How does one read aloud the Vinogradov's notation $\ll$ and $\ll_{\epsilon }$ as in
$f(x)\ll g(x)$
and
$c\ll_{\epsilon }\left( \prod\limits_{p\mid abc}p\right) ^{1+\epsilon}.$
Is the first one “is very much less than”? (This is a direct translation from Portuguese as was used informally in engineering formulae).
Added: I am not asking the meaning of this notation, rather how it is read.
This answer to the question mentioned in Unreasonable Sin's comment points to this Wikipedia entry. According to it in Analytic number theory the symbol $\ll$ in $f(x)\ll g(x)$ is to be read as “is of smaller order than”.
Added 2. As a response to LVK's comment:
"I am not asking the meaning of this notation, rather how it is read." But the way to spell out notation depends on what it means in the particular text. If Vinogradov used it to mean $f=O(g)$ in one of his papers, then in that paper the symbol should be read differently.
Let me give a specific example. I would like to know how to read $\ll _{\epsilon }$ in the following conjecture
ABC Conjecture. Suppose $A,B,$ and $C$ are positive integers, suppose $\gcd (A,B,C)=1,$ and suppose $A+B=C.$ Then $C\ll _{\epsilon }\left( \prod\limits_{p\mid ABC}p\right) ^{1+\epsilon}.$
Source: Arithmetic Algebraic Geometry, Brian Conrad, Karl Rubin, ch. 5, p.123.
The meaning as I understand it is the same as:
Suppose:
i) $A,B$ and $C$ are positive integers,
ii) $\gcd (A,B,C)=1,$
iii) $A+B=C,$
iv) $\epsilon >0$ is a positive real number.
Then there is a constant $K_{\epsilon}$ such that $C\leq K_{\epsilon}\left( \prod\limits_{p\mid ABC}p\right)^{1+\epsilon }.$
P.S. The symbol $\ll_{\epsilon}$ is also used in Terence Tao's post The probabilistic heuristic justification of the ABC conjecture.