Timeline for Lower/Upper bounds for $ \sum\limits_{i=0}^k \binom ni x^i $
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2018 at 13:24 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 3 characters in body
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| Jan 6, 2018 at 12:58 | answer | added | Max Alekseyev | timeline score: 2 | |
| Jan 6, 2018 at 12:41 | answer | added | Brendan McKay | timeline score: 7 | |
| Jan 6, 2018 at 12:40 | comment | added | zeraoulia rafik | related to :mathoverflow.net/q/55585/51189 | |
| Jan 6, 2018 at 12:27 | answer | added | zeraoulia rafik | timeline score: 0 | |
| Jan 6, 2018 at 10:26 | comment | added | Fedor Petrov | (Sorry, I was thinking about a different sum $\sum \binom{n}i x^i (1-x)^{n-i}$, which is reduced to this sum and vice versa via introducying $x/(1-x)$ as a new variable). | |
| Jan 6, 2018 at 10:19 | comment | added | Fedor Petrov | @AlexeyUstinov I doubt about geometric progression. Say, if $x=k/n$, the sum is approximately 1/2, but each summand is much less. | |
| Jan 6, 2018 at 10:06 | comment | added | Martin Sleziak |
BTW you can typeset binomial coefficient as \binom ni $\binom ni$. If you want bigger size, you can use \dbinom ni $\dbinom ni`$. (But I would not recommend the latter in the title.
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| Jan 6, 2018 at 10:05 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
edited title
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| Jan 6, 2018 at 10:01 | comment | added | Alexey Ustinov | The ratio of two consequtive summands is $\frac{a_i}{a_{i-1}}=\frac{n-i+1}{i}x$ is unimodular. So you'll easily find the largest summand. And it is an upper bound for the whole sum (up to some constant), because another terms decrease not slower than geometric progression. | |
| Jan 6, 2018 at 9:41 | review | Close votes | |||
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| Jan 6, 2018 at 9:21 | review | First posts | |||
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| Jan 6, 2018 at 9:18 | history | asked | user119319 | CC BY-SA 3.0 |