Timeline for The missing link: an inequality
Current License: CC BY-SA 3.0
28 events
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| Jan 23, 2017 at 0:28 | comment | added | Iosif Pinelis | @Suvrit : On the one hand, all we needed to prove here was the positivity of a smooth function of two variables, $x$ and $n$. As I noted before, in principle such problems can always be done by partitioning the domain -- which is what I did. This approach is only limited by the computer power, and it may take some effort to design a way to stay within such limits -- when at all possible. However, here an amazing amount of positivity was discovered by Peter Mueller. It would indeed be great to have a theory (if there is one) behind this pervasive positivity phenomenon. | |
| Jan 22, 2017 at 19:53 | comment | added | Suvrit | I really hope that there is a more elegant way to prove this inequality! | |
| Jan 22, 2017 at 19:32 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 20, 2017 at 21:18 | comment | added | T. Amdeberhan | I've not yet found a chance to go through this hard work. | |
| Jan 18, 2017 at 20:42 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 18, 2017 at 14:51 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 18, 2017 at 14:43 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 18, 2017 at 6:20 | comment | added | Iosif Pinelis | Indeed, multiple uses of Mathematica command Reduce[] in the last version of this answer are now replaced by showing that the coefficients of the mentioned relevant polynomials are all nonnegative. | |
| Jan 18, 2017 at 5:44 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 17, 2017 at 23:05 | comment | added | Iosif Pinelis | @PeterMueller : It looks like all or most instances of Reduce[] in my answer can be replaced by showing (similarly to what you suggested for $a$) that all the coefficients in the numerator become positive after changing variables $x,y$ to certain other appropriate variables. I am working on this. Positivity is indeed quite pervasive here! | |
| Jan 17, 2017 at 20:26 | comment | added | Iosif Pinelis | [previous comment continued:] This may have to do with the very naturally sounding result by Polya that for any polynomial $p=p(x_1,\dots,x_n)$ such that $p>0$ on $[0,\infty)^n\setminus\mathbf0$, there is some natural $k$ such that all coefficients of polynomial $(x_1+\dots+x_n)^k p(x_1,\dots,x_n)$ are nonnegative. | |
| Jan 17, 2017 at 20:25 | comment | added | Iosif Pinelis | @PeterMueller : Thank you for this nice comment. Indeed, this seems the simplest way to show that $a\ge0$. The amount of positivity that you discovered in this problem seems staggering. So far, my only attempt at understanding this phenomenon is to note that all the coefficients of the polynomials $1+x^k$ and $(1-x^k)/(1-x)=1+x+\dots+x^{k-1}$ are nonnegative for natural $k$. | |
| Jan 17, 2017 at 18:48 | comment | added | Peter Mueller | @IosifPinelis: As an alternative argument to show $a(x,y)\ge0$, one can use that all the coefficients of the numerator of $a(1/(1+x),1/(1+y))$ are nonnegative. As you remarked previously (in a comment to my answer), it is really amazing that positivity of certain polynomials related to the question is a result of positive coefficients. | |
| Jan 17, 2017 at 14:42 | comment | added | Iosif Pinelis | As an illustration, I have added the proof of $a\ge0$ using resultants. | |
| Jan 17, 2017 at 14:37 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 16, 2017 at 17:25 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 16, 2017 at 1:00 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 16, 2017 at 0:54 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 15, 2017 at 23:00 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 15, 2017 at 22:35 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 15, 2017 at 22:22 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 15, 2017 at 22:10 | comment | added | Iosif Pinelis | Now a complete proof, using some of the suggestions in the previous version of this answer. | |
| Jan 15, 2017 at 22:09 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2017 at 2:30 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2017 at 2:08 | comment | added | Iosif Pinelis | Only the coefficient of $\ln^2 x$ is (rather, seems, and that should not be hard to check) positive on the entire big square $(0,1)^2. | |
| Jan 11, 2017 at 1:36 | comment | added | T. Amdeberhan | Are the coefficients of $\log^2 x, \log^2t, \log x\log t$ positive individually? | |
| Jan 10, 2017 at 23:32 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 10, 2017 at 23:14 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |