Timeline for On the sum of uniform independent random variables
Current License: CC BY-SA 3.0
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| Jul 4, 2016 at 5:47 | vote | accept | CommunityBot | ||
| Jul 3, 2016 at 12:51 | comment | added | Iosif Pinelis | [Continuation of the previous comment:] So, $D'_n$ must have at least $5-1=4=3+1$ roots in $(0,1)$. Continuing thus and using the fact that $D_n$ is $n-1$ times continuously differentiable, we see that $D_n^{(n-1)}$ must have at least $3+n-1=n+2$ roots in $(0,1)$. Please let me know if you have more questions about this answer. | |
| Jul 3, 2016 at 12:50 | comment | added | Iosif Pinelis | Here are details on the previous comment. On supposing that your conjecture is false, we saw that $D_n$ must have at least $3$ roots in $(0,1)$. Plus, $0$ and $1$ are roots of $D_n$ (and of its derivatives of orders up to $n-1$). So, $D_n$ must have at least $5$ roots in $[0,1]$. By Rolle's theorem, strictly between any two roots of a differentiable function there is at least one root of its derivative. | |
| Jul 3, 2016 at 12:50 | comment | added | Iosif Pinelis | There are two typos here -- sorry. Of course, I meant to say "the derivative $D_n^{(n-1)}$ of $D_n$ of order $n-1$ has at least $3+n-1=n+2$ roots in $(0,1)$". | |
| Jul 3, 2016 at 5:45 | comment | added | user31317 | Hi Iosif,can you explain this: "Then, by the Rolle theorem, the derivative $G^{(n−1)}_n$ of $G_n$ of order $n−1$ has at least $n+2$ roots" | |
| Jul 1, 2016 at 21:24 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jul 1, 2016 at 21:09 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jul 1, 2016 at 20:58 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jul 1, 2016 at 20:51 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |