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,
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![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,](/uploads/image/z/2812608-0-8.jpg?t=f%28x%29%3Df1%28x%29%3D%28x-1%29%2F%28x%2B1%29%2Cf%28n%2B1%29%E2%86%90%E4%B8%8B%E6%A0%87%3Df%5Bfn%28x%29%5D%2C%E8%BF%99%E4%B8%AA%E5%87%BD%E6%95%B0%E5%91%A8%E6%9C%9F4%2C%E6%B1%82f2%2Cf3%2Cf4%E6%8E%A8%E5%AF%BC%E8%BF%87%E7%A8%8B%2C)
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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)