Timeline for Approximation of the form $\frac{1}{u}\pm\frac{1}{v}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 5, 2014 at 10:26 | vote | accept | Dominic van der Zypen | ||
| Sep 3, 2014 at 7:55 | answer | added | Aaron Meyerowitz | timeline score: 3 | |
| Sep 2, 2014 at 23:26 | comment | added | Gerry Myerson | See also mathoverflow.net/questions/179574/… from the same source. | |
| Sep 2, 2014 at 14:39 | comment | added | Dominic van der Zypen | @OwenBiesel Yes, because $z_1, z_2$ are (positive or negative) integers. | |
| Sep 2, 2014 at 14:10 | comment | added | Willie Wong | The solution to the best approximation problem may not be unique. $5/6 = 1/2 + 1/3 = 1/1 - 1/6$. But in any case, the search space is finite if you only want an algorithm. For the "addition" problem, let $w$ be the natural number such that $1/w$ is the largest one less than $m/n$. Then clearly $z_1 \in \{w, \ldots, 2w - 1\}$. For each candidate $z_1$ just compute the ceiling and floor of $1 / (m/n - 1/z_1)$ and find the best. A similar search algorithm can be given for the case where $z_1> 0$ and $z_2 < 0$. (The search space has size $O(n/m)$, btw.) | |
| Sep 2, 2014 at 14:05 | history | edited | Yemon Choi |
replaced a tag
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| Sep 2, 2014 at 13:47 | comment | added | Owen Biesel | Are approximations of the form $(1/z_1) - (1/z_2)$ also allowed, as the question title seems to indicate? | |
| Sep 2, 2014 at 13:20 | comment | added | Douglas Zare | Why would you look for an approximation like that? Few rational numbers have good approximations of that form. For example, you can't approximate $5$ very well like that. Anyway, there is an obvious algorithm starting with the larger piece, which can't be much smaller than $\frac{m}{2n}$ unless it is $1$, checking for the best smaller piece. Is this algorithm unsatisfactory? | |
| Sep 2, 2014 at 13:02 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |