Xn+1=2*Xn*(1-Xn)通项公式如何求,

网友回答

x(n+1)=-2[x(n)]^2+2x(n)=-2[x(n)-1/2]^2+1/2.x(n+1)-1/2=-2[x(n)-1/2]^2=(-2)^2*[x(n-1)-1/2]^(2^2)=(-2)^3*[x(n-2)-1/2]^(2^3)=.=(-2)^n*[x(1)-1/2]^(2^n)x(n+1)=1/2+(-2)^n*[x(1)-1/2]^(2^n).
======以下答案可供参考======
供参考答案1:
x(n+1)=-2[x(n)]^2+2x(n)=-2[x(n)-1/2]^2+1/2.
x(n+1)-1/2=-2[x(n)-1/2]^2
=(-2)*[(-2)(x(n-1)-1/2)^(2)]^(2)
=......
=(-2)^(2^(n)-1)*[x(1)-1/2]^(2^n)
x(n+1)=1/2+(-2)^(2^(n)-1)*[x(1)-1/2]^(2^n)
供参考答案2:
难道是传说中的黑客?