———— I 3
n \ A H .‘m ‘ .w... u u...” .

 

 

........

THE onmoeomi. POLYNOMIALSAASSQQATEEW: j

ms mama»: or ASECOND "DEGREEIEPQLYNQMMLETW Ti"

. TheiSi's fdr the’Degreevfbf‘lPhLDQ ;ii * “
,‘ MICHIGAN STATE Unwmsm : .. L
.‘DANIELARAC NUSSBAUM‘ V -'
' -. 1971 ' _ ’

 

 

      
     

37"”? !

LIP"
Michigan its at:
University

«a 911.1“

This is to certify that the
thesis entitled

The orthogonal polynomials
associated with the iteration

of a second degree polynomial.
presented by

Daniel Arac Nussbaum
has been accepted towards fulfillment

of the requirements for

PhD Jggee in mathematic S

in MM

Major professor

Date ”It“ AZ; /i7/

0-7639

/.A he] ff

#7 ’3
(7 -¥

0

ABSTRACT

THE ORTHOGONAL POLYNOMIALS ASSOCIATED WITH
THE ITERAIION OF A SECOND DEGREE POIXNOMIAL

BY

Daniel Arac Nussbaum

We consider the polynomial P(z) a 22 - p and its iterates
Pn(z) defined by Ph(z) - P(Ph_1(z)). Define F as the set of
points for which {Pn(z)} is not a normal family and p as the
equilibrium distribution for F. These definitions are due to
Brolin [1].

Kinney and Pitcher have shown in [4] that the family of
iterates of P(z) form an orthogonal family over F with respect
to the measure u in the sense that long) 635’ sz) - o
if n # m. Since Pn(z) has degree 2“, we have a lacunary set
of orthogonal polynomials. In this paper we attempt to complete
the set by finding polynomials Quiz) of degree n, with leading
coefficient 1, for which i Qn(z) 6;?Ey'dp(z) = 0 for n ¢ m

and so that Q n(z) = Ph(z). As a matter of convenience we

2 m
introduce z<n>’ where z<n> = n (P1(z))e(i’n), where
a i=0
2 e(i,n)21 is the binary expansion of n.

1-0
We achieve a matrix representation for Qn(z) and obtain

results about the matrix entries. Specifically, we obtain a

closed form for calculating the matrix entries.

We show that the Qn(z) satisfy a linear difference equa-
tion which depends only on n, n-1, n-2 and a constant, K(n),
which depends only on n and p. In the case that F is a sub-
set of the reals (p 2 2 is sufficient) we obtain non-linear
recursion formulas for K(n). We also relate these constants to
a function which represents the ratio of successive Q's.

We also show that Qn(z) can be expanded as a linear
combination of z<0>, z<1>, z<2>,..., and we obtain results about
the coefficients in this expansion.

Finally, we extend the orthogonality of the family
{Qn(z)} to certain sets which contain F. These sets are level

curves of a Green's function, and the measure we use is the normal

derivative of i log ‘2 - w‘du(w)\dz|.

THE ORTHOGONAL POLYNOMIALS ASSOCIATED WITH
THE ITERATION OF A SECOND DEGREE POIXNOMIAL

BY

Daniel Arac Nussbaum

A THESIS
Submitted t 0
Michigan State University

in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1971

 

TO BEVERLY

ii

ACKNOWIEDGEMENTS

I am deeply grateful to Dr. John R. Kinney for his sugges-
tion of this investigation and for his patience during the comple-
tion of this paper.

I also wish to express my gratitude to my wife and my

parents for their encouragement and support throughout my studies.

iii

TABLE OF CONTENTS
Page
INTRODUCTION ..................................... 1
Chapter
I COMPUTATION OF MOMENTS ........................... 4

II DERIVATION OF THE DIFFERENCE EQUATION OF THE
ORTHOGONAL POLYNOMIALS ........................... 22

II I COM HITATION OF mLYNOMIAI-S O O O O O O O O O O O O O O O O O O O O O O O 3 2
IV ORTHOGONALITY WITH RESPECT TO DIFFERENT MEASURES . 38
V EWPIES 00....0.0000000000000000000.0.0....00.... 42

BIBLIOGRAHIY O0.0...00......OOOOOOOOOOOOOOOOOOOOOO 48

iv

INTRODUCTION

This paper deals with the set of polynomials which are
orthogonal with respect to a certain measure.

Consider a polynomial P(z) of degree N and its iterates
Pn(z) given by Pn(z) = P(Pn_1(z)). Brolin [1] has discussed
the set of points F where the family {Pn(z)} is not a normal
family. Thus, for z é F and for every neighborhood, N, of 2,
there is a subsequence of {Pn(z)} which converges uniformly to
an analytic function on compact subsets of N. Brolin shows that
F is compact, contains no open set, is invariant under P and
P_1, and has capacity one. Thus
exp{-inf(l[1£ 10g Tfil dv(z)dv(w))} = 1 where the infiuum is taken
over all measures v for which £ do 8 1. We denote the measure
which minimizes the above energy integral by p. It turns out
that p is invariant under P_1 (the inverse of P). That is,
du<z> = du(P_1(Z))o

In [4], Kinney and Pitcher show that {Ph} form an
orthogonal family in the sense that i Ph(z) P;?;7'dp(z) - 0 if
n and m are unequal.

In this paper we seek methods to complete this lacunary
set of orthogonal functions. That is, we seek polynomials, Quiz),

with leading coefficient 1 so that Q n(z) - Ph(z) and
N

fiQn(z) Qm‘z) dp(z) B 0 for n f m.

While we could attempt the above problem for an arbitrary
polynomial P(z), for the sake of brevity we restrict our attention
to the case N I 2. For reasons given in Brolin [1] and Kinney-
Pitcher [4], we may take P(z) B 22 - p. We know from Brolin [1]
that F is a real set if p 2 2.

In the first chapter we derive a representation of our
orthogonal polynomials. Using a technique developed by Walsh

[9] (see also Fine [2]) in his completion of the Rademacher func-

on
tions, we introduce z:l = H (Pi(z))e(1’n), where
Q i=0
n = 2 e(i,n)2i is the binary expansion of n. The powers z<n>

i=0
give rise to an inner product matrix, Ck where c(m,n) =

 

i z<m>'z<n> dp(z) for m,n = 0,1,... .

We derive some results about the matrix CL In particular
we find a closed form for evaluating c(m,n) (Theorem 1.12), and
a result about entries c(m,n) where ml+ n is constant (Theorem
1.36).

We use c(m,n) to construct a matrix representation of our
orthogonal polynomials (Theorem 1.57) essentially as Szego does in
[5] and [7]. We do not use true moments in our matrix, and we do
not use arc length in our measure until Chapter IV.

In Chapter II we approach the problem of completing the
orthogonal family differently. We find that the Qn(z)'s satisfy
a linear difference equation (Theorem 2.8), and that the coefficients
in the equation satisfy non-linear recursion formulae (Theorem 2.9)
in the case that F is a subset of the reals. We also obtain
continued fraction representations for the coefficients (Theorem

2.13), and a link between the methods of Chapters I and II

(Theorem 2.16).

In Chapter II we also consider Tn(z), the ratio of Qn(z)
to Qn-1(z) and show it is related to the coefficients of the
linear difference equation (Lemma 2.20).

In Chapter III we construct Qn(z) as a linear combina-

tion of the z: '

s. We establish recursion formulae among the
coefficients and in so doing we link this method with Chapter II.

In Chapter IV we extend the orthogonality of the family
{Qn(z)} to sets other than F. Specifically we show in Theorem
4.9 that Qn(z) and Qm(z) are orthogonal on the level curves
of a Green's function with reSpect to the measure
31:]: log \z - w‘dp,(w)-‘dz\.

Using the results of the previous chapters, we present
two examples. The first, P(z) = 22, for which the F set is
the unit circle.

The second example we present is P(z) = 22 - 2 for which
the real interval [-2,2] is the F set. We show that the
Tchebysheff polynomials Tn(z) = 2 Cos n Cos.1 z/2 are orthogonal
on [-2,2] with respect to p, and we derive a 1934 result of
Walsh's on the orthogonality of Tchebysheff polynomials on certain

confocal ellipses.

CHAPTER I”

Let P(z) = 22 - p and its iterates {Pn(z)] be defined
as follows: Po(z) = z, P1(z) = P(z), and the iterates Pn(z) are

recursively defined by

(1.1) Pn(Z) = P(Pn_1(2)>o

Throughout, F denotes the set of points in the plane at
which {Pn(z)} is not a normal family. Brolin shows in [1] that
when P(z) is as above than F is compact, has capacity one, is
invariant under P(z) and P_1(z), and often has 2-dimensional
measure zero.

As Brolin does in [l], we introduce the measure p on F
as follows:

(1.2) Let 20 be any point in the plane with at most 2 excep-
tions. We define ”n to be the discrete measure which plaCes

weight 2-n at each of the roots Pn(z) = 20; that is, at the

2n points if if : ... i /p + 20 . Then pn converges

weakly to u, where p is independent of 20 [1; Theorem 16.1].

 

That is, for f continuous and zero outside a compact set con-
taining F, 11ml; f dun = g f du. We know that p is concentrated
on F and qup = 1, [1]. Moreover, p is symmetric with respect
to the origin by its construction. Thus for f(z) an odd function

of z,

(1.3) g f(2)du(2) = o .

It is known [4; pg. 25] that u is invariant under P 1(z).
Together with the invariance of F under P_1(z), this invariance

yields the following relation:

(1.4) i f<z>ds<z> = ] f(P_1(Z))du(Z) .

We will use this "shift" extensively.
Walsh [9] completes the Rademacher functions with what
have come to be known as Walsh (or Walsh-Rademacher) functions.

His technique, described also by Fine in [2], leads us to make the

Q
following definition: for n = 2 e(i,n)21, where e(i,n) = 0
i=0
or 1 we take
<n> m (i n)
(1'5) 2 = n (Pi(z))€ , 0
i=0
We have
(1.6) (P<z>)<“> = z<2n> .
When we multiply z<n>' by 2 we obtain:
Lemma 1.7.

(1) If e(0,n) = 0, then z-z<n> - z<n+1>

(11) If 3(0,n) = ... = 3(S,n) 3 1 and e(8+1,n) = 0
8+1 k
where s z 0, then z-z<n> - i<n+r>r+~p Z z<n+1-2 > .
<n> on (k ) k'l
Proof. By definition, 2.2 a z 20(Pk(z))e ,n , which by
k

CD
assumption equals 2 II (Pk(z))€(k’n). Since P0(z) = 2, we have
k=1
G
D
z.z<n> = Po(z) n1(Pk(z))€(k’n), which equals z<n+ . Thus, (i)
k:
is true.
3
To prove (ii), we have z-z<n> = )(z II Pk(z), where
k=0
°° (k n) <n> S 2
x = n (Pk(z))€ ’ . Since 90(2) = z, 2.2 = x n z pk<z).
k=s+2 k=1
2 <n> 3
Since 2 = P(z) + p, z-z = 7. 1'1 (P(z) + p)Pk(z) =
k=l
Q
x II P2(z)P (z) + pz<n-1>. Since P2(z) = P (z) + p,
k=2 1 k l 2
> m <n-l>
z-z<n = t n (P (2) +1»)? (2) + p z
2 k
k=2
oo
- -1
=). n P202)? (2) +pz<n3>+pz<n >-
2 k
k=3
Repeating this argument 8 times, we have
3-1 k
2'z<n> = x P:(z) + p z z<n-2 >
k=0
s-l k
(1'8) = (Ps+1(2) + I)” + p 2 z<n 2 >
k=0
s-l k
-2
= z<n+1> + p E z<n > ,

k=0
This proves (ii).
It will be necessary to form inner products of these
"powers". We make the following definition:

Definition. For m,n = 0,1,...

(1-9) c(m,n) = <2<m> z<n>> a i z<m> z<n> dlJ-(z)-

 

It follows inmediately from (1.9) that c(m,n) '3 c(n,m).
We would like to have a procedure for computing c(m,n). To that

end, we assume that p 2 2. Then F is a subset of the real

axis [1], p, is a symmetric real measure, c(m,n) = c(n,m), and we have
lermna 1.10. For p 2 2,

(i) c(2n,2m) = c(n,m)

(ii) c(2n+1,2m) = 0

(iii) c(4n+1,4m+1) I p-c(n,m)

(iv) e (4n+1 ,4m+3) c (2n+1 , 2m-l-1)

(v) c (4n+3 ,4m+3)

p°c(2n+1,2m+1) .

Proof. The first assertion follows from (1.4). Since

c(2n+1,2m) I <z<2n>-z,z<2m>>, the integrand is odd. Thus, applying

(1.3) gives c(2n+1,2m) I 0, so the second assertion follows.

<4n> <4m> 2 <4n> <4m>

Since c(4n+l,4m+1) I <2 oz, 2 oz) I <2 .2 ,z >,
we use 22 = P(z) + p to get c(4n+l,4m+1) I <P(z)z<4n>,z<4m>> +
p<z<4n>,z<4m>>. Using (1.7) and (1.3) we see that the first inner

product equals 0. Applying (1.7) twice we see that the second

inner product equals p-c(n,m). Thus, (iii) is proved.

Since c(4n+1,4m+3) I <22.z<4n>’z<4m+2>>’ we may use

<4n>, z<4m+2>> +

P(z) + p I 22 to obtain c(4n+1,4m+3) I <P(z)z

p<z<lm>,z<4m+2>>. Using (1.7) and (1.4), we see that the first

inner product is <z z<2n>,z<2m+l>>. Applying Iema 1.8 we see

z<2n+1>,z<2m+1>>, which is c(2n+1,2m+1). From

(1.4) we see that p<z<4n>,z<4m+2>> =- p<z<2n>,z<2m+1>>, which is

that this is <

0 from (1.3). Thus (iv) follows.

To show (v) we use P(z) + p = 22 to get c(4n+3,4m+3) =

<p(z)z<““+2>,z<l‘m+2>> + p c(4n+2,4m+2). Using (1.7) and (1.4)
we see that the first inner product is <z-z<2n+1>,z<2m+1>> =

<23 z<2n>,z<2m>>, which is 0, by using (1.3). Thus (v) follows.

Before proceeding to the main theorem on c(m,n), we need
the following definition:

Definition 1.11. For i,n,m I 0,1,...

A(i; n,m) = (€(i+1:n)a C(ian)s e(i+1;m). e(i.m))
and f(i; n,m) I f(A(i; n,m)) I 0 if A(i; n,m)
I (0,0,1,0) or (1,0,0,0) and equals 1 otherwise.
We may now state

Theorem 1.12. For p 2 2,

E €(ian)€(ism)
i=0

CD
c(n,m) = H f(i; nsm)'p
i=0
if m+m is even and 0 if n+m is odd.
Proof. We may assume that m s n since c(m,n) I c(n,m).

If n+m is odd the theorem follows from Lemma 1.10,

part 2.

For n+m even, we use induction on n. For n I 0,
c(0,0) I 1. For n I 1, c(1,1) I £ zzdu I I(P(z) + P>du I
I2 du +-p Idp I p, by applying (1.4) to the first integral, which
is 0 by (1.3). We now assume that the theorem is true for n s k.
For n I k+l we distinguish four cases. Case (i). For n I 23
and m I 2t we have e(0,n) I 3(0,m) I 0, and for k 2 0 we
have e(k+l,n) I e(k,s), e(k+1,m) I e(k+1,t) and
f(k+1; n,m) I f(k; s,t). If we apply Lemma 1.10, part 1 we obtain

a
m 2 6(1,S)'e(i,t)
c(n,m) = H f(i; Sst)'Pi=0
i=0 ..
w .2 c(isn)°e(i.m)
= II f(i; um) 91:1 .
i=1

If e(1,n) I e(1,m), then f(O; n,m) I 0 and s +-t is odd. The
theorem holds in this case. On the other hand, if e(1,n) I c(l,m),

then f(0; n,m) I 1 and c(n,m) I c(s,t)

m 2 e(i.n).e(i.m)
= H f(i; n.m)pi=0
i=0

Case (ii). If n I 48 + 1 and m.I 4t + 1, then e(0,n) I e(1,n) I
1 I f(0; n,m), and for k 2 0 we have e(k,28) I e(k+l,n),
e(k,2t) I e(k+1,m) and f(k; 23,2t) I f(k+l; n,m). Using Lemma

1.10, part 3, we obtain

co 2 6(1,n)e(i,m)
“mm = P C(ZSaZC) = p ll f(i; n,m)Pig=1

i=1
a, E c(i.n)°e(i.m)
, -1 i=0
= p H f(l; n,m)p p .
i=0

and the theorem.holds in this case.

Case (iii). If n I 43 + 1 and m I 4t +'3 then
1 I e(0,28+1) I e(0,n) I e(0,2t+1) I e(0,m) I e(1,m),

while 3(1,n) = o, f(O; n,m) = £(1; n,m) = f(0,28+1,2t+1) = 1,
and for k 2 1 g(k,2t+l) = c(k+l,m) and f(k; 2s+1,2t+1) =

f(k+1; n,m). Thus, using Lemma 1.10, part 4, we obtain

3 e<i.n>e(1.m>

c(n,m) - c(28+1,2t+1) a n f(i; n m)p p1-1

i=2
a, E e(i.n)e(i.m)

i=0
= n f(i; n,m)p 3
i=0

a:

and the theorem is valid for this case.
Case (iv). If n I 4sa+ 3 and m I 4t +-3, then e(0,n) I e(l,n) I

6(0,m) - e(1,m) - f(O; n,m) . 1, and for k 2 0 we have e(k,23+1) .

10

c(k+1,n), e(k,2t+l) - c(k+l,m), and f(k; 2s+1,2t+1) = f(k+1; n,m).

ThuS, from Lemma 1.10, part 5, we obtain

G
a, 2: eck.n>e<k.m)
kIl
c(n,m) I p c(25+1,2t+l) I p n f(k; n,m)p

k=1
on
a) Z C(ksn)€(kam)
-1 k=o
I p H f(k; n.m)p p .
k=o

and the theorem holds in this case. This completes the proof of
the theorem.

The following results follow from the above theorem:
Corollary 1.13. If p 2 2, then
(i) For k I 0, c(k,0) I 0.

(11) For 0 s k s 2n + 2n+1, c(2“,k) = 0 unless k a 2n in which

n n on
case c(2 ,2 ) p. Z e(l,k)
(111) For 0 s k s 2“-1, c(2n - l,k) - pi‘o .
E 8(1sk)
(iv) For all k, c(k,k) = pi=0 .
(v) For 2 s j s n < a, c(2n - 1, 2j - 1) I pj.

(vi) For 3 2 0 and for all n, c(2n,2n +2n+1 +...+ 2n+s) I p.
(vii) For all k, except as noted in (vi) above, c(2n,k) I 0.
(viii) For all n and for k s 2“, c(2n + 1,k) = p if k = 2n - 1
and 0 otherwise.
(ix) For n < 28+1 5 m and t > 8+1, c(n,mfi2t) I c(n,m) or 0.
Thus for n,m,s as above at X 2 28+1, c(n,mfix) I c(n,m) or 0.
(x) For 2mn4 < 2n - 2 and for c(2n +-2,2m) - 0, we have
c(zn +-2, 2m'+ 4) . 0.
(xi) For 2m+4 < 2n - 2 and for can + 2,2m) n 0, we have

c(2n +-2, 2m +12) - o.

(xii)

We now construct the matrix of these inner products.

11

For 2m+4 < 2n - 2 and for c(2n - 2,2n) 15 0, we have

c(2n - 2,2m +12) = 0.

will use this matrix to construct the orthogonal polynomials.

We

Definition 1.14. G is the matrix whose entries are c(m,n) for

m,n

=0,l,... O

In what follows, we will assume that p 2 2. Thus F is

real, F : [-% -- /% + p, +3: +/% + p] [1; Theorem 12.1], and G

is a symnetric matrix.

(1.15)
(1.16)
(1.17)

(1.18)

(1.19)
(1.20)

(1.21)

(1.22)

(1.23)

(1.24)

s
We note: For 0 s n,m S 2 - l, we have

e(k,n) c(k,n + 25) for o s k S 8-1

e(k,m) e(k,m + 28) for o s k S 8-1
6(81n) = 6(21111) = 0

8
3(S,n+2 ) I 5(S,m+28) I"- 1 .

For 285n528+1-1 and OSmSZS-l wehave

e(k,n) I e(k,n + 28) I e(k,n + 28'”) for 0 s k S 8-1

1

g(s,n) = e(s,n + 25) = e(s,n + 28+1) = e(s+1,n + 28+') = 1

c(m,n) = e(s.n + 28> = c(sm) = e<s+1,m) = o .

For 28sn523+1-l and m+n$28+1-l,wehave

e(k,n) I e(k,n + 23) I e(k,n + 28+1) for 0 S k S 8-1

g(s,n) I e(s,n + 28-”) I €(s+1,n + 28) I 3(s+1,n + 28+

e(s+1,n) = e(s,n + 28) = e(s,m) = e(s+1,m) - o .

1

>=1

12

For 0 s n S 28 - 1, 28 s m 3 28+1 - 1

5+1 +1
(1.25) e(k,n) = e(1t,n + 23) = e(k,n + 2 ) = e(k,n + 28 + 28 )

for k I 0,1,...,s-1 .

(1.26) g(s,n) = c(s+1,n) = e(s+l,n + 25) = e(s,n + 28‘”) = o .

+1 s+1
(1.27) e(s,n + 25) = e(s+l,n + 28 ) = 3(s,n + 23 + 2 ) e

e(s+l,n + 2S + 28‘”) = 1.

Thus, the following formulae are valid for s 2 0:

8
Theorem 1.28. (1) For 0 s n,m s 28 - 1, c(n + 23, m + 2) p c(n,m).

l

S
(ii-iii). For 28 s n s 28+ - 1 and O s m s 2 - 1,

c(n + 28+1,m) I c(n,m) and c(n + 28,m) I 0 .

+1
(iv-v). For ZS s n s 28+1 - l and m + n 5 2S - l,

c(n +-28+1,m) I c(n,m) and c(n + 28,m) I 0 .

s
(vi-vii). For 0 S n s 28 - l and 2 s m 5 28+1 - 1,

+
c(n + 28+1,m) I c(n,m) and c(n + 28 +2S 1,m) I c(n + 2S,m).

Proof. To show (i) we apply (1.15-1.18) to (1.12). To show (ii)
and (iii) we apply (1.19-1.21) to (1.12). To show (iv) and (v)

we apply (1.22-1.24) to (1.12). To show (vi) and (vii) we need some

G
additional remarks: Using (1.25-1.27) we obtain 2 e(k,n)-e(k,m) I
a s+l kggl
2-g(k,n +12 )e(k,m) and A(k; n,m) I A(k; n +.2 ,m), for
kI0
l

o s k s s. Therefore f(k; n,m) = f(k; n + 28+ ,m) for o s k s s.

.1.
80, using (1.12), c(n,m) I c(n + 28 1,m), which proves (vi). Again,
(D co
4.
using (1.25-1.27), 2 e(k,n + 25)e(k,m) I Z e(k,n + 28 + 28 1)e(k,m)
kIO kI0
and A(k; n + 28,m) I A(k; n + 28 + 28+1) for 0 s k s 3. There-

fore, f(k; n + 28,111) - f(k; n + 28 + 28+1,m) for o s k s 8.

So, by applying (1.12) we have c(n + 28,m) I c(n +28 + 28+1,m),

which proves (vii).

13

Corollary 1.29. For m,n s 28 - 1 and m +~n 2 28, c(n,m) I

 

c(n + 28,m).
Proof. We distinguish 3 cases. In the first case we have
3-1 3-1
n,m s 2 - 1 and n +-m 2 2 . Under the assumption of the

s
corollary m 2 2 - n; under the assumption of the first case

-n 2 4“"1 + 1. Thus, 11: 2 28 - 28'1 + 1 - 26"1 + 1 2 28'1. We
have
(1.30) 23'1 s m s 25'1 - 1 s 28 and o s n s 2""1 - 1 .

We apply (1.28 (vi)) to get c(n,m) a c(n + 28,m).

1

In the second case we take n 2 28- , m 5 ZS - 1. Let

N I n - 28-1. Then

(1.31) o S'N s 23‘1 and
3-1 s
(1.32) 2 SmSZ -1.
, 8-1 3-1 s
We may apply (1.28 (Vii)) to get CON +-2 ,m) I c(N +12 +-2 ,m).

That is, c(n,m) I c(n +-28,m).
S 8-1 s
In the final case, n s 2 - l, m s 2 - 1, and n +-m 2 2 .
By assumption in this case, n 2 ZS - m 2 2S - 28-1 +11 I 23-1 +.1 2 28-1.

Thus
8- ..
(1.33) O s m s 2 1 - 1 and ZS 1 s n 5 2s - 1 .

We apply (1.28 (ii)) to get c(n,m) I c(n +~28,m). This completes
the proof of the corollary.

Theorem 1.34.

 

1 1

c(n,m) I 0 for 28 S n 3 28+. - 1 and n +'m S 28+. - 1.

14

Proof. We prove the theorem by induction on 8. For 8 I 0 and
8 I 1, we have, using (1.12), c(0,1) I c(0,2) I c(0,3) I c(2,1) I 0.

We assume the theorem is true for 8 s k-l. That is,
(1.35) c(n,m) I 0 for 2k.1 s n 3 2k - 1 and n+m.s 2k - 1.

For 8 I k we have 3 cases.

In the first case, we take m,n in the rectangle defined
by 2k 5 n 3 2k +210“1 - 1 and 0 S m 5 2k-1 - 1. We take
8 I k-l and use (1.28 (iii)) and (1.35) to get our result.

In the second case we take m,n in the triangle defined
by 2k +21“1 5 n S 2k+1 - 1 and m +1n S 2k+1 - 1. We take
8 I k-l and use (1.28 (iv)) and (1.35) to get our result.

'In the third case, we take m,n in the triangle defined
by 2k 5 n s 2k +21“1 and m +-n s 2k+1 - 1. We use (1.28 (vi)),
symmetry of the matrix C. and (1.35) to get our result. Thus the
theorem is proved.

The next theorem tells us that certain diagonals in C.

can be classified by a single parameter.

Theorem 1.36 (Diagonal Theorem). If n +~m I constant, then

(1.36) c(n,m) I 0 or SKI“ +-m)

G
Where XKn +-m) I 2 c(t.n)-e(t,m).
tIO
Proof. If n +-m is odd the theorem follows easily since c(n,m) I 0

by using (1.12).
If n +-m is even we let 8 and x be defined by
l x+l
n +-m I 28 and 2 s 28 s 2 - 2. We proceed by induction on

x. Using (1.12), c(0,0) I l, c(2,0) I c(0,2) I c(1,1) I p.

15

Thus we may assume 1 2 2. Since c(m,n) I c(n,m), we may assume

m S n. Using (1.34), c(n,m) I 0 for n > 21,

We partition 6. and show that the theorem is true in each
member of the partition. We will then show that the members of

the partition give the same answers. We partition c. as follows:

2’“1

For S m S n S 21 - l, we have 0 S m - ZX-l S n - ZX-l S ZX-l - 1.

Therefore, using (1.28 (i)) we obtain c(n,m) I c(n - 21-1 + 21-1,

111 - 2x-1 + 21-1 271-1, m - 27‘4”

) ‘ P c(n ' ). Thus, if we restrict

2’"1

our attention to (the triangle) S m S n S 21 - l, the relation

(1.36) holds by induction. On the other hand, for m S 2""1 -1 S

n S 21 - l, we have, using (1.29), c(n,m) I c(n - 2x-1,m). Thus,

if we restrict our view to (the triangle) m S 2’61 - 21

1 S n S - 1,

the relation (1.36) is true by induction.

It is now sufficient to show that one entry from

2k1 S m S n S 21 - 1 equals one entry from m S ZX-l - 1 S n S

2)\ - 1. Moreover, we may assume

(1.37) 8S2 +2 --1,

for if 8 > 21-1 +_21'2 - 1, then we would have m +1n I 28 2 21 -
ZK-l. But for m S 2k1 - l S n S 21 - 1 we have m.< 2k-1 and
n S 21 - 1, whereby m + n I 28 < 21 +2)“1 - 1. Thus (1.37) holds.

2x-l-l

let k I s - 23-1. Using 21 S 28 S +-2 (as stated

at the beginning of this proof), we have
(1.38) o s k s 2*“1 - 1 .

Using (1.37) we have

16

(1.39) 0 s k s 2"”2 - 1 .

x-l

We choose c(2)‘-1 +~k, 2 +-k) as an entry for which

2)“1 S m S n S 21 - 1. We choose c(2)‘-1 +_21'2 + k, 2)"-2

S 21-1

+ k)

as an entry for which m S 2k1 - l S n . It is sufficient

to show c(2’"1 +2”2 +»k, 21'2 +1k) - c(2 +1k, 2”1 + k).

We may use (1.38) and (1.28 (1)) to get

(1.40) 12(2)"1 +'k, 2’“1 +1k) = p c(k,k) .
Adding 21-2 to (1.39) we get
(1.41) 2’1'2 s k +2’V2 s 2’"1 - 1 .
Thus, applying (1.28 (vii)),
(1.42) c(k + 21'2, k + 21-2) = c(k +21"1 + 21‘2, k + 21-2).
Using (1.39), 0 s k s 2"“2 - 1. Applying (1.28 (1)),
(1.43) p c(k,k) a c(k + 21’2, k + 21'2) .
Combining (1.40), (1.42), and (1.43), we have

ea”1 +1k, 2""1 + k) = c(k +2)“l +-2*'2, k +-2*'2),

and this completes the proof.

We are now prepared to construct the monic polynomials
Qn(z) of degree n which are orthogonal with re8pect to u.
That is, let Qo(z) I l and Qn(z) be a polynomial of degree n

with leading coefficient 1 such that

(1.44) i Qn(z) Qm(z) du(z) I 0 if n I m .

17
For example, if Q1(z) I z +1A, then
0 = i 00(2) 01(2) an =£ 1-2 on +X£ 1.8“. .

The first integral is 0 since its integrand is odd, while (1.2)

tells us that the second integral is 1. Therefore A I 0 and
(1.45) Q1(z) I z .

Kinney and Pitcher [4; pg. 27] find
(1.46) Q2n(z) I Pn(z) .

We complete this lacunary set of orthogonal polynomials to the

complete set.

SzegB uses GramrSchmidt Orthogonalization on linearly
independent functions to obtain orthogonal polynomials in [5;
pp. 227-228]. We use a slight variation, suggested by the

technique of Walsh, to obtain Qn(z).

Since the degree of z: is n, {z<n>} form a linearly

independent set of functions. For 8 S n I 0,1,2,... , let
o(n,s) be defined by

n <8)
(1.47) Qn(2) - 2 s<n,s)z .

8I0

Then for 0 S k1< n,

 

0 a <Qn(z) , z<k>> - IE Qn(2) 24‘) 611(2)

n s> k>
e]: [ an(n’8)z< 32‘ du(z).
8

Interchanging integration and summation, we get

18

n

0 - z a(n,s) z<s> z<k> du(z)
sIO
n
= z a(n,s)c(8,k) .
8I0

Also, Since i<n> can be expressed as a linear combination of the Qn's,

n
<n>
42,02). 2 > i[Qm(z)1:23:06ij(2)1180?)

If \Qn(2)l2du ~

We denote this by “Qn(z)“2. Thus we have

n
(1.48) 0 I 2 a(n,s)c(8,k) for 0 S k S n-l
8I0
and
2 n
(1.49) Mona)“ = :3 a(n,s)c(s,n) .

8I0
(1.48) and (1.49) represent n+1 linear, non-homogeneous
equations in the n+1 unknowns a(n,0), a(n,1),..., c(n,n). Thus,

as in SzegB's paper [5; pp. 227-228],

(1.50) a(n,s) I A;1 c(0,0) ... c(0,8-1) 0 c(0,s+l) ... c(0,n)
c(1,0) ... c(1,s-1) 0 c(1,s+l) ... c(1,n)

c(n.0> c(n,s-1)\\Qnu2c(n,8+1)-oo c(n,n)

 

 

where

(1.51) A0 I 1 and An I c(0,0) .... c(0,n)

 

c(n,0) .... c(n,n)

 

19

According to Theorem 1.12, c(0,8) I c(8,0) and both are

68 0; that is, 0 if 8 I 0 and 1 if 8 I 0. Thus,
3

c(1,1) ... c(1,s-1) c(1,s+l) ... c(1,n)

(1°52) O’O‘ss) - A;158,OHQn(z)“2

c(n,1) ... c(n,8-1) c(n,s+l) ... c(n,n)

 

In particular,
(1.53) c(n,0) B O for n I 1,2,... 0

Combining (1.47) with (1.52), we have shown that the follow-

ing representation holds for Qn(z):

c<1.1) .... e(1,n)
(1.54) Qn(z>=u<2n<z>\\21;1 :

c(n-1,1) .... c(n-l,n)

z<1> .... Z

 

 

Since the leading term of Qn(z) is 2“, we have
n
Qn(z) I z + lower order terms. But (1.54) indicates that

(211(2) " “Qn(Z)H2 AglAn_12<n> + lower order terms. Thus,

(1.55) \\Qn(z)\\2 Amlbgl a 1 or An = 11ml Honor)“2 .

Thus, by induction we have

n 2
(1.56) an - kno Hoke)“ .

We have thus established the following:

Theorem 1.57. For n 2 1,

 

20

Qn(z) I Dn c(l,1) .... c(1,n)

c(n-1,1) .... c(n-1,n)

<1>' .... z:

 

 

n
where on a “(gummy .- 131 = (kl) \\Qk<z)u2>'1 .

In Special cases we can be more specific about An, as

the next lemma shows.

Lemma 1.58. (1) A = P A

(11) uoznmn2 = p .

Proof. To show (i) we note that using (1.13 (2)) we obtain

c(2n,k) - p 5 (where 6, I 0 for i I j and l for
k,2“ 1’3
1 u j), When we expand A n by the row containing c(2n,k),
2
the result follows.

To show (ii) we recall that (1.46) implies Q n(z) I Pn(z).
2
2
Therefore, “Q n(2)“2 I HPn(z)“ I c(2n,2n) I p when we use (1.13
2
(2)) or (1.13 (4)). This completes the proof of the lemma.

In [5], SzegS uses I an zm ldz‘ as the matrix entries.
(11) m>
a< du

 

We have used I z . In the case that F is a subset
of the real line (this occurs for p 2 2), we might have con-

sidered I szu. If N is odd, we apply (1.3) to find this in-
tegral equals 0, but if N is even, then the integral is a poly-

nomial in p of degree N/2. Moreover, the coefficients in this

polynomial have a complex pattern. As examples, we note that

21
jzzdu-fluz) +p>du=fzdu+jpdu=p
j‘zl’du =f(1>(z) + p)2dp. =f 22:1,. + 2pjz on, + p2] dp, = p2 + p
fzédp, - 2p2 + p3

S
I zzsdu = I<P<z> + mad» - 2 (DPS-xi ‘de

1‘0 7“
[8’2] _2
z (3)398 1 j‘ 221d“ .
1,-0

The use of c(n,m) for our matrix entries seems to be
simpler since c(n,m) is 0 or a power of p (Theorem 1.12) and

c(n,m) I c(s,t) or 0 if n + m I 8 + t (Theorem 1.36).

CHAPTER II

We develop a difference equation for our orthogonal poly-
nomials as SzegS does in [1].

To insure that divisions occurring later are legitimate
we need the following lemma:
Lemma 2.1.

HQn(z)H2 I 0 for n I 0,1,2,...

‘ggggf: £ ‘Qn(z)‘2du(z) I 0 if and only if p is concentrated
on the roots of Qn(z), n I 0,1,2,... . Since p is concentrated
on F [1; Lemma 15.2], we would have F consisting of the roots
of Qn(z), n I 0,1,2,... . Then F would be a denumerable set
and have capacity 0 [8; Theorem III.8]. But F has capacity 1
[1; Lemma 15.1]. This contradiction completes the proof.

Since Qn(z) are orthogonal polynomials, the difference

equation

(2.2) one) = z Qn_1(2) + An-2Qn-2<z) + An_10n_1<z>

holds for some An-Z’ An-l [7; Theorem 3.2.1]. We can improve
the above difference equation; after a few lemmas, we will be able

to delete the middle term.

Lama 2.3. an(z) I Qn(P(z)) for n I 0,1,...

Proof. It is sufficient to show that £Qn(P(11))zs dp,(z) I 0

for 0 S 8 < 2n.

22

23

If 8 is even we let 8 I 2m. Then 0 S m< n, and

 

an(P(2))zs du == an(P(z)>zZ“‘ cl» = an(P(z))(P<z) + p)“‘dn

Shifting, the last integral is J‘Qn(z) (z + p)m dp.

Using the binomial theorem on (2 + p)!n and interchanging
integration and summation we get IQn(P(z))zs (1p, I jgo(?)pm-1J‘Qn(z);j-du.
But j S m < n and thus the integral is 0 by induction. Thus
we have our result if s is even.

If s is odd, then we claim that Qn(P(z))-z: is an odd
function of z. This is true because P(z) is even, hence
Qn(P(z)) is even and z—8 is odd. Thus IQn(P(z))z—8- dp. I 0 by
(1.3). This concludes the proof of the lam.

We have, iumediately, the following corollary:

Corollary 2.4. IF|QZn(z)‘2dp. I IF‘Qn(z)|2dp, .

31:92:. Use the previous lemma and shift (1.4).
Our next result tells us about the evenness or oddness
of the Qn(z).
Lama 2.5.
(i) an(z) is an even function of z for n I 0,1,...
(ii) Q2n+1(z) is an odd function of z for n I 0,1,...
M. Since 0211(2) IQn(P(z)) by lemma 2.3, we obtain (i).
To prove (ii), we note that by orthogonality,

fQ2n(Z)Q2n'+_-1—-(Z) as = 0. By (2.2).
(2.5) 102,12) (121.1175) due) =J02n12) 20:12) .1,

+ KZn-l IQZn“) Q2n-1(z) d“ + X2: IQ2n(z) Q2n(z) dp' '

24

Thus,
(2'6) 0 a PQ2n(z) lenEZE d“ +X2n-l jQ2n<z) Q2n-1(z) d“

2
+A2n1HQ2n(z)‘ d“ ‘

The first integral on the right hand side of (2.6) is zero
because its integrand is an odd function of z. The second integral
is zero by orthogonality of the Qn(z). Thus, since

2 ._

“one” an I 0 by learna 3.1, we must have A2“ - 0. Thus Azn = 0.

We have shown that

(2'7) Q2n+1(z) "' zan(z) + AZn-l Q2n-1(z)°

Therefore, by induction on n and the facts that
zQ2(z) I z(z2 - P) I z3 I 92 and Q1(z) I 2, we have our result.
We combine (2.2) with lemma (2.5), and we write -K(n)
instead of An-Z' Then we have shown that the following theorem
holds:

Theorem 2.8.

Q0(z) B 1 9 (21(2) 3 z , and
(2.8) Qn(z) I ZQn_1(z) - K(n) Qn_2(z) for n 2 2.

We have reduced the problem to finding relationships
among the K(n). This problem admits a solution in some cases.
As the next theorem shows, when our F set is a real set, the
K(n) satisfy a pair of non-linear recursion relations.

Theorem 2.9. If p 2 2, then

(i) K(Z) " P

25

(11) K(2n+l)IK(n+1)/K(2n) for n 1,2,...

(iii) K(2n) I p - K(2n-l) for n I 2,3,...

2
Proof. Since Qo(z) I 1, 01(2) I 2, 02(2) I z - p, we may take
2
n I 2 in Theorem 2.8. Thus 2 - p I 2(2) - K(2). We then
have K(Z) I p, giving us (i). To show (ii), we multiply (2.8)

by Qn_2(z) and integrate:

(2.10) o ==json_1(z)0n_2<z>dn - Kn HQn_2(2)H2 .

but an_2(z) IQn_1(z) +K(n-1)Qn_3(z) from (2.8), so
szn_1(z)Qn,2<z>dn = [\Qn_1<z)|2du + K(n-1)an_1(2)Qn_3(2)du

which equals “Qn_1(z)“2.

Thus, (2.10) becomes
2 2
0 = HQn_1(Z)H - K(n) HQn_2(Z)H

or

(2.11) K(n) = H0n_ltz>n2/Hon_2(z)n2 for n = 2,3,...

Thus using Corollary 2.4 and taking n 2 1, we have

1122,12 19,12
K(2n+l) = —— = ——

1122,4112 1192,4112

Dividing numerator and denominator by “Qn_1“2, we have

a \1Q,\12/\\o,,-112 Mont/11.112 green

“an-1\\2/\\Qn-1\\2 g \lQ2n_1\\2/nqzn_2“2 K(Zn)

 

 

K(2n+l

which proves (ii).

26

To show (iii), we write (2.8) as an+1(z) IzQ2n(z) -

K(2n+l)Q2n_1(z) and note that
(2-12) zQZn_1(2) = Q2n(2) +K<2n)Q2n_2(2) -
Thus,
2
\lQ2n+1(7-)H = [Q2n+1(2)[202r‘1(z) - K(2n+l)02n_1(z)]dp,
= sz2n(z)QZn+1(z)dn - K(2n+l) -0
= fzozn(z)[zqzn(z) - K(2n+1)Q2n_1(z)]du
2 2 .
= $2 Q2n(2)du - K(2n+1>j202n_1<z>02n(z>du -

Using (1.6) in the first integral and (2.12) in the second integral,

we get
uqzmenlz = f(Pczmmfinem - K(2n+1>jm§n<z> +
+ K(2n)Q2n_2(Z)Q2n(7-)]du .
Using (1.3) and (1.4), we have
HQ (MHZ = 2 (2)d + 2 ( )d - K 2 +1) 2(2)11 - x 2 +1)-x 2 0
2n+l IZQII P PJQZn Z W ( n szn u ( n ( n). .

The first integral above is zero because its integrand is

an odd function of 2. Thus
HQ (2)12 - 2 - 2 .1 2
2n+l — pHQn(Z)H 1“ n ) “(2,101)“ °
That is,

v - K<Zn+1> = \\92n,1<z>\\2/\\Qn<z>n2 .

27
2 2
But by (2.4), “Qn(z)“ = HQ2n(z)H and by (2.11)
K(2n+2) = \\02n+1\\2/\\02nl12 .
Therefore,

K(2n+2) I p - K(2n+1)
or

K(2n) = P ‘ K(Zn-l) a

which completes the proof of (iii).

Since K(2n+l) = 512111. and K(2n) = p - K(Zn-l), we

K(2n)
have K(2n+l) I Eiflill——- . Continuing this way we can develop
p-K(2n-l)
a continued fraction expansion for K(2n+l). Similarly, since
K(n)
K(2n) p - K(2n-l) and K(Zn-l) K(2n-2) , we have
K(2n) I p - %%%%—§;-, and we can develop a continued fraction

expansion for K(2n). We exhibit these continued fractions in
our next theorem.

Theorem 2.13.

 

 

 

 

(2.14) K(2n+l) I K(n+l)
p - K(n)
p - K(n-l) for n 2 3 and
'. p 2 2
p - K(3)
(2.15) K(2n) I p - K(n)
p - Kin-1)
p - K(n-Z) for n 2 2

 

p - K(3)

28

We have so far developed two approaches to the problem of
finding the orthogonal polynomials Qn(z). One method uses a
matrix representation (Theorem 1.57) in which the determinant An
appears. The other approach is by difference equation techniques
(Theorem 2.8) in which we use the coefficients K(n). The next
theorem relates these two methods.
Theorem 2.16.

-2
K(n) = An_1 An_2 An_3 for n 2 3 .
Proof. By (2.11) we have

2
K(n) = HQn_1(Z)H2/HQn_2(Z)H for n 2 2 .

We also have, by (1.55)

2 -1
Home)“ _ An A“ for n 2 1 .
-1
An-l An-2 -2
Therefore K(n) I _ I n-l An-Z An-3 , which is what we
An-z An-3

wish to show.

Another approach to computing Qn(z) for p 2 2, using the
coefficients K(n) of the linear difference equation is to consider
ratios of the Qn(z).

Definition.
(2.17) Tn(z) I=Qn(z)flQn_1(z) for n 2 1 and Qn_1(z) I 0 .
If we divide (2.8) by Qn_1(z) we have

=_§.£E1L_
(2.18) Tn(2) z Tn-l(z)

29

Since, by definition, T1(z) I z, T2(z) = z - E', we have the

following continued fraction representation for Tn(z):

Tn(z) I z - K(n)
(2.19) z - tin-11

z - K(n-Z)

2-2
2

This continued fraction reminds us of the continued fraction for
K(2n) in (2.15). In fact, if we replace 2 in (2.19) by p,

the two continued fractions (2.15) and (2.19) are identical. Thus
we have shown the following lemma.

lemma 2.20. For p 2 2 and n 2 l

Tn(p) I K(2n) .

We are in the following situation:

If we know K(2),...,K(n) then we can compute, by (2.19),
T1(z), T2(z),...,Tn(z). Then, by lemma 2.20, we get K(2n) by
evaluating Tn(p). Moreover, since K(2n-l) I p - K(2n), by
(2.9 (iii)), we have K(3), K(5),..., K(2n-l). Thus, starting
with the first n-l K's, we readily get the first 2n-l K's.
Repeating this argument gets us as many coefficients K(n) as we
want.
Example. we know (2.9 (1)) that 1((2) - p. Thus T2(z) - z - £- .

Therefore K(4) I T2(p) I p - l and K(3) I p - K(4) I l.

T3(Z) I z - fi-- f%- and

Continuing,

from which K(6) I T3(P) =

30

-l l
Tum-9199

1253111111 and K(S) — p - K(6) =-—I-.

Therefore

one).

(2.21)

rn<z>

Tn_1(z).

(2.22)

 

9-1 p-1
._1_|
T5(Z)IZ-p1-f;—L-%i-flam
2(2- 12- -1

was—4 19‘9- I 9 9

A simple observation allows us another approach to computing

Namely
=Qn (Z)

Since Qo(z) l we have observed that

n

Qn(2) = jn T (2) .

J
Using the Tn(z) we computed above we have, as examples:
01(2) = 211(2) = 2

02(2) = 11(z>12(z) = z<z - f) = 22 - p

03(2) (11(2)12<z>>r3<z) =02<z>13<z>= (22-12) (2 - —>=

2.2

23 - (p+l)z.

We already have one continued fraction representation for

in (2.19). Another may be obtained by solving (2.18) for

We have

Tn-l(z) 2 5122... .

z-Tn(z)

31

Since Tn(2) in the denominator of (2.22) can be expressed
K n+1
‘géE-izzy', we obtain the following continued fraction re-

n+

presentation for Tn(z):

Tn(z) I K(n+l)
(2-23) 2 - K(n+2)

z - K(n+3li

(2.24) z a as:

 

CHAPTER III

In this chapter we use the linear independence of the set
{z<n>} to find representations for Qn(z).
As in (1.47) let

n

<k>
(3.1) Qn(z) = z a(n,k) z .
k=0
. <n> . .
We remark that Since z 18 an n-th degree polynomial,
the z<n> are linearly independent; thus the coefficients q(n,k)

are unique.
We obtain some relationships among the q(n,k) in the
following lemmas.

Lemma 3.2.
3.2 (i) c(2n,2m) = c(n,m) for O s m S n < m
3.2 (ii) q(2n,2m+l) = O for O s m s n < m .

Proof. Since Q2n(z) ==Qn(P(z)) (Lemma 2.3) we can write Q2n(z)

2n n
as 2 a(2n,k)z<k> or as 2 a(n,k)(P(z))<k>. But
k=0

k-O
(P(z))<k> a z<2k> by (1.7). Thus

2n n
(3.3) 2 01(2n,k)z<k>= 2 c(n,k)z<k>.

k=0 k=0

Comparing coefficients in (3.3) we obtain (3.2).

We next prove:

32

33

IBmma 3.4.

(3.4) c(2n+1,2m) I 0 for 0 s m s n < m

(3.5) a(2k+3,2k+3) I a(2k+2,2k+2)

(3.6) a(2kfi3,25+1) I a(2k+2,23) - K(2kfi3)a(2k+l,28+l)

for 0 s s s k .

lgzggf. To show (3.4) we use induction on n. For n I O and 1
we use Qo(z) I l to obtain a(0,0) I l; we use Q1(z) I Pb(z) I z
to obtain a(l,1) I l and a(l,0) I 0; we use Q2(z) I P1(z) I
22 - p to obtain a(2,2) I 1, c(2,1) I O, and a(2,0) I -p.

We assume (3.4) is true for n s k. That is
a(2n+l,2m) I 0 for 0 s m s n1< R. For n I k+l, we use n I k+1

in (2.8), and we get

Q2(k+l)+1(z) = ZQ2(1<+1)(Z) ' K(2(1“"'1)+1)Qz(1<+1)-1(‘”‘) °
So,
2k+3 <8) 2k+2 > 2k+1
)3 q(2k+3,s)z = z z oz(2k+2,s)z<s - K(2k+3) z a(2k+1,s)z<8> .
SIO sIO s=0

Applying (3.2 (ii)) to the first expression on the right hand side

and the inductive assumption to the second expression, we may write

2k+3 2k+2 2k+1
z a(2k+3,s)z<s> = z z a(2k+2,s)z<s> - K(2k+3) z: a(2k+l,s)z<s>.
830 8.0 3:0

3 even 3 odd

So, by (1.8 (i)),

2k+3 s) 2k+2 (8+1) 2k+1 <8)
(3.7) 2: a(2k+3,s)z< - z a(2k+2,s)z -K(2k+3) z o,(2k+1,s)z .
8=0 s-O SIO

8 odd

34

Comparing coefficients in (3.7), we obtain a(2k+3,28) I 0
for s I 0,1,...,k+1. Thus (3.4) is true for n I k+1 and thus
true by induction.

Comparing coefficients in (3.7) also establishes (3.5)
and (3.6).

We are now able to show that when we expand Qn(z) in

<o>

> .
powers of z<n , the coefficient of z is always zero.

Lemma 3.8.
a(n,0) I 0 for n I 1,2,...

£222£° If n is odd, this follows from Lemma 3.4. If n is even,
we use (3.2 (i)) to write a(n,0) I c(n/2,0). Our result follows
by induction.
Having shown that there is no "constant coefficient", we
n

<k>
now show that Z c(n,k)z is a "monic" expansion. That is
RIO

Lemma 3.9.
c(n,n) I 1 for n I 0,1,...

Proof. We prove the lemma by induction on n. We have already
shown, in the proof of Lemma 3.4, that the result is true for
n I 0,1,2.

We assume that the theorem is true for n s'N. That is,
(3.10) a(n,n) I l for n s N .

If n I=N+1, we have to distinguish between N+l being
even or odd. If N+1 is even we take N+1 I 28. Then

am+1,n+1) - a(2s,2s), which by (3.2 (1)) equals 0103.8). which

35

by (3.10) equals 1.
If N+1 is odd, we let N+1 I Zs+1.
Then writing the difference equation for Qn(z) as expansions
<

in z n> and utilizing (1.8 (i)) we obtain:

8

8 .
(3.11) z; a(28+l,2j+l)z<2j+1> -= z c(n,m).-523“)
i=0 1‘0
5'1 2-+1>
- K(ZS-l-l) z a(2s-l,2j+l)z< J .
J=0

The second sum on the right hand side of (3.11) has no
terms involving z<23+1>. Thus, comparing coefficients in (3.11)

we obtain

a(N+1,N+1) I a(28+1,28+1) I a(28,23) I a(S,s) = 1 .

This completes the proof of the lemma.
We now show:

Lemma 3.12.

 

c(2n+1,2mfi1) I a(n,m) - K(2n+l)a(2n-1,2m+1)

Proof. We use, again, the difference equation for the Qn (2.8)

written as expansions in z<n>.

2n+1 <s> 2n <s> 2n-1 3)
z a(2n+l,s)z =2}: a(2n,s)z -K(2n+l) z a(2n-l,s)z< .
BIO sIO SIO

By (3.2 (11)) and (3.4), c(2n,28+l), a(2n-1,23) and a(2n+1,28)

are zero for O s s < n.

Thus
“ <2s+1> “ <2s>
(3.13) E a(2n+1,29+1)z I z 2 c(2n,28)z
8'0 sIO

n-l
2 +
- K(2n+1) £0 Q,(2n-1,2s+1)z< 8 1) .
8

36

<28> <Zs+1>
z - z I z

Recalling that , and conbining like

terms we have

n “-1 .
Z“ c(2n+1,28+1)z<28+1> - z [c(2n,28)-K(2n+l)a(2n-l,23+l)]z<28+1>
8'0 sIO

+ (1r(2n,2n)z<2n+]'> .

By Lemma 3.9, a(2n,2n) I 1. Thus,

n 11-1
2 +1> 2 +1>
(3-14) 2 or<2n+1.28+1)z< 8 - 2 [c(2n,23)-K(2n+1)a(2n-1,28+1)]2< ‘3
sIO sIO
+ z<2n-i-1> .

Comparing coefficients in (3.14),

a(2n+1,28+l) - a(2n,2s) - K(2n+1)a(2n-l,28+l) for n 2 o .

Since a(2n,28) I a(n,s), we obtain our result:

a(2n+1,2m+1) - c(n,m) - K(2n+1)a(2n-l,2m+l) .

Combining our results from (3.2), (3.4), and (3.12) we have:

Theorem 3 .15 .

c(n,m) I 0 if n-im is odd

a(n,m) (3621-, £22) if n and m are even
n-l m-l
c(n,m) I (AT, T) - K(n)a(n-2,m) if n and m are odd
c(n,m)Il if nImIl.
As corollaries to this theorem we have
Corollary 3.16.

1. c(2n+1,2m+I) a -x(2“+1)a(2“-1,2m+1) for o s m s 2“"2

37
“'1-1,m) - K(2“-1)a(2“-3,2m+1)

for 0 s m 5 2n-1_1

a(2“-1,2m+1) . a(2

a (2n+1, Zn- 1) I -I((2n+1)

a(2n+l,l) I (-1)n r1'11((2j+l)
i=1

a(2n+l,2n-3) I c(n,n-Z) + K(2n+l)K(2n-1)
a(2n+l,2n-5) I -K(2n+1)K(2n-1)K(2n-3) if n is odd and

K(2n+1)[K(n-1) - K(2n-3)K(2n-1)] if n is even.

CHAPTER IV

We have developed representations for Qn(x), the monic
orthogonal polynomials over F with respect to L». We now
extend the orthogonality to sets other than F by using a measure
p* induced by use of the Green's function.

We let G(z,zo.,C'/) be the Green's function for the region
bounded by the closed analytic curve 0 with pole at 20. We
recall that G(z,zo,c) I 0 on O, G(z,zo,c) > 0 in the region
bounded by c, and G(z,zb,c) + log ‘2 - 20‘ is harmonic in the
region bounded by 6.

Theorem 4.1. Let G be any closed, analytic curve containing

F in its interior. Let W E F. For 2 E c we take
<p(z) - g C(z.w.o>du<w)
* d
and dp, (z) uEE-cp(z)‘dz\ where n is normal to c at 2. Then

1‘2““) Qua) du*(2) = o if n ,1 m .
Proof. By definition we have
£Qn(2) Qm(2) d11*(2) I ions) Qm(2) (3'31: G(Z.W.G)du(w))\dZ\
=£<1£Qnm Qm(2) c<z.w.c>d..(w))\dz| .

By the Fubini Theorem, the last expression equals

38

39

H one) one) §;G(Z:Waa)‘dz‘)dp(w) .

In [8] we find the following result:
Theorem [8; Theorem 1.21]. Let f(g) be a bounded measurable
function defined on the closed analytic curve F. For g E P,

we take n to be the inner normal to F at Q. We put
u(z) =I[ f(z) g;G(Z.C.I‘) \ch .

Then u(z) is a bounded harmonic function in the region bounded
by P, and if f(g) is continuous at go 6 P, then
11m u(z) g f(go) 9

Zago

for z belonging to the region bounded by F. Thus

_" L C(Z,‘I,O) dz )dI-3(w)
H: one) Qm(2) an 1 |

= i once) 0mm duo») .

which equals zero by orthogonality. This completes the proof of
the theorem.

We examine the equilibrium measures concentrated on the
equipotential curves associated with the measure u concentrated

on F. We take

(4.2) or I [2 :£ log ‘2 - w‘dp.(w) I r} .

We always assume that r is large enough so that 6% is
simply connected and contains F in its interior.

We have the following relationship between Green's functions

and i log ‘2 - w‘dp,(w):

40
Lemma 4.3.
C(z,m,Cr) I i log ‘2 - w‘dp(w) - r .

Proof. In order to prove the lemma we must show two things. First
we must prove that as 2 tends to infinity, log ‘z-w‘dp(w) - r
tends to log ‘z‘. This is so because as 2‘» a, log ‘z-w‘ a log ‘z‘;

thus,

i log ‘2 - w‘du(w) ~ E log ‘z‘du(w) I log ‘2‘ .

Therefore as 2 a m,
i log Iz - W‘dp(W) - r ~ log ‘2‘ - r ~ log ‘2‘ .

Secondly, we must show that for z 6 CE’ i log ‘z-w‘dp(w) - r = O.
This is true by definition of C%. Thus,the lemma is proved.
There is another representation for G(z,«,Cr), as follows:

Lemma 4.4.

0(2 :maar) B 'i C(2 awscr)dIJ-(w) 0
Proof. By definition of G(z,w,c%), we know that
(4.5) G(z,w,(‘,r) + log ‘2 - w‘ .. h(z,w,cr)

where h is a harmonic function of z throughout the region

bounded by CE. Thus,

(4.6) 1‘ G(z,w,(‘,r)dp,(w) .1 -log ‘z-w‘dp(w) +£ h(z,w,(',r)dp,(w) .

The first integral on the right hand side of (4.6) equals

-G(z,m,c%) - r by Lemma 4.3. The second integral on the right

41

hand side of (4.6) is a harmonic function of 2 inside cg since
the integrand is a harmonic function of 2 inside 6%. By defini-
tion of G(z,zogC9, h(z,w,C§) I 410g ‘2 - w‘ for z 6 CE' Thus,

for z 6 Ci,
‘2 h(z,w,(‘,r)dp.(w) =+l[ log ‘2 - W‘dpfifl) 2+r.

Since i h(z,w,c%)du(w) is harmonic throughout the region

bounded by C? and constant on C4, I h(z,w,c%)dp(w) I-+r . Thus,
£G(Z,W,Gr)dp,(w) B 'G(Z,Q,Cr) + r " r = ‘G(Z,°,Cr)

Since G(z,m,c%) is unique, we combine the results of

Lemma 5.3 and Lemma 5.4 to get

(4.7) £ log ‘2 - W‘dp(W) - r I -; G(z,w,Cr)dp(w) .

Therefore

d d '
(4.8) 3:“: G(Z.W.Gr)du(W) = - 35‘ log \z - w‘de)

*
and dp , which we have previously defined as

d d
33-; G(z,w,ci)du(w)-‘dz‘ may now be taken as 33-; log ‘z-w‘du(w)-‘dz‘.
We have proved the following theorem:

Theorem.4.9.

£Qn(z) Qm(z) [gr-1'; log ‘z-w‘dp(w)] ‘dz‘ I 0 if m If n.
r

CHAPTER V

We conclude with two examples; namely, P(z) I z2 in which
case the F set is the unit circle, and P(z) I z2 - 2 in which
case the F set is the interval [-2,2] on the real axis.
Example 1. P(z) I 22.

We will show that the F set is {2: ‘2‘ I 1}, that dp
is arc length, and that the orthogonal polynomials are powers of
2. We observe that P2(2) I 24 and Pn(z) I 22“.

If ‘20‘ > 1, then for all 2 sufficiently close to 20,
{Pn(z)} converges uniformly to m. Thus {Ph(z)} is normal at
20, so that no point of ‘2‘ > 1 belongs to F.

If ‘20‘ < 1, then Ph(z) converges uniformly to O for

all z sufficiently close to Thus {Ph(z)} is normal at

20.
20, so that no point of ‘z‘ < 1 belongs to F.

0n the other hand Pn(eie) is not normal for each point
e19 of the unit circle. This is clear since each open neighbor-
hood of e19 contains points whose absolute value is greater
than one and points whose absolute value is less than one. Under
iteration the former points converge to a while the latter
points converge to 0.

Thus F I {2: ‘z‘ I 1}.

To construct p we take 2 I l and we let p“ be the

0

discrete measure placing weight 2-n at each root of Pn(z) I l.

42

43

Since these roots are the roots of unity, and since they are dis-
tributed symmetrically around the unit circle, we have

lim u I Lebesgue measure (=.l_ ° arc length). That is,
n—m n 2n

dun I %; - ‘dz‘ as n'I m.

We now use the method of Chapter I to construct Qn(z).

Lemma 5.1. 2<n> I zn I Qn(z) for n I 0,1,... .

as
Proof. For n I 2 e(s,n)28 we have

SIO I s
2 3(S,n)2
m S €(Ssn)
z<n> a H (138(2))5;(8,n)I=B II (22 ) g zsIO a zn .
SIO s=0

Thus we have the first equality.

<m>’i<n>

We have c(m,n) .{i Z dfl(z)‘ 1H‘2‘_1 Zn ‘dz‘ =

if m I n and 1 if m I n. Hence An I l for all n. Hence

Qn(z) B 1 0 ...... 0 o
0 l ...... 0 0
o . n
. . . . . I 2
0 0 '.l :
1 2 2“”1 2n

 

 

and this illustrates the second equality.

Now that we have F (I unit circle), p (I %; - arc length),
and Qn(2) (I 2“) we can use the results of Chapter IV to extend
orthogonality. For CE I {2: ‘2‘ I r} we have G(2,m,c%) I

log'% ° ‘2‘. In this case g;- is taken in the radial direction,

d 1 1
sothat a—logF- ‘2‘I—log‘z‘I—long;. Therefore
du* (2) I- - ‘dz‘. We _have obtained the well known result that
n

for z I r e19, I 2nz m--‘dz‘ I 0 if m I n .
o

Example 2. P(z) I z2 - 2 .

We will use the results of Chapter II to show that Qn(z)
is the Tchebycheff polynomial 2 Cos n arc Cos (x/Z). F will be
the real segment [-2,2] and dp,(z) '% - (4 - 22)-%d2. We con-
clude with remarks about the orthogonality of Qn(z) on certain
confocal ellipses of which F is a limitingcase.

Brolin [1; Theorem 12.1] shows that F I [-2,2]. Kinney
and Pitcher [4; pg. 27] remark, without proof, that
du(x) I - (4 - x 2) a‘dx dx. We show this in the following lemma.
Leanna 5.2. dp,(x) I -2- ° (4 - x2)-%dx.

Proof. We must show that u is invariant under P1(x). This
follows from du(P_1(x)) I l -[_4 - ((x + 2)}5)2 1 3"‘d(x + 2);} I - du(x).
But when we consider that there are two branches of P_1(x), each
contributing equally to our measure, and we have considered only
the positive branch, we get our result: dp.(P_1(x)) I dp.(x).

We use our non-linear recursion relations to find k(n)
and thus to find Qn(x).

From the recurrence formulae of Theorem 2.9 we have
k(2) I 2, k(3) I k(4) I ... I 1. Thus, Qo(x) I 1, Q1(x) I x,

Q2(x) 3x2 - 2 and
(5.3) for n 2 3, Qn(x) I XQn_1(X) ' Qn_2(x)

That Qn(x) I 2 Cos n arc Cos (x/2) satisfies (5.3) may
be seen by setting A I n are Cos (x/2), B I arc Cos (x/2) in the
identity Cos (A +B) + Cos (A - B) I 2 C08 A . Cos B.

Now we use the results of Chapter IV to extend the ortho-

gonality of the Tchebycheff polynomials to other curves.

45
We define 6R as the ellipse
-1 -l 2 -l - 2
[x<R+R ) 1 +1y<R-R )1] =1.

where z I x + iy, and R > 1. We note that 6R has x I i 2
as foci , and that as R -o l, 6R approaches the segment [-2,2].
Since 2 I w + w.1 maps ‘w‘ > R conformally onto the

exterior of 6R with the point at I going into the point at on,

log .11.. - ‘w‘ I G(z,co,6R). That is,
(5.4) 108 % ' ‘(Z ...-“22 _ 4)‘ - G(zaQ96R) 0

According to our work at the beginning of Chapter IV,

 

*
(5.5) g 2 Cos n arc Cos (z/2)-2 Cos m are Cos (2/2) d... (z) I 0
R

9:
if m 9‘ n where (1‘; (2) I %EG(z,oo,C’/r)-‘d2‘. In our next lemma

*
we compute dp, .

i.

Lanna 5.6. dn

 

G(Z,Q,6R) . ‘ 1 ‘ o
4-2
Proof. By (5.4), we must show

.3; 103 ‘2 44:21—44 . _L_ ,
VII?!
If w I r e19 (r > R) corresponds to 2, then C(2,o,6R) I log r -
log R and a; G(z,oo,6R) '% - g—E- . It is thus sufficient to show
that for z 6 6R ,
(5.7) R'1 - 53- - —-1——- .

an " 4-2 |

For 2 Ix + iy 6 6R, we have

(5.8) xIACose, yIBSine

46

where
(5.9) A=R+R' , BIR-R- .

We note that

(5.10) 3% I 1 - R"2 = R'l-B and $21; - 1 + R'2 = R'lA .

Taking the partial derivative with respect to x in both equations

of (5.8) and solving for 23', we get

R -2
(5-11) 3;: R-B-D Cos e ,

2 2 2 2 2
where D I B C08 9 +'A Sin 9.
Similarly, taking the partial derivative with re8pect to

y and solving for gg', we obtain
(5.12) 53- - R-AoD-ZSin e .
ay
Therefore,
(5.13) grad R I R-D-2(B Cos e, A Sin 9) .

We still need to find n, the unit normal to 3R. Since

-2 -2 -2 2 -
(2A x, ZB y) is normal to A x +-b 2y2 I l, we have

- 2 - - - -
n I (4x2A A +'4y B 4) % (2A 2x, 2B 2y). Using (5.8) we have

2 2 - - -
Sin 9) k(A 1Cos e. B 1Sin 9)

n a (A'ZCosZe + B-
= (BZCOSZQ +-AZSin29)-%(B C08 9, A Sin 9)

- D‘las Cos e, A Sin 9) .

47

-2 -1
Thus ill-:- Ii]: ~RoD (B Cos e, A Sin e)-D (B cos 9, A Sin 9)

= D'3(BZCos29 +-A231n29) = D'3-D2 = D"1 .

Now we show —— I D .
‘V 4-: ‘
Since 2 I w +w-1, we have
1"].

- - 2 - - - -
‘4-22‘ 35 I ‘4-(w-lw 1') ‘ 3 I ‘(w-w 1)2‘ I I ‘w-w
I |Reie - R'le'ie"1 = ‘(R-R-1)Cos e +-i(R+R-1)Sin e|-1,

which is ‘A Cos e +-i B Sin e‘-1, which equals D.1 when
i
w I R e 9. This proves the lemma.
Hence, (5.5) reduces to the following relation, first

obtained by Walsh [10]:

n m

(5.14) ng(z)T(z)-—Ldfl—IO if msén,

where Tn(z) is the n-th Tchebycheff polynomial 2 Cox n arc Cos (2/2).

10.

BIBLIOGRAPHY

Brolin, H., Invariant sets under iteration of rational functions.
Ark. Mat. 6, No. 6 (1965), 103-144.

Fine, N.J., On the Walsh functions. Trans. Amer. Math. Soc.
(1939), 372-414.

Hille, E., Analytic Function Theory, Vol. 2, Ginn and Company,
(1962).

Kinney, J.R. and Pitcher, T.S., Some connections between
ergodic theory and the iteration of polynomials. Ark.‘Mat.
8, NO. 4 (1969), 25-320

SzegB, G., fiber orthogonale polynome, die 2u einer gegebenen
kurve der komplexen Ebenen gehBren. Math. Z. 9 (1921),
218-270.

Szeg6, G., A prdblem concerning orthogonal-polynomials. Trans.
Amer. Math. Soc. 37 (1935), 196-206.

SzegB, G., Orthogonal polynomials. Amer. Math. Soc. Colloquium
Publications, Vol. 23 (1967).

Tsuji,iM., Potential Theory in Modern Function Theory.
Maruzen, Tokyo (1959).

Walsh, J.L., A closed set of normal orthogonal functions.
Amer. J. Math. 45 (1923), 5-24.

Walsh, J.L., Note on the orthogonality of Tchebycheff

polynomials on confocal ellipses. Bull. Amer. Math. Soc.
40 (1934), 84-88.

48