Solutions
463
e
xiyj /(2t)
−
N
p=1
(xiyj )
p−1
(p−1)!(2t) p−1
N
p=1
(xiyj ) p−1
(p−1)!(2t) p−1
≤ 5 (|x|/2)
N t
−N/4 /N !
det
(e
xiyj /(2t) ) 1≤i,j≤N
− det (A(t))
≤
σ∈∈G N
sgn(σ)
1≤i≤N e
xiy σ(i) /(2t)
× sup i,j
1 −
1≤i≤N
1 +
e
−xiyj /(2t)
N
p=1
(xiyj )
p−1
(p−1)!(2t) p−1 − 1
En utilisant le fait que
1 −
1≤i≤N
(1 + O()) = O()
on en conclut que
det
(e
xiyj /(2t) ) 1≤i,j≤N
− det (A(t))
≤ |det (A(t))| × O(t
−N/4 )
5. On a
det (A 1 (t), . . . , A N (t))
= det
N
p=1 b 1,p (t) B p , . . . ,
N
p=1 b N,p (t) B p
=
σ∈G N
1≤i≤N b i,σ(i) (t)
× det
B σ(1) , . . . , B σ(N )
=
σ∈G N
1≤i≤N b i,σ(i) (t)
× sgn(σ) det (B 1 , . . . , B N )
Le dernier terme est donn´ e par le d´ eterminant de Vandermonde
det (B 1 , . . . , B N ) =
1≤i (x j − x i )
De plus, nous avons
1≤i≤N
b i,σ(i) (t) =
1≤i≤N
(y i )
σ(i)−1
(σ(i) − 1)!(2t) σ(i)−1
=
(2t)
−N (N +1)/2−N
1≤k 1≤i≤N
(y i )
σ(i)−1
et