Generating functions and hypergeometric representations of ...

Introduction Generating functions of classical continuous orthogonal polynomials Hypergeometric representations of classical orthogonal polynomials

Generating functions and hypergeometricrepresentations of classical continuous orthogonal

polynomials

Maurice Kenfack Nangho

University of Dschang

[email protected]

October 5, 2018

AIMS-Volkswagen Stiftung Workshop on Introduction toOrthogonal polynomials and Application

Generating functions and hypergeometric representations of classical continuous orthogonal polynomials


Overview

1 Introduction

2 Generating functions of classical continuous orthogonalpolynomials

3 Hypergeometric representations of classical orthogonalpolynomials



Let Fn, n ∈ N, be the Fibonacci numbers defined by

Fn+1 = Fn + Fn−1, (n ≥ 1; F0 = 0; F1 = 1).

Multiplying both sides with tn and summing from n = 2 to ∞, weobtain

+∞∑n=1

Fn+1tn =+∞∑n=1

Fntn ++∞∑n=1

Fn−1tn

=+∞∑n=0

Fntn ++∞∑n=0

Fntn+1

= F (t) + tF (t), F (t) =+∞∑n=0

Fntn.



Observing that

+∞∑n=1

Fn+1tn =+∞∑n=2

Fntn−1 = t−1

( ∞∑n=1

Fntn − t

),

we obtainF (t)− t

t= F (t) + tF (t).

That is

F (t) =t

1− t − t2.

The function F is called a generating function of the series

+∞∑n=0

Fntn

.



Definition

Let {pn(x)}∞n=0 be a sequence of polynomials. A genratingfunction of {pn(x)}∞n=0 is the function G (x , t) defined by

G (x , t) =∞∑n=0

cnPn(x)tn, (1)

where {cn}∞n=0 is a sequence of real or complex numbers.



Example

Find Generating function of the sequence {xn}∞n=0 in the followingcases

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

n!, n = 0, 1, 2, ...



Solution

1 Taking Pn(x) = xn and cn = 1, n = 0, 1, 2, ... into (1) weobtain

∞∑n=0

xntn =1

1− xt, |xt| < 1.

2 Taking Pn(x) = xn and cn = 1n! , n = 0, 1, 2, ... into (1) we

obtain∞∑n=0

xn

n!tn = exp(xt).



Generating functions have many applications in mathematics. Forinstance they can be used to

1 Find an exact formula for the members of a sequence

2 Find a recurrence formula

3 Find asymptotic formula for a sequence

Objective

In this part, we show how to obtain a generating function of asequence of classical continuous orthogonal polynomials and derivetheir hypergeometric representation.



Let us recall that a sequence {pn}n of continuous polynomials,orthogonal with respect to a weight function ρ is classical if andonly if there exists a polynomial φ of degree at most two and asequence {An}n of numbers such that

pn(x) =An

ρ(x)

dn

dxn[φ(x)nρ(x)]. (2)

Definition

Generating function of classical continuous orthogonal polynomialsis the function G(x,t) for which the series expansion in aneighbourhood of t = 0, is

G (x , t) =∞∑n=0

pn(x)

Ann!tn, (3)

where An is the coefficient in (2).



Theorem

Let {pn}n be a sequence of continuous orthogonal polynomialsthat satisfies the Rodriguess formula (2) . The function

G (x , t) =ρ(z)

ρ(x)

1

1− φ′(z)t

∣∣∣∣z=ζ(x ,t)

(4)

is a generation function of {pn}n, where ζ(x , t) is the zero ofz − x − φ(z)t satisfying lim

t→0ζ(x , t) = x.



Proof

Since pn, n = 0, 1, 2, ..., is continuous and satisfies the Rodriguessformula

pn(x) =An

ρ(x)

dn

dxn[φ(x)nρ(x)]

the function given by

G (x , t) =∞∑n=0

pn(x)

Ann!tn

is a generating function of {pn}n. Let x in the interval oforthogonality of pn and let C be a circle containing x . We obtainfrom the Cauchy’s formula

dn

dxn[φ(x)nρ(x)] =

n!

2iπ

∫C

φ(z)nρ(z)dz

(z − x)n+1,

where C is a circle containing x .



Proof

So, the Rodriguess formula of pn becomes

pn(x) =An

ρ(x)

n!

2iπ

∫C

φ(z)nρ(z)dz

(z − x)n+1

and the generating function of {pn} read

G (x , t) =∞∑n=0

1

2iπρ(x)

∫C

(φ(z)t)nρ(z)dz

(z − x)n+1.



Proof

The function f : z 7→ φ(z)z−x is bounded on the compact set C for is

continuous there. So, for |t| < 13M , where M is an upper bound of

f ,∣∣∣φ(z)tz−x

∣∣∣n < 13n for all z ∈ C. Therefore, we can interchange the

summation and integral to obtain

G (x , t) =1

2iπρ(x)

∫C

∞∑n=0

(φ(z)t)nρ(z)dz

(z − x)n+1

=1

2iπρ(x)

∫C

ρ(z)dz

z − x − φ(z)t. (5)



Proof

To end the proof, we evaluate the above integral by using theResidues formula.If φ is a polynomials of degree two, one of the zeros of thedenominator p(z) = z − x − φ(z)t in the integrand tends to ∞when t tends to 0 and the other one, ζ(x , t) tends to x when ttends to 0.If the polynomial φ is of degree at most one, the zero of p(z)tends to x when t tends to 0. So, for |t| sufficiently small, theintegrand in

G (x , t) =1

2iπρ(x)

∫C

ρ(z)dz

z − x − φ(z)t

has a single pole, ζ(x , t), inside the circle C. Therefore, using theresidues formula, we obtain



Proof

G (x , t) =1

ρ(x)lim

z→ζ(x ,t)

(z − ζ(x , t))ρ(z)

z − x − φ(z)t,

=ρ(z)

ρ(x)

1

1− φ′(z)t

∣∣∣∣z=ζ(x ,t)

.



Generating function of Hermite polynomials

The Hermite polynomials {Hn}n are orthogonal polynomialsassociated with the weight ρ(x) = exp(−x2) on the real lineR = (−∞, +∞). They are known to satisfy the Rodriguessformula

Hn(x) = (−1)n exp(x2)dn

dxn

[exp(−x2)

].

Identifying with the formula (2), we obtain φ(x) = 1,ρ(x) = exp(−x2) and An = (−1)n.Therefore z − x − φ(z)t = z − x − t has only one zero z = x + t.



Generating function of Hermite polynomials

Hence, from (3) and (4), we have

∞∑n=0

Hn(x)

(−1)nn!tn =

ρ(z)

ρ(x)

∣∣∣∣z=x+t

= exp(−2xt − t2).

Taking −t for t, we obtain

G (x , t) =∞∑n=0

Hn(x)

n!tn, G (x , t) = exp(2xt − t2). (6)



Generating function of Laguerre polynomials

The Laguerre polynomials {Ln}n are orthogonal polynomialsassociated with the weight ρ(x) = xα exp(−x) on the half-lineR+ = (0, +∞) (α > −1). They are known to satisfy theRodriguess formula

L(α)n (x) =

1

n!x−α exp(x)

dn

dxn[xnxα exp(−x)],

which is of the form (2) with ρ(x) = xα exp(−x), φ(x) = x andAn = 1

n! . Therefore the polynomialp(z) = z − x −φ(z)t = z − x − zt has only one zero ζ(x , t) = x

1−t .



Generating function of Laguerre polynomials

Hence, from (3) and (4)

∞∑n=0

L(α)n (x)tn =

ρ(z)

ρ(x)

1

1− t

∣∣∣∣z= x

1−t

=ρ( x

1−t )

(1− t)ρ(x).

Taking into account the fact that ρ(x) = xα exp(−x), we obtain

G (x , t) =∞∑n=0

L(α)n (x)tn, G (x , t) = (1− t)−α−1 exp

(xt

t − 1

).

(7)



Generating function of Bessel polynomials

The Bessel polynomials {yn(x ; a)}n are orthogonal with respect tothe weight function ρ(x) = xa exp(− 2

x ) on the interval (0, +∞).These polynomials are known to satisfy the Rodriguess formula

yn(x ; a) = 2nx−a exp(2

x)

dn

dxn[x2n+a exp(−2

x)],

which is of he form (2) with ρ(x) = xa exp(− 2x ), φ(x) = x2 and

An = 2−n. Therefore, the polynomialp(z) = z − x − φ(z)t = z − x − z2t has two zeros

ζ1(x , t) =1 +√

1− 4tx

2tand ζ2(x , t) =

1−√

1− 4tx

2t.




Since limt→0

ζ1(x , t) =∞ and limt→0

ζ2(x , t) = x , we deduce from (3)

and (4)

∞∑n=0

2−nyn(x ; a)tn =ρ(z)

ρ(x)

1

1− 2zt

∣∣∣∣z=ζ2(x ,t)

,

=ρ(ζ2(x , t))

ρ(x)

1

1− 2ζ2(x , t)t.




Noting that ζ2(x , t) = 2x1+√1−4tx , we obtain

ρ(ζ2(x , t)) =

(2x

1 +√

1− 4tx

)a

exp

(−1 +

√1− 4tx

x

)and

1

1− 2ζ2(x , t)t=

1√1− 4tx

.

Therefore,∞∑n=0

2−nyn(x ; a)tn = (1− 4tx)−12

(2

1 +√

1− 4tx

)a

exp

(1−√

1− 4tx

x

),

= (1− 4tx)−12

(2

1 +√

1− 4tx

)a

exp

(4t

1 +√

1− 4tx

).

Take t2 for t to obtain




G (x , t) =∞∑n=0

yn(x ; a)tn (8)

with

G (x ; t) = (1− 2tx)−12

(2

1 +√

1− 2tx

)a

exp

(2t

1 +√

1− 2tx

).



Generating function of Jacobi polynomials

The Jacobi polynomials {P(α,β)n }n( α > −1, β > −1) are

orthogonal with respect to the weight functionρ(x) = (1− x)α(1 + x)β on the interval (−1, 1). Thesepolynomials are known to satisfy the Rodriguess formula

P(α,β)n (x) =

(−1)n

n!2n(1− x)−α(1 + x)−β

dn

dxn[(1− x)α+n(1 + x)β+n],

(9)which is of the form (2), with ρ(x) = (1− x)α(1 + x)β,

φ(x) = (1− x2) and An = (−1)nn!2n . Therefore the polynomials

p(z) = z − x −φ(z) = z − x − (1− x2)t is of degree two with zeros

ζ1(x , t) =−1−

√1 + 4tx + 4t2

2tand ζ2(x , t) =

−1 +√

1 + 4tx + 4t2

2t.

Noting that limt→0

ζ1(x , t) =∞ and limt→0

ζ2(x , t) = x ,




we deduce from (3) and (4) the following

∞∑n=0

(−2)nP(α,β)n (x)tn =

ρ(z)

ρ(x)

1

1 + 2zt

∣∣∣∣z=ζ2(x ,t)

,

=ρ(ζ2(x , t))

ρ(x)

1

1 + 2ζ2(x , t)t.

Since ζ2(x , t) = −1+√1+4tx+4t2

2t , we obtain after simplification

ρ(ζ2(x , t)) =2α+β(1− x)α(1 + x)β(

1 + 2t +√

1 + 4tx + 4t2)α (

1− 2t +√

1 + 4tx4 + 4t2)β ,

1

1 + ζ2(x , t)=

1√1 + 4tx + 4t2

.




Therefore∞∑n=0

(−2)nP(α,β)n (x)tn

= 2α+β

(1+2t+√1+4tx+t2)

α(1−2t+

√1+4tx+t2)

β√1+4tx+t2

.

Taking − t2 for t, we obtain

G (x , t) =∞∑n=0

P(α,β)n (x)tn (10)

with

G (x , t) =2α+β

(1− t + R)α (1 + t + R)β R, R =

√1− 2tx + t2.



Hypergeometric function pFq is define by the series

pFq

(a1, ..., as+1

b1, ..., bs; z

)=∞∑k=0

(a1)k ...(ap)k(b1)k ...(bq)k

zk

k!,

where (a)n is the factorial function defined as follow

(a)n = a(a + 1)(a + 2)...(a + n − 1), n ≥ 1

and(a)0 = 1.

The parameters must be such that the denominator factors in theterms of the series are never zero.



When one of the numerator parameters, let us say a1, equals −n,where n is a nonnegative integer, this hypergeometric function is apolynomial in z .

pFq

(a1, ..., ap

b1, ..., bq; z

)=

n∑k=0

(−n)k ...(ap)k(b1)k ...(bq)k

zk

k!,

Otherwise the radius of convergence R of the hypergeometricseries is given by

R =

∞ if p < q + 11 if p = q + 10 if p > q + 1.



We are going to use Generating functions or Sturm-Liouvilleequation to derive hypergeometric representations of Hermite,Laguerre, Bessel and Jacobi polynomials.



Hypergeometric representation of Hermite polynomials

Generating function of Hermite polynomials is

exp(2xt − t2) = exp(2xt) exp(−t2) =∞∑n=0

(2xt)n

n!

∞∑k=0

(−t2)k

k!

By means of the Rainville relation

∞∑n=0

∞∑k=0

A(k , n) =∞∑n=0

[ n2]∑

k=0

A(k , n − 2k), (11)

where [n2 ] denotes the greatest positive integer less than or equalto n

2 , we obtain




exp(2xt − t2) =∞∑n=0

[ n2]∑

k=0

(−1)k(2x)n−2k

(n − 2k)!k!tn.

Comparing coefficients of tn in this result and in

exp(2xt − t2) =∞∑n=0

Hn(x)

n!tn, (12)

we obtain

Hn(x) = n!

[ n2]∑

k=0

(−1)k(2x)n−2k

(n − 2k)!k!.




Using the relations

(n − k)! =n!

(−1)k(−n)kand (a)2n = 22n

(a

2

)n

(a + 1

2

)n

,

(with a = −n), we obtain after simplification

Hn(x) = (2x)n[ n2]∑

k=0

(−n

2

)k

(−n − 1

2

)n

(− 1

x2

)kk!

,

= (2x)n2F0

(−n

2 ,−n−12

−;− 1

x2

).



Hypergeometric representation of Laguerre polynomials

Factorial function is an extension of the ordinary factorial since(1)n = n!. It is particularly convenient to use the factorial functionin the binomial expansion

(1− t)−a =∞∑n=0

(−a)(−a− 1)...(−a− n + 1)

n!(−t)n,

=∞∑n=0

(a)nn!

tn. (13)

The generating function (1− t)−α−1 exp(

xtt−1

)of the Laguerre

polynomials can be written as follow

(1− t)−α−1 exp

(xt

t − 1

)=∞∑k=0

(−xt)k

k!(1− t)−1−k−α.




By means of the binomial expansion (13),

(1− t)−α−1 exp

(xt

t − 1

)=∞∑k=0

(−xt)k

k!

∞∑n=0

(1 + k + α)nn!

tn.

Using the Rainville formula

∞∑n=0

∞∑k=0

A(k, n) =∞∑n=0

n∑k=0

A(k, n − k), (14)

we obtain

(1− t)−α−1 exp

(xt

t − 1

)=∞∑n=0

n∑k=0

(−x)k(1 + k + α)n−kk!(n − k)!

tn.

Comparing coefficients of tn in this result and in (7), we obtain




L(α)n (x) =

n∑k=0

(−x)k(1 + k + α)n−kk!(n − k)!

.

Noting that (a)n = (a + k)n−k(a)k and using the relation(n − k)! = n!

(−1)k (−n)k, we obtain

L(α)n (x) =

(α + 1)nn!

n∑k=0

(−n)k(α + 1)k

xk

k!=

(α + 1)nn!

1F1

(−n

α + 1; x

).



Hypergeometric representation of Bessel polynomials

Use the power series expansion of exp(z) in a neighbourhood ofz = 0 to transform the generating function

G (x ; t) = (1− 2tx)−12

(2

1 +√

1− 2tx

)a

exp

(2t

1 +√

1− 2tx

)of the Bessel polynomials into

G (x , t) =∞∑n=0

tn

n!(1− 2tx)−

12

(2

1 +√

1− 2tx

)a+n

.

Use the Rainville identity

(1− z)−12

(2

1 +√

1− z

)a

= 2F1

(a+12 , a+2

2

a + 1; z

)with a + n taken for a and z = 2xt, to obtain




G (x , t) =∞∑n=0

∞∑k=0

(a+n+1

2

)k

(a+n+2

2

)k

(a + n + 1)k

(2xt)k

k!

tn

n!.

Take into account the relation (14) to obtain

G (x , t) =∞∑n=0

n∑k=0

(a+n−k+1

2

)k

(a+n−k+2

2

)k

(a + n − k + 1)k

(2x)k

(n − k)!k!tn.

Compare coefficients of tn in this result and in (8) to obtain

yn(x ; a) =n∑

k=0

(a+n−k+1

2

)k

(a+n−k+2

2

)k

(a + n − k + 1)k

(2x)k

(n − k)!k!.




Use the relations (13) and(a + n − k + 1)2k = (a + n − 1 + 1)k(a + n + 1)kto obtain

yn(x ; a) =n∑

k=0

(−n)k(a + n + 1)k

(− x

2

)kk!

= 2F0

(−n, a + n + 1

−;−x

2

).



Hypergeometric representation of Jacobi polynomials

We obtain series expansion of Jacobi polynomials by means of theirSturm-Liouville equation. Jacobi polynomials are known to satisfythe Sturm-Liouville equation

(1−x2)y ′′(x)+[β−α−(α+β+2)x ]y ′(x)+n(n+α+β+1)y(x) = 0.

Take

y(x) =∞∑k=0

ak(1− x)k

and use the fact that

y ′(x) =∞∑k=1

−kak(1− x)k−1, y ′′(x) =∞∑k=2

−k(k − 1)ak(1− x)k−2,

1− x2 = 2(1− x)− (1− x)2, x = 1− (1− x)

to obtain



∞∑k=0

[Akak + Bkak+1](1− x)k = 0,

where

Bk = kα + kβ + k + k2 − n2 − nα− nβ − n,

Ck =(−2α− 4 k − 2− 2 kα− 2 k2

).

Identify coefficients of (1− x)k , k = 0, 1, ... to obtain therecurrence relation

ak+1 =(−n + k)(α + β + 1 + n + k)

2(α + 1 + k)(k + 1)ak .

Iterate this relation and take k + 1 for k to have

ak =(−n)k(α + β + n + 1)k

2k(α + 1)kk!a0.

Note that (−n)k = 0, k > n, to get




y(x) = a0

n∑k=0

(−n)k(α + β + n + 1)k(α + 1)kk!

(1− x

2

)k

.

Therefore there is a constant C not depending on x such that

P(α, β)n (x) = C

n∑k=0

(−n)k(α + β + n + 1)k(α + 1)kk!

(1− x

2

)k

.

P(α,β)n (1) = C .

By means of Leibnizs rule, the Rodriuess formula (9) of Jacobipolynomials read




P(α,β)n (x) =

(−1)n(α + 1)n(β + 1)nn!2n

n∑k=0

(−1)k(1− x)n−k(1 + x)k

(α + 1)n−k(β + 1)k(n − k)!k!.

Hence P(α,β)n (1) = (α+1)n

n! and we obtain

P(α, β)n (x) =

(α + 1)nn!

n∑k=0

(−n)k(α + β + n + 1)k(α + 1)kk!

(1− x

2

)k

,

=(α + 1)n

n!2F1

(−n, α + β + n + 1

α + 1;

1− x

2

).



THANK YOU



M.E.H. Ismail, and W. Van Assche: Classical and QuantumOrthogonal Polynomial in One Variable. Cambridge UniversityPress, (2005).

McBride Elna B.: Obtaining generating functions. Vol. 21.Springer Science and Business Media, (2012).

Beals Richard, and Roderick Wong: Special functions andorthogonal polynomials. Vol. 153. Cambridge University Press,(2016).

Nikiforov A.F., and Uvarov V.B.: Special functions ofmathematical physics. Vol. 205. Basel: Birkhuser (1988).

Rainville Earl D.:Special functions, (1971).


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