Revista Colombiana de Matemáticas Volumen 32 (1998), páginas 133–142

Two new conjectures concerning positive Jacobi polynomials sums

Dimitar K. Dimitrov^{∗} & Clinton A. Merlo^{†} Universidade Estadual Paulista, Brasil

Abstract. A refinement of a conjecture of Gasper concerning the values of (α, β), −1/2 < β < 0, −1 < α + β < 0, for which the inequalities

∑_{k=0}^{n} P_{k}^{(α,β)}(x)/P_{k}^{(β,α)}(1) ≥ 0, −1 ≤ x ≤ 1, n = 1, 2, …

hold, is stated. An algorithm for checking the new conjecture using the package Mathematica is provided. Numerical results in support of the conjecture are given and a possible approach to its proof is sketched. Keywords and phrases. Jacobi polynomials, positive sums, Bessel functions, discriminant of a polynomial.

1991 Mathematics Subject Classification. Primary 33C45.

1. Introduction

The Jacobi polynomials are defined in terms of the hypergeometric function 2F_{1} by

P_{n}^{(α,β)}(x) = \frac{(α + 1)_{n}}{n!}_{2}F_{1}(−n, n + α + β + 1; α + 1; (1 − x) /2),

∗Research supported by Brazilian Science Fundation CNPq under Grant 300645/95-3.

†Research supported by a fellowship of the Brazilian Science Fundation CAPES.

where (a)_{k} = Γ (a + k) /Γ (a) is the Pochhamer symbol and

2F_{1}(a, b; c; z) = ∑_{k=0}^{∞} \begin{matrix}(a) & k & (b) & k \\ (c) & k \end{matrix} \frac{z^{k}}{k!} .

Various special cases of the inequalities

S_{n}^{(α,β)}(x) := ∑_{k=0}^{n} P_{k}^{(α,β)}(x)/P^{(β,α)}(1) ≥ 0, −1 ≤ x ≤ 1, n = 1, 2, … (1)

have been proved. Fejér [11, 12] was the first to establish inequalities of this form for α = 1/2, β = −1/2 and for α = β = 0. Fejér conjectured that (1) also hold for α = β = 1/2 and this was proved independently by Jackson [16] and Gronwall [15]. Feldheim [13] proved (1) for α = β ≥ 0. Some special cases of these inequalities were considered by Askey [1, 2] and Askey and Gasper [4] proved (1) for β ≥ 0, α + β ≥ −2. The importance of the latter result is justified by the fact that de Branges [7] used (1) for β = 0, α = 2, 4, 6, …, in the final step of his proof of the celebrated Bieberbach conjecture. Gasper [14] proved inequalities (1) for β ≥ −1/2, α + β ≥ 0.

Note that Bateman’s integral formula (Bateman [6])

\frac{P_{n}^{(α−μ,β+μ)} (x)}{P_{n}^{(β+μ,α−μ)}(1)} = \frac{Γ (β + μ + 1)}{Γ (β + 1) Γ (μ)} ∫_{−1}^{x} \frac{P_{n}^{(α,β)} (t)}{P_{n}^{(β,α)} (1)} \begin{matrix}(1 + t) & β \\ (1 + x) & β+μ \end{matrix} (x − t)^{μ−1} dt, (2)

which holds for μ > 0, and β > −1, implies the following result.

Lemma 1. If the inequalities (1) holds for (α, β), they hold for (α − μ, β + μ), μ > 0 as well. Hence, if (1) fail for some (α, β) they fail for (α + μ, β − μ) , μ > 0.

On the other hand S_{1}^{(α,β)} (x) = (α + β + 2) (1 + x) / (2 (β + 1)) . Having in mind these observations, the above mentioned results of Askey and Gasper [4] and of Gasper [14] yield: Inequalities (1) hold for α ≤ 0, β ≥ max{0, −α − 2} and α ≥ 0, β ≥ max{−1/2, −α} , and fail for β < max{−1/2, −α − 2}.

In 1993 Askey [3] drew attention to (1) for the rest of the (α, β) −plane, namely, for (α, β) in the parallelogram D_{1} = {−1/2 ≤ β < 0, −2 ≤ α + β < 0} . It was proved in [10] that (1) fail for x = 1 and for sufficiently large n, if |α − 3/2| − 1/2 ≤ β < 0. The latter and Bateman’s integral (2) disprove inequalities (1) for the left hand half of D_{1} and n large enough. Thus the only region in the (α, β) −plane for which inequalities (1) is still to be proved or disproved is the parallelogram

D = {(α, β) : −1/2 < β < 0, −1 ≤ α + β < 0} .

On the other hand, (1) hold for the upper boundary {β = 0, −1 ≤ α < 0} and fail for the lower boundary{β = −1/2, −1/2 ≤ α < 1/2} of D. Hence, by Bateman’s integral, for any θ ∈ (−1, 0) there exists an (α′, β′) ∈ D with α′ + β′ = θ such that (1) hols for {α + β = θ, β ≥ β′} and fail for {α + β = θ, β < β′} . The curve formed by the points (α′, β′) with this property will be denoted by γ. Also, denote by J_{α}(x) the Bessel function of the first kind with parameter α and let j_{α,2} be the second positive zero of J_{α}(x). The following conjecture is due to Gasper [14, p. 444].

Conjecture 1. The subregion Δ of D for which the inequalities (1) holds is given by

Δ = {(α, β) ∈ D : β ≥ β (α) , where ∫_{0}^{j_{α,2}} t^{−β(α)}J_{α}(t) dt = 0} . (3)

It may be pointed out that Gaspers’s conjecture is equivalent to the statement that

γ = {(α, β (α)) ∈ D : ∫_{0}^{j_{α,2}} t^{−β(α)}J_{α}(t) dt = 0} .

The conjecture is based on the well-known formula (see (1.8) in [3])

lim_{n→∞} (\frac{θ}{n})^{α−β+1} ∑_{k=0}^{n} \frac{P_{k}^{(α,β)} (cos (θ/n))}{P_{k}^{(β,α)} (1)}

= 2^{α}Γ (β + 1) ∫_{0}^{θ} t^{−β}J_{α} (t) dt, β < α + 1,

and on the following theorem.

Theorem 1. Let −1 < α < 1/2 and β > −1/2. Then the inequality

∫_{0}^{θ} t^{−β}J_{α} (t) dt ≥ 0

holds for any nonnegative θ if and only if

∫_{0}^{j_{α,2}} t^{−β}J_{α} (t) dt ≥ 0.

The proof of this theorem for α ∈ (−1, −1/2) is due to Askey and Steinig [5] and the case α ∈ (−1/2, 1/2) was proved by Makai [17].

Very recently Brown, Koumandos and Wang [8, 9] verified Gasper’s conjecture for the case when (α, β) lies on the lines α = β or α = −1/2.

The objective of the present paper is to state a slight refinement of Conjecture 1 and to give numerical evidence of its truth.

2. The new conjecture

For any positive integer n, set

Δ_{n} = {(α, β) ∈ D : S_{n}^{(α,β)} (x) ≥ 0 for x ∈ [−1, 1]} .

Then Gasper’s conjecture can be formulated in the equivalent form

⋃_{n=1}^{∞} Δ_{n} = Δ,

where Δ is defined by (3).

We state

Conjecture 2. For any positive integer n, Δ_{n+1} ⊂ Δ_{n}.

Denote by γ_{n} the boundary of Δ_{n} which passes through D:

γ_{n} = {(α, β) ∈ D : S_{n}^{(α,β)} (x) ≥ 0 for all x ∈ [−1, 1] and every (α, β) with α + β = α_{n} + β_{n}, β ≥ β_{n}, and for some x ∈ [−1, 1] , S_{n}^{(α,β)} (x) < 0 for (α, β) with α + β = α_{n} + β_{n}, β < β_{n}}.

The curve γ_{n} is well defined because of Lemma 1.

An equivalent formulation of Conjecture 2 is that γ_{n+1} lies above γ_{n} for any positive integer n. The latter conjecture implies that of Gasper, because of (4) and Theorem 1.

In the next section we give explicit expresions for Δ_{2} and Δ_{3} or, equivalently, for γ_{2} and γ_{3}. In Section 3 an algorithm to trace the curves γ_{n} is developed. Tables for the curves γ_{n} for n = 4 and 5 are given and the graphs of γ_{n} for n = 2, 3, 4, 5 are drawn. In Section 4 we discuss an idea of how Conjecture 2 might be proved.

3. The cases n = 2 and n = 3

In what follows we suppose that (α, β) ∈ D. First we consider the case n = 2. Straightforward calculations show that

4 (β + 1) (β + 2) S_{2}^{(α,β)} (x) = a_{2}x^{2} + 2a_{1}x + a_{0,}

where

a_{2} = (α + β + 3) (α + β + 4) ,
a_{1} = 2 (α + 2) (α + β + 3) + (α + β + 2) (β + 2) − (α + β + 3) (α + β + 4)
= (α + 1) (α + β + 4) ,
a_{0} = 2 (α + β + 2) (β + 2) + 4 (α + 1) (α + 2) + (α + β + 3) (α + β + 4)
− 4 (α + 2) (α + β + 3) = α^{2} + 3β^{2} + 3α + 7β + 4.

Obviously S_{2}^{(α,β)} (x) is convex and its minimum value is attained at x_{min} = −a_{1}/a_{2} = − (α + 1) / (α + β + 3). Observe that −1 < x_{min} < 0. Hence, S_{2}^{(α,β)} (x) ≥ 0 for x ∈ [−1, 1] if and only if it is non-negative for any real x. Since its leading coefficient is positive, then S_{2}^{(α,β)} (x) is non-negative if and only if its discriminant

(α + 1)^{2} (α + β + 4)^{2} − (α + β + 3) (α + β + 4) (α^{2} + 3β^{2} + 3α + 7β + 4)

is non-positive. Thus,

Δ_{2} = \{(α, β) ∈ D : β ≥ \frac{−3α − 10 + \sqrt{9α^{2} + 36α + 52}}{6}\} .

The case n = 3 may be treated similarly because S_{n}^{(α,β)} (−1) = 0 for any odd n. Set u = (x + 1)/2. Staightforward calculations show in fact that

\overline{S}_{3}^{(α,β)} (u) = \frac{S_{3}^{(α,β)} (x)}{u} = b_{2}u^{2} − 2b_{1}u + b_{0}

where

b_{2} = (α + β + 4)(α + β + 5)(α + β + 6)/(β + 1)(β + 2)(β + 3),
b_{1} = (α + β + 4)(α + β + 6)/(β + 1)(β + 2),
b_{0} = 2(α + β + 4)/(β + 1),

and we have to characterize the values of (α, β) in D for which \overline{S}_{3}^{(α,β)} (u) ≥ 0 for each u ∈ [0, 1] . Since \overline{S}_{3}^{(α,β)} (u) attains its minimum at u_{min} = b_{1}/b_{2} = (β + 3)/(α + β + 5) and u_{min} ∈ [0, 1] , then \overline{S}_{3}^{(α,β)} (u) ≥ 0 for u ∈ [0, 1] and those (α, β) for which the discriminant

(\frac{(α + β + 4) (α + β + 6)}{(β + 1)(β + 2)})^{2} − 2\frac{(α + β + 4)^{2}(α + β + 5)(α + β + 6)}{(β + 1)^{2}(β + 2)(β + 3)}

of \overline{S}_{3}^{(α,β)} (u) is non-negative. Therefore

Δ_{3} = \{(α, β) ∈ D : β ≥ \frac{−α − 5 + \sqrt{α^{2} + 6α + 17}}{2}\} .

4. An algorithm to find Δ_{n}

The algorithm for tracing the curves γ_{n} is based on the following simple fact.

Lemma 2. If (α_{n,}β_{n}) ∈ γ_{n}, then there exists ξ ∈ (−1, 1) for which

S_{n}^{(α_{n},β_{n})} (ξ) = \frac{d}{dx}S_{n}^{(α_{n},β_{n})} (ξ) = 0.

Proof. Assume that for some (α_{n}, β_{n}) the polynomial S_{n}^{(α_{n},β_{n})} (x) is positive at the points of local extrema in (−1, 1) . Then a continuity argument implies that there exists a neighborhood U of (α_{n}, β_{n}) such that for every (α, β) in U and for every x ∈ (−1, 1) the polynomial S_{n}^{(α,β)} (x) is positive. The latter contradicts the definition of γ_{n}. □ ✓

A well known necessary condition for a polynomial

p(x) = ∑_{ν=0}^{n} a_{ν}x^{n−ν}

to have a double root is stated in the following lemma. We recall that the discriminant D(p) of p is

D(p) = a_{0}^{2n−2} ∏_{1≤i<j≤n}(x_{i} − x_{j})^{2},

where x_{1}, …, x_{n} are the roots (zeros) of p.

Lemma 3. The discriminant D(p) of the polynomial p can be represented as a (2n − 1) × (2n − 1) determinant in the form

\frac{a_{0}D(p)}{(−1)^{n−1}} = |\begin{matrix}a_{0} & a_{1} & ⋯ & a_{n−1} & a_{n} \\ na_{0} & (n − 1) a_{1} & ⋯ & a_{n−1} \\ & ... & ... & ... & ... \\ & & a_{0} & a_{1} & ⋯ & a_{n−1} & a_{n} \\ & & na_{0} & (n − 1) a_{1} & ⋯ & a_{n−1} \\ & & & na_{0} & (n − 1) a_{1} & ⋯ & a_{n−1} \end{matrix}|

Moreover, D(p) = 0 if and only if p(x) has at least one root of multiplicity at least two.

We refer to [18, Section 1.3.3] and the references therein for the proof of this lemma and for additional information about discriminants.

Lemmas 2 and 3 immidiately yield the following result.

Theorem 2. Let S_{n}^{(α,β)}(x) = ∑_{k=0}^{n} a_{k} (α_{n}, β_{n}) x^{n−k}. If (α_{n}, β_{n}) ∈ γ_{n}, then

D(α_{n}, β_{n}) := D(S_{n}^{(α_{n},β_{n})})= 0.

The basic steps of the algorithm to construct an approximation to the curve γ_{n} are:

1. Choose k ∈ ℕ.

2. Divide the interval [−2, 1/2] into k subintervals by the mesh points α_{n}^{(i)} = −2 + 2.5i/k, i = 0, k.

3. For any fixed α_{n}^{(i)} find all the solutions β_{n,1}^{(i)}, …, β_{n,p}^{(i)} ∈ (−1/2, 0) of the

equation D (α_{n}^{(i)}, β) = 0.

4. Find that s, 1 ≤ s ≤ p, for which

S_{n}^{(α_{n}^{(i)},β_{n,s}^{(i)})}(x) ≥ 0 for x ∈ [−1, 1]

and

S_{n}^{(α_{n}^{(i)},β_{n,s}^{(i)})}(ξ) = \frac{d}{dx}S_{n}^{(α_{n}^{(i)},β_{n,s}^{(i)})}(ξ) = 0 for some ξ ∈ (−1, 1) .

5. Choose β_{n}^{(i)} = β_{n,s}^{(i)}.

6. Approximate the data (α_{n}^{(i)}, β_{n}^{(i)}) by a smooth curve.

Table 1 in the next page contains the results of the algorithm for n = 4 and n = 5, for k = 50. The values of β_{4}^{(i)} and β_{5}^{(i)} which correspond to α_{n}^{(i)} = α^{(i)} = −2 + 0.05i, i = 0, …, 50, are:

The graphs of the approximations to the curves γ_{n} for n = 2, 3, 4 and 5 are drawn in Figure 1 at the end of the paper.

5. An idea for proving Conjecture 2

The graphs of the curves γ_{2}, γ_{3}, γ_{4} and γ_{5} show that Conjecture 2 holds for n = 2, 3 and 4. It is clear that Conjecture 2 would be proved if one proves that S_{n}^{(α,β)} is nonnegative on [−1, −1] for any (α, β) for which S_{n+1}^{(α,β)} is nonnegative there. Another possible idea to prove Conjecture 2 is to show that for any (α_{n}, β_{n}) ∈ γ_{n} the inequality S_{n+1}^{(α_{n},β_{n})}(x) ≥ 0 fails for some x ∈ [−1, 1] . It turns out that for n = 2, 3 and 4 such x exists. Based on the graphs of S_{n}^{(α_{n},β_{n})}(x) and S_{n+1}^{(α_{n},β_{n})}(x) for various (α_{n}, β_{n}) ∈ γ_{n} we may state an additional conjecture which implies the truth of Conjecture 2, and thus, of Conjecture 1.

Conjecture 3. Let (α_{n}, β_{n}) ∈ γ_{n}. Then there exists a unique ξ_{n} ∈ (−1, 1) such that

S_{n}^{(α_{n},β_{n})}(ξ_{n}) = \frac{d}{dx}S_{n}^{(α_{n},β_{n})}(ξ_{n}) = 0.

i α^{(i)}  β_{4}^{(i)}  β_{5}^{(i)}  i α^{(i)}  β_{4}^{(i)}  β_{5}^{(i)}
0 −2.00  0  0
1 −1.95 −0.0124665 −0.0100482 26 −0.70 −0.29347 −0.271235
2 −1.90 −0.0248627 −0.020186 27 −0.65 −0.303304 −0.281463
3 −1.85 −0.0371837 −0.0304035 28 −0.60 −0.313026 −0.291642
4 −1.80 −0.0494251 −0.0406914 29 −0.55 −0.322637 −0.30177
5 −1.75 −0.0615829 −0.051041 30 −0.50 −0.332137 −0.311845
6 −1.70 −0.0736534 −0.0614439 31 −0.45 −0.341526 −0.321856
7 −1.65 −0.0856334 −0.0718924 32 −0.40 −0.350807 −0.331828
8 −1.60 −0.0975197 −0.0823791 33 −0.35 −0.359997 −0.341732
9 −1.55 −0.10931 −0.0928969 34 −0.30 −0.36904 −0.351576
10 −1.50 −0.121001 −0.103439 35 −0.25 −0.377995 −0.361359
11 −1.45 −0.132592  −0.1114  36 −0.20 −0.386843 −0.371079
12 −1.40 −0.144079 −0.124573 37 −0.15 −0.395585 −0.380734
13 −1.35 −0.155462 −0.135135 38 −0.10 −0.404222 −0.390324
14 −1.30 −0.166739 −0.145734 39 −0.05 −0.412754 −0.399847
15 −1.25 −0.177909 −0.156312 40 0.00 −0.421183 −0.409303
16 −1.20 −0.18897 −0.166881 41 0.05 −0.429509 −0.418691
17 −1.15 −0.199922 −0.177438 42 0.10 −0.437734 −0.428009
18 −1.10 −0.210763 −0.110763 43 0.15 −0.445858 −0.437258
19 −1.05 −0.221493 −0.198469 44 0.20 −0.453883 −0.446436
20 −1.00 −0.232112 −0.208989 45 0.25 −0.46181 −0.455544
21 −0.95 −0.242619 −0.219454 46 0.30 −0.469638 −0.464579
22 −0.90 −0.253014 −0.229886 47 0.35 −0.477371 −0.473543
23 −0.85 −0.263296 −0.240284 48 0.40 −0.485008 −0.482435
24 −0.80 −0.273467 −0.250643 49 0.45 −0.49225 −0.491254
25 −0.75 −0.283524 −0.260961 50 0.50  −0.5  −0.5
Table 1. The curves γ_{4} and γ_{5}

Moreover, there exist η′_{n} and η′′_{n} with −1 < ξ_{n} < η′_{n} < ηn^{′′} < 1 such that

S_{n+1}^{(α_{n},β_{n})}(x) < 0 for x ∈ (η′_{n}, η′′_{n}).