- 37 名前:132人目の素数さん mailto:sage [2009/11/06(金) 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)},
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