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I would like to know what is the actual meaning of the $\sim$ symbol in asymptotic analysis. Specifically wherever I look it seems to mean the following: $$\lim\limits_{n\to \infty}\dfrac{f(n)}{g(n)} = 1$$

But if one looks at this question there are people that use the symbol to mean $$\lim\limits_{n\to \infty}\dfrac{f(n)}{g(n)} = C$$ for some constant $C$. I assume this second case is something that big-theta $\Theta$ would be better used for than tilde $\sim$.

I am looking for a solid reference where I could see the usage of tilde symbol as presented in the second case.

Calvin Khor
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Michael Munta
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  • You're righ, it is used for asymptotic equivalence. The most general definition is that there exists a function $\varepsilon(n)$ with $\lim_{n\to\infty}\varepsilon(n)=0$ such that $f(n)=\bigl(1+ \varepsilon(n)\bigr)g(n)$ . – Bernard Apr 22 '20 at 08:58
  • You mean $\lim\limits_{n\to \infty} f(n) = \bigl(1+ \varepsilon(n)\bigr)g(n)$ with your last equality? – Michael Munta Apr 22 '20 at 10:05
  • Not at all, it is supposed to be valid for all $n$. It is a more explicit way to write $f(n)-g(n)=o(g(n))$. – Bernard Apr 22 '20 at 10:25
  • Can you provide a reference for the general definition you specify in your first comment? – Michael Munta Apr 22 '20 at 10:50
  • If you can read French, it is mentioned in the dedicated page Fonctions équivalentes. Other than that, The English page for Asymptotic Analysis explains why the definition in terms of little o (which is fundamentally the same). – Bernard Apr 22 '20 at 11:06
  • Thank you. In addition, have you ever seen the usage of tilde for the case when they tend to some constant instead of $1$? Is the usage of tilde in that case incorrect? – Michael Munta Apr 22 '20 at 11:19
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    No. The standard definition, as you know, uses $c=1$ (with some drawbacks, as explained in the English Wikipedia notice. I think – but I'm no specialist – that, used in this extended sense, the notation has to be explained before. – Bernard Apr 22 '20 at 11:26
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    I think it is worth mentioning that $f(x) = \Theta(g(x))$ does not imply $f(x)/g(x) \to C$ for some constant $C \neq 0$. The quotient might not converge – Sofía Marlasca Aparicio Jul 02 '20 at 07:51
  • In the theory of asymptotic expansions, $\sim$ has two meanings. It either means that $f(z)=g(z)+o(g(z))$ as $z$ tends to a critical point, or it is used for asymptotic expansions. Asymptotic expansions are usually divergent, hence you cannot put an equality sign between a function and its asymptotic expansion. One uses $\sim$ instead. I would never use $\sim$ in the second context you gave. – Gary Jul 02 '20 at 08:45
  • @Gary Can you take a look at the first comment on the question? I am confused with $f(n)=\bigl(1+ \varepsilon(n)\bigr)g(n)$. How can this be true for all $n$ if we are concerned with limiting behavior? Maybe you can provide an example? – Michael Munta Jul 03 '20 at 13:59
  • @MichaelMunta by redefining $\epsilon(n)$ for the first few values of $n$. This doesn't change the fact that $\epsilon(n)\to 0$ – Calvin Khor Jul 03 '20 at 14:06
  • @CalvinKhor Is it possible to show it by example? I don't understand what you mean by redefine. – Michael Munta Jul 03 '20 at 14:18
  • @MichaelMunta I just mean, if you think it can be done for all large enough $n$, then (as long as $g(n)$ isn't $0$) you can just define $\epsilon(n):=f(n)/g(n)-1$ for all $n$ until it kicks in – Calvin Khor Jul 03 '20 at 14:32
  • @CalvinKhor I like your answer, but I am still not understanding the stuff with $\epsilon(n)$ here... – Michael Munta Jul 04 '20 at 07:27
  • $f(n)/g(n)\to 1$ is equivalent to (if say $f,g$ never vanish) $f(n)=(1+\epsilon(n))g(n)$ where $\epsilon(n)\to 0$. This $\epsilon(n)$ can be written explicitly by the formula $\epsilon(n) = f(n)/g(n)-1$. @MichaelMunta – Calvin Khor Jul 04 '20 at 07:30
  • $\epsilon(n)$ is just a "fudge factor". Slightly simpler variant - two arbitrary quantities $a,b$ might not be equal, but you can always find an equal sign by introducing a remainder term, $a=b+R$. Explicitly $R=a-b$. – Calvin Khor Jul 04 '20 at 07:33

1 Answers1

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I found the notation you're asking for. From this and the various other known interpretations*, it should be clear that "the actual meaning of the $\sim$ symbol", even when dealing with asymptotic results, does not exist. Rather, like many notations, it is a convention that depends on the author. I'd guess one reason is that although "asymptotic analysis" is arguably its own field, its results are particularly suited to being applied in other fields, and of course different fields want different things from their notation.

* I mean of course the much more common notation $a_n\sim b_n \iff a_n/b_n\to 1$, readily found in Wikipedia and sources therein, although even this has minor variants.

1 $a\sim b$ in the sense of the linked question's answer Relation between $m$th Fibonacci number and Golden Ratio

This can be found in the following 2015 book "Stochastic Partial Differential Equations" by Sergey V. Lototsky and Boris L. Rozovsky, page 2 (Springerlink)(Google Books Preview):

Notation $a_{k} \sim b_{k}$ means $\lim _{k \rightarrow \infty} a_{k} / b_{k}=c \in(0, \infty),$ and if $c=1,$ we will emphasize it by writing $a_{k} \simeq b_{k} .$ Notation $a_{k} \asymp b_{k}$ means $0<c_{1} \leq a_{k} / b_{k} \leq c_{2}<$ $\infty$ for all sufficiently larger $k .$ The same notations $\sim, \simeq,$ and $\asymp$ can be used for functions. For example, as $x \rightarrow \infty,$ we have $$ 2 x^{2}+x \sim x^{2}, x+5 \simeq x, x^{2}(2+\sin x) /(1+x) \asymp x $$

Below I also give two "near misses".

2 $f\sim Ag$ instead of $f\sim g$

I found this a while back in this over 100 years old paper "Oscillating Dirichlet's Integrals" by G.H. Hardy (Quarterly Journal of Pure and Applied Mathematics, v.44 (1912)). Hardy was among the first few who began to use asymptotic notation. You can see it here:

The case (iii) includes certain special cases of importance. It may happen, for example, that $f / \phi$ tends to a definite limit: we then write $$ f \mathbin{\style{display: inline-block; transform: rotate(90deg)}{)|(}} \phi $$ Finally, it may happen that this limit is unity: we then write $$ f \sim \phi $$ It will be convenient, in order to avoid the frequent use of a rather inelegant symbol, to write $$ f \sim A \phi $$ instead of $f \mathbin{\style{display: inline-block; transform: rotate(90deg)}{)|(}} \phi.$ The notation implies that " there is a constant $A,$ not zero, such that $f \sim A \phi$". There is, of course, no implication that the various values of $A$ are the same;

3 $a\sim b$ in the sense of $a=\Theta(b)$

From Terry Tao's 'Compactness and Contradiction', page xii (which can be found in this extract):

I will, however, mention a few notational conventions that I will use throughout. The cardinality of a finite set $E$ will be denoted $|E| .$ We will use the asymptotic notation $X=O(Y), X \ll Y,$ or $Y \gg X$ to denote the estimate $|X| \leq C Y$ for some absolute constant $C>0 .$ In some cases we will need this constant $C$ to depend on a parameter $(\mathrm{e} . \mathrm{g} ., d),$ in which case we shall indicate this dependence by subscripts, e.g., $X=O_{d}(Y)$ or $X \ll_{d} Y$ We also sometimes use $X \sim Y$ as a synonym for $X \ll Y \ll X$.

That is, he is using $X\sim Y$ the way you'd use $Y=\Theta(X)$. Note that $a\sim b$ in the sense of 1 above implies $a=\Theta(b)$, and the implication is not reversible, since the limit might not exist.

Finally I should point out what someone correctly commented: there is also the asymptotic series notation, $f\sim\sum_{n=0}^\infty a_n \phi_n$, but its quite hard to mistake the two from context.

Calvin Khor
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  • What about the answers in the related question where people are using it to mean that the limit tends to a constant instead of $1$? Specifically where the user Rocherz says in comment "$F_m \sim \phi^m$ actually means $\displaystyle \lim_{m\to\infty} \dfrac{F_m}{\phi^m}$ is constant." Are they wrong? – Michael Munta Jul 03 '20 at 14:04
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    @MichaelMunta That's not $\Theta$ at all, that's stronger. I've seen that as well, in some old papers. I know an old paper where Hardy wrote $f\sim Ag$, for this meaning, but $A$ here was not a number, rather the notation $\sim A$ somewhat abusively means that there existed some number $A$ such that $f/g\to A$. But if you want exactly that notation in a paper, I don't recall which paper... Theres many different 'related' and 'roughly the same' meanings of $\sim$ unfortunately – Calvin Khor Jul 03 '20 at 14:10
  • @MichaelMunta I might have time (but not now). to look for more sources for other variants, for the Hardy's paper its in this question I asked https://math.stackexchange.com/questions/3318508/what-does-hardy-mean-in-this-lemma – Calvin Khor Jul 03 '20 at 14:33
  • @MichaelMunta updated my answer – Calvin Khor Jul 04 '20 at 03:54