Timeline for Background on the functional equation $F(x+1)+F(x)=f(x)$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2018 at 13:26 | comment | added | Fedor Petrov | Your $h$ equals $F$, remove one of multiples (any of two). | |
| Sep 18, 2018 at 13:03 | comment | added | soodehMehboodi | Now, could you explain that how you obtain your answer or introduce me some related books or references to do that. | |
| Sep 18, 2018 at 13:00 | comment | added | soodehMehboodi | Henec, if we denote $h(x)=F(x)\varepsilon(x) (-1)^{\lfloor x\rfloor} \quad\text{where}\quad \varepsilon(x) = \begin{cases} 1 & \text{if } \lfloor x\rfloor\text{is odd}, \\ -1 & \text{if }\lfloor x\rfloor \text{is even}. \end{cases} $ then $h$ works for all $x$. | |
| Sep 18, 2018 at 12:59 | comment | added | soodehMehboodi | @Fedor Petrov: Thanks for your answer firstly, but I checked it again and found that this $g$ doesn't work when $\lfloor x\rfloor$ is even. | |
| Sep 17, 2018 at 15:36 | comment | added | Fedor Petrov | Yes, it turns out into (1) | |
| Sep 17, 2018 at 15:31 | comment | added | soodehMehboodi | Or you give just an special solution. | |
| Sep 17, 2018 at 15:19 | comment | added | soodehMehboodi | Do you mean that by applying this $g$, the new functional equation turns into the functional equation (1)? | |
| Sep 16, 2018 at 7:55 | history | answered | Fedor Petrov | CC BY-SA 4.0 |