设F(X)=AX+B
则有F(F(X))=F(AX+B)=A(AX+B)+B=A^2X+AB+B=2X-1
等式恒等
必有
A^2=2,AB+B=-1
A=+/-根下2,对应B=-/+1/(根号2+1)
所以F(X)=(根下2)X-1/(根号2+1)
或者F(X)=-(根下2)X+1/(根号2+1)