Solving functional equations using recursion

Question

If $f:\mathbb N \rightarrow \mathbb N$ such that $f(f(f(n)))+f(f(n))+n=3f(n) \; \forall \; n\in \mathbb N,$ Then Find $f$

Solution Provided in book:

Replace $n$ with $f(n)$ successively in parent functional equation $k$ times. We get

$\underbrace {fofof...of(n)}_\text{k+3 times }+\underbrace {fofof...of(n)}_\text{k+2 times }+\underbrace {fofof...of(n)}_\text{k times }=3 \underbrace {fofof...of(n)}_\text{k+1 times }$ $\cdots \cdots(1)$

Let $a_0=n$ for some fixed $n$ and $a_{k+1}=f(a_k) \; \forall \;k\geq 0$

My doubt: How did he assume $a_0=n$ and How it became $a_{k+1}=f(a_k)$

Please Help In this. I also checked for duplicate but i couldn't find it.

Also Any other way to solve this kind of problem?

Answer 1

The sequence goes like this:

$$ \{ n, f(n) , f \circ f(n)... \}$$

The action of 'composition' with $f$ generates the next element in the sequence. Note that we are not making any assumptions about the nature of $f$ when we say that the first element of the said sequence is $n$, so everything is still done in total generality.

Answer 2

Both of these statements are just definitions, we define a new sequence $(a_i)_i$ with

$$ a_0 := n $$

and

$$ a_{k+1} = f(a_k)$$

This is a recursive definition of a sequence given by an anchor ($a_0$) and a recursion, so no assumptions have been made yet.