- 298 名前：１３２人目の素数さん mailto:sage [2011/08/24(水) 01:43:28.81 ]
- >>286 の続き
(A[n] - B[n])/√2 = D[n],
(1/2){A[n] + B[n] +(√2)C[n]} = E[n],
(1/2){A[n] + B[n] -(√2)C[n]} = F[n],
とおくと
D[n+1] = −D[n],
E[n+1] = (1+√2)E[n],
F[n+1｣ = (1-√2)F[n],
より等比数列で
D[n] = (-1)^(n-1)･D[1],
E[n] = (1+√2)^(n-1)･E[1],
F[n] = (1-√2)^(n-1)･F[1],
本問では、D[1] = 0, E[1] = (1 +√2)/√2, F[1] = -(√2 - 1)/√2,
|F[n]| = (1/√2)(√2 -1)^n < (1/√2)(1/2)^n,
A[n] = B[n] = [ (1/√8)(1+√2)^n + 1/2 ],
C[n] = [ (1/2)(1+√2)^n + 1/2 ],
