Special Functions of Mathematical (Geo-)Physics_M Gutting_2011

7/28/2019 Special Functions of Mathematical (Geo-)Physics_M Gutting_2011

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Special Functions of Mathematical (Geo-)Physics

Dr. M. GuttingTU Kaiserslautern - FB Mathematik

May 3, 2011


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Preface

These are the lecture notes of the lecture about special functions and their applicationsin mathematical (geo-)physics that took place in the summer term 2010 at the universityof Kaiserslautern. The contents can be summarized as follows.

The lecture gives an elementary approach to the theory of special functions in mathemat-ical physics with special emphasis on geophysically relevant aspects. The essential topicsof the lecture are in chronological order: the Gamma function, orthogonal polynomials,spherical polynomials (scalar, vectorial, and tensorial case), and Bessel functions. Allfields will be assisted by geophysically relevant applications.The lecture is a good preparation for further activities in the field of geomathematics suchas Constructive Approximation, Potential Theory, Inverse Problems, etc.

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Contents

1 Introduction 71.1 Example: Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Example: Geomagnetics (Maxwells Equations) . . . . . . . . . . . . . . . 8

1.3 Example: Euler Summation Formula Involving the Laplace Operator . . . 10

2 The Gamma Function 21

3 Orthogonal Polynomials 313.1 Properties of Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . 383.2 Quadrature Rules and Orthogonal Polynomials . . . . . . . . . . . . . . . 463.3 The Jacobi Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.4 Ultraspherical Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.5 Application of the Legendre Polynomials in Electrostatics . . . . . . . . . . 663.6 Hermite Polynomials and Applications . . . . . . . . . . . . . . . . . . . . 703.7 Laguerre Polynomials and Applications . . . . . . . . . . . . . . . . . . . . 73

4 Spherical Harmonics 774.1 Spherical Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2 Polynomials on the Unit Sphere in R3 . . . . . . . . . . . . . . . . . . . . . 79

4.2.1 Homogeneous Polynomials . . . . . . . . . . . . . . . . . . . . . . . 794.2.2 Harmonic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 854.2.3 Harmonic Polynomials on the Sphere . . . . . . . . . . . . . . . . . 90

4.3 Closure and Completeness of Spherical Harmonics . . . . . . . . . . . . . . 954.4 The Funk-Hecke Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

4.5 Greens Function with Respect to the Beltrami Operator . . . . . . . . . . 1004.6 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5 Vectorial Spherical Harmonics 1075.1 Notation for Spherical Vector Fields . . . . . . . . . . . . . . . . . . . . . . 1075.2 Definition of Vector Spherical Harmonics . . . . . . . . . . . . . . . . . . . 1125.3 The Helmholtz Decomposition Theorem . . . . . . . . . . . . . . . . . . . 1135.4 Closure and Completeness of Vector Spherical Harmonics . . . . . . . . . . 116

6 Bessel Functions 1216.1 Derivation and Definition of Bessel Fucntions . . . . . . . . . . . . . . . . . 121

6.2 Some Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . 127

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6 Contents

6.3 Bessel Functions with Integer Index . . . . . . . . . . . . . . . . . . . . . . 128

7 Summary 135


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Chapter 1

Introduction

The main topics of this lecture can be briefly summarized: Finding orthogonal/orthonormal basis systems in (weighted) Hilbert spaces. They

are directly related to (partial) differential equations and many (geo-)physical ap-plications.

Greens functions and corresponding integral formulas which are also related tospecific (P)DEs.

Euler summation formulas which result from the conversion of elliptic (P)DEs tointegral equations using Greens functions as a bridging tool.

We will consider 1D systems (and some generalizations to Rn) as well as systems on thesphere = S2 R3 (and some generalizations to Sn Rn+1).

1.1 Example: Gravitation

In first approximation the surface of the Earth is a sphere. If we assume that all mass iscontained within this sphere of radius R, i.e. R, we can model the gravitational field asa function U which is harmonic in the exterior of R, i.e.

xU(x) = 2

x21 +

2

x22 +

2

x23U(x) = 0 x extR , i.e. |x| = x x > R.Moreover, it has to fulfill the following decay conditions:

|U(x)| = O

1

|x|

for |x| ,

|U(x)| = O

1

|x|2

for |x| .

Potential theory tells us that it suffices to know the function U on the boundary, i.e. onthe sphere R, in order to completely determine U in the exterior

extR .

Therefore, we need a complete orthonormal basis on R (or on the unit sphere = 1)

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8 Chapter 1. Introduction

to expand U|R in a Fourier series. This is also the prerequisite for spherical wavelets andit will lead us to spherical harmonics.Further applications can be found e.g. in quantum mechanics, in many other geomathe-

matical problems, in crystallography (using higher dimensions).

1.2 Example: Geomagnetics (Maxwells Equations)

The basis of all electromagnetic considerations is the full system of Maxwells equationsgiven by

x e(x, t) + t

b(x, t) = 0

x

d(x, t) = Ff(x, t)

x h(x, t) t

d(x, t) = jf(x, t)

x

b(x, t) = 0 ,

where the unknowns are defined as follows (note that capital letters are used for scalars,lower-case letter for vectors in R3):

d electric displacemente electric fieldh magnetic displacementb magnetic field

Ff density of free chargesjf density of free currents.

All quantities are understood as averages over a unit volume in space. The electric andmagnetic displacement, d and h, can be written as

d(x, t) = 0e(x, t) + p(x, t) (1.1)

h(x, t) =1

ob(x, t) m(x, t) , (1.2)

where p is the averaged polarization, m is the (averaged) magnetization, 0 is the permit-tivity of the vacuum and 0 is the permeability of vacuum.The total charge and current density, respectively, can be written as the sum of the free

charges and currents and the bounded ones, i.e.

F(x, t) = Ff(x, t) + Fb(x, t), j(x, t) = jf(x, t) + jb(x, t)

where it is well-known that

x=div

p(x, t) = Fb(x, t), x jb(x, t) = t

Fb(x, t)

and

x=curl

m(x, t) =

tp(x, t) + j

b(x, t) .


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Chapter 1. Introduction 9

We can now reformulate Maxwells equations.

x

e(x, t)

(1.1)=

1

0 x

(d(x, t)

p(x, t)) =

1

0(Ff(x, t) + Fb(x, t))

x e(x, t) = 10

F(x, t) (1.3)

x e(x, t) = t

b(x, t) (1.4)

x b(x, t) = 0 (1.5)x b(x, t) (1.2)= 0 (x h(x, t) + x m(x, t))

= 0

jf(x, t) +

td(x, t) + x m(x, t)

x b(x, t) = 0jf(x, t) + x m(x, t) + 0 t

e(x, t) + t

p(x, t)

(1.6)

In most geomathematical problems this system of equations is too detailed to describe theoccurring phenomena. They have to be reduced as follows. Let L be the typical lengthscale of the discussed geomathematical problem and T be the typical time scale. In mostproblems we have L = 102 km 103 km and T = hours - days, such that we get for thetypical velocity of the system

L

T c

where c is the speed of light (c = 299 792 458 m/s). Thus, it can be shown, that the terms0

t

e(x, t)+ t

p(x, t) can be neglected. Hence, Maxwells equations partially decouple andthe resulting equations for the magnetic field are

x b(x, t) = 0x b(x, t) = 0 (jf(x, t) + x m(x, t)) .

Since div curl = 0 we furthermore can conclude that (by applying to the secondequation)

x (0 (jf(x, t) + x m(x, t))) = 0 . (1.7)For solving this system of equations, data of the magnetic field of the Earth are pri-marily available in the exterior of the Earth, i.e. at the Earths surface or at satellitealtitude. Thus, we can assume that the magnetization m of the surrounding medium canbe neglected. We, therefore, arrive at the Pre-Maxwell equations

x b(x, t) = 0 (1.8)x b(x, t) = 0jf(x, t) . (1.9)

Furthermore, we have due to (1.7)

x jf(x, t) = 0 .


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In earlier concepts, geoscientist assumed that the current density j is also negligible inthe spherical shell (1,2) = {x R3 | 1 < |x| < 2}, in which the magnetic field data ismeasured. This yields

x b(x, t) = 0, x b(x, t) = 0 .

Hence, the magnetic field b can be written as the gradient of a scalar potential U, i.e.

b(x, t) = xU(x, t), x (1,2) ,where U fulfills the Laplace equation

xU(x, t) = 0, x (1,2) .

This so-called Gauss-representation yields a spherical harmonic expansion of the scalarpotential U which is similar to the modeling of the gravitational field of the Earth.

Modern satellite missions like CHAMP, which is measuring the Earths magnetic field, arelocated in the ionosphere, a region where the assumption jf = 0 is not valid. Therefore,we have to deal with the Pre-Maxwell equations

x b(x, t) = 0, x b(x, t) = 0jf(x, t), x (1,2).A new concept of modeling this situation has to be applied which is the so-called Mie-representation. This also yields the need for basis systems for vector-valued functions on

the sphere, i.e. for vector spherical harmonics.

1.3 Example: Euler Summation Formula Involvingthe Laplace Operator

First we give a brief introduction to some parts of the theory of one-dimensional periodicalfunctions. These results will be used to derive the Euler summation formula in its classicalform.

-5 -3 -1 0 1 2 3 4 5-2-4

Figure 1.1: The integer lattice = Z.

Let (= Z) denote the additive group of points in R having integral coordinates (theaddition being, of course, the one derived from the vector structure ofR).The fundamental cell Fof is given by

F=

t R1

2 t < 1

2

.

Note that it is an half-open interval.


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-1 0 1

[ )-1/2 1/2

Figure 1.2: The fundamental cell F.

Definition 1.1A function F : R C is called -periodical if

F(x + g) = F(x) for all x Fand g .Example 1.2The function h : R C given by

x h(x) = e2ihx

, h ,is -periodical:

h(x + g) = e2ih(x+g) = e2ihx e2ihg = e2ihx 1 = h(x)

for all x Fand all g .

Function spaces of -periodical functions

The space of all F

C(m)(R) that are - periodical is denoted by C

(m) (R), 0

m

.

L2(R) is the space of all F : R C that are - periodical and are Lebesgue - measurableon Fwith

FL2(R) =

F|F(x)|2dx

12

< .

Clearly, the space L2(R) is the completion of C(0) (R) with respect to the norm L2(R):

L2(R) = C(0) (R)

||||L2(R)

.

Remark 1.3

An easy calculation shows us that the system{h}h

is orthonormal with respect to the L2(R) - inner product

h, hL2(R) =F

h(x)h(x)dx =

12

12e2ihxe2ih

xdx = hh =

1 , h = h

0 , h = h.

Note that (with x = d/dx)

xh(x) = ddxh(x) = ddxe2ihx = 2ihh(x)


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such that

xh(x) = ddx2

h(x) = (2ih)2h(x) =

42h2 (h) h(x), h , x R.

From classical Fourier analysis we know that has a half-bounded and discrete eigen-spectrum {(h)}h such that

(x + (h))h(x) = 0, x F

with eigenvalues (h) given by

(h)(= (h)) = 42h2, h

and eigenfunctions h(x) = e2ihx, h

, x

F. The orthonormal system

{h

}h

of eigenfunctions h : x h(x), x R, is closed in the space C(0) (R), i.e., for every > 0 and every F C(0) (R) there exist an integer N(= N()) and a linear combination

h|h|N

ahh such that supxF

F(x) h|h|N

ahh

.Since

FL2(R) = F|F(x)|2 dx

1/2

supxF

|F(x)| = FC(0) (R)

, F C(0) (R),

the closure of{h}h in (C(0) (R), C(0) (R)) implies the closure in (C(0) (R), L2(R)).

Since C(0) (R) is dense in L2(R) with respect to the norm L2(R), we find that {h}h

is a complete orthonormal system in L2(R).

Euler (Green) Function for the Laplace Operator

In this section we introduce the -Euler (Green) function ( = Z) with respect to theone-dimensional Laplacian = 2, = d

dx, i.e, Greens function with respect to

corresponding to -periodical boundary conditions. Based on the constituting propertiesof this function the Euler summation formula can be developed by integration by parts.

The formal concept of Greens function G : R R with respect to the operator andperiodic boundary conditions:

(i) (Periodicity) G is continuous in R, and for all x / and g G(x + g) = G(x).

(ii) (Differential equation) G is twice continuously differentiable for all x / with

xG(x) = 0.


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(iii) (Characteristic singularity)

G(x) 12

x sign (x)

is continuously differentiable for all x F.We want to (formally) obtain the Dirac functional by application of the differential op-erator, i.e. xG(x) = (x). However, the conditions above lead to a contradiction forh = 0. Therefore, such a G does not exist. We have to modify the conditions for theGreen function.

Definition 1.4A function G(; ) : R R is called -Euler (Green) function with respect to the operator, if it satisfies the following properties:

(i) (Periodicity) G(; ) is continuous in R, and for all x / and g G(; x + g) = G(; x).

(ii) (Differential equation) G is twice continuously differentiable for all x / with

G(; x) = 1.

(iii) (Characteristic singularity)

G(; x) 1

2x sign(x)

is continuously differentiable for all x F.

(iv) (Normalization) F

G(; x)dx = 0.

Lemma 1.5G(; ) is uniquely determined by the properties (i)-(iv).

Proof. See Exercise 1.1.

Theorem 1.6The Green function G(; ) is the negative Bernoulli function of degree 2, i.e.

G(; x) = (x x)2

2+

x x2

112

where for x R the symbol x means that integer for which holds x x < x + 1(floor operation).


Figure 1.3 gives an illustration of the -Euler (Green) function with respect to .


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-1/2 1/2x

y

Figure 1.3: Graphical illustration of the -Euler (Green) function G(; ).

Lemma 1.7For all x R the Green function possesses the Fourier series representation

G(; x) = hh=0

1

42h2h(x) =

hh=0

1

42h2h(x).

Proof. We have to compute the Fourier coefficients. For h = 0 we know that

G(; ), 0L2(R) =F

G(; x) 1dx = 0.

For h = 0:

G(; ), hL2(R) = FG(; x)h(x)dx=

F

G(; x)e2ihxdx

= lim10

1 12

G(; x)e2ihxdx + lim20+

12

2

G(; x)e2ihxdx

Consider now (with xh(x) = (2ih)2h(x):

12

2

G(; x)e2ihxdx =1

(2ih)2 12

2

G(; x)x

h

(x)dx


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=1

42h2

G(; x)xh(x)

12

21

2

2

xG(; x)xh(x)dx

=1

42h2

G(; x)xh(x) 122

G(; x)h(x) 12

2+12

2

xG(; x) =1

h(x)dx

Analogously,1 12

G(; x)e2ihxdx =1

42h2

G(; x)xh(x)

1 12

G(; x)h(x)112

+

1 12

xG(; x)

=1h(x)dx

Taking the limits we obtain (note that G(; ) is continuous on R and h C()(R)):

G(; ), hL2(R) =1

42h2

G(; 1

2)xh( 12 ) G(;0)h(0) G(; 12 )h( 12 )

+ lim20+

G(; 2) h(0) =1

1

2

0

h(x)dx + G(;0)h(0)

G(; 12

)xh(12 ) lim10G(; 1) h(0)

=1+ G(; 1

2)h(12 )

0 12

h(x)dx

=1

42h2

lim

20+G(; 2) lim

10G(; 1)

12

12

h(x)dx

=1

42h2

lim

20+G(; 2) lim

10G(; 1)

=1

42h2= 1

(h)

due to the characteristic singularity of the Green function. In more detail:

(G(; x) 12

x sign(x))

=

G(; x) 12

, x > 0G(; x) + 1

2, x < 0

which has to be continuous. Therefore, we obtain

lim20+

G(; 2) 12 = lim10G(; 1) +12

,

which gives us

lim20+

G(; 2) lim10

G(; 1) = 12 (12 ) = 1 .

This concludes our proof.


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Remark 1.8Reconsider now the first definition (with G(x) = 0):

122

G(x)

=0

h(x) G(x)h(x)dx + 1 12

G(x)

=0

h(x) G(x)h(x)dx=G(x)h(x) G(x)h(x)

122

+G(x)h(x) G(x)h(x)

1 12

Take the limits 1 0 and 2 0+ for both sides of the equation. The left hand sidebecomes:

1020+

F

G(x)h(x)dx = 42h2

F

G(x)h(x)dx ,

which is 0 for h = 0.The right hand side becomes:

1020+

h(0)(G(0) G(0+)) = 1 (1) = 1 = 0.

Thus, we have a contradiction.There is no Fourier series representation of G in the form

G(x) =h

1

(h) h(x) ,

such that (formally) G(x) =

hh(x) = (x) (denotes the Dirac distribution). There-

fore, such a Greens function is not possible.

Remark 1.9For all x R, we have

G(; x) =

n=1

1

42n2(

e2inx + e2inx)

=

n=1

1

22n2 cos(2nx)

= 122

n=1

1

n2cos(2nx).

The Fourier series ofG(; ) converges absolutely and uniformly for all x R. Elementarydifferentiation yields

G(; x) = xG(; x) = G(; x) = (x x) + 12

, x R \ .

G(; ) is called Bernoulli functionof degree 1.


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The Fourier series of G(; ) reads as follows

G(

; x) = hh=0

1

2ihh(x) =

n=11

2i

e2inx e2inx

n=

n=1sin(2nx)

n,

where the equality is understood in the L2(R)-sense.

Lemma 1.10The series

n=1

sin(2nx)n

converges uniformly in each compact interval I (g, g + 1)with g .Proof. Let 1(x) = sin(2x), x R, and k(x) =

kn=1 sin(2nx), x R, k N. By

induction on k we find that for k N

k(x) =

sin(2x) + sin(2kx)

sin(2(k + 1)x)

2(1 cos(2x)) ,for all x R \ . This gives us

|k(x)| 32(1 cos(2x)) , x R \ .

Choose a compact interval I (g, g + 1), g . Then there is a constant C (dependingon I) such that

|k(x)| C, x I.Now, for all x I and all N 2, partial summation yields

Nn=2

sin(2nx)

2n=

Nn=2

n(x) n1(x)2n

=N(x)

2N+

N1n=2

n(x)

2n(n + 1) a1(x)

4

such that

Nn=2

sin(2nx)

2n

C

2N+

C

2

N1n=2

1

n(n + 1)+

1

4.

This assures the required result.

Definition 1.11A function G(; ) : R R is called -Euler/Green function with respect to the operator if

(i) for all x R \ and for all g holds G(; x + g) = G(; x),(ii) G(; ) is continuously differentiable for all x / and xG(; x) = 1,

(iii) G(; x) 12

sign(x) is continuous for all x F,(iv) FG(; x)dx = 0.Note that G(; x) = (x x) + 12 (see Figure 1.4 for an illustration).


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-1/2

1/2

1/2

-1/2

x

y

Figure 1.4: The derivative of the -Euler (Green) function G(; ).

Classical Euler Summation FormulaTheorem 1.12Let F : [a, b] R, a < b, be a twice continuously differentiable function. Then

g

g(a,b)

F(g) =

ba

F(x)dx +

ba

G(; x)F(x) dx

+

F(x)(xG(; x)) (xF(x))G(; x)b

a,

i.e. for a, b /

:

g

g(a,b)

F(g) =

ba

F(x)dx +

ba

G(; x)F(x) dx

+

F(x)(xG(; x)) (xF(x))G(; x)b

a,

for a, b :

gg[a,b]F(g) =

b

a

F(x)dx +

b

a

G(; x)F(x) dx

+ 12

(F(b) + F(a)) G(; 0) = 1

12

(F(b) F(a)) ,

and for a , b / (analogously for a / , b ):g

g[a,b]

F(g) =

ba

F(x)dx +

ba

G(; x)F(x) dx

+1

2 F(a) + F(b)G(; b) G(; x)F(x)ba .


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Proof. First we are concerned with the case that both endpoints a, b are non-integers(cf. Figure 1.5). By partial integration (similar to the proof of Lemma 1.7) we obtain forevery (sufficiently small) > 0:

Figure 1.5: Illustration of the integration interval

x[a,b]|xg|

g

(G(; x)xF(x) F(x)xG(; x)) dx = (G(; x)xF(x) F(x)xG(; x)) ba+

g(a,b)

g

(G(; x)xF(x) F(x)xG(; x))g

g+.

By virtue of the differential equation xG(; x) = 1, x R \ , it follows that

x[a,b]|xg|g

F(x)xG(; x) dx = x[a,b]|xg|

g

F(x) dx.

By letting 0 and observing the (limit) values of the -Euler (Green) function and itsderivatives in the lattice points we obtain for the left hand side:

0b

a

G(; x)xF(x)dx +

ba

F(x)dx ,

and for the right hand side:

G(; x)

F(x)gg+ 0 G(; g)F(g) G(; g)F(g) = 0F(x)G(; x)

gg+

0 F(g) (G(; g) G(; g+)) = F(g)

Summarizing:ba

G(; x)xF(x)dx +

ba

F(x)dx + (F(x)G(; x) G(; x)F(x))b

a=

g(a,b)

g

F(g)

If a and/or b are elements of = Z, we note that

G(; g) = G(; 0) = 112 g .


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Thus, if a, b , G(; a) = G(; b) = 112

. Moreover,

F(x)

G(; x)ba = G(; b) =1

2if b

F(b)

G(; a+) =

12

if a

F(a) .

The Euler summation formula compares a weighted sum of function values at lattice pointswith the corresponding integral plus remainder term. Applications are the computationof slowly converging infinite series and numerical integration where the error given by thecorresponding estimate is optimized.


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Chapter 2

The Gamma Function

In what follows, we introduce the classical Gamma function. Its essential properties willbe explained (for a more detailed discussion the reader is referred, e.g., to [10].For real values > 0 we consider the integrals

()1

0ett1dt and

()

1ett1dt.

In order to show the convergence of () we observe that 0 < ett1 t1 holds true forall t (0, 1]. Therefore, for > 0 sufficiently small, we have

1

ett1dt 1

t1dt = t

1

= 1

.

Consequently, for all > 0, the integral () is convergent.To assure the convergence of () we observe that

ett1 =1

k=0tk

k!

t1 1tnn!

t1 =n!

tn+1

for all n N and t 1. This shows us thatA

1

ett1dt n!A

1

1tn+1

dt = n! tn+

nA

1

= n! n

1An

1

provided that A is sufficiently large and n is chosen that n + 1. Thus, the integral() is convergent.

Lemma 2.1For all > 0, the integral

0

ett1dt

is convergent.

21


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22 Chapter 2. The Gamma Function

Definition 2.2The function (), > 0, defined by

() = 0

ett1dt

is called Gamma function.

Obviously, we have the following properties:

1. is positive for all > 0,

2. (1) =

0etdt = 1.

We can use integration by parts to obtain for > 0:

(+ 1) = 0

ettdt

= ett0

0

(et)t1dt

=

0

ett1dt

= ().

Lemma 2.3The Gamma function satisfies the functional equation

(+ 1) = (), > 0.

Moreover,

(+ n) = (+ n 1) (+ 1)() =n

i=1

(+ i 1)() for > 0, n N

(n + 1) =n

i=1

(i)(1) = n! for n N0.

In other words, the Gamma function can be understood as an extension of factorials.

Lemma 2.4The Gamma function is differentiable for all > 0 and we have

() =

0

et(ln(t))t1dt.

Proof. See Exercise 2.1.An analogous proof can be used to show that is infinitely often differentiable for all > 0 and

(k)

() =

0 et

(ln(t))

k

t

1

dt, k N

.


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Chapter 2. The Gamma Function 23

Lemma 2.5 (Gau Expression of the Second Logarithmic Derivative)For > 0

(())2

()().

Equivalently, we have d

d

2ln(()) =

()()

()()

2> 0,

i.e. ln(()), > 0, is a convex function or is logarithmic convex.Proof.

(())2 =

0

et(ln(t))t1dt2

=

0

et2 t

12 (ln(t))e

t2 t

12 dt

2.

The Cauchy-Schwarz inequality yields (note that equality cannot occur since the twofunctions are linearly independent):

(())2

0

e

t2 t

12

2dt

0

e

t2 t

12 (ln(t))

2dt

=

0

ett1dt

0

ett1(ln(t))2dt

= () ().

Moreover,

d2

d2ln(()) =

d

d

()()

=()() (())2

(())2 >0

=()()

()()

2

> 0.

Note: ln(()) is convex, i.e. for t (0, 1)ln((tx + (1 t)y)) t ln((x)) + (1 t) ln((y))

= ln(

t(x))

+ ln(

1t(y))

= ln(

t(x) 1t(y))which is equivalent to (tx + (1 t)y) t(x) 1t(y) with x,y > 0.

Next we are interested in the behavior of the Gamma function for large positive values

x.


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Theorem 2.6 (Stirlings Formula)For x > 0

(x)2xx1/2ex 1 2x .Proof. Regard x as fixed and substitute

t = x(1 + s), 1 s < , dtds

= x.

We get

(x) =

0

ettx1dt =1

exxsxx1(1 + s)x1xds = xxex1

(1 + s)x1exsds

=I(x).

Our aim is to verify that I(x) satisfiesI(x)

2

x

2x.Then we have

(x)

xxex

2

x

2

x, i.e.

(x)

xx1/2ex

2 1

2

x.

For that purpose we write

(1 + s)xexs = exp (x(s ln(1 + s))) = exu2(s)

where

u(s) =

|s ln(1 + s)| 12 , s [0, )|s ln(1 + s)| 12 , s (1, 0).

Taylor expansion of u2 for s (1, ) at 0:

u2(s) = u2(0) + dds

u2(0)s + d2

ds2u2(s)s

2

2

= 0 +

1 1

1 + 0

s +

1

(1 + s)2s2

2

Therefore,

u2(s) =s2

2

1

(1 + s)2

with 0 < < 1. We interpret as a uniquely defined function of s, i.e., : s (s),such that

u(s)

s =1

21

(1 + s(s))


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is a positive continuous function for s (1, ) with the property

u(s)s 1

2 = 1

2 1

1 + s(s) 1 = 12

s(s)

1 + s(s) =

s(s)u(s)s = |(s)| |u(s)| |u(s)|.

From u2(s) = s ln(1 + s) follows that2u du =

s

1 + sds.

Obviously, s : u s(u), u R is of class C(1)(R) and thus,

I(x) =

1

(1 + s)x

1

exs

ds = 2

+

exu2 u

s(u) du.

We are able to deduce thatI(x) 22

0

exu2

du

=2

exu2 u

s(u)du

2

+

exu2

du

=

2

exu2

u

s(u) 1

2

du

2

exu2 us(u) 12

du 2

exu2 |u| du

= 4

0

exu2

u du.

Note that we have the following integrals (for , > 0)0

et

dtu=t

=1

0

euu1/1du =1

( 1

) = ( +1

),

0

t1et

dtu=t

=1

0

euu1 u

1 du =

1

(

),

0

t1et2

dtu=t2

=1

2

0

eu1u

u

12

du =1

2/2

0

euu/21du =1

2/2(

2).

Therefore, for = 1 and = x:

0

exu2

du = 12 x1/2 (12) =

2x ,


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for = 2 and = x:

0

exu2

u du = 12

x1(1) =1

2x.

This yields: I(x) 2 2 12

x

=I(x)

2

x

4 12x = 2x.This leads to the desired result.

Remark 2.7Stirlings formula can be rewritten in the form

limx

(x)2xx1/2ex

= 1 .

An immediate application is the limit relation

limx

(x + a)

xa(x)= 1, a > 0. (2.1)

This can be seen from Stirlings formula by

limx

(x + a)2(x + a)x+a

12 exa

= 1 (2.2)

due to the relation(x + a)x+a 12 = xx+a12 (1 + a

x)x+a 12

and the limits

limx

(1 + ax

)x

ea= 1, lim

x(1 +

a

x)a

12 = 1.

Lemma 2.8 (Duplication Formula)For x > 0 we have

2x1x

2

x + 1

2

=

(x).

Proof. We consider the function x (x), x > 0, defined by(x) =

2x1( x2

)( x+12

)

(x)

for x > 0. Setting x + 1 instead of x we find with the following functional equation forthe numerator

2x

x + 1

2

x

2+ 1

= 2x1x

x2

x + 1

2

,

such that the numerator satisfies the same functional equation as the denominator. This

means (x + 1) = (x), x > 0. By repetition we get for all n N and x fixed (x + n) =


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(x). We let n tend toward . For the numerator of (x + n) we then find by use of theresult in Remark 2.7, i.e. by using twice (2.1) that

limn

2x+n

1

(x+n

2 )(x+n+1

2 )2x+n1

(n2

) x2 ( n

2)x+12

(( n

2))2 = 1.

For the denominator we consider

limn

(x + n)

2x+n1(

n2

)x2 ( n

2)x+12

(( n

2))2 = limn (x + n)

2x+n1(

n2

)x+ 12 ( n

2)n1en2 (

(n2 ))2

(n2

)n1en2

= limn

(x + n)

2x+n (n2)

n+x 12 en

(( n

2))2

( n2

)n1en2

1

= limn

(x + n)2nn+x

12en

(( n

2))2

( n2

)n1en2

1=

1

since Stirlings formula yields that

limn

( n2

)

2( n2

)n21

2 en2

= 1,

i.e.

limn

(( n

2))2

2( n2

)n1en= 1,

and by the same arguments as in Remark 2.7 (set a in (2.2) to x and x in (2.2) to n) wefind that

limn

(x + n)2nn+x

12en

= 1.

We therefore get for every x > 0 and all n N(x) = (x + n) = lim

n(x + n) =

.

A periodical function with this property must be constant. This proves the lemma.

Thus far, the Gamma function is defined for positive values, i.e., x R>0 (cf. Figure2.1). We are interested in an extension of to the real line R (or even to the complexplane C) if possible.

Definition 2.9The so-called Pochhammer factorial (x)n with x R, n N is defined by

(x)n = x(x + 1) . . . (x + n

1) =

n

i=1(x + i 1).


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For x > 0 it is clear that

(x)n =(x + n)

(x)

or (x)n(x + n)

=1

(x).

The left hand side is defined for x > n and gives the same value for all n N withn > x. We may use this relation to define 1

(x)for all x R, and we see that this

function vanishes for x = 0, 1, 2, . . . (cf. Figure 2.1).

Figure 2.1: The Gamma function on the real line R.

This leads us to the following conclusion: The Gamma integral is absolutely convergentfor x C with (x) > 0, and represents a holomorphic function for all x C with(x) > 0. Moreover, the Pochhammer factorial (x)n can be defined for all complex x.Therefore we have a definition of 1

(x)for all x C.

Lemma 2.10The -function is a meromorphic function that has simple poles in 0, 1, 2, . . .. Thereciprocalx 1

(x), is an entire analytic function.

Lemma 2.11For x C,

1

(x)= lim

nnxx

n1k=1

(1 + x

k

).

Proof. The identity(x)n(n)

(n)

(x + n)=

1

(x)

is valid for all x C and all n > (x). Furthermore it is easy to see that(x)n

(n) = x

(x + 1)(x + 2) . . . (x + n

1)

1 2 . . . (n 1) = xn1

k=1

1 + xk .


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Stirlings formula tells us that for positive x R

limn

1

nx

(n)

(x + n)= 1.

and1

(x)= lim

nnxx

nk=1

1 +

x

k

(1 + x

n

)1 1 as n

= limn

nxxn

k=1

1 +

x

k

which proves the lemma for x > 0. To determine if this limit is also defined for x 0 weconsider once again s ln(1 + s) with 1 < s < :

0 s ln(1 + s) = s2

2

1

(1 + s)2, = (s) (0, 1).

Therefore, we can put s = 1k

and estimate the right hand side with its maximum ( = 0):

0 1k

ln (1 + 1k

) 12k2

.

This immediately proves that limn

nk=1

(1k

ln (1 + 1k

))exists and is positive. Moreover,

limn

nk=1 (

1k

ln

(1 + 1

k

))= lim

n

nk=1 (

1k

ln(k + 1) + ln(k)

)= lim

n

nk=1

1k

ln(n + 1)

= limn

n1k=1

1k

ln(n)

= C

where C denotes Eulers constant

C = limm

m1

k=11

k ln m

0, 577215665 . . . .

Assume now that x R. If k 2|x|, then

0 xk

ln (1 + xk

)< x

2

k2(2.3)

and

nk=1

1 +

x

k

=

nk=1

1 +

x

k

e

xk e

xk

=n

j=1 exj

n

k=11 +x

k ex

k .


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For k k0 = 2|x| we obtain by multiplying (2.3) with 1 and applying the exponentialfunction to it:

1 (1 + xk) e

xk > e

x2

k2

which shows that

limn

nk=1

(1 + x

k

)e

xk =

k=1

(1 + x

k

)e

xk

exists for all x. Moreover,

nj=1

exj = exp

x

nj=1

1j

= expxn

j=1

1

j x ln(n) + x ln(n)

= nx exp

x

nj=1

1j

x ln(n)

= nxeCx .

Therefore, limn

nxxn

k=1

(1 + x

k

)exists for all x R and it holds that

limn nx

x

n

k=1

(1 + xk) = exCx

k=1

(1 + xk) exk

where the infinite product is also convergent for all x R. By similar arguments theseresults can be extended for all x C.The proof of the previous lemma also shows us the following.

Lemma 2.12For x C,

1

(x)= xeCx

k=1

1 +x

k

e

xk .


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Chapter 3

Orthogonal Polynomials

We consider weighted Hilbert spaces on intervals in R. Those are denoted by L2

[a, b] orL2(a, b) (a, b can be , , respectively) and have the scalar product

F, Gd =b

a

F(x)G(x)d(x).

is a nondecreasing function on R which has finite limits as x and whose inducedpositive measure d has finite moments of all orders, i.e.

r = r(d) = Rxrd(x) <

for r N0 with 0 > 0. If no confusion is likely to arise, we just use the notation L2[a, b]or L2(a, b) for the spaces.If is absolutely continuous, the scalar product becomes

F, Gd =b

a

F(x)G(x)w(x)dx.

Here w is a non-negative function which is Lebesgue-measurable and for which

b

aw(x)dx > 0. It is called weight function.

Definition 3.1Two functions F, G L2(a, b) are orthogonal if F, Gd = 0.

Example 3.2The trigonometric functions Fn(x) = cos(nx) form a system of orthogonal functions onthe interval (0, ) with the weight function w(x) = 1 since

0

cos(mx) cos(nx)dx = 0

for n = m.

31


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32 Chapter 3. Orthogonal Polynomials

Definition 3.3Let {n}nN0 be a given orthonormal set (either finite or infinite). To an arbitrary real-valued function F let there correspond the formal Fourier expansion

F(x)

i=0

Fii(x)

with the coefficients Fi defined by

Fi = F, id =b

a

F(x)i(x)d(x) , i N0.

These coefficients are called Fourier coefficients ofF with respect to the given orthonormalsystem.

Theorem 3.4Let n, Fn as before with F L2(a, b). Let l 0 be a fixed integer and ai R,i = 0, . . . , l.

If G(x) =l

i=0

aii(x) and the coefficients ai are considered variable in the following inte-

gral, the integral ba

(F(x) G(x))2 d(x)

becomes a minimum if and only if ai = Fi for i = 0, . . . , l.

Proof.

F G2d = F G, F G= F, F 2(F, G) + G, G

= F2 2

li=0

aiF, i

+l

i,j=0

aiaj i, j i,j

= F2 +l

i=0|ai|2 2

l

i=0aiFi

= F2 l

i=0

|Fi|2 +l

i=0

|ai Fi|2

which is minimal if and only if ai = Fi for all i = 0, . . . , l. Therefore, the truncatedFourier series is the best approximating element.

Remark 3.5We also find Bessels equality

F l

i=0Fii

2

= F2

l

i=0|Fi|

2


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Chapter 3. Orthogonal Polynomials 33

and Bessels inequality (left hand side above is 0):l

i=0 |Fi|

2

F2

. (3.1)

Lemma 3.6The Fourier series

iN0

Fii converges in theL2-norm (in theL

2-sense) to an L

2-function

S and (F S)i for all i N0.Proof. For n m holds:

n

i=mFii

2

=n

i=mFii2 =

n

i=m|Fi|2

due to the Theorem of Pythagoras (and the orthogonality of the functions i).Bessels inequality (3.1) gives us the convergence of the series

iN0

|Fi|2. Thus, for all > 0there exists an m N such that for n m

ni=m

|Fi|2 < .

Therefore,

iN0Fii is a Cauchy sequence with respect to the L

2-norm. Since L

2(a, b) is

complete, there exists S

L2(a, b) which is the limit of this sequence.

It remains to show the orthogonality of S F: let j {0, . . . , l}S F, j = S, j F, j

= iN0

Fii, j F, j

=iN0

Fi i, j i,j

Fj = 0,

since for Fn F, Gn G holds that Fn, Gn F, G.Now we only have to show completeness of the system

{i

}i. Then F

S = 0 (in the

L2-sense), i.e. the limit of the Fourier series is the function F. We now consider propertiesof Fourier series in a more general setting.

Theorem 3.7Let {xk}kN X be a sequence of orthonormal elements of the inner product space X.Consider the following statements:

(A) The xk are closed in X, i.e. for any element x X and for all > 0 there exist ann N and coefficients a1, . . . , an K such that

x n

k=1 akxkX .


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(B) The Fourier series of anyy X converges to y (in the norm of X), i.e.

limny n

k=1y, xkXxkX = 0 .

(C) Parsevals identity holds, i.e. for all y X,

y2X = y, yX =

k=1

|y, xkX|2 .

(D) The extended Parseval identity holds, i.e. for allx, y X,

x, yX =

k=1

x, xkXxk, yX .

(E) There is no strictly larger orthonormal system containing the set {xk}kN.(F) {xk}kN is complete, i.e. if y X and y, xkX = 0 for allk N, then y = 0.(G) An element ofX is determined uniquely by its Fourier coefficients, i.e. ifx, xkX =

y, xkX for all k N, then x = y.Then holds:

(A) (B) (C) (D) = (E) (F) (G) . (3.2)If X is a complete inner product space, then also (E) = (D) and all statements areequivalent.

Proof. Assume (A). Due to the minimizing property of the truncated Fourier series (seeTheorem 3.4 which holds for any ONS {xk}k)

y

nk=1

y, xkxk

y n

k=1

akxk

,

where the last estimate is provided by (A). If on the other hand (B) holds, we can ap-proximate any y by its truncated Fourier series which shows closure. Thus, (A) (B).

By orthogonality,x

nk=1

x, xkxk, y n

k=1

y, xkxk

= x, y n

k=1

x, xkxk, y .

Using the Schwarz inequality,

x, y n

k=1

x, xkxk, y x n

k=1

x, xkxk y n

k=1

y, xkxk .


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If (B) holds, the right-hand members both tend to 0, hence (B) = (D).

Selecting x = y in (D) shows (C), i.e. (D) =

(C).

Since

0 y

nk=1

y, xkxk

2

= y2 n

k=1

|y, xk|2 ,

we see (C) = (B) and thus, (A) (B) (C) (D).

Now assume (A) and suppose {xk}kN {w}, w = xk, is also an ONS. This system is alsoclosed in X. Since (A) = (C)

w2

=

k=1

|w, xk|2 + |w, w|2 , and w2

=

k=1

|w, xk|2

Thus, w, w = 0 which contradicts w = 1. This means that (A) = (E).

Suppose there is a 0 = y X such that y, xk = 0 for all k. Then, {xk}kN {y/ y}would be a strictly larger ONS than{xk}kN. Therefore, (E) (F).

Suppose w, xk = y, xk, k N. Then w y, xk = 0, k N. Assuming (F), w y = 0and (F) = (G).

If (F) were false, we could find z = 0 with z, xk = 0, k N. For any y, y, xk =y + z, xk, k N. So y and y + z would be two distinct elements with the same Fouriercoefficients. Thus, (G) would be false and we obtain (G) = (F). This completes thechain of implications (3.2).

Assume now that additionally X is complete. We want to show (G) = (B) which is themissing implication.Let w X and consider

Sn =n

k=1w, xkxk .

For n > m we find that

Sn Sm2 =n

k=m+1

|w, xk|2 .

Bessels inequality gives us

k=1

|w, xk|2 < , thus for a given > 0 we can find an N()

such thatn

k=m+1

|w, xk|2 for all m, n N(). Thus, {Sn} is a Cauchy sequence andsince X is complete, it converges to an element S X.Let v be fixed and n v.

S Sn, xv = S, xv Sn, xv = S, xv w, xv


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and by the Schwarz inequality

|S, xv w, xv| = |S Sn, xv | S Sn xv = S Sn .Together with the convergence of Sn to S this gives us

S, xv = w, xv for v N.By (G), this implies that S = w such that the convergence reads as follows

limn

w n

k=1

w, xkxk = 0 .

This is precisely (B).Now we start to consider polynomials.

Definition 3.8The space of real polynomials up to degree n is denoted by n. The space of real poly-nomials of all degrees is , n for all n N0. For P, Q we use the scalarproduct

P, Qd =R

P(x)Q(x)d(x) (3.3)

and the induced norm.

Note that by definition of (finite moments) these integrals exist.

Definition 3.9The scalar product (3.3) is called positive definite on ifPd > 0 for all P , P 0.It is called positive definite on n ifPd > 0 for all P n, P 0.Theorem 3.10The scalar product (3.3) is positive definite on if and only if

det Mk =

0 1 k11 2 k...

......

k1 k 2k2

> 0, k N.

It is positive definite on n if and only if det Mk > 0 for k = 1, 2, . . . , n + 1.

Proof. See tutorials (Exercise 4.1).

Definition 3.11Polynomials whose leading coefficient is 1, i.e. Pk(x) = x

k + . . . with k N0, are calledmonic polynomials. The set of monic polynomials of degree n is denoted by n.Monic real polynomials Pk, k N0, are called monic orthogonal polynomials w.r.t. themeasure d if

Pk, Pld = 0 for k = l, k, l N0 and Pkd > 0 for k N0.Normalization yields the orthonormal polynomials Pk(x) = Pk(x)/ Pkd.


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Lemma 3.12Let {Pk}k=0,1,...,n be monic orthogonal polynomials. If Q n satisfies Q, Pk = 0 fork = 0, 1, . . . , n, then Q 0.

Proof. Write Q as Q(x) =n

i=0

aixi. Then

0 = Q, Pn =n

i=0

aixi, Pn

and

xi, Pn = Pi + Ri1i1

, Pn where Pi(x) = xi + Ri1(x)

=

Pi, Pn

+

Ri

1, Pn

= i,n Pn, Pn + ri1xi1 + Ri2, Pn= i,n Pn, Pn + ri1Pi1 + Si2, Pn= = i,n Pn, Pn

This gives us 0 = Q, Pn = anPn, Pn. Since Pn, Pn > 0, we obtain an = 0. Similarly,we can show that an1 = 0, an2 = 0, . . . , a0 = 0.

Lemma 3.13A set P0, . . . , P n of monic orthogonal polynomials is linearly independent. Moreover, any

polynomial P

n can be uniquely represented in the form P =

n

k=0 ckPk for some realcoefficients ck, i.e. P0, . . . , P n is a basis of n.Proof. If

nk=0

kPk 0, taking the scalar product on both sides of the equation with Pj,j = 0, . . . , n, yields that j = 0. This gives us linear independence.

If we take the scalar product with Pj on both sides of P =n

k=0

ckPk we find that

P, Pj =n

k=0

ckPk, Pj = cjPj, Pj , j = 0, . . . , n .

The difference Pnk=0 ckPk is orthogonal to P0, . . . , P n and by the previous lemma hasto be identically zero.

Theorem 3.14If the scalar product , d is positive definite on , there exists a unique infinite sequence{Pk}kN0 of monic orthogonal polynomials.Proof. The polynomials Pk can be constructed by applying Gram-Schmidt orthogonal-ization to the sequence of powers {Ek}kN0 where Ek(x) = xk. Therefore, we chooseP0 = 1 and for k N we use the recursion

Pk = Ek k1

l=0

clPl , cl = Ek, Pl

Pl, Pl . (3.4)


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Since the scalar product is positive definite, Pl, Pl > 0. Thus, the monic polynomial Pkis uniquely defined and by construction orthogonal to all Pj , j < k.The prerequisites of this theorem are fulfilled if has many points of increase, i.e. points

t0 such that (t0 + ) > (t0 ) for all > 0. The set of all points of increase of iscalled support of the measure d, its convex hull is the support interval of d.

Theorem 3.15If the scalar product , d is positive definite on n, but not on m for allm > n, thereexist only n + 1 orthogonal polynomials P0, . . . , P n.

Proof. See tutorials (Exercise 4.2).

Theorem 3.16If the moments of d exist only for r = 0, 1, . . . , r0, there exist only n + 1 orthogonal

polynomials P0, . . . , P n, where n = r0/2.Proof. The Gram-Schmidt procedure can be carried out as long as the scalar productsin (3.4) including Pk, Pk exist. This is the case as long as 2k r0, i.e. k n = r0/2.

3.1 Properties of Orthogonal Polynomials

For this section we assume that d is a positive measure on R with an infinite number ofpoints of increase and with finite moments of all orders.

Definition 3.17An absolutely continuous measure d(x) = w(x)dx is symmetric (with respect to theorigin) if its support interval is [a, a] with 0 < a and w(x) = w(x) for all x R.Theorem 3.18If d is symmetric, then

Pk(x) = (1)kPk(x) , k N0.Thus, Pk is either an even or an odd polynomial depending on the degreek.

Proof. Set Pk

(x) = (

1)kPk

(

x). We compute for k= l

Pk, Pld = (1)k+lR

Pk(x)Pl(x)d(x)

= (1)k+lR

Pk(x)Pl(x)d(x)

= (1)k+lR

Pk(x)Pl(x)d(x)

= (1)k+lPk, Pld = 0.

Since all Pk are monic, Pk(x) = Pk(x) by the uniqueness of monic orthogonal polynomials.


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Theorem 3.19Let d be symmetric on [a, a], 0 < a , and

P2k(x) = P+

k (x2

) P2k+1(x) = xPk (x2

).

Then {Pk } are the monic orthogonal polynomials with respect to the measure d(x) =x

12 w(x

12 )dx on [0, a2].

Proof. We only prove the assertion for P+k , the other case follows analogously.Obviously, the polynomials P+k are monic. By symmetry holds that

0 = P2k, P2ld = 2a

0

P2k(x)P2l(x)w(x)dx

for k = l. Therefore, we obtain for k = l

0 = 2

a0

P+k (x2)P+l (x

2)w(x)dx =

a20

P+k (t)P+l (t)t

12 w(t

12 )dt.

Theorem 3.20All zeros of Pk, k N, are real, simple, and located in the interior of the support interval[a, b] of d.

Proof. SinceR

Pk(x)d(x) = 0 for k 1, there must exist at least one point in theinterior of [a, b] at which Pk changes sign. Let x1, x2, . . . , xn, n k, be all these points.If we had n < k, then due to orthogonality

R

Pk(x)n

j=1

(x xj )d(x) = 0,

which is impossible since the integrand does no longer change sign. Therefore, we obtainthat n = k.

Theorem 3.21The zeros of Pk+1 alternate with those of Pk, i.e.

x(k+1)k+1 < x(k)k < x

(k+1)k < x

(k)k1 < < x(k)1 < x(k+1)1 ,

wherex(k+1)j , x

(k)i are the zeros in descending order of Pk+1 and Pk, respectively.

Proof. Later (uses the Christoffel-Darboux formula which we will get very soon).

Theorem 3.22In any open interval (c, d) in which d 0 there can be at most one zero of Pk.


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Proof. We perform a proof by contradiction. Suppose there are two zeros x(k)i = x(k)j in

(c, d). Let all the other zeros (within (c, d) or not) be denoted by x(k)n . Then holds

R

Pk(x)

n=i,j

(x x(k)n )d(x) =R

n=i,j

(x x(k)n )2(x x(k)j )(x x(k)i )d(x) > 0 ,

since the integrand is non-negative outside of (c, d). This is a contradiction to the orthog-

onality of Pk to polynomials of lower degree such as

n=i,j

(x x(k)n ).

Theorem 3.23For any monic polynomial P n holds

R P

2

(x)d(x) R P2n (x)d(x)where equality is only achieved for P = Pn. In other words, Pn minimizes the integral onthe left hand side above over all P n, i.e.

minPn

R

P2(x)d(x) =

R

P2n (x)d(x).

Proof. Due to Lemma 3.13 the polynomial P can be represented as follows

P(x) = Pn(x) +n1

k=0

ckPk(x).

Therefore, R

P2(x)d(x) =

R

P2n (x)d(x) +n1k=0

c2k

R

P2k (x)d(x).

This shows the desired inequality. Equality holds if and only if c0 = c1 = . . . = cn1 = 0,i.e. for P = Pn.

Remark 3.24If we consider the function

(a0, a1, . . . , an1) =R

xn +

n1k=0

akxk

2d(x) ,

we can compute the partial derivatives and set them equal to zero which yieldsR

P(x)xkd(x) = 0 , k = 0, 1, . . . , n 1.

These are exactly the conditions of orthogonality that Pn has to satisfy.Furthermore, the Hessian of is 2Mn of Theorem 3.10 which is positive definite. This

confirms that Pn gives us a minimum.


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Theorem 3.25Let 1 < p < . Then, the extremal problem of determining

minPnR

|P(x)|pd(x)

possesses the unique solution Pn.

Proof. The search for the desired minimum is equivalent to the problem of best-approximation of xn by polynomials of degree n 1 in the Lp-norm.The problem of best approximation in normed spaces is uniquely solvable if the space isstrictly convex, i.e. if from x = y = 1 and x = y we can conclude that x + y < 2.A normed space X is strictly convex if and only if x + y = x + y yields x = y ory = x with an 0. The Minkowski inequality guarantees this for the Lp-norm.

For further details see e.g. [1, 4, 9, 13, 14, 17] or other books on functional analysis.

Orthogonal polynomials fulfill a three-term recurrence relation which can be used for:

generating values of the polynomials and their derivatives, computation of the zeros as eigenvalues of a symmetric tridiagonal matrix via the

recursion coefficients,

normalization of the orthogonal polynomials,

efficient evaluation of expansions in orthogonal polynomials.The reason for the existence of these three-term recurrences is the shift property of thescalar product, i.e.

xU,Vd = U,xVd U, V .Note that there are other scalar products that do not possess this property (even thoughthey are positive definite).

Theorem 3.26Let Pk, k N0, be the monic orthogonal polynomials w.r.t. the measured (see Definition3.11). Then,

P1(x) = 0, P0(x) = 1, Pk+1(x) = (x k)Pk(x) kPk1(x), k N0, (3.5)where

k =xPk, PkdPk, Pkd , k N0,

k =Pk, Pkd

Pk1, Pk1d , k N.

The index range is infinite (k ) if the scalar product is positive definite on . Itis finite (k

d

1) if the scalar product is positive definite on d, but not on n with

n > d.


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Proof. Since Pk+1 xPk is a polynomial of degree k, we can write

Pk+1(x) xPk(x) = kPk(x) kPk1(x) +k2

j=0

k,j Pj(x)

for certain constants k, k, k,j , where P1(x) = 0 and empty sums are also zero.We take the scalar product with Pk on both sides which gives us (using orthogonality):

Pk+1, Pk xPk, Pk = kPk, Pk kPk1, Pk +k2j=0

k,jPj, Pk

= xPk, Pk = kPk, Pk=

k =

xPk, PkPk, Pk

This proves the relation for k. For k we need to take the scalar product with Pk1where k 1:

Pk+1, Pk1 xPk, Pk1 = kPk, Pk1 kPk1, Pk1 +k2j=0

k,jPj, Pk1

= xPk, Pk1 = kPk1, Pk1Since xPk, Pk1 = Pk, xPk1 = Pk, Pk + Rk1 with Rk1 k1, we find that

xPk, Pk

1

=

Pk, Pk

which provides us with

k =Pk, Pk

Pk1, Pk1 .

As a last step we take the scalar product on both sides with Pi, i < k 1, and obtain

Pk+1, Pi xPk, Pi = kPk, Pi kPk1, Pi +k2j=0

k,jPj, Pi

= xPk, Pi = k,iPi, Pi

Here we make use of the shift property of the scalar product, i.e. xPk, Pi = Pk, xPi = 0since xPi k1. Therefore, k,i = 0 for i < k 1 which finally proves (3.5).Remark 3.27If the index range in Theorem 3.26 is finite (k d 1), the relations for d and d stillmake sense, d > 0, but the polynomial Pd+1 that results from (3.5) has norm Pd+1 = 0(see also Theorem 3.15 or Exercise 4.2).

Remark 3.28Although 0 in (3.5) can be arbitrary since it is multiplied with P1 0, we define it forlater purposes as

0 = P0, P0 = R

d(x) .


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Note that k > 0 for all k N0 and for n N0Pn2 = nn1 . . . 10.

There is a converse result to Theorem 3.26 saying that any sequence of polynomials whichsatisfy a three-term recurrence relation of the form (3.5) with all k positive is orthogonalw.r.t. a positive measure with infinite support.

Theorem 3.29Let the support interval [a, b] of d be finite. Then,

a k b , k N0,0 < k max{a2, b2} , k N,

where the index range is k or k d (d as in Theorem 3.26), respectively.

Proof. Since for x in the support of d we know that a x b, the definition of k inTheorem 3.26 immediately yields the desired estimates.By definition 0 < k and it remains show the upper bound. We notice that

Pk2 = Pk, Pk = |xPk1, Pk|and apply the Cauchy-Schwarz inequality to obtain

Pk2 max{|a|, |b|} Pk1 Pk .Therefore,

PkPk1

max

{|a

|,

|b

|}and k =

Pk2

Pk12

max

{a2, b2

}.

Definition 3.30If the index range in Theorem 3.26 is infinte, the Jacobi matrix associated with themeasure d is the infinite, symmetric, tridiagonal matrix

J =

0

1 01 1

2

2 2

3. . . . . . . . .

0

.

Its leading principal minor matrix of size n n is denoted by

Jn = [J][1:n,1:n] =

0

1 01 1

2

2 2

3. . . . . . . . .

n2 n2

n10

n1 n1

.

If the index range in Theorem 3.26 is finite (k

d

1), then Jn is well-defined for

0 n d.


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Theorem 3.31The zerosx

(n)i ofPn (the orthonormal version Pn) are the eigenvalues of the Jacobi matrix

Jn of order n. The corresponding eigenvectors are given by Px(n)i whereP(x) =

P0(x), P1(x), . . . , Pn1(x)

T.

Proof. We know from Exercise 5.1 that the following system of equations holds

xP(x) = JnP(x) +

nPn(x)en ,

where en = [0, 0, . . . , 1]T is the n-th unit vector in Rn.

If we put x = x(n)i in this equation, the second summand on the right hand side drops

out. Since the first component of the vector P(x(n)i ) is 1/

0, the vector cannot be 0 and

is indeed the eigenvector to the eigenvalue x(n)i .

Theorem 3.32 (Christoffel-Darboux Formula)Let Pk denote the orthonormal polynomials with respect to the measure d. Then,

nk=0

Pk(x)Pk(t) =

n+1Pn+1(x)Pn(t) Pn(x)Pn+1(t)

x t

andn

k=0 Pk(x)2

= n+1 P

n+1(x)Pn(x)

Pn(x)Pn+1(x) .

Proof. We use the recurrence relation for the orthonormal polynomials (see Exercise5.1), i.e.

k+1Pk+1(t) = (t k)Pk(t)

kPk1(t) , k N0, P1(t) = 0, P0(t) = 10 ,

and multiply this by Pk(x) to obtain:k+1Pk+1(t)Pk(x) = (t k)Pk(t)Pk(x)

kPk1(t)Pk(x).

Now we interchange the roles of t and x and subtract the first relation, i.e.k+1Pk+1(x)Pk(t)

k+1Pk+1(t)Pk(x)

= (x k)Pk(x)Pk(t)

kPk1(x)Pk(t)

(t k)Pk(t)Pk(x)

kPk1(t)Pk(x)

= (x t)Pk(x)Pk(t)

k

Pk1(x)Pk(t) Pk1(t)Pk(x)

.

Therefore,

(x t)Pk(x)Pk(t) =

k+1

Pk+1(x)Pk(t) Pk+1(t)Pk(x)

k Pk1(t)Pk(x) Pk1(x)Pk(t) .


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Now we sum up both sides from k = 0 to k = n and make use of the teleskop sum on theright (note that P1 = 0):

nk=0

Pk(x)Pk(t) =n+1 Pn+1(x)Pn(t) Pn(x)Pn+1(t)

x t .

For the second part we take the limit t x on both sides and on the right hand side wefind

Pn+1(x)Pn(t) Pn(x)Pn+1(t)x t = Pn+1(x)

Pn(t) Pn(x)x t + Pn+1(x)

Pn(x)

x t Pn(x) Pn+1(t) Pn+1(x)

x t Pn(x)Pn+1(x)

x t

=

Pn+1(x)

Pn(t)

Pn(x)

x t

Pn(x)

Pn+1(t)

Pn+1(x)

x ttx Pn+1(x)

Pn(x)

Pn(x)

Pn+1(x)

= Pn+1(x)Pn(x) Pn(x)Pn+1(x).

Corollary 3.33Let Pk denote the monic orthogonal polynomials with respect to the measure d. Then,

n

k=0 n

i=k+1 iPk(x)Pk(t) =n

k=0 n n1

. . .

k+1Pk(x)Pk(t)

=Pn+1(x)Pn(t) Pn(x)Pn+1(t)

x t .

Proof. Put Pk = Pk/ Pk in the first formula of Theorem 3.32 and use from Theorem3.26 that

n+1 = Pn+1 / Pn together with

Pn2 = nn1 . . . 10 =n

i=0

i.

This provides us with

nk=0

Pk(x)Pk(t) =n

k=0

1

Pk2Pk(x)Pk(t)

=Pn+1Pn

Pn+1(x)Pn(t) Pn(x)Pn+1(t)Pn+1 Pn (x t)

nk=0

Pn2Pk2

Pk(x)Pk(t) =Pn+1(x)Pn(t) Pn(x)Pn+1(t)

x tn

k=0

n

i=k+1i

Pk(x)Pk(t) =

Pn+1(x)Pn(t) Pn(x)Pn+1(t)x t


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Remark 3.34From the second part of Theorem 3.32 we obtain the inequality

Pn+1(x)

Pn(x)

Pn(x)

Pn+1(x) > 0.

This can be used to prove Theorem 3.21 the following way.Let and be consecutive zeros of Pn, such that P

n()P

n() < 0 (which holds since all

zeros are simple). Then, the inequality above tells us that

Pn()Pn+1() > 0 and Pn()Pn+1() > 0.This implies that Pn+1 has opposite signs at and . Therefore, there is at least one zeroof Pn+1 between and . In this way we find at least n 1 zeros of Pn+1.For the largest zero of Pn, i.e. for

(n)1 , holds P

n(

(n)1 ) > 0 and by the inequality above

Pn+1((n)1 ) < 0. The polynomial Pn+1 has another zero at the right of

(n)1 since Pn+1(x) >

0 for x sufficiently large.A similar argument holds for the smallest zero of Pn, i.e. for

(n)n . This proves Theorem

3.21.

3.2 Quadrature Rules and Orthogonal Polynomials

Let in this section d be a measure with bounded or unbounded support, positive definiteor not. An n-point quadrature rule for d is a formula of the type

R F(t)d(t) =

n

=1

F() + Rn(F) (3.6)

The mutually distinct points are called nodes, the numbers are the weights of thequadrature rule. Rn(F) is the remainder or error term.

Definition 3.35The quadrature rule (3.6) possesses degree of exactness d if Rn(P) = 0 for all P d. Ithas precise degree of exactness d if it has degree of exactness d, but not d + 1.A quadrature rule (3.6) with degree of exactness d = n 1 is called interpolatory.

Regarding interpolatory quadrature rulesA quadrature rule is interpolatory if and only if it is obtained by integration from theLagrange interpolation, i.e. from

F(t) =n

=1

F(t)L(t) + rn1(F; t)

where L(t) =n

=1=

tttt . Thus, we obtain

= R

L(t)d(t) , = 1, 2, . . . , n and Rn(F) = R

rn1(F; t)d(t) .


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It is well-known that for P n1 holds that rn1(P; t) 0, i.e. Rn(P) = 0 and d = n1.Given any n distinct nodes an interpolatory quadrature can always be constructed.

Theorem 3.36Let 0 k n be an integer. The quadrature rule (3.6) has degree of exactness d =n 1 + k if and only if the following two conditions are both fulfilled:

(i) (3.6) is interpolatory,

(ii) The node polynomialn(t) =n

=1

(t ) satisfiesR

n(t)P(t)d(t) = 0 for all P k1.


Remark 3.37If d is positive definite, then k = n is optimal. k = n + 1 requires that n is orthogonalto all elements of n, i.e. also to itself, which is not possible.

Gauss quadratures

Definition 3.38The quadrature rule (3.6) with k = n is called Gauss quadrature rule with respect to themeasure d. Its degree of exactness is d = 2n 1.Remark 3.39The second condition in Theorem 3.36 shows that for a Gauss quadrature (i.e. k = n) wehave n = Pn. Therefore, the nodes

G are the zeros of the n-th orthogonal polynomial

with respect to d. The weights G can be found by interpolation.Note that for the Gauss quadratures we assume that d is positive definite.

Theorem 3.40All nodes G of the Gauss quadrature rule are mutually distinct and contained in theinterior of the support interval [a, b] of d. All weights G are positive.

Proof. Since the nodes G are the zeros of Pn, the first part follows from Theorem 3.20.

Now we consider the weights:

0 , b , it can be desired to have 0 = a. To achieve this we replace n byn + 1 in (3.6) and write n+1(t) = (t a)n(t). The optimal formula (see Theorem 3.36requires n to satisfy

R

n(t)P(t)(t a)d(t) = 0 , P n1.


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Therefore, n(t) = Pn(t; da) with da(t) = (t a)d(t).The remaining n nodes have to be zeros of Pn(; da). The resulting rule is called Gauss-Radau rule.

If also b < and a as well as b are both desired as nodes, we find the (n + 2)-pointGauss-Lobatto rule similarly with the measure da,b(t) = (t a)(b t)d(t).These rules have the degrees of exactness equal to 2n and 2n + 1, respectively.

3.3 The Jacobi Polynomials

Definition 3.45Let d(x) = w(x)dx with w(x) = (1 x)(1 + x), , > 1, with the support interval[1, 1]. The corresponding orthogonal polynomials are the Jacobi polynomials P(,)nwhich are normalized by the condition

P(,)n (1) =

n +

n

=

(n + + 1)

(n + 1)( + 1).

(Note that we generalize the notation of the binomial coefficients naturally via the Gammafunction.)

Remark 3.46Similar to Theorem 3.18 we find the identity

P(,)n (x) = (1)nP(,)n (x)and so we obtain

P(,)n (1) = (1)n

n +

n

.

Theorem 3.47For > 1 hold

P(,)2n (x) =

(2n + + 1)(n + 1)

(n + + 1)(2n + 1)P

(,12

)n (2x

2 1)

= (

1)n

(2n + + 1)(n + 1)

(n + + 1)(2n + 1)

P(1

2,)

n (1

2x2) ,

P(,)2n+1 (x) =

(2n + + 2)(n + 1)

(n + + 1)(2n + 2)xP

(,12

)n (2x

2 1)

= (1)n (2n + + 2)(n + 1)(n + + 1)(2n + 2)

xP(

12

,)n (1 2x2) .

Proof. We consider the first relation and use the notation d1(x) = (1 x)(1 + x)dx =(1 x2)dx and d2(x) = (1 x)(1 + x)

12 dx. It suffices to prove that

I = R P(,12

)n (2x

2

1)P(x)d1(x) =

1

1

P(,1

2)

n (2x2

1)P(x)(1

x2)dx = 0,


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where P 2n1.If P is an odd polynomial, this is fulfilled. Therefore, let P be even, i.e. P(x) = R(x2)with R

n

1. Then

I =

11

P(,1

2)

n (2x2 1)R(x2)(1 x2)dx

= 2

10

P(,1

2)

n (2x2 1)R(x2)(1 x2)dx

=

10

P(,1

2)

n (2x 1)R(x)(1 x)x12 dx

= 212

11

P(,1

2)

n (x)R( x+12 )(1 x)(1 + x)12 dx

= 212

R

P(,1

2)

n (x)R( x+12 )d2(x) = 0.

A similar argument can be used to prove the second relation of the theorem.

Remark 3.48 (Special Cases)(a) For = we are dealing with the special case of ultraspherical polynomials (or

Gegenbauer polynomials) which due to the previous theorem are even or odd poly-nomials (depending on n being even or odd).

C()n (x) =( + 1)(n + 2 + 1)

(2 + 1)(n + + 1)P(,)n (x)

=( + 1

2)(n + 2)

(2)(n + + 12

)P

(12

,12

)n (x) ,

where = = 12

, > 12

since > 1. If = 12

(or = 0), the polynomial

C(0)n vanishes identically for n 1.Another consequence of that theorem is that Jacobi polynomials with =

1

2 or = 12

can be expressed by ultraspherical polynomials.

(b) For = = 12

we obtain the Chebyshev polynomials of first kind Tn, i.e.

P(1

2,1

2)

n (x) =

ni=1(2i 1)

2nn!Tn(x) =

ni=1(2i 1)

2nn!cos(n),

where x = cos().

(c) For = = 12

we obtain the Chebyshev polynomials of second kind Un, i.e.

P(

1

2

,1

2

)

n (x) = n

i=0

(2i + 1)

2n(n + 1)! Un(x) =ni=0(2i + 1)2n(n + 1)! sin((n + 1))sin() ,


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where x = cos().

(d) The mixed variants = 12

, = 12

and = 12

, = 12

yield the Chebyshev

polynomials of third kind Vn and of forth kind Wn, respectively, i.e.

P(1

2,

12

)n (x) =

ni=1(2i 1)

2nn!Vn(x) =

ni=1(2i 1)

2nn!

cos( (2n+1)2

)

cos( 2

),

P(

12

,12

)n (x) =

ni=1(2i 1)

2n(n + 1)!Wn(x) =

ni=1(2i 1)

2n(n + 1)!

sin( (2n+1)2

)

sin( 2

),

where x = cos().

(e) For = = 0 we find the Legendre polynomials Pn, i.e.

P(0,0)n (x) = C(

12

)n (x) = Pn(x).

Note that in this case the weight function is w(x) = 1, x [1, 1].Theorem 3.49The Jacobi polynomialsy = P

(,)n (x) satisfy the following linear homogeneous differential

equation of the second order

(1 x2)y + ( ( + + 2)x)y + n(n + + + 1)y = 0

ord

dx

((1 x)+1(1 + x)+1y) + n(n + + + 1)(1 x)(1 + x)y = 0.

Proof. We note that since y n we have thatd

dx

((1 x)+1(1 + x)+1y) = (1 x)(1 + x)z

where z n. To show that z is a constant multiple of y we have to prove the orthogo-nality to any P n1, i.e.

11

ddx

((1 x)+1(1 + x)+1y)P(x)dx = 0.

We use integration by parts on the left hand side which yields (since + 1 > 0 and+ 1 > 0):1

1

d

dx

((1 x)+1(1 + x)+1y)P(x)dx = 1

1(1 x)+1(1 + x)+1yP(x)dx

=

11

yd

dx ((1 x)+1(1 + x)+1P(x)

) C(1x)(1+x)R(x)dx


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where we performed another integration by parts and R n1. Therefore, the integralvanishes. The constant factor can be calculated by comparing the highest terms, i.e.

y = knxn + . . . , y = nknxn1 + . . . , y = n(n 1)knxn2 + . . . ,and

0 =d

dx

((1 x)+1(1 + x)+1y) + C(1 x)(1 + x)y

= ( + 1)(1 x)(1 + x)+1y + (+ 1)(1 x)+1(1 + x)y+ (1 x)+1(1 + x)+1y + C(1 x)(1 + x)y

= ( + 1)(1 x)(1 + x)(1 + x)y (+ 1)(1 x)(1 + x)(x 1)y

(1

x)(1 + x)(x2

1)y + C(1

x)(1 + x)y .

Thus,

C = (n( + 1) n(+ 1) n(n 1)) = n(n + + + 1) .

Theorem 3.50Let , > 1. The differential equation

(1 x2)y + ( ( + + 2)x)y + y = 0,

where is a parameter, has a polynomial solution not identically zero if and only if = n(n + + + 1), n N0. This solution is C P(,)n (x), C = 0, and no solution whichis linearly independent of P

(,)n can be a polynomial.

Proof. Substitute y =

=0 a(x 1) in the differential equation. This gives us

0 = (x + 1)

=2

( 1)a(x 1)1 (2( + 1) + ( + + 2)(x 1))

=1

a(x 1)1

+

=0 a(x 1)

= (x 1 + 2)

=2

( 1)a(x 1)1 2( + 1)

=1

a(x 1)1

( + + 2)

=1

a(x 1) +

=0

a(x 1)

=

=2

( 1)a(x 1) 2

=1

(+ 1)a+1(x 1) 2( + 1)

=0

(+ 1)a+1(x 1)

( + + 2)

=1 a(x 1)

+

=0 a(x 1)


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Thus, the coefficients have to fulfill the relation

(( 1) ( + + 2)+ ) a 2(+ 1)a+1 2( + 1)(+ 1)a+1 =( (+ + + 1)) a 2(+ 1)(+ + 1)a+1 = 0

for N0. If we assume that y is a polynomial, we can suppose that an denotes the lastnonzero coefficient, i.e. an+1 = 0. Therefore, the factor in front of an in the recurrencerelation above has to vanish, i.e.

= n(n + + + 1).

On the other hand, if this condition for holds, we find that ai = 0 for i n + 1 sincethe factor of a+1 = 0.

Let = n(n + + + 1) and let z be a second solution of the differential equation, i.e.

ddx

((1 x)+1(1 + x)+1y) + n(n + + + 1)(1 x)(1 + x)y = 0,

d

dx

((1 x)+1(1 + x)+1z) + n(n + + + 1)(1 x)(1 + x)z = 0.

Multiply the first equation by z and the second by y and subtract them:

0 =d

dx

((1 x)+1(1 + x)+1y) z d

dx

((1 x)+1(1 + x)+1z) y

=d

dx

((1 x)+1(1 + x)+1(yz zy))

Thus, for all x [1, 1](1 x)+1(1 + x)+1(yz yz) = c = const.,

If we let x 1, we see that y and zcannot both be polynomials unless c = 0. Therefore,yz = zy for all x (1, 1), i.e. y and z are linearly dependent, z(x) = cP(,)n .Remark 3.51The Jacobi polynomials can also be defined as the polynomial solutions of the correspond-ing differential equation that additionally take the value

P(,)n (1) = n +

n .

Definition 3.52The hypergeometric function F (sometimes denoted by 2F1) is defined by

F(a, b; c; x) =

k=0

(a)k(b)k(c)k

xk

k!

=(c)

(a)(b)

k=0

(a + k)(b + k)

(c + k)

xk

k!

or its analytic continuation with a, b R, c R \ {N0}, x (1, 1).If a =

n or b =

n, n

N0, the hypergeometric function reduces to a polynomial in x

whose degree is n.


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Theorem 3.53For the Jacobi polynomials hold

P(,)n (x) = n + n F(n, n + + + 1; + 1; 1x2 )=

k=0

(n + + 1)

(n + 1)( + 1)

(n + k)(n)(k + 1)

(n + + + k + 1)( + 1)

(n + + + 1)( + 1 + k)(1)k

x 1

2

k

=(n + + 1)

n!(n + + + 1)

nk=0

(n + + + 1 + k)

( + 1 + k)

n

k

x 1

2

kThis can be reformulated into

P(,)n

(x) =1

2nn!

n

k=0

nk(n + + 1)

(n + + 1 k)(n + + 1)

(+ 1 + k)(x

1)nk(x + 1)k

=1

2n

nk=0

n +

k

n +

n k

(x 1)nk(x + 1)k

Proof. This can be shown via the properties of the hypergeometric function, in particularits differential equation. Another way uses the Rodriguez representation of the Jacobipolynomials which will follow soon.

Corollary 3.54The leading coefficient of the Jacobi polynomial P

(,)n of degree n is given by

k(,)n = 2n

2n + +

n

.

Proof. Consider the representation

P(,)n (x) =(n + + 1)

n!(n + + + 1)

nk=0

(n + + + 1 + k)

( + 1 + k)

n

k

x 1

2

kof Theorem 3.53 in the following limit:

k(,)n = limx

xnP(,)n (x)

=(n + + 1)

n!(n + + + 1)

(n + + + 1 + n)

( + 1 + n)

n

n

1

2n

= 2n

2n + +

n

.

Corollary 3.55For the derivative of the Jacobi polynomials holds

d

dxP(,)

n (x) =

n + + + 1

2 P(+1,+1)

n1 (x).


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Proof. See Exercise 6.2. (This follows immediately if both sides are expanded accordingto Theorem 3.53.)

Theorem 3.56 (Rodriguez Formula)Let , > 1. Then

(1 x)(1 + x)P(,)n (x) =(1)n2nn!

d

dx

n ((1 x)n+(1 + x)n+)

Proof. We use Leibniz rule for the differentiation of products to find

d

dx

n ((1 x)n+(1 + x)n+) = n

k=0

n

k

d

dx

k(1 x)n+

d

dx

nk(1 + x)n+

=n

k=0

n

k

(1 x)Rnk(x)(1 + x)Sk(x)

= (1 x)(1 + x)(x)

where Rj, Sj j, deg(Rj ) = deg(Sj) = j, j = 0, . . . , n, and n, deg() = n. Indetail (we will need that later):

(1 x)(1 x)(x)

=n

k=0

n

k

k1i=0

(n + i)

(1)k(1 x)n+k

nk+1i=0

(n + i)

(1 + x)n+(nk)

=n

k=0

n

k

(n + + 1)

(n + k + 1)(n + + 1)

(+ k + 1)(1)k(1 x)n+k(1 + x)+k.

Therefore,

(x) =n

k=0

n

k

(n + + 1)

(n + k + 1)(n + + 1)

(+ k + 1)(1)k(1 x)nk(1 + x)k

We now have to show that = C P(,)n with a constant C. It suffices to show that

1

1

(1

x)(1

x)(x)R(x)dx = 0


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for all R n1. We use partial integration:1

1 (1 x)

(1 x)

(x)R(x)dx =

1

1

ddxn

((1 x)n+(1 + x)n+)R(x)dx=

d

dx

n1 ((1 x)n+(1 + x)n+)R(x)

1

1 =0

1

1

d

dx

n1 ((1 x)n+(1 + x)n+)R(x)dx

= . . . = (1)n1

1

((1 x)n+(1 + x)n+)R(n)(x)dx = 0since deg(R) n 1, i.e. R(n) 0. Now we just have to determine the constant C.Consider x = 1 in , i.e. only the summand k = n remains, i.e.

(1) =

n

n

(n + + 1)

( + 1)

(n + + 1)

(n + + 1)(1)n2n = (1)n2nn!

n +

n

= (1)n2nn! P(,)n (1).

Thus, C = 2nn!

(1)n .

Remark 3.57The previous theorem also gives us the explicit representation of the second part of The-orem 3.53, i.e.

P(,)n (x) =1

2nn!

nk=0

n

k

(n + + 1)

(n + + 1 k)(n + + 1)

(+ 1 + k)(x 1)nk(x + 1)k

=1

2n

nk=0

n +

k

n +

n k

(x 1)nk(x + 1)k .

Theorem 3.58

Let , > 1. Then for n N

h(,)n =P(,)n 2 = 1

1

(P(,)n (x)

)2(1 x)(1 + x)dx

=2++1

2n + + + 1

(n + + 1)(n + + 1)

(n + 1)(n + + + 1)

and

h(,)0 =

P(,)0

2

=

11

P

(,)0 (x)

2(1 x)(1 + x)dx

= 2

++1 ( + 1)(+ 1)

( + + 2) .


57/137


Proof. Let n N. We have from the proof of Theorem 3.56

1

1 (P(,)

n (x))2

(1

x)(1 + x)dx = k(,)n

1

1P(,)n (x)x

n(1

x)(1 + x)dx

=(1)n2nn!

k(,)n

11

d

dx

n ((1 x)n+(1 + x)n+) xndx

= 2nk(,)n

11

(1 x)n+(1 + x)n+dx

= 22n

2n + +

n

11

(1 x)n+(1 + x)n+dx

where we used partial integration and Corollary 3.54.

1

1 (P(,)n (x))2 (1 x)(1 + x)dx = 22n2n + +

n 1

1(1

x)n+(1 + x)n+dx

= 2+

2n + +

n

11

1 x

2

n+ 1 + x2

n+dx

= 2++1

2n + +

n

10

yn+(1 y)n+dy

= 2++1(2n + + + 1)

(n + 1)(n + + + 1)

(n + + 1)(n + + 1)

(2n + + + 2)

=2++1

2n + + + 1

(n + + 1)(n + + 1)

(n + 1)(n + + + 1).

The case n = 0 follows similarly.

Theorem 3.59The Jacobi polynomials fulfill the following three term recurrence relation:

2(n + 1)(n + + + 1)(2n + + )P(,)n+1 (x) =

(2n + + + 1)(

(2n + + + 2)(2n + + )x + 2 2)P(,)n (x) 2(n + )(n + )(2n + + + 2)P(,)n1 (x)

for n N with P(,)0 (x) = 1 and P(,)1 (x) = 12 ( + + 2)x + 12 ( ).


3.4 Ultraspherical Polynomials

In Remark 3.48 we have already presented the ultraspherical (or Gegenbauer) polynomials

C()n and given their connection to the general Jacobi polynomials, i.e.

C()n (x) =( + 1)(n + 2 + 1)

(2 + 1)(n + + 1)P(,)n (x)

=( + 1

2)(n + 2)

(2)(n + + 12 )P

(12

,12

)n (x) ,


58/137


where = = 12

, > 12

since > 1. This relation gives us the following propertiesfor = 0:

1. The value at 1 is C()

n (1) = (n+21n ), the symmetry relation is given byC()n (x) = (1)nC()n (x),

the polynomials are either even or odd (depending on n) due to Theorem 3.47.

2. The differential equation (whose solution is y = C()n ) becomes

(1 x2)y (2 + 1)xy + n(n + 2)y = 0. (3.7)

3. We know the explicit representation, i.e.

C()n (x) =n + 2 1

n

F(n, n + 2; + 1

2; 1x

2)

=( + 1

2)

(2)n!

nk=0

n

k

(n + 2 + k)

( + 12

+ k)

x 1

2

k.

4. The leading coefficient is

k()n = limx

xnC()n (x) = 2n

n + 1

n

.

5. The derivative is again an ultraspherical polynomial, i.e.

d

dxC()n (x) = 2C

(+1)n1 (x).

6. The Rodriguez representation for the Gegenbauer polynomials reads as follows

C()n (x) =(1)n( + 1

2)(n + 2)

2nn! (2)(n + + 12

)(1 x2) 12

d

dx

n(1 x2)n+12

=(2)n(n + )(n + 2)

n! ()(2n + 2)(1

x2)

12 ddx

n

(1 x2)n+12 .

7. Their norm is computed to be

h()n =C()n 2 = 212(n + 2)n!(n + )(())2 .

8. The corresponding three term recurrence is given by

(n + 1)C()n+1(x) = 2(n + )xC()n (x) (n + 2 1)C()n1(x) (3.8)

for n N with C()0 (x) = 1 and C()1 (x) = 2x.


59/137


Lemma 3.60If n 1, C(0)n 0, but

lim0C

()n (x)

=

2

nTn(x)with the Chebyshev polynomials of first kind Tn.

Proof. We remember that

Tn(x) =2nn!n

i=1(2i 1)P

(12

,12

)n (x).

Now we consider

C()n (x)

=( + 1

2)

(2)

(n + 2)

(n + +1

2 )

P(1

2,1

2)

n (x)

=2( + 1

2)

(2 + 1)

(n + 2)

(n + + 12

)P

(12

,12

)n (x)

0 2 (12

)

(1)

(n)

(n + 12

)P

(12

,12

)n (x)

=2 n!

n

( 12

)

(n + 12

)P

(12

,12

)n (x)

=2 n!

n

1

ni=1(i 1/2)P

(12

,12

)n (x)

=2

n

n! 2nni=1(2i 1)

P(1

2,1

2)

n (x) =2

nTn(x).

Note that C()0 = 1, also as 0. Combining Theorem 3.47 and Theorem 3.53 for theultraspherical polynomials we obtain the following representations.

Lemma 3.61For n N0 hold

C

()

2n (x) = 2n + 2

1

2n F(n, n + ; + 12 ; 1 x2) ,C()2n+1(x) =

2n + 2

2n + 1

xF(n, n + + 1; + 1

2; 1 x2) .

Proof. Consider the even case first. We start with the relation between Gegenbauer andJacobi polynomials and apply Theorem 3.47.

C()2n (x) =

( + 12

)(2n + 2)

(2)(2n + + 12

)P

(12

,12

)

2n (x)

=( + 1

2)(2n + 2)

(2)(2n + + 12 )

(2n + + 12

)(n + 1)

(n + + 12 )(2n + 1)P

(12

,12

)n (2x

2

1)


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Now we use the representation of Theorem 3.53:

C()2n (x) =

( + 12

)(2n + 2)(n + 1)

(2)(n + + 12 )(2n + 1)n + 1

2

n F(n, n + ; + 12 ; 1 x2)=

2n + 2 1

2n

F(n, n + ; + 1

2; 1 x2).

The odd case can be shown analogously.

Lemma 3.62For n N0 hold

C()2n (x) = (1)n

n + 1

n

F(n, n + ; 1

2; x2)

C()2n+1(x) = (1)n2

n + n

x F(n, n + + 1; 3

2; x2)

Proof. Consider the even case. This time we apply the second variant of Theorem 3.47.

C()2n (x) =

( + 12

)(2n + 2)

(2)(2n + + 12

)P

(12

,12

)

2n (x)

=( + 1

2)(2n + 2)

(2)(2n + + 12

)(1)n (2n + +

12

)(n + 1)

(n + + 12

)(2n + 1)P

(12

,12

)n (1 2x2)

=( + 1

2)(2n + 2)(n + 1)

(2)(n + + 12 )(2n + 1)(

1)nn 12

n F(n, n + ; 12 ; 1(12x2)

2)

=( + 1

2)(2n + 2)(n + 1)

(2)(n + + 12

)(2n + 1)(1)n (n +

12

)

(n + 1)( 12

)F(n, n + ; 1

2; x2)

Note that

(n + + 12

)

( + 12

)=

n1i=0

(i + + 12

) ,

(2n + 2)

(2)= 2n

n1

i=0 (i + )2n

n1

i=0 (i + +12

) ,

such that( + 1

2)

(n + + 12

)

(2n + 2)

(2)= 22n

(n + )

().

Moreover,

2n(n + 1

2)

( 12

)=

n1i=0

(2i + 1),

2n(n + 1) =n

i=1(2i),


61/137


such that

22n(n + 1

2)

( 12

)(n + 1) = (2n + 1)

and finally

C()2n (x) = (1)n(n + )

()(n + 1)F(n, n+; 1

2; x2) = (1)n

n + 1

n

F(n, n+; 1

2; x2).

The odd case can be shown analogously.

Corollary 3.63For N N0 holds the explicit representation

C()N (x) =

N/2m=0

(1)m (N

m + )

()(m + 1)(N 2m + 1) (2x)N

2m

.

Proof. Let N = 2n, i.e. N/2 = n. we use the definition of the hypergeometric functionF (Definition 3.52) in Lemma 3.62.

C()N (x) = (1)n(n + )

()(n + 1)F(n, n + ; 1

2; x2)

= (1)n (n + )()(n + 1)

nk=0

(n)k(n + )k( 1

2)

Documents

Special Functions of Mathematical (Geo-)Physics_M Gutting_2011