−i (x)

=

n X

(a−1)

Pi

(a−1)

(x)Qn+1−i (x).

i=1

For a = 0 and b = k + 1, from equalities (1.3.6) and (1.3.6’) we get the formula k+1 X k+1 (k) (0,k+1) (k) k+1 πn (x) = Qn (x) = 2 Pn−i (x), i i=0

56

CHAPTER 2. HORADAM POLYNOMIALS (k)

which is one representation of polynomials Qn (x). In the next section we investigate generalizations of Pell and Pell–Lucas polynomials. We also mention ordinary Pell and Pell–Lucas polynomials. Notice that the Fibonacci polynomial Fn (x) is a particular case of the polynomial Pn (x), i.e., the following equality holds: x Pn = Fn (x). 2 Also, it is easy to see that the Lucas polynomial Ln (x) is a particular case of the polynomial Qn (x), i.e., x = Ln (x). Qn 2 However, for x = 1 we obtain well-known numerical sequences: Pn (1) = Pn ,

Pell sequence,

Qn (1) = Qn , Pell–Lucas sequence, 1 = Fn , Fibonacci sequence, Pn 2 1 Qn = Ln , Lucas sequence. 2

2.1.4

Generalizations of the Fibonacci and Lucas polynomials (k)

In the note [49] we consider two sequences of the polynomials, {Un,m (x)} (k) and {Vn,m (x)}, where k is a nonnegative integer and m is a positive integer. Some special cases of these polynomials are known Fibonacci and Lucas (k) (k) polynomials, for m = 2, and polynomials Un,3 (x) and Vn,3 (x), which are considered in the papers [39] and [45]. In [45] we consider the polynomials (k) (k) Un,m (x) and Vn,m (x), at first for m = 4 and then for arbitrary m. The polynomials Un,m (x) and Vn,m (x) are defined by recurrence relations ([39], [45]): Un,m (x) = xUn−1,m (x) + Un−m,m (x),

n ≥ m,

(1.4.1)

with U0,m (x) = 0, Un,m (x) = xn−1 , n = 1, 2, . . . , m − 1; and Vn,m (x) = xVn−1,m (x) + Vn−m,m (x),

n ≥ m,

(1.4.2)


57

with V0,m (x) = 2, Vn,m (x) = xn , n = 1, . . . , m − 1, m ≥ 2 and x is a real variable. In this case corresponding generating functions are given by: ∞

X t U (t) = = Un,m (x) tn 1 − xt − tm m

V m (t) =

2 − xt = 1 − xt − tm

n=0 ∞ X

Vn,m (x) tn .

(1.4.3) (1.4.4)

n=0

It is easy to get the next equality Vn,m (x) = Un+1,m (x) + Un+1−m,m (x), n ≥ m − 1. (k)

(k)

We denote by Un,m (x) and Vn,m (x), respectively, derivatives of the k th order of polynomials Un,m (x) and Vn,m (x), i.e., (k) Un,m (x) =

dk {Un,m (x)} dxk

and

(k) Vn,m (x) =

dk {Vn,m (x)}. dxk

For given real x, we take complex numbers α1 , α2 , . . . , αm , such that they satisfy: m X

αi = x,

i=1

X

αi αj = 0,

i m, and nhan,m (x) = x(m − 1)han−1−m,m (x) − xhan−1,m (x) − (2n + am − 2m)han−m,m (x) − (n + am − 2m)han−2m,m (x), for n ≥ 2m. Expanding the function Ga (x, t) (given with (3.4.2)) in powers of t, and then comparing coefficients, we get the formula [n/m]

han,m (x)

=

X (−1)n−(m−1)i (a + n − mi)i xn−mi , i!(n − mi)! i=0

which is the explicit representation of polynomials han,m (x). Several equalities of the convolution type will be proved in the following section.

190

CHAPTER 4. HERMITE AND LAGUERRE POLYNOMIALS

4.3.5

Convolution type equalities

In this section we shall prove the following statement. Theorem 4.3.3. The following equalities hold: n X

a a 2a gn−i,m (x)gi,m (y) = gn,m (x + y);

(3.5.1)

i=0 [n/m] n−mj

X X y n−i−mj (n − i − mj)j 2a gi,m (x + y); j!(n − mj)!

(3.5.2)

2a+2s a a (x + y), n ≥ 2s; Ds gn−i,m (x) Ds gi,m (y) = gn−2s,m

(3.5.3)

2a gn,m (x)

=

j=0

n X

i=0

i=0 n X

a+k a (2x), n ≥ 2k; Dk gn−i,m (x) Dk hai,m (x) = gn−2k,2m

(3.5.4)

i=0 [(n−k)/m]

X i=0 [(n−k)/m]

X i=0

(k)i a gn−k−mi,m (2x) (−1)i i!

k

= (−1)

n X

a+k gn−i−k,m (x)hai,m (x); (3.5.5)

i=0

(k)i a gn−k−mi,m (2x) (−1)i i!

k

= (−1)

n X

a ha+k n−i−k,m (x)gi,m (x); (3.5.6)

i=0 n X

a b a+b gn−i,m (x)gi,m (x) = gn,m (2x).

(3.5.7)

i=0

Proof. From (3.4.1) we get m)

F a (x, t) · F a (y, t) = (1 − tm )−2a e−(x+y)t/(1−t =

+∞ X

2a gn,m (x + y)tn .

n=0

On the other hand, we have a

a

F (x, t) · F (y, t) =

+∞ X n X n=0 i=0

Hence, we conclude that (3.5.1) is true.

a a (x)gi,m (y)tn . gn−i,m

(3.5.8)

4.3. GENERALIZATIONS OF LAGUERRE POLYNOMIALS

191

Now, starting from (3.5.8), we obtain m)

(1 − tm )−2a e−xt/(1−t

m)

= eyt/(1−t

+∞ X

2a gn,m (x + y)tn ,

n=0

and we get immediately the following 2a gn,m (x) =

+∞ n n X y t n=0

n!

!

+∞ X −n (−tm )k k k=0

!

+∞ X

! 2a gn,m (x + y)tn

.

n=0

First we multiply series on the right side. Comparing coefficients with tn we obtain the equality (3.5.2). Next, differentiating the function F a (x, t) (given with (3.4.1)) in x, one by one s-times, we get ∂ s F a (x, t) m = (−t)s (1 − tm )−a−s e−xt/(1−t ) . s ∂x So, it follows ∂ s F a (x, t) ∂ s F a (y, t) m · = t2s (1 − tm )−2a−2s e−(x+y)t/(1−t ) ∂xs ∂y s +∞ X 2a+2s = gn,m (x + y)tn+2s . n=0

Since +∞ n

∂ s F a (x, t) ∂ s F a (y, t) X X s a a · = D gn−i,m (x) Ds gi,m (y)tn , ∂xs ∂y s n=0 i=0

then we get the equality n X

2a+2s a a Ds gn−i,m (x) Ds gi,m (y) = gn−2s,m (x + y),

i=0

which represents the equality (3.5.3). Similarly, differentiating (3.4.1) and (3.4.2) in x, one by one k-times, we obtain ∂ k F a (x, t) ∂ k Ga (x, t) and , ∂xk ∂xk

192


whose product is equal to +∞

∂ k F a (x, t) ∂ k Ga (x, t) X a+k · = gn,2m (2x)tn+2k , ∂xk ∂xk n=0

i.e., +∞ X n X

+∞ X

a Dk gn−i,m (x) Dk hai,m (x)tn =

n=0 i=0

a+k gn,2m (2x)tn+2k .

n=0

Hence, we get the equality (3.5.4). Similarly, we can prove (3.5.5) and (3.5.6). Multiplying functions F a (x, t) and F b (x, t) (given in (3.4.1)), we get m)

F a (x, t) · F b (x, t) = (1 − tm )−(a+b) e−2xt/(1−t

=

+∞ X

a+b gn,m (2x)tn .

n=0

Since a

b

F (x, t) · F (x, t) =

∞ X

a gn,m (x)tn

∞ X

n=0

=

∞ X n X

b gn,m (x)tn

n=0 b a (x)tn , (x)gk,m gn−k,m

n=0 k=0

it follows that ∞ X n X

a b gn−k,m (x)gk,m (x)tn

=

n=0 k=0

∞ X

a+b gn,m (x)tn .

n=0

The required equality (3.5.7) is an immediate consequence of the last equality. If m = 1, then equality (3.5.2), (3.5.3) and (3.5.7), respectively, become (see [38]): L2a−1 (x) n n X i=0 n X i=0

=

n−j n−i−j n X X n y (n − i − j)i

j=0 i=0 a−1 s Ln−i (x) s

D

(n − i)!

D

j

i!

Li2a−1 (x + y),

L2a+2s−1 (x + y) La−1 (y) i = n−2s , i! (n − 2s)!

b−1 La−1 (x) La+b−1 (2x) n−i (x) Li = n , (n − i)! i! n!


193

where Lna−1 (x) is the Laguerre polynomial. Similarly, we can prove the following statement: Theorem 4.3.4. The following equalities hold: X a1 +···+ak (x1 + · · · + xk ); gia11,m (x1 ) · · · giakk,m (xk ) = gn,m

(3.5.9)

i1 +···+ik =n

X

1 +···+ak (x1 + · · · + xk ); hai11,m (x1 ) · · · haikk,m (xk ) = han,m

i1 +···+ik =n n X

X

X

gia1 ,m (x1 ) · · ·giak ,m (xk )

s=0 i1 +···+ik =n−s

haj1 ,m (x1 ) · · · haik ,m (xk )

j1 +···+jk =s

=

X

gia1 ,2m (2x1 ) · · · giak ,2m (2xk ).

i1 +···+ik =n

In the case of Laguerre polynomials the equality (3.5.9) becomes X i1 +···+ik =n

Lai k −1 (xk ) Lai11 −1 (x1 ) La1 +···+ak (x1 + · · · + xk ) ··· k = n . i1 ! ik ! n!

If x1 = · · · = xk = x and a1 = a2 = · · · = ak = a, then the equality (3.5.9) reduces to X ka gia1 ,m (x) · · · giak ,m (x) = gn,m (kx). i1 +···+ik =n

Moreover, for m = 1 this equality becomes X i1 +···+ik =n

4.3.6

La−1 (x) La−1 Lka−1 (kx) i i1 (x) ··· k = n . i1 ! ik ! n!

Polynomials of the Laguerre type

c,r In the note Djordjević, [50] we shall study a class of polynomials {fn,m (x)}, where c is some real number, r ∈ N ∪ {0}, m ∈ N. These polynomials are defined by the generating function. Also, for these polynomials we find an explicit representation in the form of the hypergeometric function; some identities of the convolution type are presented; some special cases are shown. The special cases of these polynomials are: Panda’s polynomials [42], [92]; the generalized Laguerre polynomials [42], [92]; the sister Celine’s polynomials [11]

194


4.3.7

c,r Polynomials fn,m (x)

Let φ(u) be a formal power–series expansion φ(u) =

∞ X

γn un ,

γn =

n=0

(−rr )n . n!

c,r We define the polynomials fn,m (x) as X ∞ −4xt c,r m −c = fn,m (x)tn . (1 − t ) φ (1 − tm )r

(3.7.1)

(3.7.2)

n=0

c,r Theorem 4.3.5. The polynomials fn,m (x) satisfy the following relations: c,r fn,m (x) =

(4rr x)n Frm−1 × A, n! (r+1)m

(3.7.3)

where # " n 1−c−rn rm−c−rn (−1)m−1 (4rr x)−m (rm)rm −m , . . . , m−1−n ; , . . . , ; m rm rm (rm−1)rm−1 A= rm−1−c−rn 1−c−rn 2−c−rn , , . . . , ; rm−1 rm−1 rm−1 (3.7.4) [n/m] X n! c,r k c + rn − rmk x = n r n (−1) fn−mk,m (x). 4 (r ) k n

(3.7.5)

k=0

Proof. Using (3.7.1) and (3.7.2), we find: ∞ X −4xt (−rr )n (−4x)n tn m −c m −c (1 − t ) φ = (1 − t ) (1 − tm )r n! (1 − tm )rn n=0

=

∞ X (4rr x)n tn n=0 ∞ X

n!

(1 − tm )−c−rn

! ∞ X −c − rn (−tm )k = k n=0 k=0 ∞ n r n−k n−k X X (4r x) t −c − r(n − k) (−1)k tmk = (n − k)! k (4rr x)n tn n!

!

n=0 k=0

(n − k − mk := n, n − k := n − mk, k ≤ [n/m])


∞ [n/m] X (−1)k (4rr )n−mk xn−mk −c − r(n − mk) X tn = (n − mk)! k n=0 k=0   [n/m] ∞ X X (4rr )n−mk xn−mk (c + r(n − mk))k   tn . = k!(n − mk)! n=0

k=0

Using the well–known equalities ([93]) (α)n+k = (α)n (α + n)k ,

(−n)k (−1)k (α)n (−1)k , = , (α)n−k = (n − k)! n! (1 − α − n)k

we find that (c)r(n−mk)+k (−1)mk (−1)mk (c + r(n − mk))k = · (n − mk)! (c)r(n−mk) (n − mk)! =

(−1)(rm−1)k (c)rn (1 − c − rn)rmk (−1)mk (−n)mk · · (1 − c − rn)(rm−1)k n! (−1)rmk (c)rn

=

(−1)(m−1)k (1 − c − rn)rmk (−n)mk (1 − c − rn)(rm−1)k n!

=

(−1)(m−1)k (rm)rmk mmk · A · B (rm − 1)(rm−1)k n! · C

where 2 − c − rn rm − c − rn 1 − c − rn · · ··· · , A= rm rm rm k k k −n 1−n m−1−n B= · · ··· · , m k m m k k 1 − c − rn 2 − c − rn rm − 1 − c − rn C= · · ··· · . rm − 1 rm − 1 rm − 1 k k k

From the other side, because of the next equality, m −c

(1 − t ) it follows

φ

−4xt (1 − tm )r

=

∞ X n=0

c,r fn,m (x)tn

195

196


(4rr x)n c,r fn,m (x) = Frm−1 × n! (r+1)m " # 1−c−rn rm−c−rn −n m−1−n (−1)m−1 (4rr x)−m (rm)rm , . . . , ; , . . . , ; rm rm m m (rm−1)rm−1 . 1−c−rn 2−c−rn rm−1−c−rn , , . . . , ; rm−1 rm−1 rm−1 These are the required equalities (3.7.3) and (3.7.4). Again, from (3.7.1) and (3.7.2), we get: ∞ X

(−4x)n tn

n=0

∞ X (−rr )n c,r = (1 − tm )c+rn fn,m (x)tn n! n=0 ! ∞ ! ∞ X c + rn X k mk c,r n = (−1) t fn,m (x)t k n=0

k=0

=

∞ [n/m] X X n=0 k=0

c,r k c + r(n − mk) (x)tn . (−1) fn−mk,m k

Hence, we get [n/m] X n! c,r k c + r(n − mk) x = n r n (x), (−1) fn−mk,m 4 (r ) k n

k=0

which yields the equality (3.7.5).

4.3.8

c,r (x) Some special cases of fn,m

If r = 1 and m > 1, then (3.7.3) and (3.7.4) become (−4x)n c,1 fn,m (x) = 2m Fm−1 × n! " # m m ( 4x ) 1−c−n 2−c−n m−c−n −n 1−n m−1−n , , . . . , ; , , . . . , ; m m m m m m (1−m)m−1 . 1−c−n 2−c−n m−1−c−n , , . . . , ; m−1 m−1 m−1 If m = 1 and r > 1, then (3.7.3) and (3.7.4) yield " # (4rr x)−1 1−c−rn r−c−rn (4rr x)n , . . . , ; −n; c,r r−1 r r (r−1) fn,1 (x) = . r+1 Fr−1 1−c−rn r−1−c−rn n! , . . . , ; r−1 r−1


197

For r = 0 in (3.7.2), and φ(u) = eu , we have (1 − tm )−c φ(−4xt) =

∞ X

c,0 fn,m (x)tn ,

n=0

and hence we get the following equalities: e

−4xt

m c

= (1 − t )

∞ X

c,0 fn,m (x)tn ,

n=0

and also ∞ X (−4x)n tn

n!

n=0

=

∞ X c (−tm )k k k=0

=

∞ [n/m] X X n=0 k=0

!

∞ X

! c,0 fn,m (x)tn

n=0

c c,0 (x)tn . (−1)k f k n−mk,m

So, we get xn =

[n/m] n! X k c (x). (−1) f c,0 (−4)n k n−mk,m

(3.8.1)

k=0

For c = 0 in (3.7.3) and (3.7.4), we get the following formula (4rr x)n Frm−1 × n! (r+1)m # 1−rn rm−rn −n m−1−n (−1)m−1 (4rr x)−m mm , . . . , ; , . . . , ; rm rm m m (rm−1)rm−1 . rm−1−rn 1−rn , . . . , ; rm−1 rm−1

0,r fn,m (x) = "

(−4)n For c = 1, m = 2 and r = 2, then, γn = and by (3.7.3) and (3.7.4), n! we get the following formula −2n 1−2n 2−2n 3−2n −n 1−n −1 (4x)n 1,2 4 , 4 , 4 , 4 ; 2 , 2 ; 123 x2 fn,2 (x) = 6 F3 −2n 1−2n 2−2n n! 3 , 3 , 3 ; Note that the generalized Laguerre polynomials are the special case of c,r c−1,1 the polynomials fn,m (x), that is, Lcn,m (x) = fn,m (x/4). So, we get the following representation

Lcn,m

xn = n!

" 2m Fm−1

2−c−n m+1−c−n −n x−m mm ; m , . . . , m−1−n ; (1−m) m−1 m ,..., m m 2−c−n m−c−n , . . . , ; m−1 m−1

# .

198


For the Laguerre polynomials Lcn,1 (x) ≡ Lcn (x), where −c

(1 − t)

exp

−xt 1−t

=

∞ X

Lcn (x)

n=0

tn , n!

the following statement holds.

Theorem 4.3.6. Let D =

d and g ∈ C ∞ (−∞, +∞) and g(x) 6= 0, then dx

n−k n X X n − k k n (−1) Dn−k−j {g −1 }Dj {g} = (−1)n n!. k j j=0

k=0

Proof. Using the known formula ([96]) DLcn (x) = DLcn−1 (x) − Lcn−1 (x),

n ≥ 1,

we get the following equalities: DLcn (x) = (D − 1)Lcn−1 (x), D2 Lcn (x) = (D − 1)2 Lcn−2 (x), ........................... Ds Lcn (x) = (D − 1)s Lcn−s (x), n ≥ s. Hence, for s = n, we obtain that n X

! n Dn Lcn (x) = (−1)k Dn−k {1} k k=0 n−k n X X n − k k n = (−1) Dn−k−j {g −1 }Dj {g}. k j k=0

j=0

Since Dn Lcn (x) = (−1)n n!, we get n X k=0

(−1)k

n−k n X n−k Dn−k−j {g −1 }Dj {g} = (−1)n n!, k j

which leads to (3.8.2).

j=0

(3.8.2)


199

Depending to the chosen functions g(x) and from (3.8.2), we get some interesting relations. 1◦ For g(x) = eax , a is any rial number, and g −1 (x) = e−ax , we get an

n n−k X a−k X

k!

k=0

j=0

(−1)j = 1. j!(n − k − j)!

2◦ If g(x) = (1 + x)α , for x > −1, α 6= 0, then we get (α)n Γ(α)

n n−k X (1 + x)k X

1

k!

(1 − α − n)k+j−2 Γ(α − j + 2)

k=0

j=0

=

(1 + x)n , αn!

or n n−k X X (−1)j (1 + x)k

Q

j i=0 (α

+ 1 − i)

Q

n−k−j−1 (α s=0

+ s) =

k!j!(n − k − j)!

k=0 j=0

3◦ For g(x) = ax , a > 0 and a 6= 1, we obtain n n−k X X (−1)j (ln a)n−k = 1. j!k!(n − k − j)! k=0 j=0

4◦ For g(x) = xα ex , we have the following formula j n n−k X X n−k−j X X (−1)j (α)j (α + 1 − l)l x−i−l k=0 j=0

i=0

k!i!l!(n − k − j)!(j − l)!

l=0

= 1,

or in the form j n n−k X X n−k−j X X (−1)k+i (−n)k+j+i (α)j k=0 j=0

i=0

xi+l k!i!l!(j − l)!

l=0

=

n! , Γ(α + 1)

hence, for α = n, n ∈ N, we get j n n−k X X n−k−j X X (−1)k+i (−n)k+j+i j! k=0 j=0

i=0

l=0

xi+l i!l!(j − l)!

= 1.

(1 + x)n . α

200

CHAPTER 4. HERMITE AND LAGUERRE POLYNOMIALS 5◦ For g(x) = xα , x ≥ 0 and x 6= 1, α 6= 0, we obtain n n−k X X (−1)j Qn−k−j−1 (α + j) Qj−1 (α − s)xk s=0

i=0

k!j!

k=0 j=0

4.3.9

= xn .

Some identities of the convolution type

Using the following equality ∞ X −4xt m c/2 c/2,r m −c/2 = (1 − t ) fn,m (x)tn , (1 − t ) φ (1 − tm )r

(3.9.1)

n=0

and by (3.7.1) and (3.7.2), we get ∞ X

c/2,r fn,m (x)tn

=

n=0

∞ X n X c/2 n=0 k=0

=

k

∞ [n/m] X X c/2

k

n=0 k=0

c/2,r

(−1)k fn−k,m (x)tn+mk−k c/2,r

(−1)k fn−mk,m (x)tn .

Hence, we get [n/m] c/2,r fn,m (x)

=

X k=0

c/2 c/2,r (−1) fn−mk,m (x). k k

(3.9.2)

Again, by (3.7.1) and (3.7.2), for φ(u) = eu we find: X ∞ −4(x1 + x2 + · · · + xs )t m −c c,r (1 − t ) φ fn,m (x1 + x2 + · · · + xs )tn , = (1 − tm )r n=0

that is, m −c/s

(1 − t )

e

−4x1 t (1−tm )r

m −c/s

· · · · · (1 − t )

e

−4xs t (1−tm )r

=

∞ X

c,r fn,m (x1 + · · · + xs ) tn ,

n=0

hence ∞ X n=0

! c/s,r fn,m (x1 )tn

∞ X

! c/s,r fn,m (x2 )tn

n=0

=

...

∞ X

! c/s,r fn,m (xs )tn

n=0 ∞ X n=0

c,r fn,m (x1 + · · · + xs )tn .


201

So, we get X

c/s,r

c/s,r

c/s,r

c,r fi1 ,m (x1 )fi2 ,m (x2 ) . . . fis ,m (xs ) = fn,m (x1 +x2 +· · ·+xs ). (3.9.3)

i1 +···+is =n

Let c = c1 + · · · + ck and x = x1 + · · · + xk , then we have, at, on the one side X ∞ −4(x1 + · · · + xk )t c1 +···+ck ,r = (1 − tm )−c1 −···−ck φ fn,m (x1 + · · · + xk )tn , (1 − tm )r n=0

and on the other side m −c1 −···−ck

(1 − t )

−4(x1 + · · · + xk )t φ (1 − tm )r ! ∞ X c1 ,r n fn,m (x1 )t · ··· · = n=0

=

∞ X

! ck ,r (xk )tn fn,m

n=0



 ,r ,r fic11,m (x1 ) . . . fickk,m (xk ) tn .

X 

n=0

∞ X

i1 +···+ik =n

Hence we get the following formula X ,r ,r c1 +···+ck ,r fic11,m (x1 ) · · · · · fickk,m (xk ) = fn,m (x1 + · · · + xk ).

(3.9.4)

i1 +···+ik =n

If x1 = x2 = · · · = xs =

x , then (3.9.3) becomes s

c/s,r

X

c/s,r

c,r fi1 ,m (x/s) . . . fis ,m (x/s) = fn,m (x),

i1 +···+is =n

where s is a natural number. For r = 0 in (3.7.2), and φ(u) = eu , we have m −c

(1 − t ) φ(−4xt) =

∞ X

c,0 fn,m (x)tn ,

n=0

whence we get e−4xt = (1 − tm )c

∞ X n=0

c,0 fn,m (x)tn ,

(3.9.5)

202


and also ∞ X (−4x)n tn n=0

n!

=

∞ X c (−tm )k k k=0

=

∞ [n/m] X X n=0 k=0

!

∞ X

! c,0 fn,m (x)tn

n=0

k c f c,0 (x)tn . (−1) k n−mk,m

So, we get xn =

[n/m] n! X k c (−1) f c,0 (x). (−4)n k n−mk,m k=0

For c = 0 in (3.7.3) and (3.7.4), we get the following formula (4rr x)n 0,r Frm−1 × fn,m (x) = n! (r+1)m   1 − rn rm − rn −n m − 1 − n (−1)m−1 (4rr x)−m mm ;  rm , . . . , rm ; m , . . . ,  m (rm − 1)rm−1  . 1 − rn rm − 1 − rn ,..., ; rm − 1 rm − 1

Bibliography [1] M. Abramowitz and I. A. Stegun’ Handbuk of Mathematical Functions, National Bureau of Standards, Washington, 1970. [2] R. André–Jeannin, A general class of polynomials, Fibonacci Quart. 32.5 (1994), 445–454. [3] R. André–Jeannin, A note on a general class of polynomials, Part II., Fibonacci Quart. 33.4(1995), 341–351. [4] R. André–Jeannin, A generalization of Morgan–Voyce polynomials, Fibonacci Quart. 32.3(1994), 228–231 [5] R. André–Jeannin, Differential properties of a general class of polynomials, Fibonacci Quart. 33.5(1995), 453–458. [6] G. Bandana, Unification of bilateral generating functions, Bull. Calcutha Math. Soc. 80 (1988), 209–216. [7] W. W. Bell, Special Functions for Scientists and Engineers, Van Nostrand, London, 1968. [8] B. Brindza, On some generalizations of the Diophantine equation 1k + 2k + · · · + xk = y z , Acta Arith. 44 (1984), 99–107. [9] L. Carlitz, Arithmetic properties generalized Bernoulli numbers, J. Reine Angew. Math. 202 (1959), 174–182. [10] S. Chandel, A note on some generating functions, Indian J. Math. 25(1983), 185–188. [11] M. Celine Fasenmyer, Some generalized hypergeometric polynomials, Bull. Amer. Math. Soc. 53(1947), 105–108. 203

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