Timeline for Prove/disprove $(\int_{0}^{2 \pi} \!\!\cos f(x) d x)^{2}+(\int_{0}^{2 \pi}\!\!\! \sqrt{(f'(x))^{2}+\sin ^{2} f(x)}dx)^{2}\ge 4\pi^{2}$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 29, 2021 at 19:27 | comment | added | Robert Bryant | @MattF.: Actually, your $f$ does not produce an actual circle on the unit sphere, i.e., a curve that is the intersection of the sphere with a plane. That's probably why your integral isn't coming out to be $(2\pi)^2$. | |
| Jan 29, 2021 at 12:00 | answer | added | Robert Bryant | timeline score: 5 | |
| Jan 29, 2021 at 11:03 | comment | added | user44143 | Let $f(x)=\pm \arccos(1/\sqrt{1+\sin^2 x})$, with $+$ for $x<\pi$, $-$ for $x>\pi$. Then $C$ is a circle but the integral comes to $1.33252 (2\pi)^2$. What gives? See wolframalpha.com/input/… | |
| Jan 29, 2021 at 2:08 | comment | added | Paata Ivanishvili | This can be technical but nevertheless: you can always try "optimal control approach". See Section 3 here arxiv.org/pdf/1712.04590.pdf how it works. In your case due to a requirement $f(0)=f(2\pi)$ you will need to introduce one more variable $\int_{0}^{t} f'(s)ds=z$ in (3.6). Eventually the question will reduce to finding a function B of 4 variables $(t,x,y,z)$ which satisfies the corresponding Hamilton--Jacobi--Bellman PDE. See this example mathoverflow.net/questions/275980/… how it works. | |
| Jan 28, 2021 at 23:10 | answer | added | Minh | timeline score: -6 | |
| Jan 23, 2021 at 13:26 | comment | added | Robert Bryant | One cannot show that 'there's no other $f$ [besides constants] that makes the left hand side equal (or be less than) $(2\pi)^2$', since it is not true. As the OP has remarked, if the corresponding closed curve $C$ is any circle on the sphere, then equality holds. Thus, there is (at least) a $3$-parameter family of functions $f$ that give equality, while the case of constant $f$ only constitutes a $1$-parameter family. | |
| Jan 21, 2021 at 11:04 | history | asked | FFjet | CC BY-SA 4.0 |