09/11/06 23:35:20
>>34
λ = -1: A[n] -3B[n] +3C[n] -D[n] = (-1)^n,
λ =-1/3: A[n] -B[n] -C[n] +D[n] = (-1/3)^n,
λ = 1/3: A[n] +B[n] -C[n] -D[n] = (1/3)^n,
λ = 1: A[n] +3B[n] +3C[n] +D[n] = 1,
よって
A[n] + D[n] = (1/4){1 - (-1/3)^(n-1)},
A[n] - D[n] = (1/4){(-1)^n + (1/3)^(n-1)},
B[n] + C[n] = (1/4){1 - (-1/3)^n},
B[n] - C[n] = (1/4){-(-1)^n + (1/3)^n},
よって
A[n] +3C[n] = (3^n){A[n] - C[n]} = (1/2){1+(-1)^n} ≡ g[n],
3B[n] + D[n] = (3^n){B[n] - D[n]} = (1/2){1-(-1)^n} ≡ u[n],
よって
A[n] = g[n]･(1/4){1+(1/3)^(n-1)},
B[n] = u[n]･(1/4){1+(1/3)^n},
C[n] = g[n]･(1/4){1-(1/3)^n},
D[n] = u[n]･(1/4){1-(1/3)^(n-1)},