$O_q(1)$ notation

Question

I cant find anywhere what this notation is supposed to mean:

$$\frac{L'(s,\chi_0)}{L(s,\chi_0)} = -\frac{1}{s-1}+O_q(1)$$

Where $\chi_0$ is the principal character mod $q$

What is $O_q(1)$ here? Im guessing it has something to do with order.

Answer 1

This is a "big Oh" notation. More specifically, we write $$ f(x) = O(g(x)),$$ where $f(x)$ is some function and $g(x) \geq 0$, if there are constants $C$ and $X$ such that $$ \lvert f(x) \rvert \leq C g(x) \quad (\forall x > X).$$ There are variants of big oh notations. It can also be used as $x \to x_0$ for some finite $x_0$; in this case, we say that $f(x) = O(g(x))$ as $x \to x_0$ if there are constants $C$ and $\delta$ such that $$ \lvert f(x) \rvert \leq C g(x) \quad (\forall x \text{ with } \lvert x - x_0 \rvert < \delta).$$

The constant $C$ is sometimes called "the implicit constant". In some cases it is useful to indicate what the implicit constant depends on. For example, for any $\epsilon > 0$, it's true that $$ \tau(n) = O_\epsilon(n^\epsilon),$$ for all $n \geq 1$ as $n \to \infty$, where $\tau(n)$ is the number of divisors of $n$. The notation $O_\epsilon(\cdot)$ means that the implicit constant depends on $\epsilon$. In practice, if $\epsilon$ is smaller, then the implicit constant is larger.

In your particular case, your asymptotic holds as $s \to 1$. The strength of the statement is that the leading growth is independent of the character, but the implicit constant in the error term depends on the conductor of the character.

This is a statement about the Laurent expansion of $L'/L$ around the value $s = 1$.