G. Gasper Using sums of squares to prove that ... - Fuchs-braun.com

Let α be real valued. Substituting the sum of squares of Laguerre polynomials
expansion (from [7, (91)])
(4.3)
n−k
L α+2k
n−k (2x) = ∑
j=0
(n − k − j)! (2k + 2j + α)(2k + α) j
j! (2k + α)(2k + α + 1) n+j−k
× (−1) j x 2j ( L α+2k+2j
n−k−j (x) ) 2
into the special case of [3, (5.4)]
(4.4)
|L α n(z)| 2 = (α + 1) n
n!
n∑
k=0
and changing the order of summation yields
(4.5) |L α n (z)|2 = (α + 1) n
n!
n∑
k=0
Then application of (3.7) gives
(4.6) |L α n (z)|2 = (α + 1) n
n!
1
k! (α + 1) k
(x 2 + y 2 ) k L α+2k
n−k (2x)
(n − k)! (2k + α)(α) k
k! α(α + 1) n+k
(−1) k x 2k
× 2 F 1 (−k, k + α; α + 1; 1 + y 2 /x 2 ) ( L α+2k
n−k (x)) 2
.
n∑
k=0
(n − k)! (2k + α)(α) k
k! α(α + 1) n+k
y 2k
× 2 F 1 (−k, 1 − k; α + 1; 1 + x 2 /y 2 ) ( L α+2k
n−k (x)) 2
.
Since L α 0 (x) ≡ 1 and the coefficients on the right hand side of (4.6) are
clearly positive when α > −1 and y ≠ 0, the expansion (4.6) proves that the
Laguerre polynomials have only real zeros when α > −1. This also follows,
in particular, from the inequalities
(4.7)
and
(4.8)
|L α n(z)| 2 ≥ (α + 1) n
n! n! (n + α) n
y 2n 2F 1 (−n, 1 − n; α + 1; 1 + x 2 /y 2 ), α > −1,
|L α n(z)| 2 ≥ |L α n(x)| 2 + (α + 1) n
n! n! (n + α) n
y 2n , α > −1, n ≥ 1,
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