\(f(10) = 1024\text{.}\) To find \(f(10)\text{,}\) we need to know \(f(9)\text{,}\) for which we need \(f(8)\text{,}\) and so on. So build up from \(f(0) = 1\text{.}\) Then \(f(1) = 2\text{,}\) \(f(2) = 4\text{,}\) \(f(3) = 8\text{,}\) .... In fact, it looks like a closed formula for \(f\) is \(f(n) = 2^n\text{.}\) Later we will see how to prove this is correct.
in-context