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I learnt big-O-notation in analytic NT books but in papers they usually use $\ll$ instead. My friend told me they are absolutely same.

1- I wanted to be sure if they are same? and

2- if so just curious why there are two different notations?

3- How to pronounce $\ll$ in talking? and $\gg$? (which has no similarity to O notation I think).

4- How to pronounce $f \asymp g$?

PS There is a similar question here but unf it doesn't answer all questions.

  • I'm afraid I can't give you a complete answer, as I am not sure about the relationship between the notation. But in terms of how to say "<<" and ">>" I'd say "much greater than" or "much less than" – MRobinson Oct 02 '18 at 10:08
  • @MRobinson, it is the same notation as "much less/greater than" but I am not sure if they called same, because the meaning of notation is different; for example $x>> \sqrt{x}$ but not much greater than. –  Oct 02 '18 at 10:13
  • If it's being used the same way then I guess using the same wording could be appropriate? "Is the order of" – MRobinson Oct 02 '18 at 10:16
  • For large enough $x$, it actually is true that $x$ is much greater than $\sqrt x.$ I suspect that that is the intended interpretation in the context where this occurred, though it is hard to be sure without actually seeing exactly what the context was. – David K Oct 02 '18 at 11:42
  • @DavidK, should I say "double less than" or "much less than" or "aymptotically much less than"? (in asymptotic limits in analytic number theory) –  Oct 02 '18 at 11:47
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    I might go with "much less than" or "asymptotically much less than," depending on context. I would only say "asymptotically" if it was not already obvious from context that we were dealing either with asymptotic or "large $x$" behavior. But others would say "less than less than": see the answers to https://math.stackexchange.com/questions/199753/how-does-one-read-aloud-vinogradovs-notation-ll-and-ll-epsilon – David K Oct 02 '18 at 12:27

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1) Your friend is right: The two statements $f = O(g)$ and $f \ll g$ mean exactly the same thing, namely that there exists some $C > 0$ such that $|f(x)|\leqslant C\cdot g(x)$ for all $x$ large enough, small enough, or...

3) I'd say this out loud as "$f$ is at most a constant times $g$."

2) In my experience the $O(\cdot)$-notation is more convenient in certain manipulations than Vinogradov's notation: One can write \begin{align*} \pi (x) = \frac{x}{\log x} \left( 1 + O\left( \frac{1}{\log x} \right) \right) \end{align*} using the $O(\cdot)$-notation, but this cannot really be written as elegantly using Vinogradov's notation. On the other hand, Vinogradov's notation has a very nice and clean look and so it's sometimes preferred for typographical reasons. So each notation has its advantages and disadvantages.

4) I'd say, "$f$ and $g$ have the same order of magnitude."

primes.against.humanity
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