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Jul 26, 2018 at 19:44 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 26, 2018 at 19:26 comment added Iosif Pinelis There was a typo in the previous version of the answer: I missed the minus in $-4/5[=-1.3+1/2]$. This is now fixed.
Jul 26, 2018 at 19:23 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 26, 2018 at 14:44 comment added Armin Weiß thanks for the change! ...In my calculation $F(u,a) = 1.06589...$ for $a=u=0.095260$ though ($a=u=0.095260$ seems to be correct)...
Jul 26, 2018 at 12:01 vote accept Armin Weiß
Jul 25, 2018 at 18:16 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 25, 2018 at 17:23 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 25, 2018 at 17:17 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 25, 2018 at 17:14 comment added Iosif Pinelis I have modified the answer, to take into account that $\theta\ge1$ and the log is base $2$.
Jul 25, 2018 at 17:12 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 25, 2018 at 15:35 comment added Armin Weiß oops... $\theta \geq 1$ (so $u\leq 1/5$)
Jul 25, 2018 at 15:25 comment added Iosif Pinelis I am still not quite sure: do you want $\theta\ge1$ or $\theta\le1$? Because $\theta\ge1$ corresponds to $u\le1/5$, not $u\ge1/5$. In any case, I think your specification the values of $\theta$ of interest would only simplify the proof. Also, in mathematical literature, $\log$ without a specification of the base usually means the natural log (I think $\ln$ is better for that purpose).
Jul 25, 2018 at 11:45 comment added Armin Weiß Thanks, a lot!! Sorry, I was not precise: $\theta \geq 1$ (so $u\geq 1/5$) and the logarithm is base 2.
Jul 24, 2018 at 18:11 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 24, 2018 at 18:05 history edited Iosif Pinelis CC BY-SA 4.0
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Jul 24, 2018 at 17:54 history answered Iosif Pinelis CC BY-SA 4.0