Timeline for Is there a definition of $\log(x)$ for quaternion/octonion $x$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2023 at 18:24 | comment | added | Dieter Kadelka | @Sidhardt Ghoshal: I computed $x^{0.1}$ as $\exp(\log(x)*0.1)$ with $\log(x)$ as in the question. | |
| Nov 3, 2023 at 15:47 | comment | added | Sidharth Ghoshal | @DieterKadelka how did you compute $x^{0.1}$ over the quarternions/complexes? There are notions of branch cut here that can mess those results up but if you look at the entire set of possible $(0.1)^{\text{th}}$ roots then one of them will indeed cancel out and give back $x$ | |
| Nov 3, 2023 at 9:48 | vote | accept | Dieter Kadelka | ||
| Nov 3, 2023 at 9:48 | comment | added | Dieter Kadelka | @DaveBenson: Thank you once again. It seems that actually no solution for this problem is possible, even for complex $x$. For instance, if $x = 1+4j$, then $(x^{10})^{0.1} \not= x$. | |
| Nov 2, 2023 at 21:49 | comment | added | Dave Benson | @OlegEroshkin Have you ever tried computing Bessel functions of the first kind $J_n(z)$? The power series converges for all $z$, but if $z$ is not small then it involves massive cancellation, and this results in poor control over rounding errors. This is why Miller's backwards recurrence algorithm was developed; this is used by Unix bc -l for example. | |
| Nov 2, 2023 at 16:09 | comment | added | Oleg Eroshkin | @DieterKadelka What do you mean "Using power series is too slow."? I am sure that library functions to calculate logs, exponential functions, trig functions and all other transcendental functions are using power series (with some clever tricks). Any formula that you will write will be calculated in the end using power series. | |
| Nov 2, 2023 at 15:33 | comment | added | Sidharth Ghoshal | It might behave well in commutative subsets of the quaternions and octonions, similar to how in matrix exponentiation $e^{A+B} = e^A e^B$ if $A,B$ commute | |
| Nov 2, 2023 at 12:57 | history | edited | Dave Benson | CC BY-SA 4.0 |
added 177 characters in body
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| Nov 2, 2023 at 10:33 | comment | added | Dieter Kadelka | Thank you, I've changed the question. Using power series is too slow. I tried the algorithm of the question for $q = (1,v)$ with $\|v\| < 1$. Here we actually have $\exp(\log(q)) = \log(\exp(q)) = q$, Which seems to indicate that power series are not needed. | |
| Nov 2, 2023 at 10:14 | history | answered | Dave Benson | CC BY-SA 4.0 |