# 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 $$1\cdot 10^{10}$$. I was wondering why this is?

• This is only 34 bits. Try 2^{128} or for RSA 2^{2048} – kelalaka Oct 14 '18 at 19:43
• I get a root @ $e^e$. Strange. Ah - natural logs. – Paul Uszak Oct 14 '18 at 21:49

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 $$f(N) for all $$N>n$$.
• And, as $n\rightarrow \infty$ equation $\rightarrow \infty$ – kelalaka Oct 14 '18 at 19:57