Timeline for Asymptotic equivalence for functions with zeros
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 18, 2012 at 12:59 | comment | added | Kevin Smith | 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. | |
| May 18, 2012 at 11:57 | answer | added | Brendan McKay | timeline score: 2 | |
| May 18, 2012 at 11:56 | comment | added | Henry Cohn | 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). | |
| May 18, 2012 at 11:06 | history | edited | Kevin Smith | CC BY-SA 3.0 |
Corrected grammar, added ``EDIT'' statement.
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| May 18, 2012 at 11:00 | history | edited | Kevin Smith | CC BY-SA 3.0 |
deleted 15 characters in body
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| May 18, 2012 at 10:44 | comment | added | Kevin Smith | 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. | |
| May 18, 2012 at 10:27 | comment | added | Feldmann Denis | 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 ? | |
| May 18, 2012 at 10:25 | comment | added | Kevin Smith | It is not the only problem, you see - multiplication by zero is also a problem. | |
| May 18, 2012 at 10:14 | history | asked | Kevin Smith | CC BY-SA 3.0 |