if f0(x)=x^2 and fn+1(x)= f0(fn(x)) for n= 0,1,2,....., find an expression for fn(x) and use mathematical induction to prove it.

Question

if f0(x)=x^2 and fn+1(x)= f0(fn(x)) for n= 0,1,2,....., find an expression for fn(x) and use mathematical induction to prove it. 

Answer 1

N=0:    f1(x) = f0(f0(x)) = (x^2)^2 = x^4

 

N=1: f2(x) = f0(f1(x)) = (x^4)^2 = x^8

 

N=2: f3(x) = f0(f2(x)) = (x^8)^2 = x^16

 

N=3: f4(x) = f0(f(3)) = (x^16)^2 = x^32

 

 

So the pattern appears to be Fi(x) = x^(2^(i+2)) for some integer N=i>0

This serves as the induction hypothesis.

 

Let z = N = i+1

 

Fz = F0(Fi(x)) = [ x^(2^(i+2)) ]^2 = [ x^(2^(i+2)) ] [ x^(2^(i+2)) ]

 

                                                    = x^ [ (2^(i+2)) + (2^(i+2))] <--- same base, adds the exponents

 

                                                    = x^ [ 2* (2^(i+2) ]  <---   adding any number to itself is the same as doubling

 

                                                    = x^ [ 2^(i+3) ] <--- same base, adds the exponents

 

                                                    = x^ [ 2^ (i+1+2) ]

 

                                                    = x^ [2^(z+2)]

 

The statement holds true for z=i+1

End of Proof