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What do I call two sequences $a, b$ such that $\lim_{n\to\infty} |a_n - b_n| = 0$? Or what do I call two functions $f, g$ such that $\lim_{x\to c} |f(x) - g(x)| = 0$? (For my purposes, these are essentially the same thing, and I will happily extend a term for one to the other.)

This seems like such a straightforward condition that it must have a standard term, but I can't find it (in either context). I looked through the Wikipedia article on all of the variations of big-$O$ and the like, but these are all too weak. If $\lim_n a_n$ (hence $\lim_n b_n$) existed, then I could call $a$ and $b$ ‘coterminal’, but that limit might not exist. In an incomplete space, I have seen $a$ and $b$ called ‘co-Cauchy’ under the weaker assumption that one (hence both) is Cauchy, but I don't want to assume that either. I could call $\exp f$ and $\exp g$ ‘asymptotic’ (as $x \to c$), but I want to refer to $f$ and $g$ directly.

Surely somebody knows a term for this?

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They are called equivalent. I am half-joking half-serious. –  Włodzimierz Holsztyński Jun 28 '13 at 7:40
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How about "asymptotically equal/equivalent"? The only problem with this term would be that you can pose a similar condition on the ratio and will then need another name for it :-) –  efq Jun 28 '13 at 11:01
    
I have the feeling that ‘asymptotic’ should mean what I want, but when I look it up, it doesn't. –  Toby Bartels Jul 1 '13 at 0:27

1 Answer 1

You could say $ a $ and $ b $ are mutually asymptotic. Example: "Since $ n -1/n$ and $n+1/n $ are mutually asymptotic. .." And usually even drop the word "mutually".

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My understanding of the accepted meaning of ‘asymptotic’ would also allow one to say that $n - 1$ and $n + 1$ are mutually aysmptotic (which I take to mean that each is asymptotic to the other), but these don't meet my requirement. That is why I'm searching for another term. –  Toby Bartels Jul 1 '13 at 0:27
    
I was thinking in terms of asymptotes in calculus. Y=1/x has y=0 as asymptote but not y=1. –  Bjørn Kjos-Hanssen Jul 1 '13 at 1:31
    
True, that does fit what I want. –  Toby Bartels Jul 1 '13 at 15:42
    
I still hold out hope that there might be a better answer? –  Toby Bartels Jul 4 '13 at 13:33

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