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Timeline for Implausible inequality

Current License: CC BY-SA 3.0

19 events
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Aug 20, 2017 at 4:55 vote accept axk
Aug 19, 2017 at 23:14 answer added Yaakov Baruch timeline score: 1
Aug 17, 2017 at 5:45 history edited axk CC BY-SA 3.0
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Aug 17, 2017 at 4:50 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 16:49 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 14:26 comment added Yaakov Baruch I would more simply write:$\quad \exists C,\epsilon >0$ constants... it holds that: LHS $\le C t^{-1-\epsilon}$
Aug 16, 2017 at 14:20 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 14:10 comment added axk @Yaskov Baruch: Thanks for the observations. $a\ge 2b$ would be perfect if it works out.
Aug 16, 2017 at 13:50 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 13:23 comment added Yaakov Baruch My previous comment is wrong: since $t\ge dC$ if $d\rightarrow\infty$ then also $t\rightarrow\infty$ and the LHS does not go to $\infty$.
Aug 16, 2017 at 7:19 comment added Yaakov Baruch Notice also that as $d\rightarrow\infty$ the LHS goes to $\infty$ too, so I think for the question to be interesting it must be that $a$ is larger than something that depends on $d$.
Aug 16, 2017 at 6:59 comment added Yaakov Baruch Without assuming $b=2$, you need $a\ge 2b$ otherwise $t^b \left(\frac{t}{d}\right)^{d/2}\left(\frac{d+a}{t+a}\right)^{(d+a)/2}\rightarrow \infty$ and cannot be bound by any $C$.
Aug 16, 2017 at 6:52 comment added Fedor Petrov hm, the limit of the quotient is 0 only if $b<a/2$, no?
Aug 16, 2017 at 6:26 comment added Yaakov Baruch Why is $b\ge2$ needed? If the inequality holds for $b>2$ doesn't it all the more so for $b=2$?
Aug 16, 2017 at 1:56 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 1:50 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 1:03 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 0:58 history edited axk CC BY-SA 3.0
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Aug 16, 2017 at 0:53 history asked axk CC BY-SA 3.0