Timeline for A third degree surface and a touching sphere [closed]
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2016 at 19:20 | comment | added | Vladimir Zolotov | My argument was wrong. It only works only for $a > (2/5)^{1/3}$, where the curvature is monotonic. | |
| Apr 8, 2016 at 10:34 | comment | added | Lev Borisov | Correct. The discriminant is $(x-y)^2(y-z)^2(x-z)^2$, so if the roots are real, then it is nonnegative. | |
| Apr 8, 2016 at 5:46 | comment | added | Sergei | @Lev Borisov - I missed your idea with discriminant. You checked it is negative, but it has to be nonnegative? | |
| Apr 8, 2016 at 5:36 | comment | added | Sergei | @VladimirZolotov - may you give more details, with proper references on connection of curvature with value of radius? I do not know much of diffgeometry... | |
| Apr 8, 2016 at 5:33 | comment | added | Sergei | @LevBorisov -so "easy"? Still no solution based on convexity, sorry. | |
| Apr 7, 2016 at 21:34 | comment | added | Vladimir Zolotov | Which has curvature $< 1$, while the circle of radius $\frac{\sqrt 3}{2}$ has curvature $> 1$. And we are done. | |
| Apr 6, 2016 at 21:29 | comment | added | Vladimir Zolotov | The closest points to $(1,1,1)$ on $xyz=1$ are of the form $(a,1/\sqrt a,1/\sqrt a)$. So, it suffices to check this curve. | |
| Apr 6, 2016 at 20:23 | comment | added | Lev Borisov | The most elegant solution is to write "This is an easy exercise which we leave to the reader." | |
| Apr 6, 2016 at 20:21 | comment | added | Lev Borisov | It is easy to see by convexity of sorts that there are no other solutions near $(1,1,1)$. Specifically, let $x=e^a$, $y=e^b$, $z=e^c$. Then you have to solve $e^{2a}-3e^{a} + .... = -6$ for $a+b+c=0$. By Jensen's you will not have nontrivial solutions in the region where $f(t)=e^{2t}-3e^t$ is convex. | |
| Apr 6, 2016 at 18:37 | comment | added | Sergei | @Vladimir Zolotov - pochtaname(c)gmail.com | |
| Apr 6, 2016 at 18:35 | comment | added | Sergei | @Lev Borisov - I solved it by Lagrange. The aim of the question is to find geometrical solution using convexity of related surface. | |
| Apr 6, 2016 at 17:34 | comment | added | Vladimir Zolotov | @Sergei Can you add a reference to the joint paper and/or motivation, please | |
| Apr 6, 2016 at 15:08 | comment | added | Lev Borisov | You can do Lagrange multiplers for $xyz$ on the sphere. This will give you three equations like $xy=(2z-3)\lambda$ and so on. By subtracting one of such equation from the other you get $x=z$ or $y=-2\lambda$. Either way you get linear relations on $x,y,z$ and it should be easy to solve. Then you will get $xyz\geq 1$ with minimum at $(1,1,1)$. | |
| Apr 6, 2016 at 15:01 | comment | added | Lev Borisov | You can try to take log coords $x=e^a,y=e^b,z=e^c$ and then see if the inequality $(e^a-3/2)^2+(e^b-3/2)^2+(e^c-3/2)^2\leq 3/4$ gives a convex shape. But I think that my approach with the discriminant is short, if not "nice". | |
| Apr 6, 2016 at 14:53 | comment | added | Sergei | Lev Borisov - I proved the underlying inequality of 3rd degree as a lemma for some result in joint paper. But the proof is not nice or short. So I am interesting to find another one using geometry and convexity. | |
| Apr 6, 2016 at 14:45 | comment | added | Lev Borisov | Certainly, computer gives accurate enough coordinates of four roots to know that they are complex. If that's not enough for you, try to write it as opposite of sum of squares of quadratic polynomials. | |
| Apr 6, 2016 at 14:40 | comment | added | Sergei | Somebody of the usual set "put on hold" may prove it? Will be thankful... | |
| Apr 6, 2016 at 14:28 | history | closed |
Ben McKay Stefan Kohl♦ Wolfgang Joonas Ilmavirta Stefan Waldmann |
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| Apr 6, 2016 at 14:21 | comment | added | Lev Borisov | I am getting discriminant to be $(1/4)(s-3)^2(-s^4 + 6s^3 - 24s^2 +56s - 60)$. The remaining polynomial of degree $4$ is negative for all $s$. | |
| Apr 6, 2016 at 14:14 | review | Close votes | |||
| Apr 6, 2016 at 14:28 | |||||
| Apr 6, 2016 at 13:56 | comment | added | Lev Borisov | Consider the polynomial of degree three with roots $x,y,z$. The information given implies that it is equal to $f(t)=t^3-st^2 +(s^2/2 -3s/2 + 3)t -1$. Three roots are real iff the discriminant is nonnegative. This gives you a polynomial inequality on $s$, which you presumably can check. | |
| Apr 6, 2016 at 13:38 | review | Low quality posts | |||
| Apr 6, 2016 at 14:29 | |||||
| Apr 6, 2016 at 13:36 | comment | added | Lev Borisov | Decent question, wrong forum. | |
| Apr 6, 2016 at 13:21 | history | asked | Sergei | CC BY-SA 3.0 |