Timeline for Mapping exponentiation onto addition
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2022 at 8:27 | history | edited | Jojo | CC BY-SA 4.0 |
deleted 1 character in body
|
| Jun 20, 2022 at 8:14 | history | edited | Jojo | CC BY-SA 4.0 |
added 166 characters in body
|
| Jun 20, 2022 at 8:11 | vote | accept | Jojo | ||
| Jun 20, 2022 at 7:52 | history | became hot network question | |||
| Jun 20, 2022 at 7:41 | answer | added | Emil Jeřábek | timeline score: 13 | |
| Jun 19, 2022 at 22:11 | comment | added | LSpice | @GeraldEdgar, such functions still have $h$ constant (take $a = 1$). Then the restriction is that $f$ is constant on powers (i.e., that $b \mapsto f(a^b)$ is constant for all $a$), and we may take $g = f$ and $h = 0$. | |
| Jun 19, 2022 at 21:02 | comment | added | Oliphaunt | @Anixx thanks for that link, that seems very apropos. Also to the original poster at math.se. | |
| Jun 19, 2022 at 19:40 | comment | added | Gerald Edgar | If we find that there is no non-trivial solution in reals, how about asking for maps $\mathbb N \to \mathbb N$ where $f(a^b) = g(a)+h(b)$. | |
| Jun 19, 2022 at 19:32 | history | edited | Jojo | CC BY-SA 4.0 |
added 11 characters in body
|
| Jun 19, 2022 at 19:30 | review | Close votes | |||
| Jun 24, 2022 at 3:01 | |||||
| Jun 19, 2022 at 19:12 | comment | added | Emil Jeřábek | It's not clear to me what is the intended domain of $a$ and $b$, as $a^b$ is not well defined for arbitrary real $a,b$. But anyway, the natural solution, which works for $a>1$ and $b>0$, is $\log\log a^b=\log\log a+\log b$. | |
| Jun 19, 2022 at 19:06 | answer | added | LSpice | timeline score: 3 | |
| Jun 19, 2022 at 19:00 | history | edited | LSpice | CC BY-SA 4.0 |
`\eqref`
|
| Jun 19, 2022 at 18:59 | comment | added | LSpice | It's clear to me that $a^b$ isn't symmetric in $a$ and $b$, but not that $f(a^b)$ isn't; and, indeed, the example of $f = 0$ shows that it can be. I guess more important is that $f(1^b)$ is constant. | |
| Jun 19, 2022 at 18:54 | comment | added | Anixx | This may be relevant: en.wikipedia.org/wiki/… | |
| Jun 19, 2022 at 18:47 | history | edited | Jojo | CC BY-SA 4.0 |
deleted 27 characters in body
|
| Jun 19, 2022 at 18:38 | history | asked | Jojo | CC BY-SA 4.0 |