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6.5 The Toda Equations 257
where A n and B n are Hankelians defined as
A n = |φ m | n , 0 ≤ m ≤ 2n − 2,
B n = |φ m | n , 1 ≤ m ≤ 2n − 1,
φ =(m +1)φ m+1 .
m
Proof.
(n) n+1 (n)
B =(−1) A 11 ,
1n
(n+1)
A 1,n+1 =(−1) B n . (6.5.10)
n
It follows from Theorems 4.35 and 4.36 in Section 4.9.2 on derivatives of
Turanians that
(n+1)
D(A n )= −(2n − 1)A ,
n+1,n
(n+1)
D(B n )= −2nB ,
n,n+1
(n) (n+1)
D(A )= −(2n − 1)A ,
11 1,n+1;1n
(n) (n+1)
D(B 11 )= −2nB 1,n+1;1,n . (6.5.11)
The algebraic identity in Theorem 4.29 in Section 4.8.5 on Turanians is
satisfied by both A n and B n .
(n) (n)
B y = B n D(B ) − B D(B n )
2
n 2n−1 11 11
(n) (n+1) (n+1)
=2n B B − B n B
11 n,n+1 1,n+1;1n
(n) (n+1)
=2nB B
1n 1,n+1
(n) (n+1)
= −2nA A .
11 11
Applying the Jacobi identity,
(n) (n+1) (n) (n+1)
A A (y 2n − y 2n−2 )= A n+1 A − A n A
11 11 11 11
(n+1) n+1 (n+1)
= A n+1 A − A A
1,n+1;1,n+1 n+1,n+1 11
(n+1) 2
= − A
1,n+1
2
= −B .
n
Hence,
y (y 2n − y 2n−2 )=2n,
2n−1
which proves the theorem when n is odd.
(n+1) 2 (n+1) (n+1)
A y = A D(A n+1 ) − A n+1 D(A )
1,n+1 2n 11 11
(n+2) (n+1) (n+2)
=(2n +1) A n+1 A 1,n+2;1,n+1 − A 11 A n+2,n+1
(n+1) (n+2)
= −(2n +1)A A .
1,n+1 1,n+2