How to solve f(f(n)) ? This is NOT calculus.

Question

I am trying to solve this

 

f(n) = n - 1         if n > 5

      = f(f(n+1))   if n < or = 5

 

What is the name of this kind of function ? f(f(n+1)). A function within another function?

 

f(1) = f(f(1+1)) = f(f(2))

f(2) = f(f(2+1)) = f(f(3))

f(3) = f(f(3+1)) = f(f(4))

f(4) = f(f(4+1)) = f(f(5))

f(5) = 5 - 1 = 4

f(6) = 6 - 1 = 5

f(7) = 7 - 1 = 6

 

I can only find the answer when n > 5. When n < or = 5, I can only see f(f(k)). How to find the answers for f(1),f(2),f(3) & f(4) ? Thank you.

Answer 1

There are two things here,

1. the piecewise function, defined as f(n)= n-1 for the domain, n>5 and.                                           f(n)=f(f(n+1)) if n≤5
2. Let's rewrite the function using other function notations, say h(n) and g(n). For the section of the domain when n≤5,

                        Let h(n)=n-1, g(n)=n+1 

f(n)=h(g(n))

This is composition of functions. Other notation is h(n)οg(n).

 

for the part of your question, solving the function for n<5

f(4)=h(g(4))

=h(4+1)

=(4+1)-1=4

 

f(3)=h(g(3))
      =h(3+1)
      =(3+1)-1=3

 

f(2)=h(g(2))
     =h(2+1)
     =(2+1)-1=2

 

f(1)=h(g(1))
     =h(1+1)
     =(1+1)-1=1