User talk:SuperJedi224/Turing Machines

Binary counter
For your information, your binary counter indeed reaches \(n\) in \(O(n)\) steps - \(\frac{n}{2^1}\) numbers below \(n\) are odd, so take \(2\cdot 1\) steps to count up. \(\frac{n}{2^2}\) numbers below \(n\) are even but indivisible by 4, so take \(2\cdot 2\) steps to count up. In general \(\frac{n}{2^k}\) numbers below \(n\) take \(2\cdot k\) steps to count up (this might actually be slightly less, because we round down to nearest integers). By summing we get that to get to \(n\) we will have to make at most \(\sum_{k=1}^\infty \frac{n}{2^k}\cdot 2k=2n \sum_{k=1}^\infty\frac{k}{2^k}=2n\cdot 2=4n=O(n)\). LittlePeng9 (talk) 19:21, April 30, 2015 (UTC)