高分问题求助,急~问题如下图,如
(1) An=f[A(n-1)],f(-1)(An)=A(n-1),A(n+1)=f(An),又 f(x)+f(-1)(x)<5x/2=5f[f(-1)(x)], ∴A(n+1)+A(n-1)<5An/2, 2A(n+1)-5An+2A(n-1)<0,即A(n+1)-2An<(1/2)[An-2A(n-1)],A(n)-2A(n-1)<(1/2)[A(n-1)-2A(n-2)],…,A3-2A2<(1/2)(A2-2A1),A2-2A1<(1/2)(A1-2A0),依次迭代,得A(n+1)-2An<(1/2)^n·(A1-2A0)=(1/2)^n·(-6)=-6/(2^n), ∴ A(n...全部
(1) An=f[A(n-1)],f(-1)(An)=A(n-1),A(n+1)=f(An),又 f(x)+f(-1)(x)<5x/2=5f[f(-1)(x)], ∴A(n+1)+A(n-1)<5An/2, 2A(n+1)-5An+2A(n-1)<0,即A(n+1)-2An<(1/2)[An-2A(n-1)],A(n)-2A(n-1)<(1/2)[A(n-1)-2A(n-2)],…,A3-2A2<(1/2)(A2-2A1),A2-2A1<(1/2)(A1-2A0),依次迭代,得A(n+1)-2An<(1/2)^n·(A1-2A0)=(1/2)^n·(-6)=-6/(2^n), ∴ A(n+1)<2An-6/(2^n)
(2) 假设存在常数A,B使A0=B0,A1=B1, ∵ B0=A+B=8,B1=2A+(B/2)=10, ∴ A=B=4, Bn=4[2^n+2^(-n)]。
且当n≥2时,An
∴ 当n≥2时,An收起