f(x)=f1(x)=(x-1)/(x+1),f(n+1)←下标=f[fn(x)],这个函数周期4,求f2,f3,f4推导过程,

f(x)=f1(x)=(x-1)/(x+1),f(n+1)←下标=f[fn(x)],这个函数周期4,求f2,f3,f4推导过程,
f(x)=f1(x)=(x-1)/(x+1),f(n+1)←下标=f[fn(x)],这个函数周期4,求f2,f3,f4推导过程,

f(x)=f1(x)=(x-1)/(x+1),f(n+1)←下标=f[fn(x)],这个函数周期4,求f2,f3,f4推导过程,
fn=f[fn-1(x)]=f{f[fn-2(x)}}=f{f{…f(x)}}即n重f(x)可记为f^n(x)
所以有,f2=f^2(x)=-1/x,
f3=f^3(x)=(1+x)/(1-x),
f4=f^4(x)=x,
故f5=f(x)
即你要求fn+1就把fn当成x代入方程f(x).即fn+1=(fn-1)/(fn+1).

f2=f[f1(x)]
=f[(x-1)/(x+1)]
=(fx-1)/(fx+1)
=--1--2/x
f3=f[f(2)]
=--1--2/f2
=2x/(x+2)--1
=1--4/(2+x)
f4=f[f3]
=1--4/(2+f3)
=1--4/[3x+2/(x+2)]
=-(x+6)/(3x+2)