Timeline for O(·) is not a function, so how can a function be equal to it?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2018 at 9:16 | comment | added | Michel Fioc | A possibility would be to use slightly different notations for the set of asymptotic functions, $O(h)$, and for an undetermined element of this set, say $\mathord{\in}O(h)$. So, if $f-g \in O(h)$, one can write $f = g+\mathord{\in}O(h)$ instead of the ambiguous $f = g+O(h)$. You can then write without problem $\mathord{\in}O(h) = f-g$. Other possible notations for an unspecified element of $O(h)$ might be $\dot O(h)$, $\hat O(h)$... | |
| Dec 10, 2018 at 12:57 | comment | added | leftaroundabout | I find computations like that jarring; those equals signs aren't bidirectional anymore. I'm not sure there's more of a problem with writing $f(x) \in e^x(e^{2x}+O(x)) \subset e^x + o(1)$. I suppose that's also abuse of notation; basically you're overloading the $=$ operator whereas I prefer to lift $+$ and $\cdot$ to also operate on sets. | |
| Dec 10, 2018 at 12:48 | comment | added | David Richerby | The rearrangement only works in standalone statements. It's much more common in the middle of calculations, where that kind of thing doesn't work, and where multiple functions get absorbed together into the Landau notation. (Stuff like $f(x) = e^{-x}(e^{2x}+O(x)) = e^x+o(1)$). | |
| Dec 10, 2018 at 12:42 | comment | added | leftaroundabout | You could also write $f(n) - 3n \in O(\log n)$. Though I admit that it can be handy to conclude a multiple-step computation with $f(n) = \ldots = 3n + O(\log n)$. | |
| Dec 10, 2018 at 12:34 | history | answered | David Richerby | CC BY-SA 4.0 |