Timeline for Does function $f(x)=f(2x)$, $f(x)$ - non const, exist? ($f(x)$ - continuous function on real numbers)
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2020 at 11:57 | comment | added | Dmitriy Shekhmatov | Ok... Thank you. | |
| Dec 2, 2020 at 11:38 | comment | added | Wlod AA | THEOREM $\ \ $ Every function $\ f:\mathbb R\to\mathbb R\ $ such that (1)$\ \forall_{x\in\mathbb R} f(2\cdot x)=f(x),\ $ and (2) $f$ is continuous at $0$, is constant. $\ \ $ PROOF $\frac x{2^n} \to 0\ $ hence $f(x)=f(\frac x{2^n})=f(0).\ \ $ End of Proof (It's trivial. End of discussion). | |
| Dec 2, 2020 at 10:17 | comment | added | Dmitriy Shekhmatov | Okay, what about this one? $$ \begin{cases} x\ =\ 0,\ f(x)\ :=\ 1;\\ x\ \neq\ 0,\ f(x)\ :=\ |\sin(\,2\cdot\pi\cdot \log_2(|x|)\,)| \end{cases} $$ | |
| Dec 2, 2020 at 6:19 | comment | added | Wlod AA | You require continuity. However, your variant is not continuous at $0$ since $f$ swings between $1$ and $-1$ again and again when the argument approaches $0$ hence your $f$ is not continuous at $0$. | |
| Dec 1, 2020 at 23:25 | comment | added | Dmitriy Shekhmatov | Thank you for shorter formula! Could you answer why that system cannot be an answer? $$ \begin{cases} x\ =\ 0,\ f(x)\ :=\ 0;\\ x\ \neq\ 0,\ f(x)\ :=\ \sin(\,2\cdot\pi\cdot \log_2(|x|)\,) \end{cases} $$ Sorry for the stupid question... | |
| Nov 28, 2020 at 6:48 | history | answered | Wlod AA | CC BY-SA 4.0 |