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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.

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  • $\begingroup$ @PeterTamaroff It sounds better! $\endgroup$ Sep 20, 2012 at 15:03
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    $\begingroup$ See math.stackexchange.com/questions/36364/what-does-ll-mean where the dual use of that notation is discussed. $\endgroup$ Sep 20, 2012 at 16:47
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    $\begingroup$ @AméricoTavares: If I had to read $X \ll_{\epsilon} Y$, I would say "$X$ is much less than a constant depending on $\epsilon$ times $Y$". $\endgroup$
    – JavaMan
    Sep 20, 2012 at 18:13
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    $\begingroup$ It is not "very much less than". The symbol is about order. $\endgroup$
    – timur
    Sep 23, 2012 at 17:47
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    $\begingroup$ "controlled by a constant multiple of ..., with the constant possibly depending on epsilon", or if you just don't want to be quite as you write it on blackboard, "less less sub epsilon". $\endgroup$
    – timur
    Sep 23, 2012 at 18:32

7 Answers 7

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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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  • $\begingroup$ 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 $\epsilon$"? $\endgroup$ Sep 21, 2012 at 14:33
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    $\begingroup$ @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. $\endgroup$ Sep 21, 2012 at 14:39
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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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  • $\begingroup$ +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. $\endgroup$ Sep 21, 2012 at 13:21
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    $\begingroup$ This is getting dangerously close to reading $X\simeq Y$ as "X-tilde-over-horizontal-line-Y". $\endgroup$
    – user31373
    Sep 21, 2012 at 17:12
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    $\begingroup$ @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. $\endgroup$ Sep 21, 2012 at 23:07
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    $\begingroup$ @LVK, I've heard X~Y read "$X$ twiddles $Y$," though I wouldn't recommend it. $\endgroup$ Sep 21, 2012 at 23:09
  • $\begingroup$ @GerryMyerson Thanks again for the clarification. $\endgroup$ Sep 22, 2012 at 7:05
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"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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    $\begingroup$ Many thanks for your answer. $\endgroup$ Sep 29, 2012 at 21:33
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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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    $\begingroup$ If you are unsure how to answer the question, why did you write this as an answer? $\endgroup$
    – JavaMan
    Sep 20, 2012 at 18:12
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I prefer to say " doubly less than ", which is more convenient than " less than less than " when speaking Chinese.

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I would say/I have heard it said "$A$ is (upper) bounded by $B$, (perhaps) up to a (big/small) constant factor (depending on ...)", or "we have a (lower/upper) bound of $A$ by $B$, [same as before]".

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