Timeline for Limit of Integration of continuous function
Current License: CC BY-SA 4.0
15 events
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| May 1 at 5:36 | history | edited | User1865345 | CC BY-SA 4.0 |
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| May 1 at 4:05 | history | edited | Zhanxiong |
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| Mar 27, 2021 at 6:00 | history | tweeted | twitter.com/StackStats/status/1375689044248641536 | ||
| Mar 23, 2021 at 4:16 | history | became hot network question | |||
| Mar 23, 2021 at 4:09 | vote | accept | edison | ||
| Mar 22, 2021 at 20:44 | comment | added | whuber♦ | Given any $\epsilon\gt 0,$ Chebychev's Inequality assures us that for sufficiently large $n,$ the volume of the region where $|(x_1+x_2+\cdots+x_n)/n-1/2|\le \epsilon$ is very close to $1.$ Assuming $|f|$ is bounded on $[0,1],$ this implies the integral is an average of the values of $f$ within this neighborhood of $1/2$ plus a vanishingly small term. Thus, provided $f$ has a well-defined limiting average value at $1/2$ (which its continuity there assures), the integrals must be approaching that average. | |
| Mar 22, 2021 at 20:20 | answer | added | Zhanxiong | timeline score: 13 | |
| Mar 22, 2021 at 20:09 | history | rollback | whuber♦ |
Rollback to Revision 2
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| Mar 22, 2021 at 19:51 | history | rollback | edison |
Rollback to Revision 1
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| Mar 22, 2021 at 19:50 | history | edited | Zhanxiong | CC BY-SA 4.0 |
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| Mar 22, 2021 at 19:42 | review | Close votes | |||
| Mar 22, 2021 at 22:09 | |||||
| Mar 22, 2021 at 19:29 | comment | added | edison | @whuber can you please explain how to apply Chebyshev's Inequality to prove the argument is close to 1/2. | |
| Mar 22, 2021 at 19:17 | review | First posts | |||
| Mar 23, 2021 at 2:24 | |||||
| Mar 22, 2021 at 19:16 | comment | added | whuber♦ | One method: use Chebyshev's Inequality to show that when $n$ is large enough, almost all the time the argument of $f$ is close to $1/2.$ It's unclear what you might mean by "any bounds," because you have explicitly given $[0,1]^n$ as the limits of the integral and obviously all these iterated integrals lie between the extremes of $f$ (which are finite because $f$ is continuous and $[0,1]$ is compact). | |
| Mar 22, 2021 at 19:09 | history | asked | edison | CC BY-SA 4.0 |