# What does $\ll$ mean?

I saw two less than signs on this Wikipedia article and I was wonder what they meant mathematically.

http://en.wikipedia.org/wiki/German_tank_problem

EDIT: It looks like this can use TeX commands. So I think this is the symbol: $\ll$

-
+1 for bringing up the German tank problem. Reminds me of a variant a prof once posed to us: when you arrive in a town, and you see a tram with number 14, how can you estimate the number of trams riding in that town. – Raskolnikov May 2 '11 at 16:45

In the occurrence of "$\ll$" you are asking about, it means "much less than". If you look at the fourth entry here, this is the first meaning listed for $\ll$.

As Charles has correctly pointed out, this symbol is also used in advanced mathematics to describe a certain relationship in the growth of two functions. That is the second meaning listed.

-
That works for me! – ssiddi38 May 2 '11 at 5:09
@ssiddi38: glad to help! – Zev Chonoles May 2 '11 at 5:41
Downvoter: Why the downvote? – Zev Chonoles May 2 '11 at 5:41
@ZevChonoles There are lot of stupids out there who downvote people for no particular reason. The best solution is to only ignore them. – user93957 Jan 3 '14 at 11:24

"$a\ll b$" can also mean "$a$ at least as smaller than $b$ as it is needed for my arguments to be true".

It is in that sense that one sometimes writes, for example, "let $x$ be such that $0< x\ll 1$" to mean "let $x$ be a positive number as small as needed for the following to hold".

-

It does not mean "much less than". It is the Vinogradov symbol, similar to the Hardy-Landau-etc. Big O notation.

$$f(x)\ll g(x)$$ means that there exists some N and k > 0 such that, for all x > N, $f(x)<k\cdot g(x).$ In slightly more informal terms, it means that the asymptotic growth of f(x) is no faster than that of g(x).

-
@Charles: That is one use of it, yes, but I don't believe it is the meaning in the occurrence that the OP is asking about. However, I will edit my answer to be less categorical in saying that $\ll$ means "much less than". – Zev Chonoles May 2 '11 at 5:44
Actually, it is. The precise version would replace $\approx$ with = and multiply by a factor of (1 + o(1)). – Charles May 2 '11 at 5:47
@Charles: Unfortunately I am inexperienced with both statistics and asymptotic analysis. Could you describe in more detail the way in which the $k$ and $N$ which occur in the OP's link are functions that are asymptotically related like this? – Zev Chonoles May 2 '11 at 6:05
@Moron: Yes, I mentioned Vinogradov and Big O. – Charles May 2 '11 at 12:40
@Charles: Yes, I was only agreeing with your answer (you have my +1 already). – Aryabhata May 2 '11 at 15:52

This is the perfect example of the overloaded symbol. In measure theory, we use $\nu << \mu$ if the measure $\nu$ is absolutely continuous with respect to $\mu$, i.e., for any measurable $E$, we have $\mu(E) = 0\Rightarrow \nu(E) =0.$ Most mathematical symbols require a context to be interpreted unambiguously.

-

Another way to think about the much less than $\ll$ is in the spirit of approx $\approx$. When you write $a \ll b$ you say that errors of size $a$ don't matter in assessing a quantity of size $b$. So in the article you reference, saying $k \ll N$ allows replacement of $N-k$ by $N$ to simplify the expression, or $N-k \approx N$. For practical purposes, such as the one in the article, an answer within $10\%$ is plenty good enough.

-

Perhaps not its original intention, but we (my collaborators and former advisor) use $X \gg Y$ to mean that $X \geq c Y$ for a sufficiently large constant $c$. Precisely, we usually use it when we write things like:

$$f(x) = g(x) + O(h(x)) \quad \Longrightarrow \quad f(x) = g(x) (1 + o(1))$$

when $g(x) \gg h(x)$.

-
I noticed you have deleted a few nice answers... Is there any particular reason for this? – Mariano Suárez-Alvarez Aug 31 '11 at 4:44
@Mariano. Thanks for your interest. Every once in a while I go through and delete some old answers which don't add to the conversation (either the answers get repeated and the latter is accepted, other answers are more inclusive and incapsulate my answer, etc.) I figured this is standard operating procedure, but based on your comment, maybe it is not. Is this acceptable behaviour? – JavaMan Aug 31 '11 at 4:56
it is quite acceptable! (It got picked by the software running the site, as it is rather unusual) I tend to prefer to keep such answers, in any case :) – Mariano Suárez-Alvarez Aug 31 '11 at 5:04
@Mar Thank you again for checking. I didn't feel the deleted answers added anything to the conversation (if they add even a little, I leave them). In the future I will use more discretion in both which questions to answer (based on the quality of my given answer) and which answers I delete after a while. – JavaMan Aug 31 '11 at 5:14

It means significantly smaller than, if I'm not mistaken.

-