# How does one read aloud Vinogradov's notation $\ll$ and $\ll_{\epsilon }$?

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).

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.

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@PeterTamaroff It sounds better! – Américo Tavares Sep 20 '12 at 15:03
@Byron Schmuland Thanks for correcting the name in the title! – Américo Tavares Sep 20 '12 at 15:10
See math.stackexchange.com/questions/36364/what-does-ll-mean where the dual use of that notation is discussed. – Unreasonable Sin Sep 20 '12 at 16:47
@AméricoTavares: If I had to read $X \ll_{\epsilon} Y$, I would say "$X$ is much less than a constant depending on $\epsilon$ times $Y$". – JavaMan Sep 20 '12 at 18:13
It is not "very much less than". The symbol is about order. – timur Sep 23 '12 at 17:47

As a default, I second Gerry Myerson's answer "less-than-less-than", although just reading "sub-epsilon" seems a little odd. In a situation where I'm trying to reinforce the meaning, I might read $A\ll B$ as "$A$ is dominated by $B$", and $A\ll_\varepsilon B$ as "$B$ dominates $A$ depending on $\epsilon$", but this may be defeating the purpose/intent of the question. (Edit: or, better, as noted by Americo Tavares, to say "$A$ is dominated by $B$ depending on $\varepsilon$", to avoid the need to read ahead...)

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Thanks. This has nothing to do with the original question, but as for $A\ll_{\epsilon}B$ instead of "$B$ dominates $A$ depending on $\epsilon$" couldn't we say with the same meaning "$A$ is dominated by $B$ depending on $\epsilon$"? – Américo Tavares Sep 21 '12 at 14:33
@Americo... oh, yes, certainly the verb can be arranged so that the $A$, $B$ are spoken in the same order as appearing on the page. Probably preferable, yes, rather than seeming to require look-ahead. – paul garrett Sep 21 '12 at 14:39
Thanks for your comment. – Américo Tavares Sep 21 '12 at 14:41

I read $\ll$ as "less than less than". I don't think I have ever put $\ll_{\epsilon}$ in words, but if I did it would probably come out "less than less than sub epsilon".

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+1 Thanks! It appears e.g. in ch.5 of Arithmetic Algebraic Geometry by Brian Conrad, Karl Rubin. And in Terence Tao's last post. – Américo Tavares Sep 21 '12 at 13:21
This is getting dangerously close to reading $X\simeq Y$ as "X-tilde-over-horizontal-line-Y". – user31373 Sep 21 '12 at 17:12
@Américo, I didn't mean to suggest I hadn't ever seen the notation with the subscript, only that I have never had occasion to read it out loud to anyone. – Gerry Myerson Sep 21 '12 at 23:07
@LVK, I've heard X~Y read "$X$ twiddles $Y$," though I wouldn't recommend it. – Gerry Myerson Sep 21 '12 at 23:09
@GerryMyerson Thanks again for the clarification. – Américo Tavares Sep 22 '12 at 7:05

"Controlled by a constant multiple of ..., with the constant possibly depending on epsilon", or if you don't want to be quiet as you write it on blackboard, "less less sub epsilon".

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From my experience, for $\ll$ we say "Less than less than" and as for $\ll_\epsilon$, I would say "Less than less than where the constant depends on epsilon."

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Many thanks for your answer. – Américo Tavares Sep 29 '12 at 21:33

Extending Charles' answer in the thread I linked to, $f(x)\ll_{\epsilon} g(x)$ means that for a given $\epsilon$ there exists some $N$ and $k$ such that for all $x > N$, $f(x) < k\cdot g(x)$. In other words, $k$ and $N$ depend on the value of $\epsilon$. As to how you would read or pronounce that, I am not sure.

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If you are unsure how to answer the question, why did you write this as an answer? – JavaMan Sep 20 '12 at 18:12