Timeline for The missing link: an inequality
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
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| Jan 18, 2017 at 17:53 | comment | added | user90369 | To simplify the problem, maybe is useful to use complementary functions, so that there are only konvex functions. I haven’t checked if this works, it’s only an example: $f_n(x):=\frac{1-x^{2n+1}}{1-x}$, $g_n(x):=\frac{1}{(1+x^{2n+1})(1+0.1/n-x)}$, $h_n(x):=\frac{(1+0.1/n-x)(1-x)}{ 1-x^{2n+2}}$ ; It follows $F_n(x)=\ln((1+x^{4n-1}) (1+x^{2n}) f_n(x)g_n(x)h_n(x))$ konvex, if $f_n, g_n, h_n$ are konvex . | |
| Jan 13, 2017 at 14:24 | answer | added | Yaakov Baruch | timeline score: 3 | |
| Jan 13, 2017 at 10:31 | answer | added | Dima Pasechnik | timeline score: 2 | |
| Jan 11, 2017 at 20:27 | answer | added | Peter Mueller | timeline score: 21 | |
| Jan 11, 2017 at 16:51 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 10, 2017 at 23:14 | answer | added | Iosif Pinelis | timeline score: 11 | |
| Jan 10, 2017 at 22:03 | comment | added | Pietro Majer | Let's recall the old version mathoverflow.net/questions/246919/… | |
| Jan 10, 2017 at 19:07 | comment | added | T. Amdeberhan | @SteveHuntsman: All terms inside the log are log-convex or can be turned around to be so, except for $1-x^{2n+1}$. | |
| Jan 10, 2017 at 18:56 | comment | added | Steve Huntsman | Since products of log-convex functions are log-convex, it might be worth looking at when $\frac{x^\alpha \pm x^{-\alpha}}{x^\beta \pm x^{-\beta}}$ is log-convex on $(0,1)$. Perhaps that problem is i) tractable and ii) after pulling out factors of $x$ you might get lucky. | |
| Jan 10, 2017 at 18:24 | comment | added | Steve Huntsman | @T.Amdeberhan- I have a passing familiarity. I suspect that in practice these methods will be hard-pressed since the degree of $V$ is so large even for small $n$. Another idea for fixed $n$ is to use en.wikipedia.org/wiki/Sturm%27s_theorem. But $V$ has repeated factors, which makes this that much more intricate. | |
| Jan 10, 2017 at 18:14 | answer | added | Matt Young | timeline score: 10 | |
| Jan 10, 2017 at 14:41 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 10, 2017 at 5:26 | answer | added | user103395 | timeline score: 5 | |
| Jan 10, 2017 at 2:17 | answer | added | Joseph O'Rourke | timeline score: 6 | |
| Jan 10, 2017 at 1:51 | comment | added | T. Amdeberhan | @SteveHuntsman: It might be useful although I'm not familiar with the methods there. Are you? | |
| Jan 10, 2017 at 1:34 | comment | added | Steve Huntsman | For $n$ fixed, we can use interval analysis a la chapter 5 of books.google.com/books?id=IEN56sqHtR8C | |
| Jan 10, 2017 at 1:32 | history | edited | GH from MO |
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| Jan 10, 2017 at 1:27 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 10, 2017 at 1:24 | comment | added | Gerhard Paseman | It might turn into a good tool if, instead of investigating convexity of log f for rational functions f, one looks at convexity of log p - log q for polynomials p and q, and finds a more complicated but computationally simpler test. Gerhard "Cross Multiplication Makes One Cross" Paseman, 2017.01.09. | |
| Jan 10, 2017 at 1:21 | comment | added | T. Amdeberhan | @NateEldredge: It would look very complicated here, but I could write in terms of $P=numerator$ and $Q=denominator$. | |
| Jan 10, 2017 at 1:08 | comment | added | Nate Eldredge | Maybe it would help people if you can write that polynomial into the question? | |
| Jan 10, 2017 at 0:36 | comment | added | T. Amdeberhan | @FanZheng: Thanks for the comment. But, we need concrete proofs if you don't mind my saying. | |
| Jan 10, 2017 at 0:35 | comment | added | T. Amdeberhan | @NateEldredge: Yes, you get positivity of a polynomial. | |
| Jan 10, 2017 at 0:24 | comment | added | Fan Zheng | While for a fixed n, such problems can in theory be solved by quantifier elimination of real closed fields, I'm not sure if this is still the case when you have an integer variable in the exponent. | |
| Jan 10, 2017 at 0:21 | comment | added | Nate Eldredge | When you take the second derivative, you get a rational function $p(x)/q(x)$ whose denominator is clearly positive on $(0,1)$. Right? So this reduces to showing that the polynomial $p(x)$ is positive on $(0,1)$. | |
| Jan 9, 2017 at 22:40 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |