The rule says that \(f(6) = f(5) + 11\) (we are using \(6 = n+1\) so \(n = 5\)). We don't know what \(f(5)\) is though. Well, we know that \(f(5) = f(4) + 9\text{.}\) So we need to compute \(f(4)\text{,}\) which will require knowing \(f(3)\text{,}\) which will require \(f(2)\text{,}\)… will it ever end?
Yes! In fact, this process will always end because we have \(\N\) as our domain, so there is a least element. And we gave the value of \(f(0)\) explicitly, so we are good. In fact, we might decide to work up to \(f(6)\) instead of working down from \(f(6)\text{:}\)
It looks that this recursively defined function is the same as the explicitly defined function \(f(n) = n^2\text{.}\) Is it? Later we will prove that it is.