RODRIGUES FORMULA FOR JACOBI POLYNOMIALS ON THE UNIT CIRCLE
MASTERS THESIS
Presented in Partial Fulfillment of the Requirements for the Degree Master of
Science in the Graduate School of the Ohio State University
By
Griffin Alexander Reiner-Roth
Graduate Program in Mathematics
The Ohio State University
2013
Master’s Examination Committe:
Professor Rodica Costin, Advisor
Professor Barbara Keyfitz c Copyright by
Griffin Alexander Reiner-Roth
2013 ABSTRACT
We begin by discussing properties of orthogonal polynomials on a Borel measur- able subset of C. Then we focus on Jacobi polynomials and give a formula (analogous to that of [5]) for finding Jacobi polynomials on the unit circle. Finally, we consider some examples of Jacobi polynomials and demonstrate different methods of discov- ering them.
ii ACKNOWLEDGMENTS
I would like to thank my advisor, Dr. Rodica Costin, for all her help on this thesis; for finding a really interesting and doable project for me, for spending so much time explaining the basics of orthogonal polynomials to me, and for perfecting this paper.
I also thank my parents for making sure I never starved.
iii VITA
2011 ...... B.A. Mathematics, Vassar College
2011-Present ...... Graduate Teaching Associate, Department of Mathematics, The Ohio State Univer- sity
FIELDS OF STUDY
Major Field: Mathematics
iv TABLE OF CONTENTS
Abstract ...... ii
Acknowledgments ...... iii
Vita...... iv
List of Tables ...... vii
CHAPTER PAGE
1 Introduction ...... 1
1.1 Historical Background ...... 1 1.2 Orthogonality in Hilbert Spaces ...... 4 1.3 Orthogonal Polynomials with Respect to Measures ...... 5 1.4 Approximation by Orthogonal Polynomials ...... 12 1.5 Classical Orthogonal Polynomials ...... 15
2 Jacobi Polynomials ...... 17
2.1 Jacobi Polynomials on the Real Line ...... 17 2.2 Jacobi Polynomials on the Unit Circle ...... 19 2.3 Properties of Jacobi Polynomials on the Unit Circle ...... 21 2.4 Connection between Jacobi Polynomials on the Real Line and on the Unit Circle ...... 22 2.5 Calculating the Inner Product ...... 26 2.6 Formulas for Cn, Dn ...... 32 2.7 Conjecture for An, Bn ...... 35 3 New Results ...... 37
3.1 Commutation Relations ...... 37 3.2 Rodrigues’ Formula ...... 39
v 4 Examples of Jacobi Polynomials on the Unit Circle ...... 41
4.1 Using the Gram-Schmidt Process ...... 41 4.2 Using Theorem 9 ...... 43 4.3 Using the New Rodrigues’ Formula ...... 44 4.4 Comparison of Formulas ...... 45
5 Codes and Computer Calculation Results ...... 47
5.1 OPRL ...... 47 5.2 OPUC ...... 48
Bibliography ...... 57
vi LIST OF TABLES
TABLE PAGE
5.1 Table 1: Comparison of Rodrigues’ formulas on the real line ..... 49
5.2 Table 2: Computation times using the new Rodrigues’ formula ... 54
vii CHAPTER 1
INTRODUCTION
1.1 Historical Background
The study of orthogonal polynomials has a rich history, which can be traced to as early as the late eighteenth century, linked to Legendre’s study of planetary motion.
About a century later, orthogonal polynomials arose (seemingly, in a more natural manner) when searching for solutions to Sturm-Liouville problems. These polynomi- als are referred to as the classical orthogonal polynomials (we discuss properties of these polynomials in Section 1.5). Some of the first mathematicians to study these polynomials are Chebyshev, Markov, and Stieltjes. These three names come up again, in connection with orthogonal polynomials, regarding the Markov-Stieltjes inequality, which “plays a fundamental role in the theory of momentum probems” [7]. Cheby- shev conjectured the inequality, and Markov and Stieltjes (independently) proved the result.
Orthogonal polynomials are defined on the real line (OPRL), or on the unit circle
(OPUC); there is far more research that has been done on orthogonal polynomials on the real line. Simon [17] conjectures one reason is “[OPRL] examples appear in so many places that most scientists are exposed to them early in their education.
1 The applications of OPUC are subtle and beautiful, but less concrete.” However,
Szeg¨oproved a relation between two special classes of these sets of polynomials, so an additional application of OPUC is that knowledge of these polynomials can illuminate
OPRL facts. In Section 2.4, we state and prove this relation.
In the early twentieth century, Szeg¨o,the father of OPUC, is responsible for the revolution in the study of orthogonal polynomials by considering a wide variety of problems. In many ways, OPUC are analogous to OPRL. As an example, polynomials from both sets possess the Christoffel-Darboux Formula; for more information on the
Christoffel-Darboux Formula in both settings, see Simon [17], Szeg¨o[19] (to whom the formula is due), and Freud [7] (whose work on the unit circle relies heavily on the minimum problem, i.e., minimizing the Christoffel function). In fact, “The study of the minimum problem is central to Szeg¨o’sinitial papers and have been a recurrent theme since” [17]. Other analogues on the unit circle include Favard’s Theorem (a recent proof is given in [6]), Gauss-Jacobi quadrature, and the fact that they all satisfy three-term recursion relations.
We conclude this section with a short summary of some areas of mathematics that have benefitted from the study of OPRL, and to a lesser extent, OPUC. We mentioned quadrature, an area that can be considered independently of orthogo- nal polynomials. Asymptotic behavior of orthogonal polynomials is a large field of study, and has applications to problems such as to Riemann-Hilbert problems. The asymptotic behavior of the ultraspherical and Laguerre polynomials (examples of clas- sical orthogonal polynomials) have been studied by Darboux and Fej´er,respectively.
Asymptotic properties of orthogonal polynomials have continued to be investigated,
2 and results have been found by several mathematicans, from Szeg¨oto Simon. In- terpolation procedures, very important in applications, have been studied by, among others, Geronimus. A solution to the momentum problem (or the Hamburger momen- tum problem, named after Hamburger for his work on the problem) was proved due in no small part to results of orthogonal polynomials. Approximation theory (with profound results due to Nevai), Lagrange interpolation, continued fractions, stochas- tic processes, and even coding theory are subjects with results relying on orthogonal polynomials.
3 1.2 Orthogonality in Hilbert Spaces
Orthogonal polynomials are the Rodney Dangerfield of analysis.
Barry Simon, OPUC on One Foot
Let (H, h·, ·i) be a Hilbert space and k · k be the associated norm, defined by kfk = phf, fi.
Definition 1. Let F ⊆ H be a countable set of functions {fk : k ≥ 0}. fi is orthog- onal to fj if hfi, fji = 0, and F is an orthogonal set if for i 6= j, hfi, fji = 0. The family F is an orthonormal set if F is an orthogonal set and, in addition, kfik = 1 for each i ≥ 0.
Let F be a countable, linearly independent set of functions. Then there exists an orthogonal set G in bijection with F such that span(F) = span(G). Such a G can be constructed using the Gram-Schmidt process:
Theorem 2. (The Gram-Schmidt Process) Let n ∈ N and {fk : 0 ≤ k ≤ n} be a linearly independent set of functions. Define g0 = f0 and for 1 ≤ k ≤ n,
k−1 X hgj, fki g = f − g . k k kg k2 j j=0 j
Then {gk : 0 ≤ k ≤ n} is an orthogonal set and span{fk} = span{gk}.
(For a proof of this theorem, see, for example, Sadun [15].) The process above can be extended to a countable family of linearly independent functions. We remark that one can obtain an orthonormal set from an orthogonal set by replacing each gk g with k . kgkk
4 1.3 Orthogonal Polynomials with Respect to Measures
Let X be an interval in R or the unit circle and µ a positive Borel measure on X. Define Z 2 2 H = L (X, dµ) = f : X → C: f is measurable and |f| dµ < ∞ , X the space of square integrable functions with respect to the measure µ. The binary operation defined by Z hf, gi = fg dµ (1.3.1) X for f, g ∈ H is an inner product on H and it is well-known that (H, h·, ·i) becomes a Hilbert space. Notice that this inner product is finite by the Cauchy-Schwarz
Inequality: Z Z
|hf, gi| = fg dµ ≤ |f| · |g| dµ X X Z 1/2 Z 1/2 ≤ |f|2 dµ |g|2 dµ < ∞. X X Z Assumptions. We assume that the moments exist, that is, zk dµ is defined for X all k ≥ 0 (later, we give a more general definition of moments, but this definition will Z be sufficient on the unit circle). Further, assume µ is normalized so that 1 dµ = 1. X The set of polynomials {zk : k ≥ 0} is linearly independent, so one may apply
Gram-Schmidt to this set to obtain the set of monic orthogonal polynomials of degree n ≥ 0. These are the orthogonal polynomials on X with respect to the measure µ.
Throughout, the set {ϕn} = {ϕn : n ≥ 0} will denote the set of monic orthogonal polynomials, where ϕn(z) is degree n (i.e., for each n ∈ N, there exist complex num- n−1 n X k bers ak such that ϕn(z) = z + akz ). In certain situations, the set of orthonormal k=0 5 polynomials is desired; denote this set by {Φn}. In more theoretical settings, {ϕn} is usually of interest simply because monic polynomials are easier to work with. In the context of applications, {Φn} may be preferred for its use in approximation.
The first apparent difference between ϕn and Φn involves the leading coefficient of Φn, which we denote by κn:
n Φn(z) = κnz + ··· .
ϕn −1 Of course, Φn = , therefore, κn = kϕnk . kϕnk We mention a few observations about orthogonal polynomials. Since whenever
0 ≤ m < n, hϕm, ϕni = 0, then
m hϕn(z), z i = 0, 0 ≤ m < n.
m Conversely, the conditions hϕn(z), z i = 0 for 0 ≤ m < n are sufficient for {ϕk(z)}
m to be an orthogonal set: if for each n, hϕn(z), z i = 0 whenever m < n, then by the linearity of the inner product, hϕn(z), p(z)i = 0 if p(z) is a polynomial of degree less than n. In particular, hϕn(z), ϕk(z)i = 0 if k < n.
Definition 3. For i, j ≥ 0, denote the moments of µ by σi,j, where Z i j σi,j = z z dµ. X n X j If we write ϕn(z) = cjz (with cn = 1 because ϕn is monic), then for 0 ≤ m < j=0 n, Z n m m X 0 = hϕn(z), z i = ϕn(z)z dµ = cjσj,m. X j=0 Therefore, n−1 X cjσj,m = −σn,m, m = 0, 1, . . . , n − 1. (1.3.2) j=0 6 So finding ϕn(z) amounts to solving the above n×n system of equations for c0, c1,..., cn−1.
Definition 4. An n×n matrix A = (ai,j) is Hermitian if ai,j = aj,i for all 1 ≤ j, k ≤ n.A quadratic form in n variables z1, . . . , zn is a polynomial in these variables of homogeneous degree two.
Note that hx, Axi is a quadratic form (generated byA).
The coefficients of (1.3.2) form a Hermitian matrix which generates the quadratic n X form zizj, and the determinant of this system is i,j=0
σ0,0 σ0,1 ··· σ0,n
σ1,0 σ1,1 ··· σ1,n D = . n ......
σn,0 σn,1 ··· σn,n
If µ is supported on the real line, then Dn is a Hankel determinant. A Hankel matrix is a matrix that satisfies ai,j = ai−1,j+1 for each i > 0 and j < n. If µ is supported on the unit circle, then Dn is a Toeplitz determinant. A Toeplitz matrix satisfies ai,j = ai+1,j+1 for 0 ≤ i, j < n. We will see later why these are such determinants for µ on the real line and on the unit circle.
The system (1.3.2) of equations has a solution by the Sylvester criterion for positive definiteness: the quadratic form generated by a Hermitian matrix A is positive definite if and only if the principal minors of A are positive. The quadratic form generated n X i j here is z z , which is positive definite, so the principal minors (that is, |D0|, i,j=0 |D1|,..., |Dn−1|) are positive (an argument explaining why this quadratic form is
7 positive definite is given in Horn and Johnson [11]). This result gives the following method of constructing orthogonal polynomials:
Proposition 5. Let
σ0,0 σ0,1 ··· σ0,n
......
pn(z) = σ σ ··· σ n−1,0 n−1,1 n−1,n
n 1 z ··· z and suppose either X ⊆ R or X = ∂D, the unit circle. Then for each n ≥ 1,
1 ϕn(z) = pn(z). Dn−1
1/2 Dn Also, kϕnk = . Dn−1
Proof. Fix n ≥ 1 and 0 ≤ m < n. Recall an equivalent condition for {pk(z)} to be
m an orthogonal set is hpn(z), z i = 0 for 0 ≤ m < n.
σ0,0 σ0,1 ··· σ0,n
. . . . Z Z . . .. . m m m hpn(z), z i = pn(z)z dµ = · z dµ X X σ σ ··· σ n−1,0 n−1,1 n−1,n
n 1 z ··· z
σ0,0 σ0,1 ··· σ0,n
. . . . Z . . .. .
= dµ. X σ σ ··· σ n−1,0 n−1,1 n−1,n
m m n m z zz ··· z z
8 Each σi,j is a constant, so the integral of the determinant may be applied to the last row of the matrix. Thus, we have
σ0,0 σ0,1 ··· σ0,n
...... m hpn(z), z i = σ σ ··· σ n−1,0 n−1,1 n−1,n
R m R m R m n X z dµ X z z dµ ··· X z z dµ
σ0,0 σ0,1 ··· σ0,n
......
= . σ σ ··· σ n−1,0 n−1,1 n−1,n
σ0,m σ1,m ··· σn,m
We claim σj,m = σm,j for 0 ≤ j ≤ n.
m If X ⊆ R, then zm = z and Z j+m σj,m = z dµ = σm,j. X
(So if X ⊆ R, then Dn is a Hankel determinant because Z Z i+j (i−1)+(j+1) σi,j = z dµ = z dµ = σi−1,j+1 X X for i > 0 and j < n.) eiθ If X = ∂ , then zm = z−m. The substitutions z = eiθ, dν(t) = dµ , and D ieiθ k = j − m yield Z Z π Z π k ikt σj,m = z dµ(z) = e dν(t) = cos(kt) + i sin(kt) dν ∂D −π −π Z π = cos(kt) dν −π Z π = cos(−kt) dν = σm,j. −π 9 (So if X = ∂D, then Z Z i−j i+1−(j+1) σi,j = z dµ = z dµ = σi+1,j+1 ∂D ∂D for 0 ≤ i, j < n.)
This claim implies we can rewrite the last row of the matrix above as
σ0,0 σ0,1 ··· σ0,n
......
. σ σ ··· σ n−1,0 n−1,1 n−1,n
σm,0 σm,1 ··· σm,n
th th Now, the m and n rows are equal, so the determinant is zero and {pn(z)} forms an orthogonal set. Also, one can see that the leading coefficient of pn is Dn−1 by expansion by minors.
Next, we find the norm of ϕn. Since ϕn(z) is orthogonal to any polynomial of degree less than n,
2 n kϕnk = hϕn(z), ϕn(z)i = hϕn(z), z i .
The calculation above with m = n gives
σ0,0 σ0,1 ··· σ0,n
n 1 n 1 . . . . Dn hϕn(z), z i = hpn(z), z i = . . .. . = . D D D n−1 n−1 n−1
σn,0 σn,1 ··· σn,n Therefore, s Dn kϕnk = . Dn−1
10 Thus, the moments help give the leading coefficient of Φn. In particular,
1/2 Dn−1 κn = . Dn
11 1.4 Approximation by Orthogonal Polynomials
What is the definition of an approximation to a number? Any number other than that number!
Arnold Ross
Orthogonal polynomials are studied for several reasons (in both the applied and pure branches of mathematics) and are of great importance due to their applications to approximating functions – this branch of mathematics is called Approximation theory.
Let f be a function on X that we wish to approximate by polynomials, and suppose {pk(z)} is a given sequence of polynomials. The goal, if possible, is to find a sequence of constants {ck} such that
∞ X f(z) = ckpk(z) k=0 as a uniformly convergent series on X. One of the most famous examples of such an expansion is the Taylor series expansion of f, but this expansion exists if and only if f ∈ C∞(X), and even then, the series may not converge. This idea motivates ap- proximation by orthogonal polynomials. One nice feature of orthogonal polynomials
12 is that the coefficients can be discovered easily: if such an expansion exists, then take the expansion, multiply through by ϕj(z), and integrate to get ∞ X f(z) = ckϕk(z) k=0 ∞ Z Z X f(z)ϕj(z) dµ = ckϕk(z)ϕj(z) dµ X X k=0 (1.4.1) ∞ X Z = ck ϕk(z)ϕj(z) dµ k=0 X
2 = cjkϕjk .
Note that, in the representation (1.4.1), the series converges in H, that is, in square average. Even if the series is not uniformly convergent on X, we may still interchange the integral with the limit in (1.4.1) because the inner product is a continuous linear
X 2 functional, so in particular, if ckϕk(z) converges to f(z) in L (which is true by DX E Parseval’s identity), then for any ϕj ∈ H, ckϕk, ϕj → hf, ϕji.
So each cj can be found with the formula
1 Z cj = 2 f(z)ϕj(z) dµ. kϕjk X Not only is this sequence of constants relatively easy to obtain, it is also the optimal set of constants to use to approximate f(z):
2 Theorem 6. Let f ∈ L (X, dµ) and n ∈ N. Then ( n )
X n+1 inf f − cjΦj : {c0, . . . , cn} ∈ C j=0 is attained if and only if cj = hf(z), Φj(z)i.
This is a key result of the theory of Hilbert spaces (for a proof, see Akhiezer and
Glazman [2]). The theorem states that the best approximation (in the L2 norm) 13 by a polynomial of degree n is determined by the orthogonal polynomials and the coefficients are given by the projections of f onto each ϕj(z).
14 1.5 Classical Orthogonal Polynomials
Some of the first classes of orthogonal polynomials appeared as eigenfunctions of
Sturm-Liouville problems, that is, as solutions of differential equations
00 0 σ(x)y (x) + τ(x)y (x) + λny(x) = 0 with appropriate boundary conditions, where σ(x) is a polynomial of degree at most two, τ(x) a linear polynomial, and λn are constants that need to be identified. These polynomials are also rich in extra properties and are now called the classial orthogonal polynomials. Some of these properties include:
1. their derivatives also form an orthogonal set,
2. they all possess a Rodrigues’ type formula:
dn P (x) = w(x)−1 (w(x)σ(x)n) n dxn with w the weight function and σ some polynomial,
3. they satisfy a differential-difference relation of the form
0 π(x)Pn(x) = (αnx + βn)Pn(x) + γnPn−1(x),
4. they satisfy a non-linear equation of the form
0 2 2 σ(x)(Pn(x)Pn−1(x)) = (αnx + βn)Pn(x)Pn−1(x) + γnPn (x) + δnPn−1(x) with σ a polynomial of degree at most two.
Not only do these properties hold for all of the classical orthogonal polynomials, the classical orthogonal polynomials are the only orthogonal polynomials that satisfy these four properties (Golinskii, [8]). 15 The three classes of the classical orthogonal polynomials are:
(α,β) 1. Jacobi Polynomials: Traditionally denoted by Pn (x), they are orthogonal in L2([−1, 1], µ) with dµ = w(x) dx, the weight function w(x) = (1 − x)α(1 + x)β
(α,β) (α > −1, β > −1). Note that the condition α, β > −1 is needed for Pn to be integrable with respect to the measure (1 − x)α(1 + x)β dx on [−1, 1]. On the other hand, many of the other properties hold for α, β ∈ C \ {−1, −2,...} (there exists a three-term relation, the Rodrigues’ formula holds, and in fact, they form a complete set with respect to the complex-valued measure (1 − x)α(1 + x)β dx).
2. Laguerre polynomials: Orthogonal in L2([0, ∞), µ) with dµ = w(x) dx, w(x) =
−x α (α) th e x (α > −1), Ln denotes the n Laguerre polynomial. 3. Hermite polynomials: These polynomials are orthogonal in L2((−∞, ∞), µ)
−x2 th with dµ = w(x) dx, w(x) = e . The polynomial Hn(x) denotes the n Hermite polynomial, which is orthogonal with respect to w(x) = e−x2 .
Many special subclasses appeared of great interest. For example, the ultraspherical
(or Gegenbauer) polynomials are the Jacobi polynomials with α = β, the Chebyshev 1 polynomials of the first kind are the Jacobi polynomials with α = β = , and the 2 Legendre polynomials satisfy α = β = 0. These polynomials occur in many settings, including those mentioned in Section 1.1.
In the following, we focus our attention on the Jacobi polynomials.
16 CHAPTER 2
JACOBI POLYNOMIALS
2.1 Jacobi Polynomials on the Real Line
Jacobi polynomials have been studied extensively both on the real line and the unit circle. We begin with a summary of results on the real line, and we focus on the scalar case.
2 α β Let Q(x) = x − 1 and wr(x) = (1 − x) (1 + x) , where α, β > −1. Let N denote the nonnegative integers: N = {0, 1, 2,...}.
For each n ∈ N, define pn(x) by the Rodrigues formula:
n 1 d n pn(x) = n [Q(x) wr(x)] . wr(x) dx
The set {pn(x): n ∈ N} is the set of Jacobi polynomials (up to constant multiplica- tion), and is an orthogonal set. A relatively new result gives a nice, recursive property of the Jacobi polynomials:
Theorem 7. (Costin [5]) Define the operator Ak by
0 Ak = kQ (x) + (α + β)x + α − β + Q(x)∂x.
Then for all n, pn(x) = A1A2 ... An1.
17 This result greatly reduces the difficulty of finding the nth Jacobi polynomial for a specified n. For example, the computation time of finding p15(x) with the Rodrigues formula is 16.718 seconds, and with Theorem 7, the computation time is only 0.522 seconds. The code used and the times for computing other Jacobi polynomials are given in Chapter 5.
18 2.2 Jacobi Polynomials on the Unit Circle
Define the inner product of functions defined on the unit circle ∂D by 1 Z dz hf, gi = f(z)g(z)w(z) , 2πi ∂D z where w is the complex Jacobi weight z + z−1 α z + z−1 β w(z) = 1 − 1 + 2 2 and α, β ≥ 0 (on the unit circle, integrability does not hold if α or β are negative).
Since w(z) = w(z) (since on the unit circle, z−1 = z), the values of w(z) are completely determined by its values for z on the upper half circle (or lower). For z on the upper √ half of the unit circle, we may write z = x+i 1 − x2 for x ∈ [−1, 1]. Since R(z) = x, √ we may write w(z) = w(x + i 1 − x2) = (1 − x)α(1 + x)β. A more useful expression for this inner product is Z Z π iθ 1 dz 1 iθ −iθ α β ie dθ f(z)g(z)w(z) = f(e )g(e )(1 − cos θ) (1 + cos θ) iθ 2πi ∂D z 2πi −π e 1 Z π = f(eiθ)g(e−iθ)(1 − cos θ)α(1 + cos θ)β dθ. 2π −π dz This change of variables explains why we write instead of dz: converting z to eiθ z yields an additional term, ieiθ.
For x ∈ [−1, 1], define
α− 1 β− 1 w1(x) = (1 − x) 2 (1 + x) 2
α+ 1 β+ 1 w2(x) = (1 − x) 2 (1 + x) 2 , and let pn(x), qn(x) be the Jacobi polynomials defined as n 1 d n pn(x) = n [Q(x) w1(x)] w1(x) dx n 1 d n qn(x) = n [Q(x) w2(x)] w2(x) dx 19 2 for n ≥ 0 and q−1(x) = 0. (Recall that Q(x) = x − 1). Define µ0 on [−1, 1] and µ on ∂D to be the measures
α β µ0 = µ0(x) = (1 − x) (1 + x) = wr(x) √ (1 − x)α(1 + x)β w(z) µ = µ(z) = µ(x + i 1 − x2) = = . 2πiz 2πiz
20 2.3 Properties of Jacobi Polynomials on the Unit Circle
Throughout this paper, we use the following facts for polynomials on the unit circle which are orthogonal with respect to the Jacobi weight.
j Useful Fact 1. For 0 ≤ j < n, ϕn(z), z = 0 (the proof is given in Section 1.3).
Useful Fact 2. Given n ∈ N, σi,j = σj,i for all 0 ≤ i, j ≤ n (the proof of which is also in Section 1.3).
−1 n −n Useful Fact 3. For all z ∈ ∂D, z = z , and zn = z = z for n ∈ N (it is evident).
Useful Fact 4. w(z) = w(z).
Useful Fact 5. A straightforward calculation gives √ Z Z π Z 1 2 1 iθ iθ 1 f(x + i 1 − x ) f(z) dµ(z) = f(e )w(e ) dθ = − √ dµ0(x) 2 ∂D 2π −π 2π −1 1 − x Z 1 √ 1 2 = − f(x + i 1 − x )w1(x) dx 2π −1
2 for all f ∈ L (∂D, dµ).
Useful Fact 6. Each ϕn(z) has real coefficients (a proof is given in Freud [7]).
21 2.4 Connection between Jacobi Polynomials on the Real Line
and on the Unit Circle
In his review of Simon’s Orthogonal Polynomials on the Unit Circle, Nevai noted
“Szeg¨odiscovered that all OPRL systems living on a finite interval can be mapped to
OPUC systems” [13]. The following theorem describes how the systems are related for Jacobi weights.
Theorem 8. (Szeg¨o,[19]) Let {pn(x)}, {qn(x)} be sets of real polynomials orthogonal
α−1/2 β−1/2 with respect to w1(x), w2(x), respectively (recall w1(x) = (1 − x) (1 + x) and
α+1/2 β+1/2 w2(x) = (1 − x) (1 + x) ). Then there exist constants an, bn, cn, dn such that
−n n −1 pn(x) = an z ϕ2n(z) + z ϕ2n(z )
−n+1 n−1 −1 = bn z ϕ2n−1(z) + z ϕ2n−1(z ) z−n−1ϕ (z) − zn+1ϕ (z−1) q (x) = c 2n+2 2n+2 n n z − z−1 z−nϕ (z) − znϕ (z−1) = d 2n+1 2n+1 . n z − z−1 Proof. We prove the first equality, and sketch the proof of the next three since they are similar.
−n n −1 Fix n ∈ N and let ψn(z) = z ϕ2n(z) + z ϕ2n(z ). We prove the first equality in the following steps:
n−j n+j Step 1. For each j < n, ϕ2n(z), z + z = 0.
n−j n+j j −j Step 2. If ϕ2n(z), z + z = 0, then ψn(z), z + z = 0.
j −j j Step 3. If ψn(z), z + z = 0 for 0 ≤ j < n, then ψn(z), x = 0 for j < n.
j Step 4. If ψn(z), x = 0 for j < n, then ψn(z) is a scalar multiple of pn(x).
22 Step 1 is easy to establish because zn−j +zn+j is a polynomial of degree n+j < 2n, and so it is orthogonal to ϕ2n(z).
Next, note that ϕ2n(z) is a polynomial with real coefficients. Thus,
−1 ϕ2n(z) = ϕ2n(z) = ϕ2n(z ), and
n−j n+j n−j n+j −1 j−n −j−n 0 = ϕ2n(z), z + z = hϕ2n(z), z + z i = ϕ2n(z ), z + z .
Therefore,
n−j n+j −1 j−n −j−n 0 = ϕ2n(z), z + z + ϕ2n(z ), z + z
−n j −j n −1 j −j = z ϕ2n(z), z + z + z ϕ2n(z ), z + z
j −j = ψn(z), z + z , proving Step 2.
To prove the next step, consider the change of variables z = eiθ. Then zj + z−j =
2 cos(jt), and cos(jt) is a polynomial (of degree j) in cos θ, say
j X k Pj(cos θ) = νk cos t. k=0 So we have
j j −j X k ψn(z), z + z = 2 hψn(z),Pj(x)i = 2 νk ψn(z), x k=0
j −j If ψn(z), z + z = 0 for j = 1, then hψn(z), xi = 0. By induction on j, we see that ψn(z) is orthogonal to each polynomial in x of degree less than n and Step 3 is proved.
23 j Finally, we must show that ψn(z) is orthogonal to x with respect to w1(x). Notice
w(eiθ) = (1 − cos θ)α(1 + cos θ)β = (1 − cos θ)α−1/2(1 + cos θ)β−1/2| sin θ|
= w1(cos θ)| sin θ|.
−n Also, ψn(z) is real-valued because it is twice the real part of z ϕ2n(z), so we may write ψn(z) = ψn(x). Using Steps 1 and 3, Z j −j dz 0 = ψn(z)(z + z )w(z) ∂D z Z π iθ = 2i ψn(e ) cos(jt)w1(cos θ)| sin θ| dθ −π Z 1 = i ψn(x)Pj(x)w1(x) dx. −1
Since ψn(x) is orthogonal to a set of polynomials whose span is the set of polynomials of degree at most n − 1, we are done. The second equality follows from this one.
For the third equality, the argument is similar if we let
z−n−1ϕ (z) − zn+1ϕ (z−1) ψ (z) = 2n+2 2n+2 n z − z−1 and use the orthogonality condition
Z π sin(j + 1)t qn(cos θ) w2(cos θ)| sin θ| dθ = 0, −π sin θ and the fourth equality follows from this one.
So we are able to write each Jacobi polynomial pn, qn in terms of orthogonal polynomials on the unit circle. The converse is also true.
Theorem 9. (Szeg¨o,[19]) There exist constants An, Bn, Cn, Dn such that
−n −1 z ϕ2n(z) = Anpn(x) + Bn(z − z )qn−1(x) (2.4.1) −n+1 −1 z ϕ2n−1(z) = Cnpn(x) + Dn(z − z )qn−1(x).
24 Proof. From the first and third equalities of the previous theorem, with n−1 in place of n in the third equality,
an −1 −n pn(x) − (z − z )qn−1(x) = 2anz ϕ2n(z) cn −n 1 1 −1 z ϕ2n(z) = pn(x) − (z − z )qn−1(x). 2an 2cn A similar argument may be used to obtain the other equality.
25 2.5 Calculating the Inner Product
This section is devoted to calculating hf, gi for real polynomials f, g. Even though we do not give explicit formulas for the orthonormal polynomials, we give the formulas for the moments, which can be used to find the norm of ϕn, which determines Φn. It suffices to consider polynomials with real coefficients becasue all Jacobi polynomials have this property.
We develop a systematic procedure for practical calculations related to the inner product. Let ψn(z), ψm(z) be real, monic polynomials of degrees n, m with n ≥ m. Write
n X (n) j (n) ψn(z) = aj z , an = 1. j=0
We know that each coefficient of ψn is real, so we may write
X (m) (n) j−i ψn(z)ψm(z) = ai aj z 0≤i≤m 0≤j≤n Z Z X (m) (n) j−i ψn(z)ψm(z) dµ(z) = ai aj z dµ(z) ∂D ∂D 0≤i≤m 0≤j≤n Z X (m) (n) j−i = ai aj z dµ(z) 0≤i≤m ∂D 0≤j≤n Z X (m) (n) |j−i| = ai aj z dµ(z). 0≤i≤m ∂D 0≤j≤n
w(z) So to find hf, gi for given polynomials f, g, it suffices to find the integral of zk on z
26 k −k the unit circle for each k ∈ N (notice that the integrals of z and z are the same). Z 1 Z π zk dµ(z) = (cos kθ + i sin kθ)(1 − cos θ)α(1 + cos θ)β dθ ∂D 2π −π 1 Z π = cos kθ(1 − cos θ)α(1 + cos θ)β dθ 2π −π 1 Z π + i sin kθ(1 − cos θ)α(1 + cos θ)β dθ. 2π −π Consider the second integral on the right in the above line.
Z π Z π sin kθ(1 − cos θ)α(1 + cos θ)β dθ = sin kθ(1 − cos θ)α(1 + cos θ)β dθ = 0 −π −π because sin kθ is odd and the weight is even. Now let {Tk(θ): k ≥ 1} denote the Chebyshev polynomials of the first kind. Z 1 Z π zk dµ(z) = cos kθ(1 − cos θ)α(1 + cos θ)β dθ ∂D 2π −π Z π 1 α β = Tk(θ)(1 − cos θ) (1 + cos θ) dθ. 2π −π θ Using Weierstrass substitution with t = tan( ), define the 0th moment to be 2 Z 1 Z π T (α,β) = dµ = (1 − cos θ)α(1 + cos θ)β dθ ∂D 2π −π 1 Z ∞ 1 − t2 α 1 − t2 β 2 dt = 1 − 2 1 + 2 2 2π −∞ 1 + t 1 + t 1 + t 1 Z ∞ 2t2 α 2 β 2 dt = 2 2 2 2π −∞ 1 + t 1 + t 1 + t 2α+β Z ∞ t2α = 2 α+β+1 dt. π −∞ (1 + t ) Maple may be used to compute this integral, but perhaps more informative is the following calculation: begin by computing the first integral above:
Z π (1 − cos θ)α(1 + cos θ)β dθ. −π 27 One may verify with Mathematica (see also Abramowitz and Stegun [1]) that Z (1 − cos θ)α(1 + cos θ)β dθ =
θ −α−1/2 1 1 3 θ sin θ sin2 (1 − cos θ)α(1 + cos θ)β F − α, β + ; β + ; cos2 2 2 1 2 2 2 2 − , 2β + 1 where
∞ n X (a)n(b)n z F (a, b; c; z) = 2 1 (c) n! n=0 n is the hypergeometric function and
n−1 Y (x)n = x(x + 1) ··· (x + n − 1) = (x + j) j=0 is the Pochhammer symbol (for x ∈ C and n ∈ N \{0}, the Pochhammer symbol is defined as above and (x)0 = 1). Now evaluate the antiderivative at the endpoints.
θ −α−1/2 lim sin θ sin2 (1 − cos θ)α = lim sin 2θ(sin θ)−2α−1(1 − cos 2θ)α θ→0+ 2 θ→0+
= lim 2 sin θ cos θ(sin θ)−2α−1(1 − cos 2θ)α θ→0+ (1 − cos 2θ)α = 2 lim θ→0+ (sin θ)2α 1 − cos 2θα = 2 lim 2 θ→0+ sin θ 2 sin2 θα = 2 lim 2 θ→0+ sin θ = 2α+1
28 and for θ → 0+, Z lim (1 − cos θ)α(1 + cos θ)β dθ θ→0+ 1 1 3 θ 2α+1(1 + cos θ)β F − α, β + ; β + ; cos2 2 1 2 2 2 2 = lim − θ→0+ 2β + 1 1 1 3 2α+β+1 F − α, β + ; β + ; 1 2 1 2 2 2 = − 2β + 1 Γ α + 1 Γ β + 3 = −2α+β+1 2 2 . (2β + 1)Γ(α + β + 1)
The antiderivative for θ → 2π− is the same except for the sign because sin θ on (π, 2π) is negative, so
Z π Γ α + 1 Γ β + 3 (1 − cos θ)α(1 + cos θ)β dθ = 2α+β+2 2 2 −π (2β + 1)Γ(α + β + 1) 2α+β+1 Γ α + 1 Γ β + 3 T (α,β) = 2 2 . π (2β + 1)Γ(α + β + 1)
Then the problem of computing hf, gi is reduced to rewriting Tk(θ) as a sum of products of powers of (1 − cos θ), (1 + cos θ) because the problem of solving the equation of orthogonality will be reduced to solving equations with terms T (α+r,β+s) for r, s ≥ 0. So the next step is to find a method of computing all the moments.
(α,β) Suppose k ≥ 0 and that we know the number mk for all α, β ≥ 0, where
Z π (α,β) k α β mk = cos θ(1 − cos θ) (1 + cos θ) dθ. −π
Then we can find
Z π (α,β) k+1 α β mk+1 = cos θ(1 − cos θ) (1 + cos θ) dθ −π
29 by
cosk+1 θ = cos θ cosk θ = (1 − (1 − cos θ)) cosk θ = cosk θ − (1 − cos θ) cosk θ Z π (α,β) k+1 α β mk+1 = cos θ(1 − cos θ) (1 + cos θ) dθ −π Z π = cosk θ(1 − cos θ)α(1 + cos θ)β dθ −π Z π − (1 − cos θ) cosk θ(1 − cos θ)α(1 + cos θ)β dθ −π Z π = cosk θ(1 − cos θ)α(1 + cos θ)β dθ −π Z π k α+1 β (α,β) (α+1,β) − cos θ(1 − cos θ) (1 + cos θ) dθ = mk − mk . −π
(α,β) (α,β) (α+1,β) (α,β) Hence, mk+1 = mk − mk , and by induction, we can find mn for all n. The first three terms are 1 Γ α + 1 Γ β + 3 m(α,β) = T (α,β) = 2α+β+2 2 2 0 2π (2β + 1)Γ(α + β + 1) (β − α − 1)Γ α + 1 Γ β + 3 m(α,β) = m(α,β) − m(α+1,β) = 2α+β+2 2 2 1 0 0 (2β + 1)(α + β + 1)Γ(α + β + 1)
(α,β) (α,β) (α+1,β) m2 = m1 − m1 (α2 + β2 − αβ + 3α − β + 1)Γ α + 1 Γ β + 3 = 2α+β+2 2 2 . (2β + 1)(α + β + 1)Γ(α + β + 1)
Recall
Z 1 Z π zk dµ(z) = cos kθ(1 − cos θ)α(1 + cos θ)β dθ ∂D 2π −π and cos kθ is a polynomial in cos θ of degree k, precisely, the kth Chebyshev polyno- mial. The first three Chebyshev polynomials in x are 1, x, and 2x2 − 1, so the first
30 three moments are Z 2α+β+1 Γ α + 1 Γ β + 3 dµ(z) = 2 2 ∂D π (2β + 1)Γ(α + β + 1) Z 2α+β+1 (β − α − 1)Γ α + 1 Γ β + 3 z dµ(z) = 2 2 ∂D π (2β + 1)(α + β + 1)Γ(α + β + 1) Z 2 (α,β) (α,β) z dµ(z) = 2m2 − m0 ∂D 2α+β+2 (α2 + β2 − 2αβ + 2α − 2β)Γ α + 1 Γ β + 3 = 2 2 . π (2β + 1)(α + β + 1)Γ(α + β + 1)
Other moments are found similarly.
31 2.6 Formulas for Cn, Dn √ Throughout this section, we identify z with x + i 1 − x2. So any polynomial in z √ can be expressed as P (x) + i 1 − x2 Q(x) for some real polynomials P , Q.
Lemma 10. For all n, Cn and Dn ∈ R.
Proof. We know that for all n, √ −n+1 2 z ϕ2n−1(z) = Cnpn(x) + 2i 1 − x Dnqn−1(x).
We also know that each ϕn(z) is a polynomial with real coefficients. Let {tj} ⊂ R n X j such that ϕn(z) = tjz with tn = 1 (because ϕn is monic). Write j=0 √ −n+1 √ −n+1 2 2 z ϕ2n−1(z) = x + i 1 − x · ϕ2n−1 x + i 1 − x
2n−1 √ −n+1 √ j 2 X 2 = x + i 1 − x · tj x + i 1 − x j=0 2n−1 √ n−1 √ j 2 X 2 = x − i 1 − x · tj x + i 1 − x j=0 √ ˜ 2 ˜ = Pn(x) + i 1 − x · Qn(x), ˜ ˜ where Pn(x) and Qn(x) are real polynomials in x that satisfy the above line. Then √ √ ˜ 2 ˜ 2 Pn(x) + i 1 − x · Qn(x) = Cnpn(x) + 2i 1 − x Dnqn−1(x) √ ˜ 2 ˜ Pn(x) − Cnpn(x) = i 1 − x (2Dnqn−1(x) − Qn(x)). √ Now 1 − x2 cannot divide the left hand side unless each side is zero. In particular,
Cn and Dn are real.
˜ ˜ n ˜ For each n ≥ 1, there exist real polynomials Rn(x), Sn(x) such that z = Rn(x) + √ 2 ˜ ˜ ˜ i 1 − x · Sn(x). Let cn and dn be the leading coefficients of Rn(x) and Sn(x), respectively. 32 ˜ ˜ n−1 Lemma 11. For all n ≥ 1, deg(Rn) = n, deg(Sn) = n − 1, and cn = dn = 2 . ˜ ˜ Moreover, the degree and leading coefficient of Pn(x) are the same as those of Rn(x), ˜ ˜ and the degree and leading coefficient of Qn(x) are the same as those of Sn(x). √ Proof. We prove the first statement by way of induction. For n = 1, z = x+i 1 − x2, ˜ ˜ giving Rn(x) = x, Sn(x) = 1, and c1 = d1 = 1. So the base case is true. Fix k ≥ 1 and suppose the claim holds for k. Consider zk+1. √ √ k+1 k 2 ˜ 2 ˜ z = z · z = x + i 1 − x Rk(x) + i 1 − x · Sk(x) √ ˜ 2 ˜ ˜ 2 ˜ ˜ = xRk(x) + x Sk(x) − Sk(x) + i 1 − x Rk(x) + xSk(x) .
Therefore,
˜ ˜ 2 ˜ ˜ Rk+1(x) = xRk(x) + x Sk(x) − Sk(x) ˜ ˜ ˜ Sk+1(x) = Rk(x) + xSk(x).
˜ 2 ˜ By the inductive hypothesis, xRk(x) and x Sk(x) are degree k + 1 polynomials with k−1 ˜ leading coefficients ck = dk = 2 . Then Rk+1(x) is a degree k + 1 polynomial with k ˜ leading coefficient ck+1 = 2 . Similarly, Sk+1(x) is a degree k polynomial with leading
k coefficient dk+1 = 2 . Next, for all n,
√ 2n 2n−1 ˜ 2 ˜ X j n X j Pn(x) + i 1 − x · Qn(x) = tjz = z + tjz . j=0 j=0
˜ ˜ ˜ ˜ n For each j < 2n, deg(Rn) > deg(Rj) and deg(Sn) > deg(Sj), so z is the function in √ x and i 1 − x2 of highest degree and we are done.
Proposition 12. (Formula) For all n,
2n−1 2n−2 Cn = ,Dn = . (α + β + n)n (α + β + n + 1)n−1 33 In particular,
D α + β + n n = . Cn 2
Proof. Since
√ √ 2 ˜ 2 ˜ Cnpn(x) + 2i 1 − x Dnqn−1(x) = Pn(x) + i 1 − x · Qn(x)
and Cn is real,
˜ Cnpn(x) = Pn(x) ˜ Pn(x) cn Cn = = (n,n) , pn(x) χp
(n,k) th where χp is the k coefficient of pn(x). Lemma 2 gives the numerator and a result
(n,n) (n) of Costin [5] gives χp = (α + β + n) . Similarly, Dn ∈ R implies √ √ 2 2 ˜ 2i 1 − x · Dnqn−1(x) = i 1 − x · Qn(x) ˜ n−2 Qn(x) dn 2 Dn = = (n−1,n−1) = . 2qn−1 2χq (α + β + n + 1)n−1
34 2.7 Conjecture for An, Bn
2n 2n−1n An = ,Bn = . (α + β + n + 1)n (α + β + n + 1)n Write
√ −n ˜ 2 ˜ z ϕ2n(z) = An(x) + i 1 − x · Bn(x).
(n,k) (n,k) th ˜ ˜ Define a and b to be the k coefficient of An(x) and Bn(x), respectively, and
(n,k) (n,k) (n,k) a (n,k) b A = (n,k) ,B = (n−1,k) . χp 2χq
Then the coefficients of ϕ2n can be discovered by solving
A(n,n) = A(n,k)
B(n−1,n−1) = B(n−1,j) Z ϕ2n(z) dµ(z) = 0 ∂D for 0 ≤ k ≤ n − 1 and 0 ≤ j ≤ n − 2. For n = 1, 2, 3, 4, we get
2 1 A = ,B = 1 α + β + 2 1 α + β + 2 4 4 A = ,B = 2 (α + β + 3)(α + β + 4) 2 (α + β + 3)(α + β + 4) 8 12 A = ,B = 3 (α + β + 4)(α + β + 5)(α + β + 6) 3 (α + β + 4)(α + β + 5)(α + β + 6) 16 A = , 4 (α + β + 5)(α + β + 6)(α + β + 7)(α + β + 8) 32 B = . 4 (α + β + 5)(α + β + 6)(α + β + 7)(α + β + 8)
At n = 5, the problem is too computationally demanding for Maple to solve (due to solving hϕ10, 1i = 0).
35 One way to determine An and Bn is to find t0 = ϕ2n(0) for all n because
(n,n) n−1 (n,n) a 2 (1 + t0) An = A = (n,n) = (n,n) χp χp n−1 2 (1 + t0) Bn = (n,n) . 2χq For 1 ≤ n ≤ 8, we have
α + (−1)nβ ϕ (0) = . n α + β + n
For ϕn to be monic, we know Cn, Dn for n odd (Section 2.6). While the values of
An, Bn are conjecture, we know that
B n n = An 2 for all n (see Ismail [12]).
36 CHAPTER 3
NEW RESULTS
3.1 Commutation Relations
0 Let Ak = kQ +xL1 +L2 +Q∂x, as in Costin [5]. In the Jacobi scalar case, L1 = α+β and L2 = α − β. Converting to the variable z, we have ∂ ∂z ∂ 2z ∂ = = ∂x ∂x ∂z z − z−1 ∂z ! z + z−1 2 2z (x2 − 1)∂ = − 1 ∂ x 2 z − z−1 z z2 − 1 = ∂ 2 z 2 Ak = k(2x) + (α + β)x + (α − β) + (x − 1)∂x
2 = (2k + α + β)x + α − β + (x − 1)∂x z + z−1 z2 − 1 A = (2k + α + β) + α − β + ∂ . k 2 2 z
Now, this expression for Ak holds for given constants α, β. However, when converting 1 1 to the unit circle, the real Jacobi polynomials depend on α − , β − (see Section 2 2 2.4). Therefore, we write
z + z−1 z2 − 1 A = (2k + α + β − 1) + α − β + ∂ . k 2 2 z
37 If r is a function of z, then
z + z−1 z2 − 1 A r = (2k + α + β − 1) + α − β + ∂ r k 2 2 z z + z−1 z2 − 1 ∂ = (2k + α + β − 1) r + (α − β)r + r 2 2 ∂z z2 + 1 z(z2 − 1) ∂ zA r = (2k + α + β − 1) r + (α − β)zr + r k 2 2 ∂z z2 + 1 z2 − 1 z ∂ A (zr) = (2k + α + β − 1) r + (α − β)zr + r + r k 2 2 ∂z z2 − 1 zA r = A (zr) − r. k k 2 Other relations are found also with straightforward calculation. A summary of some important relations is given below.
n znA r = A (znr) − (z2 − 1)zn−1r 1 1 2 2 + α + β ∂ A r = [A + z] r0 + (1 − z−2)r z 1 1 2 n n n−1 z ∂zr = ∂z(z r) − nz r 1 1 2zr ∂ r = ∂ r + . z2 − 1 z z z2 − 1 (z2 − 1)2
These relations are useful when writing differential equations for Jacobi polynomials on the unit circle.
38 3.2 Rodrigues’ Formula
Our new result is the following Rodrigues’ formula for Jacobi polynomials on the unit circle.
Proposition 13. For each n ≥ 1,
n ϕ2n(z) = AnA1(z A2 ... An1) α + β + 1 ϕ (z) = C A zn−1 + zn−2(z2 − 1) A ... A 1. 2n−1 n 1 2 2 n
n−1 Note that the term A1z is an operator, not a polynomial.
Proof. We know for all n,
z + z−1 z + z−1 z−nϕ (z) = A p + B z − z−1 q 2n n n 2 n n−1 2 z + z−1 p = A ... A 1 n 2 1 n z + z−1 q = A ... A 1 n−1 2 2 n B n n = An 2 n znA r = A (znr) − (z2 − 1)zn−1r. 1 1 2
Write r = A2 ... An1. This information gives
−n −1 z ϕ2n(z) = AnA1 ... An1 + Bn z − z r n = A A ... A 1 + z − z−1 r n 1 n 2 n ϕ (z) = A znA ... A 1 + z2 − 1 zn−1r 2n n 1 n 2 n n = A A (znr) − zn−1(z2 − 1)r + zn−1 z2 − 1 r n 1 2 2 n = AnA1(z r).
39 For the second equality, we use
z + z−1 z + z−1 z−n+1ϕ (z) = C p + D z − z−1 q 2n−1 n n 2 n n−1 2 D α + β + n n = . Cn 2 We get
α + β + n ϕ = C zn−1A ... A 1 + C (z2 − 1)zn−2r 2n−1 n 1 n 2 n n − 1 α + β + n = C A (zn−1r) − zn−2(z2 − 1)r + zn−2(z2 − 1)r n 1 2 2 α + β + 1 = C A (zn−1r) + zn−2(z2 − 1)r . n 1 2
40 CHAPTER 4
EXAMPLES OF JACOBI POLYNOMIALS ON THE UNIT
CIRCLE
In this chapter, we find a few orthogonal polynomials in specific cases, using three different approaches, and we will compare each approach. The first method is the classical one: the Gram-Schmidt process. Then we will use the connection between
OPRL and OPUC (Theorem 9). The final method will utilize our new result (Propo- sition 13).
4.1 Using the Gram-Schmidt Process
1 Example 1. α = β = . 2 The inner product here is
Z π √ √ Z π 1 iθ −iθ 1 iθ iθ hf, gi1 = f(e )g(e ) 1 − cos θ 1 + cos θ dt = f(e )g(e )| sin(t)| dθ. 2π −π 2π −π
For each n, ϕn(z) is found by computing
n−1 n n−1 it int n X hϕk(z), z i n X hϕk(e ), e i1 ϕn(z) = z − 2 ϕk(z) = z − 2 ϕk(z). kϕkk kϕkk k=0 k=0
41 Then the first six polynomials are
ϕ1(z) = z 1 ϕ (z) = z2 + 2 3 1 ϕ (z) = z3 + z 3 3 2 1 ϕ (z) = z4 + z2 + 4 5 5 2 1 ϕ (z) = z5 + z3 + z 5 5 5 3 9 1 ϕ (z) = z6 + z4 + z2 + . 6 7 35 7 Example 2. α = β = 1.
The inner product is
Z π Z π 1 iθ iθ 1 iθ −iθ 2 hf, gi2 = f(e )g(e )(1 − cos θ)(1 + cos θ) dt = f(e )g(e ) sin t dθ, 2π −π 2π −π and the first six polynomials are
ϕ1(z) = z 1 ϕ (z) = z2 + 2 2 1 ϕ (z) = z3 + z 3 2 2 1 ϕ (z) = z4 + z2 + 4 3 3 2 1 ϕ (z) = z5 + z3 + z 5 3 3 3 1 1 ϕ (z) = z6 + z4 + z2 + . 6 4 2 4 3 1 Example 3. α = , β = . 2 2 The inner product is
Z π 1 iθ −iθ hf, gi3 = f(e )g(e )(1 − cos θ)| sin θ| dθ. 2π −π 42 The first six polynomials are
1 ϕ (z) = z + 1 3 1 1 ϕ (z) = z2 + z + 2 2 2 3 3 1 ϕ (z) = z3 + z2 + z + 3 5 5 5 2 4 2 1 ϕ (z) = z4 + z3 + z2 + z + 4 3 5 5 3 5 6 18 3 1 ϕ (z) = z5 + z4 + z3 + z2 + z + 5 7 7 35 7 7 3 27 9 9 9 1 ϕ (z) = z6 + z5 + z4 + z3 + z2 + z + . 6 4 28 14 14 28 4
4.2 Using Theorem 9
Example 1. Finding ϕ1(z) reduces to finding C1, D1, p1(x), and q0(x). If C1 and
D1 are known, then an easy way to arrive at ϕ1(z) is to use (2.4.1) (in Theorem 9).
And if C1, D1 are not known, then some elementary algebra may be used to discover
ϕ1. For n = 1, we have √ 2 ϕ1(z) = C1p1(x) + 2i 1 − x D1q0(x) √ 2 = C1(2x) + 2i 1 − x D1.
We can rewrite everything in terms of z and write ϕ1(z) = z + a0 for some constant a0.
−1 −1 −1 z + a0 = C1(z + z ) + D1(z − z ) = (C1 + D1)z + (C1 − D1)z .
1 Equating coefficients gives C = D = , and therefore, a = 0 and ϕ (z) = z. In 1 1 2 0 1 general, finding the orthogonal polynomials of even degree (using this approach) is
43 much more difficult. Consider ϕ2(z), so we must find A1, B1 (we already have p1(x)
2 and q0(x)). Write ϕ2(z) = z + b1z + b0.
−1 −1 −1 z ϕ2(z) = z + b1 + b0z = (A1 + B1)z + (A1 − B1)z .
The difficulty of finding orthogonal polynomials of even degree is that there is one fewer equation given than there is when finding those of odd degree. Finding ϕ1 amounted to solving three equations with three unknown variables, but finding ϕ2 involves solving two equations with three variables. In particular, ϕ1(z) required
−1 −1 C1 − D1 = 0 because the left hand side had no z term. But because z ϕ2(z) does, we would have to know b0 to find ϕ2(z). So we must add an equation for which to solve without introducing new variables. We may use the orthogonal condition, that is, solve for hϕ2, 1i = 0 (we also have the option of considering hϕ2, ϕ1i = 0). Z π 2iθ iθ 4 hϕ2, 1i = e + b1e + b0 | sin θ| dθ = 4b0 − = 0 −π 3 1 b = . 0 3 1 Then ϕ (z) = z2 + . In this case, the only way to find A , B (which are desired 2 3 1 1 to find ϕ2(z)) is first to find ϕ2. For n ≥ 2, at least some knowledge of An, Bn is crucial for finding ϕ2n(z) if we want to use this approach. Note that knowledge of A n suffices, and we know this ratio, given in Section 2.7. Bn
4.3 Using the New Rodrigues’ Formula
1 Example 1. If α = β = , then 2 z + z−1 z2 − 1 A = 2k + ∂ . k 2 2 z 44 Then the first three orthogonal polynomals are
1 z + z−1 ϕ (z) = C A + z−1(z2 − 1) 1 = 2 + z − z−1 = z 1 1 1 2 2 2 z + z−1 z2 − 1 1 ϕ (z) = A A z = 2z + = z2 + 2 1 1 3 2 2 3
2 ϕ3(z) = C2 A1A21 + (z − 1)A21 1 = A 2z + 2z−1 + (z2 − 1)(2z + 2z−1) . 6 1 Even by the third polynomial, the calculation becomes tedious by hand, but with the assistance of a computer program, we see that this approach is much more efficient
(results are included in Chapter 5).
4.4 Comparison of Formulas
Consider first the first two methods. An advantage of the method in 4.2 is that ϕ2n−1 is no more difficult to find for general α, β because we never need to use the inner product. However, the even case is different: the easiest way of finding ϕ2n (with this method) is by setting up appropriate equalities with An, Bn, the coefficients of ϕ2n, and solving hϕ2n, 1i = 0. But the results of Sections 2.6, 2.7 (which give formulas for
An, Bn, Cn, Dn) yield the orthogonal polynomials with much more ease. Revisiting
Example 1, ϕ2 can be found by using A1, B1, a conjecture for which was given in Section 2.7, giving
2n 2 2n−1n 1 A1 = = ,B1 = = (α + β + n + 1)n 3 (α + β + n + 1)n 3 and
1 ϕ (z) = (A + B )z2 + (A − B ) = z2 + . 2 1 1 1 1 3 45 Using the Gram-Schmidt Process requires various integrals to be computed. We saw in Section 2.5 that for arbitrary α, β, the formula for the integral involves the hypergeometric function. Therefore, the methods outlined in 4.2 and 4.3 are desir- able to find ϕn. If An, Bn, Cn, Dn are known, then the latter two methods seem comparable. However, from a programming perspective, the new Rodrigues’ Formula is preferable, as will be seen in Chapter 5.
46 CHAPTER 5
CODES AND COMPUTER CALCULATION RESULTS
5.1 OPRL
The Rodrigues formula gives the Jacobi polynomials:
n 1 d n pn(x) = n [Q(x) wr(x)] . (5.1.1) wr(x) dx The code used is
> QQ := x2 − 1 : WW := (1 − x)α(1 + x)β: diff (QQ nWW , x$n) > pp := (n, x) → collect simplify , x ; WW
Theorem 7 gives the formula
pn(x) = A1 ... An1. (5.1.2)
The code used is
> xA := (k, f ) → simplify (2k + α + β) · x · f + (α − β) · f + (x 2 − 1 ) · diff (f , x);
> xA1 := proc(k, f)
description “apply recursive operator to obtain Jacobi polynomial (RL)”;
option remember:
if k = 1 then 47 return xA(1 , f );
else
return xA1 (k − 1 , xA(k, f ));
end if: k = 1;
end proc;
The times (in seconds) to compute the nth Jacobi polynomial (2 ≤ n ≤ 25) with each method are given in Table 1. The table shows that using xA1 is computation- ally less expensive. pp is relatively inefficient to use because the formula requires several applications of differentiation of (x2 − 1)2(1 − x)α(1 + x)β, which requires the product rule of differentiation. Alternatively, equation (5.1.2) involves differentiating polynomials, requiring only the power rule (along with multiplication by polynomials, which are not differentiated). Therefore, the disparity in computation times among the first few Jacobi polynomials is minor because the product rule is applied only a few times. But for larger values of n, we see that computing pn(x) via equation (5.1.2) is preferable.
5.2 OPUC
As we mentioned, the problem of determining ϕn for general α, β using the Gram- Schmidt Process is very difficult due to the hypergeometric function, so consider 1 α = β = (as in Example 1). Finding each ϕ means computing 2 n
n−1 n X hϕj(z), z i ϕ (z) = zn − ϕ (z) (5.2.1) n kϕ k2 j j=1 j
48 Table 1: Comparison of Rodrigues’ formulas on the real line
Polynomial Degree Using pp Using xA1 Degree Using pp Using xA1
2 0.026 < 0.001 14 10.961 0.447
3 0.027 0.001 15 16.718 0.522
4 0.039 0.003 16 25.005 0.734
5 0.083 0.008 17 36.911 0.900
6 0.181 0.016 18 52.623 1.110
7 0.338 0.031 19 73.989 1.427
8 0.608 0.046 20 97.395 1.796
9 1.003 0.070 21 122.850 2.179
10 1.640 0.124 22 159.331 2.637
11 2.725 0.164 23 >240 3.153
12 4.290 0.269 24 >240 3.728
13 6.785 0.311 25 >240 4.480
49 with the inner product
1 Z π f(eiθ)g(eiθ)| sin(t)| dθ. 2π −π 1 > inprod := · int(f(t) · conjugate(g(t)) · abs(sin(t)), t = −Pi..Pi); 2 · Pi > c0p := subs(f(t) = 1, g(t) = 1, inprod);
Here, c0p = kϕ0k. Then tp1 = ϕ1 and c1p = kϕ1k are
subs(f(t) = exp(I · t), g(t) = 1, inprod) > tp1 := z − ; c0p > c1p := subs(f(t) = exp(I · t), g(t) = exp(I · t), inprod);
In this way, we can find all the orthogonal polynomials. But this method is very inefficient (we do not consider a closed form of this program or computation times of obtaining these polynomials with this method).
Recall the new formula we found is
n ϕ2n(z) = AnA1(z A2 ... An1) α + β + 1 ϕ (z) = C A zn−1 + zn−2(z2 − 1) A ... A 1, 2n−1 n 1 2 2 n where
2n−1 2n Cn = ,An = . (α + β + n)n (α + β + n + 1)n
These constants may be coded as
2k−1 > CC := (α, β, k) → factor expand ; pochhammer(α + β + k, k) 2k > AA := (α, β, k) → factor expand ; pochhammer(α + β + k + 1, k)
Then each ϕn can be written with the procedures 50 > # Defining the operator to find the monic orthogonal polynomials
> zA := (α, β, k, f ) → z + z−1 z2 − 1 simplify (2 · k + α + β) · · f + (α − β) · f + · diff (f , z) 2 2
> # The operator A1 ... Ak > zA1 := proc(k, f)
description “apply recursive operator to obtain Jacobi polynomial (UC,
starting with A1)”; option remember:
if k = 1 then
return zA(α, β, 1, f);
else
return zA1(α, β, k − 1, zA(k, f ));
end if: k = 1;
end proc;
> # The operator A2 ... Ak > zA2 := proc(k, f)
description “apply recursive operator to obtain Jacobi polynomial (UC,
starting with A2)”; option remember:
if k = 1 then
return zA(α, β, 2, f);
else
return zA2(α, β, k − 1, zA(k, f ));
end if: k = 2;
51 end proc;
> # Define ϕ2k for k ≥ 2 > PHIEVEN := (α, β, k) → 1 1 1 1 collect AA(α, β, k) · zA α − , β − , 1, zk · zA2 α − , β − , k, 1 , 2 2 2 2 z, factor)
> # Define ϕ2k−1 for k ≥ 2 > PHIODD := (α, β, k) → 1 1 1 1 collect CC (α, β, k) · zA α − , β − , 1, zk−1 · zA2 α − , β − , k, 1 2 2 2 2 α + β + 1 1 1 + · zA2 α − , β − , k, 1 , z, factor 2 2 2
> # Define ϕk > PHI := proc(k)
description “gives phi k”;
option remember:
if k = 1 then α − β return z + ; α + β + 1 elif k = 2 then 2(α − β)z α + β return z2 + + ; α + β + 2 α + β + 2 elif type(k, even) then k return PHIEVEN α, β, ; 2 else k + 1 return PHIODD α, β, ; 2 end if;
end proc;
52 For simplicity, we explicitly define ϕ1 and ϕ2 above. Table 2 displays computation times needed to compute ϕn for 1 ≤ n ≤ 25 using the program PHI . First, the first two polynomials are found immediately because they were hard- coded into PHI . Another note to consider is that not all computation times increase.
For example, ϕ14 is discovered in less time than ϕ13 is. But it is reasonable to expect the even degree polynomials are easier to compute because PHIEVEN is a simpler function.
Table 2 demonstrates that the new Rodrigues’ formula is very efficient for symbolic calculation. In fact, OPUC of a given degree seem to be simpler to compute than
OPRL, as the times in Table 2 are much faster than those in Table 1. The reason for this difference is, for a given n, xA1 executes n steps. By contrast, if n is even, n then PHI uses PHIEVEN , which executes − 1 steps (similarly, if n is odd, then 2 n + 1 − 1 steps are executed), so even though PHI is a more intricate program, it 2 applies fewer recursive steps.
As a result of PHI , discovering a list of some monic orthogonal polynomials is an easy task. A few results are shown after Table 2.
53 Table 2: Computation times using the new Rodrigues’ formula
Polynomial Degree Using PHI 13 0.132
1 <0.001 14 0.126
2 <0.001 15 0.214
3 0.005 16 0.207
4 0.004 17 0.300
5 0.013 18 0.245
6 0.017 19 0.472
7 0.029 20 0.371
8 0.029 21 0.682
9 0.050 22 0.450
10 0.052 23 0.967
11 0.086 24 0.656
12 0.111 25 1.203
54 ϕ0(z) = 1
α − β ϕ1(z) = z + α + β + 1
2 (2α − 2β) z α + β ϕ2(z) = z + + α + β + 2 α + β + 2 2 2 2 3 (α − β) z 3α + 3α − 2αβ + 3 β + 3β z α − β ϕ3(z) = z + 3 + + α + 3 + β (α + β + 3) (α + β + 2) α + β + 3 3 2 2 2 4 (α − β) z 3α + 3α − 2αβ + 3β + 3β z (α + β + 1) (α − β) z α + β ϕ4(z) = z + 4 + 2 + 4 + α + 4 + β (α + β + 4) (α + β + 3) (α + β + 4) (α + β + 3) α + β + 4 4 2 2 3 2 2 2 2 2 5 (α − β) z 5α − 6αβ + 5α + 5β + 5β z (α − β) 5α + 15α + 2αβ + 5 β + 10 + 15β z 5α − 6αβ + 5α + 5β + 5β z α − β ϕ5(z) = z + 5 + 2 + 2 + + α + 5 + β (α + β + 4) (α + β + 5) (α + β + 4) (α + β + 5) (α + β + 3) (α + β + 4) (α + β + 5) α + β + 5 5 2 2 4 2 2 3 2 2 2 6 (α − β) z 5α − 6αβ + 5α + 5β + 5β z (α − β) 5α + 15α + 2αβ + 5 β + 10 + 15β z (α + β + 2) 5 α − 6αβ + 5α + 5β + 5β z ϕ6(z) = z + 6 + 3 + 4 + 3 α + 6 + β (α + β + 5) (α + β + 6) (α + β + 5) (α + β + 6) (α + β + 4) (α + β + 5) (α + 6 + β)(α + β + 4)
55 (α + β + 1) (α − β) z α + β + 6 + (α + 5 + β)(α + β + 6) α + β + 6 6 2 2 5 2 2 4 7 (α − β) z 7α − 10αβ + 7α + 7β + 7β z (α − β) 7α + 21α − 2αβ + 14 + 7β + 21β z ϕ7(z) = z + 7 + 3 + 5 α + 7 + β (α + β + 6) (α + β + 7) (α + β + 5) (α + β + 6) (α + β + 7) 210α + 210β + 385α2 − 62αβ + 385β2 − 42α2β − 42α β2 − 20α3β + 18α2β2 − 20 αβ3 + 35β4 + 210α3 + 210β3 + 35α4 z3 + (α + β + 5) (α + β + 6) (α + β + 7) (α + β + 4) (α − β) 7α2 + 21α − 2αβ + 14 + 7β2 + 21β z2 7α2 − 10αβ + 7α + 7β + 7 β2 z α − β + 3 + + (α + β + 5) (α + β + 6) (α + β + 7) (α + β + 6) (α + β + 7) α + β + 7 7 2 2 6 2 2 5 8 (α − β) z 7α − 10αβ + 7α + 7β + 7β z (α − β) 7α + 21α − 2αβ + 14 + 7β + 21β z ϕ8(z) = z + 8 + 4 + 8 α + 8 + β (α + β + 7) (α + β + 8) (α + β + 6) (α + β + 7) (α + β + 8) 210α + 210β + 385α2 − 62αβ + 385β2 − 42α2β − 42αβ2 − 20α3β + 18α2β2 − 20 αβ3 + 35β4 + 210α3 + 210β3 + 35α4 z4 + 2 (α + β + 5) (α + β + 8) (α + β + 7) (α + β + 6) (α − β)(α + β + 3) 7α2 + 21α − 2 αβ + 14 + 7β2 + 21β z3 (α + β + 2) 7α2 − 10αβ + 7 α + 7β + 7β2 z2 (α + β + 1) (α − β) z α + β + 8 + 4 + 8 + (α + β + 5) (α + β + 8) (α + 7 + β)(α + β + 6) (α + β + 6) (α + β + 7) (α + β + 8) (α + β + 7) (α + β + 8) α + β + 8 8 2 2 7 2 2 6 9 (α − β) z 9α + 9α − 14αβ + 9β + 9β z (α − β) 3α + 9α − 2αβ + 6 + 3 β + 9β z ϕ9(z) = z + 9 + 4 + 28 α + 9 + β (α + β + 8) (α + β + 9) (α + β + 8) (α + β + 9) (α + β + 7) 21α4 + 126α3 − 28α3β − 54α2β + 231α2 + 30α2β2 + 126α − 54αβ2 − 82αβ − 28α β3 + 126β + 231β2 + 126β3 + 21β4 z5 + 6 (α + β + 8) (α + 9 + β)(α + β + 6) (α + β + 7) (α − β) 63 α4 + 630α3 + 28α3β + 2205α2 + 58α2β2 + 450α2β + 28α β3 + 3150α + 1850αβ + 450αβ2 + 2205 β2 + 63β4 + 630β3 + 1512 + 3150β z4 + 2 (α + β + 5) (α + β + 6) (α + β + 7) (α + β + 9) (α + β + 8) 21 α4 + 126α3 − 28α3β − 54α2β + 231α2 + 30α2β2 + 126α − 54 αβ2 − 82αβ − 28αβ3 + 126 β + 231β2 + 126β3 + 21β4 z3 + 4 (α + β + 8) (α + β + 9) (α + β + 6) (α + β + 7) (α − β) 3α2 + 9α − 2αβ + 6 + 3β2 + 9β z2 9α2 + 9α − 14αβ + 9β2 + 9β z α − β + 12 + + (α + β + 8) (α + β + 9) (α + 7 + β) (α + β + 8) (α + β + 9) α + β + 9 9 2 2 8 2 2 7 10 (α − β) z 9α + 9α − 14αβ + 9β + 9β z (α − β) 3α + 9α − 2αβ + 6 + 3β + 9β z ϕ10(z) = z + 10 + 5 + 40 α + 10 + β (α + β + 9) (α + β + 10) (α + β + 8) (α + β + 9) (α + β + 10) 21α4 + 126α3 − 28α3β − 54α2β + 231α2 + 30 α2β2 + 126α − 54αβ2 − 82αβ − 28αβ3 + 126β + 231β2 + 126 β3 + 21β4 z6 + 10 (α + β + 8) (α + β + 9) (α + β + 7) (α + β + 10)
56 (α − β) 63α4 + 630α3 + 28α3β + 2205α2 + 58α2β2 + 450α2β + 28αβ3 + 3150α + 1850αβ + 450αβ2 + 2205β2 + 63β4 + 630β3 + 1512 + 3150β z5 + 4 (α + β + 6) (α + β + 10) (α + β + 9) (α + β + 8) (α + β + 7) (α + β + 4) 21α4 + 126 α3 − 28α3β − 54α2β + 231α2 + 30α2β2 + 126α − 54αβ2 − 82αβ − 28αβ3 + 126β + 231β2 + 126β3 + 21β4 z4 + 10 (α + 6 + β)(α + β + 10) (α + β + 9) (α + β + 8) (α + β + 7) (α − β)(α + 3 + β) 3α2 + 9α − 2αβ + 6 + 3 β2 + 9β z3 (α + 2 + β) 9α2 + 9α − 14αβ + 9β2 + 9 β z2 (α + β + 1) (α − β) z α + β + 40 + 5 + 10 + (α + β + 8) (α + β + 9) (α + β + 7) (α + β + 10) (α + β + 8) (α + 9 + β)(α + β + 10) (α + 9 + β)(α + β + 10) α + β + 10 BIBLIOGRAPHY
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