Timeline for Sum of subset of geometric series: a^2^n
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2013 at 21:16 | history | edited | user9072 |
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| May 11, 2010 at 3:37 | comment | added | Qiaochu Yuan | You can truncate the series at any finite point and assume that it continues like a geometric series and that gives you a sequence of upper bounds a + a^2 + ... + a^{2^n}/(1 - a^{2^n}). For moderately large n and moderately small a the error in this approximation will be pretty small. | |
| May 11, 2010 at 3:25 | vote | accept | Henry Yuen | ||
| May 11, 2010 at 3:25 | comment | added | Henry Yuen | Well, in particular I'm looking for an upper bound that's tighter than 1/(1-a). | |
| May 11, 2010 at 3:17 | vote | accept | Henry Yuen | ||
| May 11, 2010 at 3:25 | |||||
| May 11, 2010 at 2:59 | comment | added | Qiaochu Yuan | The partial sums grow really, really, really slowly. What exactly do you want to know? | |
| May 11, 2010 at 2:55 | answer | added | Wadim Zudilin | timeline score: 16 | |
| May 11, 2010 at 2:41 | comment | added | Henry Yuen | Interesting. Are there any characterizations about how slowly the partial sums grow with respect to the growth of the partial sums of the geometric series? | |
| May 11, 2010 at 1:44 | comment | added | Qiaochu Yuan | There really isn't one. This is an example of a lacunary function (en.wikipedia.org/wiki/Lacunary_function), and it's known more for its undesirable analytic properties than anything else. (Admittedly, it satisfies a nice functional equation.) | |
| May 11, 2010 at 1:37 | history | asked | Henry Yuen | CC BY-SA 2.5 |