Numer AlgorDOI 10.1007/s11075-013-9807-7

ORIGINAL PAPER

Inequalities for zeros of Jacobi polynomials via Sturm’stheorem: Gautschi’s conjectures

Yen Chi Lun ·Fernando Rodrigo Rafaeli

Received: 26 April 2013 / Accepted: 25 November 2013© Springer Science+Business Media New York 2013

Abstract Let x(α,β)n,k , k = 1, . . . , n, be the zeros of Jacobi polynomials P

(α,β)n (x)

arranged in decreasing order on (−1, 1), where α, β > −1, and θ(α,β)n,k =

arccos x(α,β)

n,k . Gautschi, in a series of recent papers, conjectured that the inequalities

nθ(α,β)n,k < (n + 1)θ

(α,β)

n+1,k

and

(n + (α + β + 3)/2)θ(α,β)

n+1,k < (n + (α + β + 1)/2)θ(α,β)n,k ,

hold for all n ≥ 1, k = 1, . . . , n, and certain values of the parameters α and β. Weestablish these conjectures for large domains of the (α, β)-plane by using a Sturmianapproach.

Keywords Gautschi’s conjectures · Jacobi polynomials · Zeros · Inequalities

Mathematics Subject Classification (2010) Primary 33C45

Research supported by FAPESP, CNPq and CAPES

Y. C. LunUniversidade Estadual de Campinas - UNICAMP, Sao Jose do Rio Preto, Sao Paulo, Brazile-mail: yen.chilun@yahoo.com.tw

F. R. Rafaeli (�)Universidade Estadual Paulista - UNESP, Sao Jose do Rio Preto, Sao Paulo, Brazile-mail: rafaeli@ibilce.unesp.br

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1 Introduction

The inequalities conjectured by Gautschi [5–7] were inspired in a paper by Leopardi[12] that deals with a problem involving error estimates of positive weight quadra-ture in the unit sphere S

d ⊂ Rd+1 when the quadrature rule possesses the property

called quadrature regularity [9, 10]. Leopardi’s approach led to a conjecture on amonotonicity property of rescaled Jacobi polynomials (see [13])

˜P (α,β)n (x) := P (α,β)

n (x)/P (α,β)n (1).

If−1 < x(α,β)

n,n < · · · < x(α,β)

n,1 < 1

denote the zeros of P(α,β)n (x), we can define θ

(α,β)

n,k = arccos x(α,β)

n,k , and therefore,

0 < θ(α,β)

n,1 < · · · < θ(α,β)n,n < π.

Leopardi suggested: for α ≥ 1, n ≥ 1 and θ ∈(

0, θ(α,α−1)1,1

)

,

˜P (α,α−1)n

(

cosθ

n

)

< ˜P(α,α−1)n+1

(

cosθ

n + 1

)

. (1)

The most natural attempt to prove the inequality (1) is to check that

nθ(α,α−1)n,1 < (n + 1)θ

(α,α−1)n+1,1 . (2)

This was exactly what Gautschi and Leopardi observed and formulated in [4]. Inaddition, they investigated the analogous problems for general Jacobi polynomials:for n ≥ 1 and α, β > −1, consider

˜P (α,β)n

(

cosθ

n

)

< ˜P(α,β)

n+1

(

cosθ

n + 1

)

(3)

for either of the two intervals

0 < θ < θ(α,β)

n,1 , 0 < θ < π,

or equivalently

nθ(α,β)

n,1 < (n + 1)θ(α,β)

n+1,1. (4)

Koumandos [11] proved (3) when n ≥ 1, θ ∈ (0, π) and (α, β) coincides withone of the points (1/2, 1/2), (1/2, −1/2) and (−1/2, 1/2).

Note that the inequality (4) involves only the largest zero, x(α,β)

n,1 = cos(

θ(α,β)

n,1

)

, ofthe n-th Jacobi polynomial. So a natural question arises: does the inequality (4) holdfor all the zeros of Jacobi polynomial? (Or does such an inequality hold for otherzeros of Jacobi polynomials?) If so, in what domains of the (α, β)-plane is it true?

That was exactly what Gautschi proposed in [6]: determine, numerically, thedomain of validity in the (α, β)-plane of the inequalities

nθ(α,β)

n,k < (n + 1)θ(α,β)

n+1,k, k = 1, 2, . . . , n. (5)

Gautschi tested the inequalities computationally using a bisection method imple-mented in Matlab. The strategy was checked when (5) holds for all n ≤ N for the

Numer Algor

cases N = 50, N = 100 and N = 200. Of course, in these cases, the domain ofvalidity of (5) depends on N . His conclusions are shown in Fig. 1:

Based on the above results, Gautschi stated that when N → ∞, there is a domainlimit in the (α, β)-plane in which the inequalities (5) hold unrestrictedly for all n. SeeFig. 2.

So, one of Gautschi’s conjectures is:

Conjecture 1 For all n ≥ 1, the inequalities

nθ(α,β)n,k < (n + 1)θ

(α,β)

n+1,k

hold for each k, k = 1, . . . , n, and (α, β) in the region S as described in Fig. 2.

Following a suggestion from Askey (see [7, Section 1]), Gautschi examined com-putationally whether similar inequalities hold when the factor n is replaced byn+ (α +β + 1)/2 because it naturally arises in the theory of Jacobi polynomials andtheir zeros. Gautschi stated the following:

Conjecture 2 For all n ≥ 1, and 1 ≤ k ≤ n, the inequalities

(n + (α + β + 3)/2)θ(α,β)

n+1,k < (n + (α + β + 1)/2)θ(α,β)n,k , (6)

hold for (α, β) in �, where

� = {(α, β) : α, β > −1, α �∈ (−1/2, 1/2) and β �∈ (−1/2, 1/2)}\ {(±1/2, ±1/2)} . (7)

Another conclusion of Gautschi was: for (α, β) ∈ R1/2, where R1/2 = {(α, β) :|α| ≤ 1/2, |β| ≤ 1/2, α2+β2 �= 1/2}, the inequality (6) holds with > replaced by <

for all n = 1, 2, 3, . . ., and 1 ≤ k ≤ n. In fact, this conjecture is already establishedby Ahmed, Laforgia and Muldoon in Theorem 3.1 (ii) of [1] by means of a new andmore general version of the Sturm comparison theorem, proved in the same paper.See also comment [8].

Fig. 1 Numerical tests made byGautschi [6] to determine thedomain of validity of (5): Hesaid that for 1 ≤ n ≤ N (eitherN = 50, N = 100 andN = 200), the domain ofvalidity of (5) is the domainbounded above by BN , on theleft by the vertical segment atα = −1 between BN andK = {(α, β) : −1 < α <

−0.5, β = −α − 1}, and belowby K followed by CN . The curveBN turns downward and thecurve CN upward as N increases

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Fig. 2 Gautschi observed alsothat the curves BN and CN inFig. 1 very likely tend towardshorizontal lines β = ±0.5 asN → ∞. If so, the inequalities(5) are expected to be valid forall n in the horizontal strip S ={(α, β) : α > −1, |β| ≤ 0.5} cutoff on the left by the linesegment K, and with the point(−0.5,−0.5) removed

We establish the Conjecture 1 for large regions of the (α, β)-plane and we provethe Conjecture 2 (Figs. 3 and 4).

2 Sturm’s comparison theorem and a descendent

The basic tool in our approach is Sturm’s comparison theorem for zeros of solutionsof Sturm-Liouville differential equations. In Szego’s book [13, p. 19] we find thefollowing version:

Theorem 1 (Sturm’s comparison theorem) Let y(x) and Y(x) be solutions of thedifferential equations

y ′′(x) + f (x)y(x) = 0 (8)

and

Y ′′(x) + F(x)Y (x) = 0, (9)

Fig. 3 The graph of �

concerning the validity of (6)

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Fig. 4 For n ≤ N (either N = 5and N = 50), the inequalitiesnθ

(α,β)n,k < (n + 1)θ

(α,β)

n+1,k ,k = 1, . . . , n, hold in the regionbounded by BN , on the left bythe vertical line α = −0.5 andbelow by the horizontal lineβ = −0.5

where f, F ∈ C(a, b) and f (x) ≤ F(x) in (a, b). Let x1 and x2, with a < x1 <

x2 < b be two consecutive zeros of y(x). Then the function Y(x) has at least onevariation of sign in the interval (x1, x2) provided f (x) �≡ F(x) there. The statementholds also:

• for x1 = a if

y(a + 0) = 0 or limx→a+

{y ′(x)Y (x) − y(x)Y ′(x)} = 0; (10)

• for x2 = b if

y(b − 0) = 0 or limx→b−

{y ′(x)Y (x) − y(x)Y ′(x)} = 0. (11)

We present a direct consequence of Sturm’s comparison theorem:

Corollary 1 Let f, F ∈ C(a, b) and y(x) and Y(x) be solutions of the differentialequations (8) and (9) and both have n and m distinct zeros in (a, b), respectively. Letthe zeros of y(x) in (a, b) be x1 < x2 < · · · < xn and those of Y(x) be X1 < X2 <

· · · < Xm.

(i) Suppose that y(x) and Y(x) satisfies

Y(a + 0) = 0 or limx→a+

{y ′(x)Y (x) − y(x)Y ′(x)} = 0, (12)

and

y(b − 0) = 0 or limx→b−

{y ′(x)Y (x) − y(x)Y ′(x)} = 0. (13)

If m ≤ n and, in addition, if there exists η ∈ (a, b) such that f (η) = F(η) andF(x) − f (x) < 0 for x ∈ (a, η) and F(x) − f (x) > 0 for x ∈ (η, b), thenxk < Xk for every k = 1, . . . , m;

(ii) Suppose that y(x) and Y(x) satisfies

y(a + 0) = 0 or limx→a+

{y ′(x)Y (x) − y(x)Y ′(x)} = 0, (14)

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and

Y(b − 0) = 0 or limx→b−

{y ′(x)Y (x) − y(x)Y ′(x)} = 0 (15)

If n ≤ m and, moreover, if there exists η ∈ (a, b) such that f (η) = F(η) andF(x) − f (x) > 0 for x ∈ (a, η) and F(x) − f (x) < 0 for x ∈ (η, b), thenxk > Xk for every k = 1, . . . , n.

Proof Without loss of generality we prove only item (i) because the arguments inthe other cases are similar.

We suppose that y(x) possesses i zeros in (a, η) and n − i zeros in [η, b), that is

a < x1 < x2 < . . . < xi < η ≤ xi+1 < . . . < xn < b.

The cases i = 0 and i = n are immediate from Theorem 1, so that we consider1 < i < n.

First we shall prove that xj < Xj for j = 1, . . . , min{i, m}. Assume the contrary,that there is a j0 with 1 ≤ j0 ≤ min{i, m}, such that Xj0 < xj0 . Since F(x) < f (x)

for x ∈ (a, xj0) then, by Theorem 1, the function y(x) would change sign at leastonce in all the intervals (a, X1), . . . (Xj0−1, Xj0). This means that xj0 < Xj0 is acontradiction. Therefore xj < Xj for j = 1, ..., min{i, m}.

Note that if min{i, m} = m, the proof is complete. Now suppose that min{i, m} =i < m. We shall show that xj < Xj for j = i + 1, . . . , m. By Theorem 1, sincef (x) < F(x) for x ∈ (xi+1, b), Y(x) changes sign at least once in (xn, b), at leasttwice in (xn−1, b), and so on, at least n − i times in (xi+1, b). Hence,

Xm ∈ (xn, b) ⇒ Xm > xn > xm,

Xm−1 ∈ (xn−1, b) ⇒ Xm−1 > xn−1 > xm−1,...

Xi+1 ∈ (xi+1, b) ⇒ Xi+1 > xi+1,

that is, Xm > xm, Xm−1 > xm−1, and so on, until Xi+1 > xi+1.

The arguments of the proof were based on [1]. The particular case n = m reducesto Theorem 4.1 in [2].

3 Preliminary technical results

In this section we provide a technical lemma. It consists of analysing a particularfunction that appear naturally in the proof of Theorem 1 present in the next section.

We begin defining the function ˜fn(θ; h, α, β) by

˜fn(θ; h, α, β) = 1/4 − α2

4h2 sin2(θ/2h)+ 1/4 − β2

4h2 cos2(θ/2h)+

(

n

h+ α + β + 1

2h

)2

. (16)

Lemma 1 Let n ≥ 1. It follows:

(i) If α ≥ 1/2 and β > 1/2 then ˜fn+1(θ; n + 1, α, β) − ˜fn(θ; n, α, β) changessign only once in the interval (0, nπ) and it is from negative to positive;

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(ii) If α ≥ 1/2 and −1/2 < β ≤ 1/2 then ˜fn+1(θ; n+1, α, β)− ˜fn(θ; n, α, β) <

0 in (0, nπ);(iii) If (α, β) ∈ {(α, β) : α, β ≥ 1/2}/{(1/2, 1/2)} then

˜fn+1(θ; γn+1, α, β) − ˜fn(θ; γn, α, β) > 0

in (0, γnπ), where γn = n + (α + β + 1)/2.

Proof We define the auxiliary function R(θ) by

R(θ) = ˜fn+1(θ; n + 1, α, β) − ˜fn(θ; n, α, β)

= Rα(n, θ) + Rβ(n, θ) − (α + β + 1)(4n2 + 2n(α + β + 3) + α + β + 1)

4n2(n + 1)2,

where

Rα(n, θ) := 4α2 − 1

16

{

1

n2

[

sin

(

θ

2n

)]−2

− 1

(n + 1)2

[

sin

(

θ

2n + 2

)]−2}

Rβ(n, θ) := 4β2 − 1

16

{

1

n2

[

cos

(

θ

2n

)]−2

− 1

(n + 1)2

[

cos

(

θ

2n + 2

)]−2}

.

In order to prove item (i), we consider α ≥ 1/2 and β > 1/2. Thus

limθ→0+ R(θ) = −9α + 12n2 + 15

4n2(n + 1)2

(

β + (α + 1)[

α + 4(

n2 + 1)]

3α + 4n2 + 5

)

< 0

and

limθ→nπ− R(θ) = +∞.

So it is sufficient to prove that R(θ) increases in (0, nπ). Let s(u) and r(u) be definedby

s(u) = u3 cos u

sin3 uand r(u) = u3 sin u

cos3 u.

Calculating the derivative of R(θ) with respect to θ we obtain

dR(θ)

dθ= 4α2 − 1

2θ3[s(u2) − s(u1)] + 4β2 − 1

2θ3[r(u1) − r(u2)],

where u1 = θ/2n and u2 = θ/(2n + 2). Observe that the positivity of dR(θ)/dθ

would follow from the inequalities

s(u2) > s(u1) (17)

and

r(u1) > r(u2). (18)

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We shall prove that s(u) decreases and r(u) increases when u increases in(0, π/2). We have

ds(u)

du= u2[3 sin u cos u − u sin2 u − 3u cos2 u]

sin4 u

anddr(u)

du= u2[3 sin u cos u − 2u cos2 u + 3u]

cos4 u.

Using the Taylor expansion of cos u and sin u around u = 0 we obtain, for 0 < u <

π/2, the following inequalities:

cos u < 1, cos u > 1 − u2

2and cos u < 1 − u2

2+ u4

24,

and

sin u > u − u3

6and sin u < u − u3

6+ u5

120.

Obviously, for 0 < u < π/2,

3 sin u cos u − u sin2 u − 3u cos2 u < 3

(

u − u3

6+ u5

120

)(

1 − u2

2+ u4

24

)

− u

(

u − u3

6

)2

− 3u

(

1 − u2

2

)2

= u5p(u2)

2880< 0,

where p(u) = −48 − 176u+ 3u2 has one positive zero given by 4(22 + √493)/3 ≈

58.9381 which is greater than π2/4, and

3 sin u cos u−2u cos2 u+3u > 3

(

u − u3

6

)(

1 − u2

2

)

−2u+3u = u(u2 − 4)2

4> 0.

Summarizing, we conclude that ds(u)/du < 0 and dr(u)/du > 0 for u ∈ (0, π/2).Hence, s(u1) < s(u2) and r(u1) > r(u2) which are exactly the inequalities (17) and(18). This completes the proof of item (i).

To prove item (ii), we need to show, for α ≥ 1/2 and −1/2 < β ≤ 1/2, thatR(θ) < 0 in (0, nπ).

Note that Rβ(0) = (4β2 − 1)(2n + 1)/n2(n + 1)2 ≤ 0 and, for θ ∈ (0, nπ),

dRβ(θ)

dθ= 4β2 − 1

2θ3[r(u1) − r(u2)] ≤ 0,

where u1 = θ/2(n + 1), u2 = θ/2n and r(u) is defined as above. Then Rβ(θ) ≤ 0em (0, nπ). On the other hand,

d

dθ

[

Rα(θ) − (α + β + 1)(4n2 + 2n(α + β + 3) + α + β + 1)

4n2(n + 1)2

]

= 4α2 − 1

2θ3 [s(u2) − s(u1)] > 0

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in (0, nπ), where s(u) is defined as above, and

Rα(nπ) − (α + β + 1)(4n2 + 2n(α + β + 3) + α + β + 1)

4n2(n + 1)2

= a2(n, α)(

β + 1

2

)2 + a1(n, α)(

β + 1

2

)

+ a0(n, α), (19)

where

a2(n, α) = − 3n + 1

2n(n + 1)2< 0,

a1(n, α) = −2α + 6nα + 5n + 3

2n(n + 1)2 < 0,

a0(n, α) = b2(n)

(

α − 1

2

)2

+ b1(n)

(

α − 1

2

)

+ b0(n),

with

b2(n) = −5n2 − 1 + n2

(

csc2(

πn2n+2

))

4n2(n + 1)2< 0,

b1(n) = −15n2 + n2

(

csc2(

πn2n+2

))

+ 6n − 1

4n2(n + 1)2< 0,

b0(n) = − 5n + 3

2n(n + 1)2 < 0.

Observe that the free coefficient a0(n, α) of the quadratic in (19) does not have anyzero bigger than 1/2, by means of Descartes’ rule of signs, so a0(n, α) < 0 forα ≥ 1/2. Using the same argument, we conclude that the polynomial (19) possessesno real zero bigger than −1/2, thus (19) is negative for β > −1/2.

Summarizing, for α ≥ 1/2 and −1/2 < β ≤ 1/2, the function Rβ(θ) decreasesin (0, nπ) and Rβ(0) ≤ 0, so Rβ(θ) ≤ 0 for θ ∈ (0, nπ), and the function

Rα(θ) − (α + β + 1)(4n2 + 2n(α + β + 3) + α + β + 1)

4n2(n + 1)2(20)

increases in (0, nπ) and assumes a negative value at the point θ = nπ , therefore (20)is negative for all θ in (0, nπ). This completes the proof of item (ii).

The proof of item (iii) follows almost immediately from some easy calculations.We have

˜fn+1(θ; γn+1, α, β) − ˜fn(θ; γn, α, β)

= 1/4 − α2

4

{

1

γ 2n+1

[

sinθ

2γn+1

]−2

− 1

γ 2n

[

sinθ

2γn

]−2}

+1/4 − β2

4

{

1

γ 2n+1

[

cosθ

2γn+1

]−2

− 1

γ 2n

[

cosθ

2γn

]−2}

.

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So the inequality ˜fn+1(θ; γn+1, α, β) > ˜fn(θ; γn, α, β) holds for (α, β) ∈ {(α, β) :α, β ≥ 1/2}/{(1/2, 1/2)} and 0 < θ < γnπ if

1

γ 2n+1

[

sinθ

2γn+1

]−2

− 1

γ 2n

[

sinθ

2γn

]−2

< 0

and

1

γ 2n+1

[

cosθ

2γn+1

]−2

− 1

γ 2n

[

cosθ

2γn

]−2

< 0.

Having in mind that cos u and sin u are positive in (0, π/2), the last inequalitieswould follow from

γn sinθ

2γn

< γn+1 sinθ

2γn+1(21)

and

γn cosθ

2γn

< γn+1 cosθ

2γn+1. (22)

Since 0 < γn < γn+1 and cos u is positive and a decreasing function in (0, π/2) theinequality (22) holds. Let g(θ) be defined by

g(θ) := γn+1 sinθ

2γn+1− γn sin

θ

2γn

.

Then g(0) = 0 and

g′(θ) = 1

2

[

cosθ

2γn+1− cos

θ

2γn

]

> 0

in (0, γnπ). Thus 0 = g(0) < g(θ) which implies the inequality (21). The proof ofLemma 1 is complete.

4 Our results on Gautschi’s conjecture

In this section we present some results in order to prove Gautschi’s conjecture,described in the introduction. We begin by finding a function whose zeros areh θ

(α,β)

n,k , h being a parameter. Recall that [13, p. 67] the function

yn(θ; α, β) =(

sinθ

2

)α+1/2 (

cosθ

2

)β+1/2

P (α,β)n (cos θ)

is a solution of the Sturm-Liouville differential equation

y ′′n(θ; α, β) + fn(θ; α, β) yn(θ; α, β) = 0,

where

fn(θ; α, β) = 1/4 − α2

4 sin2(θ/2)+ 1/4 − β2

4 cos2(θ/2)+

(

n + α + β + 1

2

)2

.

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Then a straightforward change of variables implies that

zn(θ; h, α, β) := yn(θ/h; α, β) =(

sinθ

2h

)α+1/2 (

cosθ

2h

)β+1/2

P (α,β)n

(

cosθ

h

)

,

whose zeros in (0, hπ) are hθ(α,β)

n,1 < hθ(α,β)

n,2 < · · · < hθ(α,β)n,n , is a solution of

z′′n(θ; h, α, β) + ˜fn(θ; h, α, β) zn(θ; h, α, β) = 0,

where˜fn(θ; h, α, β) = fn(θ/h; α, β)

h2

coincides with (16), i.e.,

˜fn(θ; h, α, β) = 1/4 − α2

4h2 sin2(θ/2h)+ 1/4 − β2

4h2 cos2(θ/2h)+

(

n

h+ α + β + 1

2h

)2

.

4.1 About conjecture 1

We state the following:

Theorem 2 The inequality

nθ(α,β)

n,k < (n + 1)θ(α,β)

n+1,k

holds for every n ≥ 1 and each k, k = 1, . . . , n, and for every (α, β) in the region

Sn = {(α, β) : −1/2 ≤ α, β ≤ 1/2} ∪ {

(α, β) : β > −1/2, α ≥ max{1/2, δn(β)}},where δn(β) is given by

δn(β) =2(β + 1)(β + 3) −

[

2n(n + β + 3) + (β + 2)(β + 3)][

1 + cos(

nπ/(n + 1))

]

(2n + β + 3)[

1 + cos(

nπ/(n + 1))

] .

Proof The function y = zn(θ; n, α, β) has exactly n zeros

nθ(α,β)

n,1 < nθ(α,β)

n,2 < · · · < nθ(α,β)n,n

in (a, b) = (0, nπ).Driver and Jordaan [3] proved that,

cos(

θ(α,β)

n+1,n+1

)

= x(α,β)

n+1,n+1 < −1 + 2(β + 1)(β + 3)

2n(2n + α + β + 3) + (β + 3)(α + β + 2).

So, for α ≥ δn(β), we obtain nπ ≤ (n + 1)θ(α,β)

n+1,n+1. Therefore, the function Y =zn+1(θ; n + 1, α, β) has at most n zeros in (0, nπ).

Obviously zn(θ; n, α, β) and zn+1(θ; n + 1, α, β) satisfy the requirements (12)and (13) for every fixed α > −1/2 and β > −1/2.

• For α ≥ 1/2 and β > 1/2, item (i) of Lemma 1 states that ˜fn+1(θ; n+1, α, β)−˜fn(θ; n, α, β) changes sign only one time in the interval (0, nπ) and it is fromnegative to positive. Then Corollary 1 implies the desired result.

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• For α ≥ 1/2 and −1/2 < β ≤ 1/2, item (ii) says that ˜fn+1(θ; n + 1, α, β) −˜fn(θ; n, α, β) < 0 in (0, nπ). Applying Theorem 1 we obtain the inequalitiesnθ

(α,β)n,k < (n + 1)θ

(α,β)

n+1,k.

• To establish nθ(α,β)n,k < (n + 1)θ

(α,β)

n+1,k for −1/2 ≤ α, β ≤ 1/2, we use part (ii) ofTheorem 3.1 by S. Ahmed, A. Laforgia and M. E. Muldoon [1] which reads asfollows: If α2 ≤ 1/4 and β2 ≤ 1/4, then

(n + (α + β + 1)/2)θ(α,β)n,k < (n + (α + β + 3)/2)θ

(α,β)

n+1,k

for every n ≥ 1 and k = 1, . . . , n. On the other hand

θ(α,β)n,k

θ(α,β)

n+1,k

<n + (α + β + 3)/2

n + (α + β + 1)/2≤ 1 + 1

n

hold for α, β ∈ [−1/2, 1/2]. This completes the proof of Theorem 1.

Remark 1 We show a figure that shows the domain Sn of the validity of the inequal-ities nθ

(α,β)n,k < (n + 1)θ

(α,β)

n+1,k. The analysis is performed for the cases N = 5 andN = 50.

Note that the region Sn approaches to

S∞ = {(α, β) : −1/2 ≤ α, β ≤ 1/2} ∪ {(α, β) : α ≥ 1/2, −1/2 < β ≤ l},

for largest values of n, where l = −2 +√

12

(

2 + π2) ≈ 0.436145 is a positive real

number. The proof consists of proving the two statements below.

(i) The function δn(β) − l is positive for β > l and goes to {(α, β) : β =0 and 1/2 > α};

(ii) δn(−1/2) < 0 for every n ≥ 1.

In fact, the numerator of δn(β) is a polynomial in β of degree 2 with positive leadingcoefficient, so by Descartes rule of sign this quadratic has two real zeros with oppo-site signs. The denominator of δn(β) is positive for all n ≥ 1 and β > −1. Sincecot(x) < 1/x for x ∈ (0, π/4) we have

δn(l) = −2n (2n + 2) + (2n + 1)

√

2(

2 + π2) + 2 − π2 cot2

(

π2n+2

)

4n +√

2(

2 + π2) + 2

< −(2n + 1)

(

−2 +√

2(

2 + π2)

)

4n + 2 +√

2(

2 + π2)

.

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On the other hand

−(2n + 1)

(

−2 +√

2(

2 + π2)

)

4n + 2 +√

2(

2 + π2)

|n=1 ≈ −0.792553 <−1

2,

and

d

dx

[

−(2x + 1)

(

−2 +√

2(

2 + π2)

)

4x + 2 +√

2(

2 + π2)

]

= −4

(

2 + π2 −√

2(

2 + π2)

)

(

4x +√

2(

2 + π2) + 2

)2< 0.

In other words, the expression above assumes a negative value less than −1/2 at thepoint n = 1 and decreases when n increases. Thus δn(l) < −1/2 which means thatthe positive zeros of δn(β) are greater than l for all positive integers n. Now, we shallprove that such a positive zero converges to l when n goes to infinity. We denote thisby

β+(n) :=8 − (2n + 5)κ(n) −

√

4 + (

4n2 + 4n − 1)

κ(n) [6 − κ(n)]

2 (−2 + κ(n)),

where κ(n) := 1 + cos(

nπ/(n + 1))

, the positive zero of δn(β). Using the limitsrelation

limn→∞ n

[

1 + cos( nπ

n + 1

)]

= 0+

and

limn→∞ n2

[

1 + cos( nπ

n + 1

)]

= π2

2

we verify easily that β+(n) goes to l when n → ∞. Summarizing, for β > l, thegraph of δn(β) is to the right-hand side of the vertical line β = l and goes to it whenn goes to infinity. Finally, δn(−1/2) < 0 follows from

δn(−1) = −2n(n + 2) + 2

2n + 2< 0.

Comparing the regions S conjectured by Gautschi showed in Fig. 2 and the regionS∞ obtained by Theorem 1 we have that S∞ ⊂ S. See Fig. 5.

Corollary 2 Let n ≥ 1. Then, for (α, β) in Sn,

˜P (α,β)n (cos(θ/n)) < ˜P

(α,β)

n+1 (cos(θ/(n + 1))) ,

holds in θ ∈(

0, (n + 1)θ(α,β)

n+1,1

)

.

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Fig. 5 The region S∞

4.2 About conjecture 2

We assert the following:

Theorem 3 The inequality

(n + (α + β + 3)/2)θ(α,β)

n+1,k < (n + (α + β + 1)/2)θ(α,β)n,k

holds for every n ≥ 1 and each k, k = 1, . . . , n, and for (α, β) in � as defined in (7)

Proof For this purpose, we apply Theorem 1 with y = zn(θ; γn, α, β) and Y =zn+1(θ; γn+1, α, β) in (0, γnπ), where γn = n + (α + β + 1)/2. For α, β ∈ �, wehave to prove that zn(θ; γn, α, β) and zn+1(θ; γn+1, α, β) satisfy

limθ→0

[

zn(θ; γn, α, β)z′n+1(θ; γn+1, α, β) − z′

n(θ; γn, α, β)zn+1(θ; γn+1, α, β)] = 0.

This is evident for α > 1/2. For −1 < α < −1/2 we use the asymptotic relations(

sin(θ/2γn))α+1/2 ∼ (

θ/2γn

)α+1/2 and(

cos(θ/2γn))β+1/2 ∼ 1,

when θ is sufficiently small. So by Lemma 1 item (iii) and Theorem 1 we concludethe proof.

Finally, we mention an interesting consequence of Theorems 1 and 2 and theasymptotic formula (see [13, Theorem 8.1.2])

limn→+∞ nθ

(α,β)n,k = j

(α)k ,

where j(α)k is the kth positive zero of the Bessel function Jα(z).

Corollary 3 Let n ≥ 1. Then, for k = 1, . . . , n, the inequalities

j(α)k

n + (α + β + 1)/2< θ

(α,β)n,k ,

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hold for (α, β) in �, and

θ(α,β)

n,k <j

(α)k

n,

hold for (α, β) in Sn.

Acknowledgment The authors kindly thank the referees for their valuable suggestions.

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