User blog comment:BlauesWasser/How do you come up with growth rates for functions?/@comment-31966679-20180824143844

For example, if I tried to calculate the growth rate of \(A(n) = n + 1\):

It looks slow growing so let's compare it to small numbers for the fast-growing hierarchy (most people call it FGH).

\(f_{0}(n) = n + 1\), by definition.

\(f_{1}(n) = f_{0}^{n}(n)\), so it would be \(n + \underbrace{1 + 1 + \ldots + 1}_{n - 1} + \underbrace{1 + 1 + \ldots + 1}{n}\)

Or the easier way to express it, roughly \(2n\)

However, \(f_{0}(n)\) works because \(A(n) = n + 1\) AND \(f_{0}(n) = n + 1\)

Let's try another one!

\(B(n) = n! = n(n - 1)(n - 2)\ldots(3)(2)(1)\)

This doesn't diagonalize over \(f_{n}(x)\), so let's test small results for it.

\(f_{0}(n) = n + 1\)

\(f_{1}(n) \approx 2n\)

\(f_{2}(n) \approx 2 ^ {n ^ {2} + 2n}\)

But wait! \(f_{2}(n)\) grows faster than n!, but f_{1} grows slower!

So now we say that \(f_{1}(n) < B(n) < f_{2}(n)\)