Ultraspherical Gauss--Kronrod Quadrature Is Not Possible for $\lambda > 3$

$: Ultraspherical Gauss--Kronrod Quadrature Is Not Possible for $\lambda > 3$$
ULTRASPHERICAL GAUSS–KRONROD QUADRATURE IS NOTPOSSIBLE FOR λ > 3∗

FRANZ PEHERSTORFER† AND KNUT PETRAS‡

SIAM J. NUMER. ANAL. c© 2000 Society for Industrial and Applied MathematicsVol. 37, No. 3, pp. 927–948

Abstract. With the help of a new representation of the Stieltjes polynomial it is shown by usingBessel functions that the Stieltjes polynomial with respect to the ultraspherical weight function withparameter λ has only few real zeros for λ > 3 and sufficiently large n. Since the nodes of the Gauss–Kronrod quadrature formulae subdivide into the zeros of the Stieltjes polynomial and the Gaussiannodes, it follows immediately that Gauss–Kronrod quadrature is not possible for λ > 3. On theother hand, for λ = 3 and sufficiently large n, even partially positive Gauss–Kronrod quadrature ispossible.

Key words. Gauss–Kronrod quadrature, Stieltjes polynomials, orthogonal polynomials, Besselfunctions

AMS subject classifications. 33C10, 33C45, 42C05, 65D32

PII. S0036142998327744

1. Introduction. Let µ be a nonnegative nontrivial measure and let pn(x, dµ) :=pn(x) = xn + . . . , n ∈ N, be the monic polynomials orthogonal with respect to µ, i.e.,

(1.1)

∫Rxjpn(x) dµ(x) = 0 for j = 0, . . . , n− 1.

To estimate the error in Gaussian quadrature, Kronrod [10] suggested the followingquadrature rule in 1964, currently known as Gauss–Kronrod quadrature formulae:

(1.2)

∫Rf(x) dµ(x) =

n∑ν=1

σν,nf(xν,n) +n+1∑κ=1

λκ,n+1f(yκ,n+1) +Rn[f ],

where xν,n are the Gaussian nodes, i.e., the zeros of the orthogonal polynomials pnwith respect to µ and the nodes yκ,n+1 and weights σν,n, λκ,n+1 are chosen so thatthey maximize the polynomial degree of exactness of the quadrature formula. Itturned out, see, e.g., Monegato [12] or Notaris [14], that (as usual, Pn denotes the setof polynomials of degree n or less):

Rn[f ] = 0 for all f ∈ P3n+1

if and only if the nodal polynomial, called the Stieltjes polynomial,

En+1(x, dµ) := En+1(x) :=n+1∏κ=1

(x− yκ,n+1)

∗Received by the editors April 24, 1998; accepted for publication (in revised form) December 29,1998; published electronically February 29, 2000. This work was supported by the Austrian Fonds zurForderung der wissenschaftlichen Forschung, project P12985-TEC. The research was initiated andpartly done during a stay of the second author at Linz, which was supported by the OsterreichischerAkademischer Austauschdienst.

http://www.siam.org/journals/sinum/37-3/32774.html†Institut fur Mathematik, Johannes Kepler Universitat Linz, Altenbergerstrasse 69, A-4040 Linz,

Austria ([email protected]).‡Institut fur Angewandte Mathematik, Technische Universitat Braunschweig, Pockelsstrasse 14,

D-38106 Braunschweig, Germany ([email protected]).

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928 FRANZ PEHERSTORFER AND KNUT PETRAS

satisfies the orthogonality property

(1.3)

∫RxjEn+1(x)pn(x) dµ(x) = 0 for j = 0, . . . , n,

i.e., En+1 is orthogonal with respect to the weight pn(x) dµ(x). Note that, sincethe weight changes sign, the orthogonality property (1.3) will not, in general, implythat the zeros of En+1 are contained in the support supp(µ) of µ; they may even becomplex. By (1.2.), however, of foremost interest are measures µ for which all zeros ofthe Stieltjes polynomial En+1 are simple and in supp(µ). If both of these conditionsare fulfilled we say that Gauss–Kronrod quadrature (1.2) with respect to µ is possiblefor n. By the way (see, e.g., Monegato [12]) if Gauss–Kronrod quadrature is possiblefor n and if the zeros of En+1 and pn strictly interlace, then the quadrature weightsλκ,n+1 in (1.2) are positive. In this case, we say briefly that µ admits partially positiveGauss–Kronrod quadrature.

It was a big surprise when it turned out (see Barrucand [2]) that for the specialcase w(x) = 1, i.e., dµ(x) = dx, on [−1, 1], and zero otherwise, the (Stieltjes) polyno-mials with the orthogonality property (1.3) have already been discovered by Stieltjes[20, pp. 439–441] in 1894. For the Legendre case w(x) = 1, Stieltjes conjectured thatthe zeros of En+1 are simple, located in [−1, 1], and strictly interlace with the zerosof the Legendre polynomial pn. Forty years later, Szego [21] proved Stieltjes’s conjec-ture. In fact he was able to prove the conjecture even for the ultraspherical weightson [−1, 1],

(1.4) w(x) = wλ(x) =(1− x2

)λ−1/2, where 0 ≤ λ ≤ 2;

or in other words, positive Gauss–Kronrod quadrature is partially possible for λ ∈[0, 2] and all n ∈ N. (Later, Monegato [11] proved the positivity of all quadratureweights in (1.2) for all λ ∈ (0, 1].) Thus, the question arises for which other weightfunctions Gauss–Kronrod quadrature is possible; in particular, what about the ultra-spherical case with parameters λ ∈ (−1/2, 0) or λ ∈ (2,∞)? In [16] and [18] the firstauthor has shown that weight functions of the form

(1.5) w(x) =√

1− x2 v(x), where v ∈ C2[−1, 1] and v > 0 on [−1, 1],

admit positive Gauss–Kronrod quadrature for sufficiently large n. In addition, anasymptotic representation of the Stieltjes polynomial En+1 is given. If v vanishes atone of the boundary points ±1, then additional assumptions (see [18, Theorem 11and Corollary 12]) have to be fulfilled such that the asymptotic representation of theStieltjes polynomials holds true on compact subintervals of [−1, 1]. However, for theultraspherical weights (1.4), these additional conditions could not be verified. By acompletely different approach, Ehrich [4] and [5] succeeded in finding an asymptoticrepresentation for the Stieltjes polynomial

(1.6) Eλn+1(x) := En+1(x,wλ)

if 0 ≤ λ ≤ 2 (we simply write En+1(x,wλ) instead of En+1(x, dµ) with µ′ = wλ).For λ > 2, based on numerical experiments, Gautschi and Notaris [8] conjecturedthat wλ admits positive Gauss–Kronrod quadrature for a fixed n ∈ N if and only if λis less than a certain number Λn. Under the assumption that this conjecture holds,they calculated Λn numerically for some n ≤ 40 and always obtained Λn ≥ 4. As the

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ULTRASPHERICAL GAUSS–KRONROD QUADRATURE 929

0.5 1

−1 −0.5x

0.5

−0.5

−1 −0.5 0.5 1

x

−1

Fig. 1. 2200(1− x2)2E4

201(x). Fig. 2. 2199(1− x2)2E4

200(x).

following results show, this conjectured bound has to be modified for large n. Moreprecisely, by deriving a new representation of the Stieltjes polynomials, we are able toprove with the help of Bessel functions that Gauss–Kronrod quadrature is not possiblefor λ > 3 and sufficiently large n. More precisely, we have the following theorem.

Theorem 1. Let λ > 3. Then, for each δ > 0, Eλn+1 has less than δ · n real zerosfor all sufficiently large n.

In section 3, a more detailed description of Eλn+1 is given for λ > 3. There, weshow that for x ∈ [ε− 1, 1− ε], ε > 0,

(1.7) 2nEλn+1(x) = c · nλ−3 ρn(x)

1− x2+ ωn, ρn(x) =

{x if n is even,

1 otherwise,

where c 6= 0 and ωn is an oscillating part of size O(1 + nλ−4) (see Figures 1 and 2).As the next theorem shows, the bound in the assumptions of Theorem 1 is sharp.Theorem 2. For sufficiently large n, all zeros of the Stieltjes polynomial E3

n+1

are simple, lie in (−1, 1), and strictly interlace with the zeros of pn(·, w3).Remarks.(1) Concerning the zeros of Eλn+1 for λ > 3, relation (1.7) shows that for given

ε > 0 and n > n0(ε) the Stieltjes polynomial Eλn+1 has no zeros in [−1+ε,−ε]∪[ε, 1−ε].(2) The case λ ∈ (2, 3) is completely open. In view of the above results, we expect

that Gauss–Kronrod quadrature is possible.(3) According to some numerical tests, we conjecture that the assertion of Theo-

rem 2 is true for all n.Let us note that for two particular sets of parameters, the Jacobi case

(1.8) w(x) = wα,β(x) = (1− x)α(1 + x)β

with corresponding orthogonal polynomials p(α,β)n (x) = xn + · · · can be reduced to

the ultraspherical case. More precisely, we have that the polynomials p(α,−1/2)n are

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related to the ultraspherical polynomials p(α,α)2n by the relations (see Szego [22, eq.

(4.1.4)])

(1.9a) p(α,−1/2)n (2x2 − 1) = (−1)np(−1/2,α)

n (1− 2x2) = cnp(α,α)2n (x)

and

(1.9b) p(α,1/2)n (2x2 − 1) = (−1)np(1/2,α)

n (1− 2x2) =cnxp

(α,α)2n+1(x),

from which it follows (see Monegato [12, eq. (32)]) that

(1.10a) E(α,1/2)n+1 (2x2 − 1) = (−1)n+1E

(1/2,α)n+1 (1− 2x2) = E

(α,α)2n+2(x)

and

(1.10b) E(α,−1/2)n+1 (2x2 − 1) = (−1)n+1E

(−1/2,α)n+1 (1− 2x2) = xE

(α,α)2n+1(x)− dn,

where

dn =

∫ 1

−1

(1− x2

)αp

(α,α)2n (x)E

(α,α)2n+1(x)x2n+1 dx

/∫ 1

−1

(1− x2

)αp

(α,α)2n (x)x2n dx.

Thus, relation (1.10a) and Szego’s result give immediately that wα,β admits Gauss–Kronrod quadrature if (α, β) ∈ {(1/2 , γ) : −1/2 ≤ γ ≤ 3/2} ∪ {(γ , 1/2) : −1/2 ≤γ ≤ 3/2}. Using (1.10b), Rabinowitz [19] has shown by calculating dn (note that it

depends only on dn whether E(α,−1/2)n+1 has all zeros in [−1, 1] or not) that Jacobi–

Gauss–Kronrod quadrature is not possible for even n (odd n) if β = −1/2 and α ∈(1/2 , 3/2) (α = −1/2 and β ∈ (3/2 , 5/2]). Applying Theorem 1 from above andrelation (1.10a), we obtain that Jacobi–Gauss–Kronrod quadrature is not possible ifα = 1/2 and β > 5/2 or if β = 1/2 and α > 5/2.

This paper is organized as follows. In section 2, we first prove a new representationof Stieltjes polynomials. With its help and the help of Bessel functions, Theorem 1 isproved in section 3. Theorem 2 is proved in section 4, where the technique might beapplied usefully in order to obtain further results for integer values of λ.

2. A new representation of the Stieltjes polynomial. It is well known thatthe sequence of polynomials pn orthogonal with respect to µ satisfies a three termsrecurrence relation of the form

(2.1) pn(x) = (x− αn)pn−1(x)− βnpn−2(x), n = 1, 2, . . . ,

where p0(x) := 1 and p−1(x) := 0. The kth monic associated polynomials p(k)n (x) =

xn + . . . , k ∈ N, are defined by the shifted recurrence relation

(2.2) p(k)n (x) = (x− αn+k)p

(k)n−1(x)− βn+kp

(k)n−2(x), n = 1, 2, . . . ,

where p(k)0 (x) := 1 and p

(k)−1(x) := 0. They satisfy the important relation (see, e.g.,

Dickinson [3] or the first author [17])

(2.3)p

(k+1)n−1 (z)

p(k)n (z)

=1

βk+1· Qk(z)

Qk−1(z)+O

(1

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as z →∞ where for z ∈ C \ supp(µ)

(2.4) Qk(z) :=

∫R

pk(x)

z − x dµ(x) = pk(z)Q0(z)− β1p(1)k−1(z), k = 1, 2, . . .

and

Q0(z) :=

∫R

dµ(x)

z − x , Q−1(z) := 1, and β1 :=

∫Rdµ(x).

In view of (2.2) and Favard’s theorem (see Freud [7, Theorem II.1.5]), the polynomials

p(k)n are orthogonal with respect to a positive measure µ(k), which is given by the

relation

(2.5)1

βk+1· Qk(z)

Qk−1(z)=

∫R

dµ(k)(x)

z − x(see Van Assche [23]). Note that in view of (2.3),

(2.6)

∫Rdµ(k)(x) = 1 and p

(k+1)n−1 (z) =

∫R

p(k)n (z)− p(k)

n (x)

z − x dµ(k)(x)

for k ≥ 1. Let us now derive the announced new representation for the Stieltjespolynomial, which is crucial in the sequence.

Theorem 3. The Stieltjes polynomial En+1(·, µ) is given by

(2.7) En+1(x, µ) = pn+1(x, µ)− βn+2

∫R

pn(y, µ)− pn(x, µ)

y − x dµ(n+1)(y).

Proof. By definition, En+1pn is orthogonal to Pn with respect to dµ and hence

En+1pn =n∑j=0

µjp2n+1−j ,

where µ0 = 1 and µj ∈ R. Thus, using the relation

(2.8) p2n+1−j = p(n+1)n−j pn+1 − βn+2p

(n+2)n−j−1pn,

which can be proved by induction arguments, it follows that

(2.9) En+1pn =

n∑j=0

µjp(n+1)n−j

pn+1 − βn+2

n−1∑j=0

µjp(n+2)n−j−1

pn.

Considering (2.9) at the zeros of pn, we obtain

(2.10) pn =n∑j=0

µjp(n+1)n−j ,

which implies again by (2.9) that

(2.11) En+1 = pn+1 − βn+2

n−1∑j=0

µjp(n+2)n−j−1.

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Hence, by (2.10) and (2.6), the assertion is proved.Because of (2.7), we are especially interested in how the kth associated orthogo-

nality measure µ(k) looks. Let us assume that Q0(z) from (2.4) is of the form

Q0(z) =

∫ 1

−1

w(x)

z − x dx for z ∈ C \ [−1, 1],

where w is nonnegative on [−1, 1] and w ∈ Lp[−1, 1], p > 1 or w lnw ∈ L1[−1, 1].Then it follows that (see Nevai [13, eq. (15)] or Peherstorfer [17, Theorem 3.9]) fork ∈ N,

(2.12)1

βk+1· Qk(z)

Qk−1(z)=

∫ 1

−1

w(k)(x)

z − x dx,

where(2.13)

βk+1w(k)(x) =

β1

(p

(1)k−1pk−1 − pkp(1)

k−2

)(x)w(x)[

πpk−1(x)w(x)]2

+[pk−1(x)

∫ 1

−1− w(t)

x− t dt− β1p(1)k−2(x)

]2=

k∏j=1

βj

w(x)

|Q−k−1(x)|2 almost everywhere (a.e.) on [−1, 1]

if Qk−1 has no zeros in C \ [−1, 1] and w(k) ∈ Lp[−1, 1], p > 1 or w lnw ∈ L1[−1, 1].As usual,

∫− denotes the Cauchy principal value and

Q±k (x) := limε→+0

Qk(x± iε)

(the notation “ε→ +0” means that ε→ 0 and ε > 0). Note that by the first relationof (2.4) and the Sokhotskij–Plemelj formula

Q−k (x) +Q+k (x)

2=

∫ 1

−1

− pk(t)w(t)

x− t dt a.e. on [−1, 1]

and

Q−k (x)−Q+k (x)

2= iπpk(x)w(x) a.e. on [−1, 1]

and moreover

(2.14) Q±k (x) =

∫ 1

−1

− pk(t)w(t)

x− t dt∓ iπpk(x)w(x) a.e. on [−1, 1].

Let us mention that at the points x, where (2.14) holds and w(x) > 0, Q±k (x) has no

zeros, since pk and p(1)k−1 have no common zero. Finally, let us point out that in the

ultraspherical case, representation (2.12) with (2.13) hold (see Nevai [13, Example 4])

for the kth associated ultraspherical weight w(k)λ (x), λ > −1/2.

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3. Proof of Theorem 1. In this section, we apply the representation fromTheorem 3 to the weight function wλ with λ > 3. Therefore, we decompose thecorresponding integral into parts each containing one of the difficulties. Namely, forx ∈ (−1, 1), write

rn(t) =w

(n+1)λ (t)√1− t2 , qx(t) =

x+ t

1− x2

and

fx(t) :=

(1−

(1− t2)λ−1(1− x2

)λ−1

)1

t− x − qx(t)

(rn is approximately constant in each compact subinterval of (−1, 1), qx interpolates1t−x at ±1, fx has simple zeros at ±1, and f ′x is a bounded function). The decompo-sition is as follows:∫ 1

−1

pn(t)− pn(x)

t− x w(n+1)λ (t) dt

=

∫ 1

−1

pn(t)qx(t)w(n+1)λ (t) dt

− pn(x)

∫ 1

−1

qx(t)w(n+1)λ (t) dt

+rn(x)(

1− x2)λ−1

∫ 1

−1

pn(t)− pn(x)

t− x(1− t2)λ−1/2

dt

+

∫ 1

−1

(pn(t)− pn(x))

(1− t2)λ−1/2(1− x2

)λ−1· rn(t)− rn(x)

t− x dt

+

∫ 1

−1

(pn(t)− pn(x))w(n+1)λ (t)fx(t) dt

≡(I1 + I2 + I3 + I4 + I5)(x).

The purpose is to show that for λ > 3 the integral I1(x) yields the main contributionto the Stieltjes polynomial if x ∈ [ε− 1, 1− ε], where ε > 0.

Notation. Throughout this section, we set

α = λ− 1

2, N = n+ λ, and t = cos θ, where θ ∈ [0, π].

Furthermore, denote by Kα the modified Bessel function and by H(1)α = Jα + iYα one

of the Bessel functions of the third kind (see Abramowitz and Stegun [1, Chap. 9]).By x → a + 0 (analogously for x → a − 0), we mean that x = a + ε with ε → 0 andε > 0.

We divide the proof into 6 lemmata and a final phase.Lemma 3.1. Let ε > 0. For θ ∈ [0, π − ε], we have

rn(t) =4

π2Nθ |H(1)α (Nθ)|2

· (1 +O(n−1)).D

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Proof. The coefficients β in the three term recurrence relation (2.1) are given by

βn =(n− 1)(n+ 2λ− 2)

4(n+ λ− 1)(n+ λ− 2)and β1 =

√π Γ(λ+ 1/2)

Γ(λ+ 1).

Therefore,

rn(t) = γnsin2α−1 θ

|Q−n (t)|2 ,

where

γn =π

βn+24n+α

Γ(n+ 2λ)Γ(n+ 1)

Γ(n+ λ)Γ(n+ λ+ 1)=

4π

4n+α· (1 +O(n−1)

).

According to Elliot [6], there are even functions As with A0(x) ≡ 1 and odd functionsBs, both being differentiable and bounded, such that |Q−n |2 has the following formalasymptotic expansion:

|Q−n (t)|2 ∼ 41+αΓ2(n+ α+ 1) · 4nN2αn!2

·(

2n+ 2α

n

)−2

· θ sin2α−1 θ

·∣∣∣∣∣Kα(−iNθ + 0)

∞∑s=0

As(−i θ2 + 0)

(2N)2s−Kα+1(−iNθ + 0)

∞∑s=0

Bs(−i θ2 + 0)

(2N)2s+1

∣∣∣∣∣2

(in Elliot’s formula, we set k = N and z = t−0i, i.e., ζ = −(iθ/2)+0). This expansionis uniformly valid except for t in a neighborhood of −1, i.e., it is uniformly valid forθ ∈ [0, π − ε], ε > 0. From

Kα(z) =iπ

2eνiπ/2H(1)

α (iz) for −π < arg z ≤ π

2,

we deduce

|Q−n (t)|2 ∼ 4−α−nπ3n · (1 +O(n−1)) · θ sin2α−1 θ

·∣∣∣H(1)

α (Nθ)∣∣∣2 ∣∣∣∣∣1 +

∞∑s=1

As(−i θ2 )

(2N)2s+θH

(1)α+1(Nθ)

H(1)α (Nθ)

∞∑s=0

Bs(−i θ2 )

iθ(2N)2s+1

∣∣∣∣∣2

for θ ∈ [0, π − ε]. Since the functions Bs are odd, we obtain

|Q−n (t)|2 = 4−α−nπ3n · (1 +O(n−1)) · θ sin2α−1 θ

·∣∣∣H(1)

α (Nθ)∣∣∣2 ∣∣∣∣∣1 +O(n−2) +

θH(1)α+1(Nθ)

H(1)α (Nθ)

O(n−1)

∣∣∣∣∣2

.

We set f(x) =xH

(1)α+1(x)

H(1)α (x)

, so thatθH

(1)α+1(Nθ)

H(1)α (Nθ)

= 1N f(Nθ) and limx→0 |f(x)| = 2α (see

Abramowitz and Stegun [1, eq. 9.1.9]). Furthermore, H(1)α has no zero and the

relation f(x) = O(x) holds for increasing x. Hence,θH

(1)α+1(Nθ)

H(1)α (Nθ)

is uniformly bounded

for θ ∈ [0, π] and all integers n. This yields the assertion.Lemma 3.2. If α ≥ 1/2, then the values rn(x) are uniformly bounded for all

x ∈ [−1, 1] and all integers n. Furthermore, rn(t)−rn(x)t−x is bounded for all n and all

x ∈ [ε− 1, 1− ε], where ε > 0.

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Proof. The function rn is symmetric. Therefore, to prove the boundedness of rn,we only have to show that

g(x) =1

x|H(1)α (x)|2

is bounded in [0,∞). The limit limx→0 |g(x)| exists, H(1)α has no zeros, and we have

limx→∞ |g(x)| = π2 (see Abramowitz and Stegun [1, eqs. 9.1.9 and 9.2.3]), yielding

the first part of the lemma.According to Olver [15, Theorem D], the main term of the derivative of Q−n

asymptotically equals the derivative of the main term of Q−n for increasing argument.This means that we may differentiate the main term on the right-hand side of theequation in Lemma 3.1 in order to obtain the main term of r′n. If t ∈ [ε/2−1, 1−ε/2],we have (see Abramowitz and Stegun [1, eq. 9.1.27])(

|H(1)α (Nθ)|2

)′= 2N (JαJ

′α + YαY

′α) (Nθ)

= −2N (JαJα+1 + YαYα+1) (Nθ) +2α

θ

(J2α + Y 2

α

)(Nθ)︸︷︷︸

=O(N−1)

,

which implies

|r′n(t)| = 2

π2

|Jα(Nθ)Jα+1(Nθ) + Yα(Nθ)Yα+1(Nθ)|θ sin θ|H(1)

α (Nθ)|4+O(1).

We have (cf. Watson [24, sect. 7.51, eq. (3)]),

Jα(z)Jα+1(z) + Yα(z)Yα+1(z) = O(z−2).

Since

|H(1)α (z)|2 =

2

πz· (1 +O(z−1)

)(see Abramowitz and Stegun [1, eq. 9.2.7]), it follows that r′n(t) is bounded indepen-

dently of n. Hence, for x ∈ [ε − 1, 1 − ε] and |t − x| ≤ ε/2 we have rn(t)−rn(x)t−x =

r′n(ξ) = O(1), where ξ ∈ [ε/2− 1, 1− ε/2]. If |t− x| ≥ ε/2, we clearly have∣∣∣∣rn(t)− rn(x)

t− x∣∣∣∣ ≤ 4

ε‖rn‖∞.

This finishes the proof of the second statement of the lemma.Lemma 3.3. Let λ ≥ 1 and let x remain in a compact subinterval of (−1, 1).

Then

2n+1Eλn+1(x) = βn+2

∫ 1

−1

pn(t)qx(t)w(n+1)λ (t) dt+O(1 + nλ−5).

Proof. Define

Mε(f) = sup|x|≤1−ε

|f(x)|.

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For fixed ε > 0, the summand I2(x) is of the same order as Mε(pn) (i.e., of Mε(pn+1))on [ε− 1, 1− ε], since the corresponding integrand is uniformly bounded.

We have

I3(x) =(1− x2

)1−λrn(x)p

(1)n−1(x).

Therefore, taking into consideration (2.14), I3(x) is also of the same order asMε(pn+1).

The terms rn(t)−rn(x)t−x and 2npn(t)

(1 − t2

)λare bounded independently of n.

Hence, also I4(x) is of the same order as Mε(pn+1).In order to estimate I5(x), we note that (see Szego [22, eq. (7.33.6)])

2n∫ 1

−1

|pn(t)fx(t)√

1− t2 | dt = O

(∫ π/2

0

min{nλ, θ−λ} · sin4 θ dθ

)= O(1+nλ−5).

Now, we estimate the integral in Lemma 3.3.Lemma 3.4. Let i ∈ {0, 1} and λ > 3, then

2n∫ 1

0

tipn(t)w(n+1)λ (t) dt =

1

2απ√πNλ−3

∫ ∞0

x1−αJα(x)

|H(1)α (x)|2

dx+O(1 + nλ−4).

Proof. Replacing pn by its asymptotic representation eq. (8.21.17) in Szego [22]and using Lemma 3.1 yields

2npn(t)rn(t) =1

2απ√πNθ sin θ sinα θ|H(1)

α (Nθ)|2·

·(Jα(Nθ) +O

( 1

N|Jα(Nθ)|+ min{θα+2Nα,

√θN−3}

)).

We furthermore replace cosi θ · sin3/2−α θ by θ3/2−α(1 +O(θ2))

and thus obtain

2n∫ 1

0

tipn(t)w(n+1)(t) dt =1

2α√π

∫ π/2

0

Jα(Nθ)√N θα−1|H(1)

α (Nθ)|2dθ + δn,

where

δn = O

(∫ π/2

0

|Jα(Nθ)|(N−1 + θ2) + min{θα+2Nα,√θN−3}√

N θα−1|H(1)α (Nθ)|2

dθ

).

We subdivide this latter integral at the point 1/N and use

Jα(Nθ) =

O((Nθ)α

)if θ < 1/N,

O

(1√Nθ

)if θ ≥ 1/N,

(see Abramowitz and Stegun [1, eqs. 9.1.7 and 9.2.1]) and

1

|H(1)α (Nθ)|2

=

O((Nθ)2α

)if θ < 1/N,

O (Nθ) if θ ≥ 1/N,

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(see Abramowitz and Stegun [1, eqs. 9.1.9 and 9.2.3]). This implies

2n∫ 1

0

tipn(t)w(n+1)(t) dt

=nα−3/2

2απ√π

∫ π/2

0

Jα(Nθ)

(Nθ)α−1|H(1)α (Nθ)|2

dθ +O(1 + nλ−4)

=Nα−5/2

2απ√πNα−5/2

{∫ ∞0

x1−αJα(x)

|H(1)α (x)|2

dx +O

(∫ ∞Nπ/2

x1−α|Jα(x)||H(1)

α (x)|2dx

) }+O(1+nλ−4).

The above estimates for |Jα| and |H(1)α |−2 yield the lemma.

For symmetry reasons, we have uniformly for x ∈ [ε− 1, 1− ε], ε > 0, that

2n∫ 1

−1

pn(t)− pn(x)

t− x w(n+1)(t) dt =2ρn(x)nλ−3

(1− x2)2απ√π

∫ ∞0

x1−αJα(x)

|H(1)α (x)|2

dx+O(1 +nλ−4)

if λ > 3, where

ρn(x) =

{x if n is even,

1 otherwise.

For ε ≤ |x| ≤ 1 − ε, ε > 0, this yields the main contribution to Eλn+1 if the integralon the right-hand side does not vanish. This will be shown in the next two lemmata.

Lemma 3.5. Let α ≥ 0 and Mα := |H(1)α |. Denote furthermore by jα,1 and j′α,1

the first positive zeros of Jα and J ′α, respectively. Then,(i) for x >

√α2 − 1/4,

1

M2α(x)

=π

2

√x2 − (α2 − 1/4) · (1− ε(x)

)−1,

where 0 ≤ ε(x) ≤ 3(α2 − 1/4)

4x4·(

1− α2 − 1/4

x2

)−3

;

(ii) for x ∈ [j′α,1 , jα,1],

J ′′α(x) ≥ −(

1− α2

x2

)Jα(j′α,1)

and

Jα(x) ≥ Jα(α)

(1 + α(x− α)− (x− α)2

2− α2 ln

x

α

);

(iii) for α ≥ 3,

jα,1 > α+3√

3α, Jα−3/2(α) ≥ 1

2Jα−3/2

(α− 3

2

),

and Jα−3/2 > 0 on [α− 3/2, α].

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Proof. (i) Since we have the formal expansion (see Abramowitz and Stegun [1,eq. 9.2.28])

M2α(x) ∼ 2

πx

∞∑k=0

{(k∏µ=1

4α2 − (2µ− 1)2

4x2

)(k∏µ=1

2µ− 1

2µ

)}

and since the remainder does not exceed the (k + 1)th term in absolute value andis of the same sign, provided that k > α − 1/2 (see Abramowitz and Stegun [1, eq.9.2.31]), we obtain

M2α(x) ≤ 2

πx

∞∑k=0

{(4α2 − 1

4x2

)k( k∏µ=1

2µ− 1

2µ

)}=

2

π√x2 − (α2 − 1/4)

and

M2α(x)− 2

π√x2 − (α2 − 1/4)

≥ − 2

πx

∞∑k=2

(4α2 − 1)k−1(4x2)k ( k∏

µ=1

2µ− 1

2µ

)( k∑µ=2

4µ(µ− 1)

)

= − 2

πx· 4α2 − 1

12x4·∞∑k=0

(4α2 − 1

4x2

)k(k + 1)(k + 2)(k + 3) ·

k+2∏µ=1

2µ− 1

2µ

≥ − 2

πx· 4α2 − 1

12x4·∞∑k=0

(4α2 − 1

4x2

)k· 9

4

k∏µ=1

2µ+ 5

2µ

= − 2

π· 3(α2 − 1/4)

4x4· 1√

x2 − (α2 − 1/4)·(

1− 4α2 − 1

4x4

)−3

.

(ii) Section 9.5.2 in Abramowitz and Stegun [1] means that j′α,1 ≥ α and thatJ ′α < 0 on [j′α,1 , jα,1]. The first statement now follows from Bessel’s differentialequation:

J ′′α(x) = −J′α(x)

x−(

1− α2

x2

)Jα(x).

We have for x ∈ [j′α,1 , jα,1] that

Jα(x) = Jα(j′α,1) +

∫ x

j′α,1

∫ t

j′α,1

J ′′α(u) du dt

≥ Jα(j′α,1)

(1−

∫ x

j′α,1

∫ t

j′α,1

(1− α2

u2

)du dt

)

≥ Jα(α)

(1−

∫ x

α

∫ t

α

(1− α2

u2

)du dt

),

which is the right-hand side of the stated inequality.(iii) In the second inequality of (ii), we set x = α+ bα1/3, where b > 0, such that

Jα(x)

Jα(α)≥ 1 + bα4/3 − 1

2b2α2/3 − α2 ln

(1 + bα−2/3

).

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Assuming bα−2/3 < 1 and taking the first three terms in the Taylor expansion ofln(1 + t), we obtain further

Jα(x)

Jα(α)≥ 1− 1

3b3.

This is not negative if 0 ≤ b ≤ 3√

3. For the proof of the second inequality in (iii),apply the second inequality in (ii) again yielding

Jµ(µ+ 3/2)

Jµ(µ)≥ 3

2µ− 1

8− µ2 ln

(1 +

3

2µ

)=: f(µ).

We have

limµ→∞ f(µ) = 1 and f ′′′(µ) =

54

µ(2µ+ 3)3> 0,

such that f is monotonically increasing. Hence, f(µ) ≥ f(3/2) > 12 .

Lemma 3.6. Let α ≥ 3; then∫ ∞0

x1−αJα(x)

M2α(x)

dx > 0.

Proof. Set µ =√α2 − 1/4 and % = α + 3

√3α. Then we have Jα(x) ≥ 0 for

0 < x < ρ. Using Jα(x) ≥ −Mα(x), we obtain∫ ∞0

x1−αJα(x)

M2α(x)

dx

>

∫ ∞µ

π

2

√x2 − (α2 − 1/4) · x1−αJα(x) dx

+

∫ ∞%

(1

M2α(x)

− π

2

√x2 − (α2 − 1/4)

)x1−α · (−Mα(x)

)dx

=π

4µ3−α

∫ ∞1

√t− 1 · t−α/2Jα(µ

√t) dt

−√π

2

∫ ∞%

4√x2 − µ2

(1√

1− ε(x)−√

1− ε(x)

)x1−α dx ≡ I − II,

where ε is defined in Lemma 3.5(i). According to Gradshteyn and Ryzhik [9, eq.(6.592.10)], we have

I =(µπ

2

)3/2

· µ−αJα−3/2(µ).

Furthermore, since Jα−3/2 is first increasing and then decreasing on [α−3/2, jα−3/2,1],we obtain by Lemma 3.5(iii) that

Jα−3/2(µ) ≥ min{Jα−3/2(α− 3/2) , Jα−3/2(α)} ≥ 1

2Jα−3/2(α− 3/2).

For the estimation of II, we first note that for x ≥ %,(1− µ2

x2

)−1

≤(

1− µ2

%2

)−1

=α

23

2 3√

3·

1 + 2 3

√3α2 + 3

√9α4

1 + 12

3

√3α2 + 1

243

√9α4

.

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The second factor in the last term is decreasing for increasing α, such that(1− µ2

x2

)−1

≤(

1 +1

2 3√

3+

3√

3

72

)−1(1 +

13√

3

)2

· α2/3

2 3√

3≤ 3

4α2/3,

which, in particular, means that ε(x) ≤ ( 34

)4. Part (i) of Lemma 3.5 now implies

II ≤√π

2

∫ ∞%

4√x2 − µ2 · 3µ2

4x4

(1− µ2

x2

)−3

· x1−α√1− (3/4)4

dx

≤ 8

7µ2

∫ ∞%

x−5/2−α(

1− µ2

x2

)−11/4

dx

≤ π

6µ2α11/6 · 1

(α+ 3/2)%α+3/2.

The lemma is therefore proved if

Jα−3/2(α− 3/2) >

√8

9π

(µ

%

)α+3/2

· α11/6

(α+ 3/2)µ.

Since α1/3Jα(α) is an increasing function of α (see Watson [24, section 8.54]), we have

Jα−3/2(α− 3/2) ≥ 3

√3

2α− 3J3/2(3/2) ≥

√3

5π(α− 3/2)−1/3.

Considering the last two inequalities we obtain by some easy estimates that we stillhave to show (

µ

ρ

)α+3/2

· α1/6

√40

27< 1.

We estimate as follows:

(µ

ρ

)α+3/2

≤(α

ρ

)α+3/2

≤(

1 +3

√3

α2

)−( 12 +

3√α2

3

)· 3√3α·δα

≤ e− 3√3n ·δα ,

where

δα =

(1 +

3

2α

)·(

1 +1

23

√3

α2

)−1

≥ 81

83> 0.975.

The lemma follows since

maxα

α1/6e−0.975 3√3α =1√

1.95e 6√

3.

Since 2npn+1(x) = O(1) uniformly if x remains in a fixed compact subinterval of(−1, 1), we have proved now that for λ > 3, given ε > 0 and n > n0(ε), the Stieltjespolynomial Eλn+1 has no zeros in Vε := [ε−1,−ε]∪ [ε, 1−ε]. The Stieltjes polynomialhas real coefficients and is symmetric. Therefore each complex zero yν + izν , yν 6= 0

and zν 6= 0 gives rise to a factor[(x− yν)2 + z2

ν

][(x+ yν)2 + z2

ν

] ≥ (x2 − y2ν

)2in the

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factorization of En+1. Now let En+1 have exactly k ≤ n+ 1 real or purely imaginaryzeros, m of them lying in (−∞, ε − 1) and (1 − ε,∞), respectively, and additionally4s = n+ 1− k ≥ 0 zeros ±yν ± izν . Then, for ε ≤ |x| ≤ 1− ε,

|Eλn+1(x)| ≥ (1− ε− x)m(x2 − ε2

)(k−2m)/2(1− ε+ x)m

s∏ν=1

(x2 − y2

ν

)2︸︷︷︸

=: q2(x2), q ∈ Ps

.

According to the above estimates, there is a positive constant c such that

c · nλ−3

2n≥∫ε≤|x|≤1−ε

[(1− ε)2 − x2

]m(x2 − ε2

)(k/2)−mq2(x2) dx

>

∫ (1−ε)2

ε2

[(1− ε)2 − t]m(t− ε2

)(k/2)−mq2(t) dt.

= · · · .

We substitute t =[(1 − 2ε)u + (1 − ε)2 + ε2

]/2 in order to obtain a Jacobi weight

function times the square of a polynomial in the integrand. This yields

· · · =(

1− 2ε

2

)(n+3)/2 ∫ 1

−1

(1− u)m(1 + u)(k/2)−mp2(u) du

with a monic polynomial p ∈ Ps. This may be estimated from below by choosing pas the monic polynomial being orthogonal with respect to the Jacobi weight function(1 − u)m(1 + u)(k/2)−m. The resulting integral can now be calculated explicitly andwe obtain by using equations (4.21.6) and (4.3.3) in Szego [22] that

c · nλ−3

2n≥ 2(1− 2ε)(n+3)/2

n+ 3Γ

(n+ 5− k

4

)Γ

(n+ 5 + k

4

)· Γ(n+ 5− k

4+m

)Γ

(n+ 5 + k

4−m

)·[Γ

(n+ 3

2

)]−2

≥ 2(1− 2ε)(n+3)/2

n+ 3Γ

(n+ 5− k

4

)Γ

(n+ 5 + k

4

)·[Γ

(n+ 5

4

)]2

·[Γ

(n+ 3

2

)]−2

=π(1− 2ε)(n+3)/2

2n(n+ 3)· Γ(n+ 5− k

4

)Γ

(n+ 5 + k

4

)·[Γ

(n+ 3

4

)]−2

.

By Stirling’s formula, this inequality implies that if k ≥ d · n, then we must have forsufficiently large n

(1− d)1−d(1 + d)1+d(1− 2ε)2 ≤ 1.

The latter inequality yields d = O(√ε). Since ε is arbitrary, the theorem is

proved.

4. Proof of Theorem 2. In order to prove Theorem 2, we state a lemma,which might, for example, also yield a more elementary proof of Theorem 1 for integerparameters λ ≥ 4. At least for given λ, we may check an additional condition, whichwould yield the assertion, by using a computer algebra package. For arbitrary λ oneneeds an extra idea.

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Lemma 4.1. Let λ = m+ 1 ∈ N,

δn(z) =m∑ν=0

µν,nz2ν , where µν,n = (−1)m−ν

(m

ν

)( m∏µ=ν+1

n+m+ µ+ 1

n+ µ

)

and denote by yk the zeros of the polynomial δn. Then we may write the Stieltjes

polynomials with respect to the weight function w(x) =(1− x2

)λ−1/2as follows:

(4.1)2n

µ0,nEλn+1(cos θ) · |δn(eiθ)|2 =

m∑ν=0

µν,nTn+1+2ν(cos θ)− gn(cos θ),

where the polynomials gn ∈ P2m−1 satisfy the interpolatory relations

(4.2) 2gn

(y2k + 1

2yk

)= δn(y−1

k )y−n−1k .

Proof. In the proof, denote by c1, c2, and c3 some nonvanishing constants. Westart with the representation in [16]; setting

Φn(z) :=c1z

n+1

Qn((z + z−1)/2

) = 1 +∞∑ν=1

dν,nzν if |z| < 1,

we have

Eλn+1(cos θ) = 2−n< (S∗n+1[Φn](eiθ)− dn+1,n/2).

Here, <(z) denotes the real part of z, Sn+1 denotes the (n+1)th partial sum operatorof the Fourier series, and

S∗n+1[Φn](z) = zn+1 · Sn+1[Φn](1/z).

We note that, for symmetry reasons, dn+1,n/2 = 0 for even n. Denote by Uk thekth Chebyshev polynomial of the second kind. Since λ is an integer, we have (seeSzego [22, eq. (4.9.22)]),

Qn((z + z−1)/2

)=

∫ 1

−1

pλn(x)wλ(x)

z − x dx = c2

m∑ν=1

(−1)m−νµν,n∫ 1

−1

Un+2ν(x)

√1− x2

z − x dx

= c2zn+1

m∑ν=1

(−1)m−νµν,nz2ν = c2zn+1δn(z),

which implies

Φn(z) =c3

δn(z).

Consider the decomposition

δn(z) =m∏ν=1

(z2 − uν).

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Showing that for each uν 6∈ R, its conjugate complex number is also in the set of alluµ, and, since the square roots of uµ are zeros of Qn

((z + z−1)/2

), we know that

|uµ| > 1. We write

(4.3)1

δn(z)=

m∑ν=1

aνz2 − uν , where aν =

m∏µ=1µ6=ν

1

uν − uµ .

Let first n be even. Then,

1

c3Sn+1[Φn](z) =

m∑ν=1

aνz2 − uν

(1− zn+2

u(n+2)/2ν

)

=1

δn(z)−

m∑ν=1

zn+2

u(n+2)/2ν

· aνz2 − uν

=1

δn(z)− 1

δn(z)·m∑ν=1

zn+2

u(n+2)/2ν

·m∏µ=1µ6=ν

z2 − uµuν − uµ

=1

δn(z)· (1− zn+2 intpol(z2)

).

Here, we denote by intpol ∈ Pm−1 the interpolating polynomial (with real coefficients)

with respect to the points (uν , u−(n+2)/2ν ). Thus,

1

c3S∗n+1[Φn](z) = zn+1 · δ−1

n (1

z) (1− z−n−2 intpol(z−2)),

i.e.,

1

c3S∗n+1[Φn](eiθ) = ei(n+1)θ · 1

δn(eiθ)

(1− ei(n+2)θ intpol(e2iθ)

)=

δn(eiθ)

|δn(eiθ)|2(ei(n+1)θ − e−iθ intpol(e−2iθ)

),

yielding

2n

c3Eλn+1(cos θ) · |δn(eiθ)|2 = <

(δn(eiθ)ei(n+1)θ − e−iθδn(eiθ) intpol(e−2iθ)

)=

m∑ν=0

µν,nTn+1+2ν(cos θ)−< (e−iθδn(eiθ) intpol(e−2iθ)).

Comparing the leading coefficients on both sides of the equation, we obtain c3 = µ0,n.Our polynomial gn ∈ P2m−1 is now given by

gn(cos θ) = < (e−iθδn(eiθ) intpol(e−2iθ)),

i.e.,

2gn

(z2 + 1

2z

)=

1

zδn(z) intpol(z−2) + zδn(z−1) intpol(z2).

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Denote by yk the zeros of δn; then

(4.4) 2gn

(y2k + 1

2yk

)= ykδn(y−1

k ) intpol(y2k) = ykδn(y−1

k )y−n−2k .

For odd n, we obtain analogously to the even case that

2n

c3Eλn+1(cos θ) · |δn(eiθ)|2 =

m∑ν=0

µν,nTn+1+2ν(cos θ)−< (e−2iθδn(eiθ) intpol(e−2iθ))

− dn+1,n

2c3δn(eiθ)δn(e−iθ).

We have, using the representation (4.3), that

dn+1,n = −c3m∑ν=1

aν

u(n+3)/2ν

.

Therefore, dn+1,n is −ρn times the main coefficient hn of intpol, which is now the

polynomial in Pm−1 interpolating the data {(uν , u−(n+3)/2ν ); ν = 1, . . . ,m}. As

above, we can show that

2gn

(z2 + 1

2z

)=

1

z2δn(z) intpol(z−2) + z2δn(z−1) intpol(z2)− hn · δn(z)δn(z−1).

We see that the coefficients of z2m and z−2m vanish, such that we have gn ∈ P2m−2.Moreover, the polynomial pn satisfies the interpolating property

(4.5) 2gn

(y2k + 1

2yk

)= y2

kδn(y−1k )y−n−3

k .

Proof of Theorem 2. With some calculation (or easier with some computer al-gebra), we obtain the following asymptotic relations for the quantities used in thelemma. First,

y1 = 1 +3− i√3

2n− 15− 3i

√3

4n2+

21− 2i√

3

2n3+O(n−4),

while y2 = y1 and yk = −yk−2, k = 3, 4. We set

t :=y2

1 + 1

2y1= 1 +

3− 3i√

3

4n2− 9− 9i

√3

2n3+

345− 351i√

3

16n4+O(n−5)

and have

δn(y−11 ) =

36− 36i√

3

n2− 180− 108i

√3

n3+

720− 720i√

3

n4+O(n−5).

Using (1 +

z

n

)n+ z2− z2

12n

= ez(

1 +O( z4

n3

)),

we obtain

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v :=1

2δn(y−1

1 )y−n−1

= e(−3+i√

3)/2

(18− 6i

√3

n2− 54− 18i

√3

n3+

126− 54i√

3

n4+O(n−5)

).

By symmetry, we have the representation

gn(z) =

{anz

2 + bn if n is odd,

anz3 + bnz if n is even.

For odd n, the parameters an and bn are determined by

ant2 + bn = v,

ant2 + bn = v,

which implies

an ==v=t2 and bn =

=(v · t−2)

=(t−2),

where =z denotes the imaginary part of the complex number z. For even n, we obtainanalogously

an ==(v · t−1)

=t2 and bn ==(v · t−3)

=(t−2).

We finally obtain

gn(cos θ) = −(an sin2 θ +

3√

3 sin√

32 + 3 cos

√3

2

n2

)·{

1 if n is odd

cos θ if n is even

}+O(n−3),

where

an = − 4

e√e

[(1 +

3

n

)(√3 sin

√3

2− cos

√3

2

)+

1

n2·(

5√

3 sin

√3

2− kn cos

√3

2

)]+O(n−3)

and

kn =

{3 if n is odd,

6 if n is even.

This means in particular that

|gn(cos θ)| ≤ (0.6 +O(n−1))

sin2 θ +5

n2+O(n−3).

Setting θ = θν = νπn+3 , 1 ≤ ν ≤ (n+ 3)/2, in

2n

µ0,nEλn+1(cos θ) · |δn(eiθ)|2 + gn(cos θ)

=

(−4

n+ 4

n+ 1sin2 θ +

12

(n+ 2)(n+ 1)cos2 θ

)cos(n+ 3)θ

− 6n+ 3

(n+ 2)(n+ 1)sin 2θ sin(n+ 3)θ,

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we have

eν :=(−1)ν+1 2n

µ0,nEλn+1(cos θν) · |δn(eiθν )|2

≥4n+ 4

n+ 1sin2 θν − 12 cos2 θν

(n+ 2)(n+ 1)− 0.6 sin2 θν − 5

n2+O

(sin2 θνn

+1

n3

)≥3.4 sin2 θν − 17

n2+O

(sin2 θνn

+1

n3

).

By sinx > 2πx for 0 ≤ x ≤ π

2 , we obtain eν > 0 for sufficiently large n. Hence, forsufficiently large n, Eλn+1 changes sign between cos θν+1 and cos θν , which yields therequired number of zeros in [0, 1] and therefore by symmetry the required number of

zeros in [−1, 1]. For ϕν := ν−1/6n+3 π, 2 ≤ ν ≤ (n + 3)/2, we obtain the same sign of

Eλn+1◦cos as for θν . Using the explicit representation of the ultraspherical polynomial

P(3)n ,

sin5 θ · P (3)n (cos θ) = const ·

(sin2 θ sin(n+ 3)θ + 3 sin θ cos(n+2)θ

n+2 − 3 sin(n+1)θ(n+2)(n+1)

)= const

(n+2)(n+1)

{ [(n2 + 6n+ 11) sin2 θ − 3

]sin(n+ 3)θ + 3(n+ 3) sin θ cos θ cos(n+ 3)θ

}(see Szego [22, eq. (8.4.13)]), we see that P

(3)n changes sign between cos θν and cosϕν ,

ν = 2, 3, . . . , (n+ 3)/2. This proves the theorem.As announced above, we want to show now how one might prove Theorem 1 for

integer λ ≥ 4Lemma 4.2. Let the assumptions of Lemma 4.1 be valid and denote by zk the

zeros of the polynomial

γ(z) =

m∑ν=0

(2m− νm

)zν

ν!.

Then, the polynomials gn in (4.1) satisfy

gn

(±(

1 +z2k

8n2

)+ o(n−2)

)= (−1)m

(±1)n+1m!

2nmγ(−zk)ezk/2 + o

(n−m

).

Proof. The polynomial δn may be rewritten as

δn(z) = (−1)m(n+m

m

)−1 m∑ν=0

(2m− νm

)(n+ ν

ν

)(1− z2

)ν,

such that

limn→∞n(1− y2

k) = zk.

Inserting this in (4.2), we obtain the assertion.Corollary 4.3. If the function γ in Lemma 4.2 has only simple zeros z1, . . . , zm,

we obtain for the polynomials gn in (4.1) that

gn(x) = n−mhn(x) + sn(x),

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where hn, rn ∈ P2m−1, rn(x) = o(nm−2) and

hn

(±(

1 +z2k

8n2

))= (−1)m

(±1)n+1m!

2γ(−zk)ezk/2, k = 1, . . . ,m.

Remark. Suppose that γ has only simple zeros z1, . . . , zm. Then,

(−1)mm!(8n2)m−1 ·m∑k=1

{γ(−zk)ezk/2

m∏ν=1ν 6=k

(z2k − z2

ν)−1 ·m∏ν=1

(2 +z2k + z2

ν

8n2)−1

}

=(−1)mm!

2(4n2)m−1 ·

m∑k=1

{γ(−zk)ezk/2

m∏ν=1ν 6=k

(z2k − z2

ν)−1

}+O(n2m−4)

is the main coefficient of hn(z), i.e., the coefficient of z2m−1 if n is even and thecoefficient of z2m−2 if n is odd. If the constant

m∑k=1

{γ(−zk)ezk/2

m∏ν=1ν 6=k

(z2k − z2

ν)−1

}

does not vanish, we obtain as at the end of section 3 that a lot of zeros of Eλn+1 arenot real for sufficiently large n. We have verified these properties for λ = 4, 5, . . . , 14.

REFERENCES

[1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, National Bureau ofStandards, Washington, DC, 1964.

[2] P. Barrucand, Integration numerique, abscisse de Kronrod-Patterson et polynome de Szego,C. R. Acad. Sci. Paris Ser. A-B, 270 (1970), pp. 336–338.

[3] D. J. Dickinson, On certain polynomials associated with orthogonal polynomials, Boll. Un.Mat. (3) Ital., 13 (1958), pp. 116–124.

[4] S. Ehrich, Asymptotic properties of Stieltjes polynomials and Gauss-Kronrod quadrature for-mulae, J. Approx. Theory, 82 (1995), pp. 287–303.

[5] S. Ehrich, Asymptotic behaviour of Stieltjes polynomials for ultraspherical weight functions,J. Comput. Appl. Math., 65 (1995), pp. 135–144.

[6] D. Elliot, Uniform asymptotic expansions of the Jacobi polynomials and an associated func-tion, Math. Comp., 25 (1971), pp. 309–315.

[7] G. Freud, Orthogonal Polynomials, Pergamon Press, Oxford, 1971.[8] W. Gautschi and S. Notaris, An algebraic study of Gauss-Kronrod quadrature formulae for

Jacobi weight functions, Math. Comp., 51 (1988), pp. 231–248.[9] I. S. Gradshteyn and I. M. Ryzhik, Table of Series, Integrals and Products, Academic Press,

New York, 1965.[10] A. S. Kronrod, Nodes and Weights for Quadrature Formulae. Sixteen-Place Tables. Nauka,

Moscow, 1964 (in Russian); Consultants Bureau, New York, 1965 (in English).[11] G. Monegato, Positivity of the weights of extended Gauss-Legendre quadrature rules, Math.

Comp., 32 (1978), pp. 243–245.[12] G. Monegato, Stieltjes polynomials and related quadrature rules, SIAM Rev., 24 (1982), pp.

137–158.[13] P. Nevai, A new class of orthogonal polynomials, Proc. Amer. Math. Soc., 91 (1984), pp.

409–415.[14] S. Notaris, An Overview of Results on the Existence or Nonexistence and the Error Term of

Gauss-Kronrod Quadrature Formulae, manuscript.[15] F. W. J. Olver, The asymptotic solution of linear differential equations of the second order

in a domain containing one transition point, Philos. Trans. Roy. Soc. London, Ser. A, 249(1956), pp. 65–97.

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[16] F. Peherstorfer, On the asymptotic behaviour of functions of the second kind and Stieltjespolynomials and on Gauss-Kronrod quadrature formulas, J. Approx. Theory, 70 (1992),pp. 156–190.

[17] F. Peherstorfer, Finite perturbations of orthogonal polynomials, J. Comput. Appl. Math.,44 (1992), pp. 275–302.

[18] F. Peherstorfer, Stieltjes polynomials and functions of the second kind, J. Comput. Appl.Math., 65 (1995), pp. 319–338.

[19] P. Rabinowitz, Gauss-Kronrod integration rules for Cauchy principle value integrals, Math.Comp., 41 (1983), pp. 63–78.

[20] T. J. Stieltjes, Correspondance d’Hermite et de Stieltjes, Vol. II, Gauthier–Villars, Paris,1905.

[21] G. Szego, Uber gewisse Polynome, die zu einer oszillierenden Belegungsfunktion gehoren,Math. Ann., 110 (1935), pp. 501–513.

[22] G. Szego, Orthogonal Polynomials, A.M.S., Providence, RI 1939.[23] W. Van Assche, Orthogonal polynomials, associated polynomials and functions of the second

kind, J. Comput. Appl. Math., 37 (1991), pp. 237–249.[24] G. N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge University Press,

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Ultraspherical Gauss--Kronrod Quadrature Is Not Possible for $\lambda > 3$

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Transcript of Ultraspherical Gauss--Kronrod Quadrature Is Not Possible for $\lambda > 3$