Timeline for Mapping exponentiation onto addition
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2022 at 14:37 | comment | added | Jojo | Thankyou. This is much simpler than trying to look at $a=b$. | |
| Jun 21, 2022 at 11:25 | comment | added | Emil Jeřábek | You showed yourself in the question a reduction to $h(1)=0$, $g=f$ by subtracting a constant. Likewise, by subtracting a constant from $f$, you can arrange $f(e)=0$, thus $f(e^b)=h(b)$. This reduces the equation to $h(b\log a)=h(b)+h(\log a)$, that is, $h(xy)=h(x)+h(y)$. Then $h(e^x)$ satisfies Cauchy’s equation, thus if continuous, there is $\alpha$ such that $h(x)=\alpha\log x$ for $x>0$. Also, $h(-x)=h(x)+h(-1)$, and $h(-1)+h(-1)=h(1)=0$, thus $h(-x)=h(x)$, i.e., $h(x)=\alpha\log|x|$ for $x\ne0$, and $f(x)=g(x)=h(\log x)=\alpha\log|\log x|$. Now, undo the constant-subtracting reductions. | |
| Jun 20, 2022 at 19:39 | comment | added | Jojo | Thanks for the link, this was very interesting and counterintuitive to me. What method would you use to show that the solution you gave is unique? My default is to try and rewrite functional equations as ODEs, but it doesn't always work very well | |
| Jun 20, 2022 at 9:30 | comment | added | Emil Jeřábek | If $f$, $g$, or $h$ is required to be continuous, then it’s not hard to show that the only solutions are $h(x)=\alpha\log|x|+\beta$, $g(x)=\alpha\log|\log x|+\gamma$, $f(x)=\alpha\log|\log x|+\beta+\gamma$ for some constants $\alpha,\beta,\gamma\in\mathbb R$. For discontinuous solutions, see e.g. the discussion in en.wikipedia.org/wiki/Cauchy's_functional_equation (this is a slightly different equation, but the same applies mutatis mutandis for the equation $h(xy)=h(x)+h(y)$). | |
| Jun 20, 2022 at 9:05 | comment | added | Jojo | Could you explain why that is? Might it be unique if $f$ is continuous? | |
| Jun 20, 2022 at 8:36 | comment | added | Emil Jeřábek | It’s not going to be unique without some kind of continuity requirement. | |
| Jun 20, 2022 at 8:29 | comment | added | Jojo | This is a very nice and simple solution. Do you have thoughts on whether it's unique as a solution to my $(3)$? I think the logarithm is the unique solution to $f(ab) = f(a) + f(b)$, up to a scaling maybe. I edited my question to specify more clearly | |
| Jun 20, 2022 at 8:11 | vote | accept | Jojo | ||
| Jun 20, 2022 at 7:41 | history | answered | Emil Jeřábek | CC BY-SA 4.0 |