Little oh notation

Question

The GNFS-function $f(n)$ with $n$ the number for which we search the factorisation shows the complexity of that factorisation:

\begin{align} f(n) = e^{\left(\sqrt[3]{\frac{64}{9}} + o(1)\right)\left(\ln n\right)^{\frac{1}{3}} \left(\ln\ln n\right)^{\frac{2}{3}}} \end{align}

What is the meaning and value of $o(1)$?

Thanks for the help!

Answer 1

It means, by definition: "which goes to $0$ as $n\to\infty$."

Now, for the value... it could be any function converging to $0$. In this particular case, you could try to get more information about this low-order term, but without more information we cannot say anything else.

Answer 2

Given two functions $h, g \colon \mathbb{N} \to \mathbb{N}$, in general $h \in o(g)$ means that for every choice of a constant $k > 0$, you can find a constant $a$ such that the inequality $0 \leq h(n) < k \, g(n)$ holds for all $n > a$.

Here, $1$ stands for the constant function $g(n) = 1$ for all $n \in \mathbb{N}$. Therefore, $o(1)$ in your expression stands for any function $h \colon \mathbb{N} \to \mathbb{N}$ such that, for every choice of a constant $k > 0$ there is a constant $a$ such that $0 \leq h(n) < k$ holds for all $n > a$. In other words, $\lim_{n \to \infty} h(n) = 0$.