Timeline for A functional equation with a quadratic solution
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2017 at 19:53 | vote | accept | user118723 | ||
| Dec 19, 2017 at 19:53 | vote | accept | user118723 | ||
| Dec 19, 2017 at 19:53 | |||||
| Dec 19, 2017 at 19:53 | vote | accept | user118723 | ||
| Dec 19, 2017 at 19:53 | |||||
| Dec 19, 2017 at 3:52 | comment | added | Alexandre Eremenko | @user118723: I posted a general solution. You may select those which are concave. | |
| Dec 19, 2017 at 3:44 | answer | added | Alexandre Eremenko | timeline score: 1 | |
| Dec 19, 2017 at 3:28 | answer | added | Somos | timeline score: 1 | |
| Dec 19, 2017 at 3:23 | comment | added | user118723 | Thanks for your comment, but your solutions are not concave, as required by (1). Is it possible to construct a concave solution using trig functions like this? | |
| Dec 19, 2017 at 2:59 | comment | added | Alexandre Eremenko | Your solutions are not the only solutions, in fact there are many, for example $\cos(x-y)=\cos x\cos y+\sin x\sin y$ ($f_3=0$). Same with $\sin(x-y)$ and there are many more. | |
| Dec 19, 2017 at 1:02 | comment | added | Somos | I think that $u(x)$ satisfies linear recurrences with constant coefficients. | |
| Dec 19, 2017 at 0:41 | history | edited | user118723 | CC BY-SA 3.0 |
added 159 characters in body
|
| Dec 19, 2017 at 0:32 | review | First posts | |||
| Dec 19, 2017 at 0:59 | |||||
| Dec 19, 2017 at 0:32 | history | asked | user118723 | CC BY-SA 3.0 |