# In Latex, what does X_n \asymp Y_n mean ?

I've tried to find the answer from various places:

https://groups.google.com/forum/#!topic/sci.math/2DIwkFqATXE

http://mathoverflow.net/questions/51761/does-fxgx-imply-fx-asymp-gx

But I somehow I find these explanations a bit confusing

In my case, I am talking about 2 sequences of numbers X_n and Y_n, so in latex, what does X_n \asymp Y_n mean ?

I really appreciate any help !

P/S Now does it make sense to have X_n \asymp Y_n, given that X_n is non-increasing and Y_n is non-decreasing ?

https://groups.google.com/forum/#!topic/sci.math/2DIwkFqATXE

http://mathoverflow.net/questions/51761/does-fxgx-imply-fx-asymp-gx

But I somehow I find these explanations a bit confusing

In my case, I am talking about 2 sequences of numbers X_n and Y_n, so in latex, what does X_n \asymp Y_n mean ?

I really appreciate any help !

P/S Now does it make sense to have X_n \asymp Y_n, given that X_n is non-increasing and Y_n is non-decreasing ?

## All Answers (8)

Anh Nguyen Duc· Oxford University Clinical Research UnitJomar Fajardo Rabajante· University of the Philippines Los BañosX_n \asymp Y_n means X_n is asymptotically equivalent to Y_n (this is Theta in BACHMANN-LANDAU NOTATIONS). In layman's words: X_n increases as fast as Y_n (not faster nor slower) --- they have the same "speed". lim X_n/Y_n=constant as n->+inf.

for example: X_n=2n^2 and Y_n=3n^2+2 are asymptotically equivalent. X_n=2n^3 and Y_n=3n^2+2 are NOT asymptotically equivalent.

Anh Nguyen Duc· Oxford University Clinical Research UnitJomar Fajardo Rabajante· University of the Philippines Los BañosHerbert H H Homeier· Universität RegensburgE.g. X_n = 1+ 1/n ; Y_n = 1 - 1/n^2

Increasing/decreasing is not the best question here. Also, the concept of "speed" as mentioned above is misleading: In the above example, Y_n converges faster to 1 than X_n. But still X_n \asymp Y_n for n \to \infty .

In this context, you should also come across the notion of asymptotic expansion, and the big-O and small-o notations.

In the example above, you could write also

X_n = 1 + O(1/n) for n \to \infty

Y_n= 1 + O(1/n^2) = 1+ o(1/n) for n \to \infty

Note that one sometimes considers n as a continuous variable, and then,

X_n \asymp Y_n can also be used for limits to n=0, say. Note also the '+' in the second line. It is correct.

X_n - 1 = O(1/n) means that X_n - 1 ~ K * 1/n for some constant K with 0 < |K| < \infty .

Y_n -1 = o(1/n) means that Y_n - 1 is "asymptotically smaller" than 1/n, thus, converges faster to zero than 1/n.

See any decent introduction to analysis for more details.

Jomar Fajardo Rabajante· University of the Philippines Los BañosAnh Nguyen Duc· Oxford University Clinical Research UnitJomar Fajardo Rabajante· University of the Philippines Los Bañosanother example: lim (2+1/n)/(1-1/n^2)=2. so they are asymp equivalent. but (2+1/n) and (1-1/n^2) approach different limits as n->inf.

counterexample: lim n^2/(n^3+1)=0. so they are NOT asymp equivalent.

another counterexample: lim 2n^5/(2n^3+4)=inf. so they are NOT asymp equivalent.

Can you help by adding an answer?