Problem: Given the linear fractional transformation of x into f1(x)=x+12x−1. Define fn+1(x)= f1(fn(x)) for n=1,2,3,⋯. Assuming that f35(x)=f5(x), it follows that f28(x) is equal to
Answer Choices:
A. x
B. x1
C. xx−1
D. 1−x1
E. None of these
Solution:
If g(x) is the inverse of the transformation f1(x), then g(fn+1(x))=fn(x). Since f35(x)=f5(x), successive application of this formula yields f31(x)=f1(x)∴f30(x) =g(f1(x))=x∴f29(x)=g(x)∴f28(x)=g(g(x)). But g(x)=2−xx+1∴f28(x)= g(g(x))=1−x1.