Negative time complexity? Just finishing an investigation into Shor's algorithm, and the following equation,
$$
O\big(\big(\log N\big)^2 \big(\log \log N\big)\big(\log \log \log N\big)\big)
$$
is given for its time complexity. However, this is negative for most small values, becoming 0 at $1\cdot 10^{10}$.
Why is this?
The big O notations describes the complexity when $N$ approaching infinity, it is not a formula giving you exact running time for all $N$. 

Roughly, let $f(N)$ be the function for the running time of the algorithm, the big O notation says that there exists $n$ and a constant $c$ such that 
 $f(N)&lt;c\cdot (\log N)^2(\log\log N)(\log\log\log N)$ for all $N&gt;n$.