if f0(x)=1/(2-x) 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)=1/(2-x) 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.

I really have no idea.

Answer 1

Let's use the information we have to see what we can find as an explicit representation for the f⁽ⁿ⁺¹⁾(x) to see how to prove it by induction...

f⁰(x) = 1/(2-x) = f(x)

f⁽ⁿ⁺¹⁾(x) = f⁰(fⁿ(x))

f¹(x) = f⁰(f⁰(x)) = f(f(x)) = f(1/(2-x)) = 1/(2 - 1/(2-x)) = 1/[ (3-2x)/2-x ] = 2-x/(3-2x)

f²(x) = f⁰(f¹(x)) = f(f¹(x)) = 1/(2 - [ (2-x)/(3-2x) ] ) = 1/[ (4-3x)/(3-2x) ] = (3-2x)/(4-3x)

f³(x) = f⁰(f²(x)) = f(f²(x)) = 1/(2 - [ (3-2x)/(4-3x) ] ) = 1/[ (5-4x)/(4-3x) ] = (4-3x)/(5-4x)

A pattern can be seen such that...

f^n-1(x) = f⁰(f^(n-2)(x)) = f(f^(n-2)(x)) = [ (n) - (n-1)x ] / [ (n+1) - nx ]

Does fⁿ(x) follow?

fⁿ(x) = f⁰(f^(n-1)(x)) = f(f^(n-1)(x)) = 1/(2 - [ (n) - (n-1)x ] / [ (n+1) - nx ])

After simplification we arrive at the above we get...

fⁿ(x) = [ (n+1) - nx ] / [ (n+2) - (n+1)x ]

This is indeed the expression we would have if we used the representation in bold based on n->n+1