Numer Algor DOI 10.1007/s11075-007-9098-y ORIGINAL PAPER
On a conjectured inequality of Gautschi and Leopardi for Jacobi polynomials Stamatis Koumandos
Received: 15 March 2007 / Accepted: 17 April 2007 © Springer Science + Business Media B.V. 2007
Abstract Motivated by work on positive cubature formulae over the spherical (α,β)
surface, Gautschi and Leopardi conjectured that the inequality (α,β) θ Pn+1 (cos n+1 ) (α,β) Pn+1 (1)
holds for α, β > −1 and n ≥ 1, θ ∈ (0, π ), where
Pn
(cos nθ )
(α,β)
Pn (1) (α,β) Pn (x)
<
are
the Jacobi polynomials of degree n and parameters (α, β). We settle this conjecture in the special cases where (α, β) ∈ ( 12 , 12 ), ( 12 , − 12 ), (− 12 , 12 ) . Keywords Jacobi polynomials · Inequalities · Trigonometric functions Mathematics Subject Classifications (2000) Primary 26D05 · Secondary 33B10 · 33C45 · 42A05 1 Introduction and results (α,β)
Let Pn (x) be the Jacobi polynomials of degree n and parameters (α, β). In their work [3], Gautschi and Leopardi considered inequalities related to the largest zeros of Jacobi polynomials and also inequalities involving the scaled Jacobi polynomials having the value 1 at x = 1. In particular, Gautschi and Leopardi conjectured in [3] that, given α > −1, β > −1, then the inequality (α,β)
Pn
(cos nθ )
(α,β)
Pn
(1)
(α,β)
<
θ Pn+1 (cos n+1 ) (α,β)
Pn+1 (1)
S. Koumandos (B) Department of Mathematics and Statistics, University of Cyprus, P. O. Box 20537, 1678 Nicosia, Cyprus e-mail:
[email protected]
,
(1.1)
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either holds for all n = 1, 2, . . . and 0 < θ < π or it is false for n = 1 and some θ. They also provided compelling evidence, both analytic and numerical, in support of this conjecture. (cf. Conjecture 4 of [3]). Inequalities such as (1.1) can be applied to establish positive cubature formulae over the spherical surface. See [3] and [4] for details. In this note our aim is to show that (1.1) is valid in the cases where (α, β) ∈
1 1 1 1 1 1 , , ,− , − , , 2 2 2 2 2 2
(1.2)
and this can be considered as a further evidence in support of the Gautschi and Leopardi conjectured inequality. Our result is mentioned in [3, p. 11]. We recall that (− 12 ,− 12 )
Pn
(cos θ)
(− 1 ,− 1 ) Pn 2 2 (1) ( 1 , 12 )
Pn2
(cos θ)
( 12 , 12 )
Pn
(1)
( 1 ,− 12 )
(cos θ)
Pn2
( 12 ,− 12 )
Pn
(− 12 , 12 )
Pn
(cos θ)
(− 12 , 12 )
Pn
(1)
(1)
= cos nθ
=
=
=
sin(n + 1)θ (n + 1) sin θ sin(n + 12 )θ
(2n + 1) sin θ2 cos(n + 12 )θ cos θ2
.
(1.3)
The normalized polynomials listed in (1.3) are the Chebyshev polynomials of the first, second, third and fourth kind, respectively. See [2, Section 2.5, pp. 101–102]. As it is already pointed out in [3], the case α = β = −1/2 of (1.1) is trivial as both of its sides are identically equal to cos θ. The cases of (1.1) given in (1.2) are not obvious and we deal with them here. It is worth noting that most of the classical inequalities for the Jacobi polynomials have been first established for the trigonometric cases that correspond to (1.3) above. See [1] for related results and background information on the subject. It is therefore reasonable to start our investigations regarding (1.1) by considering the cases (1.2) in which it reduces to a trigonometric inequality. We note also that for the cases listed in (1.2), the inequality for the zeros given as Conjecture 3 of [3] holds as an immediate consequence of the location of the zeros of the trigonometric functions listed in (1.3). This can also be verified via the cases of equality in Theorems 6.21.2 and 6.21.3 of [5]. Our result can be stated as follows.
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Theorem 1 Inequality (1.1) holds for all n = 1, 2, 3 . . . and 0 < θ < π, when (1) α = β = 1/2,
(2)
α = 1/2, β = −1/2,
and (3) α = −1/2, β = 1/2.
In the next section we give a proof of this theorem. In Section 3 we provide some additional comments on related inequalities.
2 Proof of Theorem 1 In the case (1) using the second of the relations (1.3) above we see that (1.1) amounts to θ sin n+1 n
(n + 1) sin nθ
<
θ sin n+2 n+1 θ (n + 2) sin n+1
(2.1)
for all n = 1, 2, 3 . . . and 0 < θ < π, which we proceed to prove. By elementary trigonometric identities we deduce that (2.1) is equivalent to −
1 θ 1 θ 1 cot θ + cot − cot > 0 . (n + 1)(n + 2) n+2 n+1 n+1 n
(2.2)
We first prove that θ 1 θ 1 cot − cot > 0 , n+2 n+1 n+1 n
(2.3)
for all n = 1, 2, 3 . . . and 0 < θ < π. For n = 1, inequality (2.3) reduces to the elementary inequality θ 1 1 cot − cot θ > 0 , 3 2 2 which is easy to prove. For n ≥ 2, inequality (2.3) is equivalent to θ θ θ θ cot − cot > 0 . n+2 n+1 n+1 n
(2.4)
We set f (x) := x cot x, a :=
θ θ , b := , n+1 n
so that the left hand side of (2.4) equals (n + 1)2 f (a) − f (b ) + f (b ) (n + 1)2 f (a) − n(n + 2) f (b ) = > 0, (n + 1)(n + 2) (n + 1)(n + 2) because the function f (x) is positive and strictly decreasing on (0, π/2). This establishes (2.3) and also shows that (2.2) holds for π/2 ≤ θ < π .
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In order to prove (2.2) for 0 < θ < π/2 we multiply both sides by θ, and using the notation above we see that this reduces to (n + 1)2 f (a) − f (b ) + f (b ) − f (θ) > 0, (2.5) (n + 1)(n + 2) which is clearly true. The proof of (2.2) is now complete. In a similar way we can handle the case (2) which reduces to showing θ sin 2n+1 2n
θ (2n + 1) sin 2n
<
θ sin 2n+3 2n+2 θ (2n + 3) sin 2n+2
for all n = 1, 2, 3 . . . and 0 < θ < π. We observe that (2.6) is equivalent to
(2n + 2)(2n + 1) f (c) − f (d) + 2 f (d) − f (θ) > 0 ,
(2.6)
(2.7)
where c :=
θ θ , d := , 2n + 2 2n
and f (x) as above. For all n ≥ 1 we have c < d < π/2, hence f (c) − f (d) > 0. When π/2 ≤ θ < π , we have f (θ) < 0 and therefore (2.7) is valid. On the other hand, when 0 < θ < π/2 we have f (d) − f (θ) > 0 and therefore (2.7) holds and in this case as well. The proof of (2.6) is complete. Finally, in the case (3) we need to show that θ cos 2n+1 2n θ cos 2n
<
θ cos 2n+3 2n+2 θ cos 2n+2
(2.8)
for all n = 1, 2, 3 . . . and 0 < θ < π, and this is equivalent to tan
θ θ > tan 2n 2n + 2
which is obvious.
3 Remarks The closely related inequality θ sin 2n+1 2n θ sin 2n
<
θ sin 2n+3 2n+2 θ sin 2n+2
,
holds true for all n = 1, 2, 3 . . . and 0 < θ < π, because it reduces to cot
θ θ < cot , 2n 2n + 2
(3.1)
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while the inequality θ cos 2n+1 2n
θ (2n + 1) cos 2n
<
θ cos 2n+3 2n+2 θ (2n + 3) cos 2n+2
(3.2)
fails to hold for all n = 1, 2, 3 . . ., because the function Ln (θ) := satisfies Ln (0) = Recalling that
θ cos 2n+3 2n+2 θ (2n + 3) cos 2n+2
−
θ cos 2n+1 2n
θ (2n + 1) cos 2n
1 1 − < 0 and Ln (π ) = −Ln (0) > 0. 2n + 3 2n + 1 ( 1 ,− 12 )
Pn2
(cos θ)
(− 12 , 12 )
Pn
(− 12 , 12 )
Pn
(cos θ)
( 12 ,− 12 )
Pn
(1)
(1)
=
=
sin(n + 12 )θ sin θ2
cos(n + 12 )θ
(2n + 1) cos θ2
,
(cf. [1]), the suggested by (3.1) inequality for Jacobi polynomials (α,β)
Pn
(cos nθ )
(β,α)
Pn
(1)
(α,β)
<
θ Pn+1 (cos n+1 ) (β,α)
Pn+1 (1)
,
(3.3)
is, in general, not true. The normalization given in (3.3) has often been used in proofs of positivity of polynomial sums, as noted in [1]. See also Sections 7.3 and 7.4 of [2], which contain related results. In particular, Section 7.4 of [2] describes the role of this normalization in the proof of the Bieberbach conjecture. Acknowledgement I would like to thank one of the referees for helpful comments and suggestions for the improvement of the presentation of the paper.
References 1. Askey, R.: Orthogonal polynomials and special functions. In: Regional Conf. Lect. Appl. Math., vol. 21. SIAM, Philadelphia, PA (1975) 2. Andrews, G.E., Askey, R., Roy, R.: Special functions. In: Encyclopedia of Mathematics and its Applications, vol. 71 (paperback edn). Cambridge University Press, Cambridge (2000) 3. Gautschi, W., Leopardi, P.: Conjectured inequalities for Jacobi polynomials and their largest zeros. In: DWCAA06 Proceedings in Numerical Algorithms, doi:10.1007/s11075-007-9067-5 (2007) 4. Leopardi, P.: Positive weight quadrature on the sphere and monotonicities of Jacobi polynomials. In: Numerical Algorithms, doi:10. 1007/s11075-007-9073-7 (2007) ˝ Orthogonal polynomials. In: Colloquium Publications, 2nd edn., vol. 23. American 5. G. Szego: Mathematical Society, Providence, RI (1975)
