04/07/03 15:28
>264-266
【263の(1)】
1≦i≦n-1 ⇒ Q_i ≡(P_i)^2 - P_{i-1}･P_{i+1} ≧0.
(略証） n(変数の数)に関する帰納法で示す。
For i=1,
Q_1 = (P_1)^2 - P_0･P_2 = (1/n)^2･(∑[j=1,n] x_j)^2 - {2/n(n-1)}(∑[j>j'] x_j･x_j')
= {1/[n^2･(n-1)]}{(n-1)∑[j=1,n](x_j^2) -2･∑[j>j'] x_j･x_j'}
= {1/[n^2･(n-1)]}∑[j>j'] (x_j-x_j')^2 ≧ 0.
n=2 ⇒ i=1 ∴ n=2のとき成立。
Consider a new member x'= x_{n+1}.
s'_i ≡ s_i + x'･s_{i-1} (1≦i≦n),
P'_i ≡ [(n+1-i)/(n+1)]･P_i + [i/(n+1)]x'･P_{i-1},
Q'_i ≡ (P'_i)^2 - P'_{i-1}･P'_{i+1},
and
(n+1)^2･Q'_i = (n-i)(n-i+2)･Q_i + (n-i)(i-1)[P_i･P_{i-1}－P_{i+1}P_{i-2}]x' +(i^2-1)･Q_{i-1}(x')^2 +[P_i-P_{i-1}x'}^2.
Provided Q_i≧0, Q_{i-1}≧0 for a certain n, then
P_i/P_{i+1} increase monotonously with i: P_i/P_{i+1} ≧ P_{i-2}/P_{i-1}.
∴ Q'_i≧0.