On a conjectured inequality of Gautschi and Leopardi for Jacobi polynomials

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)


−1, β > −1, then the inequality (α,β)

Pn

(cos nθ )

(α,β)

Pn

(1)

(α,β)


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 .

Numer Algor

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


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


tan 2n 2n + 2  

which is obvious.

3 Remarks The closely related inequality θ sin 2n+1 2n θ sin 2n