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 ?
i dont think researchgate currently supports latex format.
X_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.
Jomar, thank you so much for a quick reply. So does it make sense to have X_n \asymp Y_n, given that X_n is non-increasing and Y_n is non-decreasing ?
hmmm... i dont think so. for example X_n=1/n (non increasing) and Y_n=n^2 (non decreasing)...
Well, there are also examples that do make sense.
E.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.
True that saying speed is not mathematically appropriate. That is why i put it in quotation marks "". But it is one of the best ways to explain it to layman. It seems the one who asked the question is still starting to understand the topic. We can define \asymp in mathematical way as what i show in the succeeding sentence - we can even make it more rigourous as we want. But we need to see who is the reader/audience.
Thank you all for the exciting discussion. In the example, X_n = 1+ 1/n ; Y_n = 1 - 1/n^2, both series converge to the same limit. So, is it also possible that X_n and Y_n go to different limit but they are still \asmp with each other ?
one way to check \asymp is lim X_n/Y_n=constant (not equal to zero) as n->+inf. for example, lim (1+1/n)/(1-1/n^2)=lim n/(n-1)=1.
another 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.
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