# Asymptotic equivalence for functions with zeros

I am considering the relative asymptotic behavior of a pair of real functions on the positive real axis, say $f$ and $g$. There is no $x_0$ such that $f$ and $g$ are non-zero for all $x>x_0$.

I need a sensible definition of asymptotic equivalence when there is not necessarily an $x_0$ such that $f$ and $g$ are non-vanishing for all $x>x_0$.

That is, I need an analogue of the statement (which only makes sense if such an $x_0$ exists):

There exists $A>0$ and $B<\infty$ such that $$A\leq \left|\frac{f(x)}{g(x)}\right|\leq B$$ for all sufficiently large $x$. Equivalently, $f(x)=O(g(x))$ and $g(x)=O(f(x))$ as $x\rightarrow\infty$.

By sensible'', I require that the analogous statement firstly implies the above whenever $f$ and $g$ are non-vanishing for all sufficiently large $x$ and, secondly, that it does not imply the stronger statement $|f(x)/g(x)|\rightarrow C$ as $x\rightarrow \infty$ whenever $f$ and $g$ are non-vanishing for all sufficiently large $x$.

EDIT: I also require that one does not need to know anything about the zeros of the pair, i.e. excluding particular subsets of the domain is not a satisfactory extension of the definition.

The application I have in mind is number theoretic. Specifically, I am interested in the relative asymptotics of $L(x)$ and $M(x)$.

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It is not the only problem, you see - multiplication by zero is also a problem. –  Kevin Smith May 18 '12 at 10:25
If the only problem you see is division by zero, why not simply use $A∣g(x)∣\le∣f(x)∣\le B∣g(x)∣$ for all $x$ large enough, or (as this would implies same zeros for $f$ and $g$, which you may not like), for all $x$ large enough and where $f$ and $g$ dont vanish ? –  Feldmann Denis May 18 '12 at 10:27
The statement and where $f$ and $g$ do not vanish'' literally does reduce to the required statement, yet it is no longer a statement about $f$ and $g$. For the purpose of doing analysis with it, one would have to replace $f$ and $g$ with some other pair that are non-zero at those points. This is unsatisfactory because one has introduced an infinite set of new values and it is not clear how these values should be chosen. –  Kevin Smith May 18 '12 at 10:44
I don't think there's any standard definition of the sort you are looking for, and I doubt there's any way to formulate a broadly useful definition. When I've run into this issue before (for example, in asymptotics where the terms had oscillatory factors that occasionally vanish), the right thing to do has just been to give explicit error terms. I.e., instead of saying something is asymptotic to $\sin(x)/x$, say that it equals $\sin(x)/x + O(1/x^2)$ (or whatever error term you can prove). –  Henry Cohn May 18 '12 at 11:56
Indeed, thank you Henry. The explicit error terms in the case I am interested are equally difficult, so I was trying to look at the problem from another angle. I was thinking that $A|f|^2|g|\leq |f||g|^2\leq B |f|^2|g|$ would work, but it does seem somewhat artificial. –  Kevin Smith May 18 '12 at 12:59

It isn't clear if you intend that $f$ and $g$ are eventually zero at the same places. Otherwise I wouldn't want to call them asymptotically equivalent. What you need is $$f(x) = (1+o(1)) g(x),$$ where as usual $o(1)$ means some function converging to 0.
In the case I have in mind they are not eventually zero at the same places, so perhaps extending the definition of asymptotic equivalence is too strong. Specifically, I am considering $L(x)$ and $M(x)$, so your suggestion would lead me to attempting to prove or disprove the statement $L(x)=(1+o(1))M(x)$. This is exactly the kind of thing I am getting at, as it seems it may not be the case if RH holds, yet the failure of RH appears to imply it. –  Kevin Smith May 18 '12 at 12:32
Maybe something like $f(x)=(1+o(1))g(x) + o(1)$ with suitably small $o(1)$ terms is what you need. –  Brendan McKay May 18 '12 at 13:18
@ Brendan: Your suggestions don't allow the cases when $f$ and $g$ don't eventually vanish at the same places, and they both amount to $|f/g|\rightarrow 1$ if $g$ doesn't vanish eventually. My reference to the possible implications of RH is not related to the zeros of $f$ and $g$, it is to do with the behavior of $\zeta (2\rho)$. Although I think RH is indirectly related to the zeros of $f$ and $g$ through some old theorems of Polya and Turan. –  Kevin Smith May 19 '12 at 11:16