Timeline for Why we can analytically define $ε$ in dual numbers so to distinguish $ε$ from $-ε$ but cannot do so in complex and split-complex numbers?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2022 at 16:11 | comment | added | Anixx | @LaurentMoret-Bailly in that comment $b$ is meant to be positive, I thought it was obvious. | |
| Jun 5, 2022 at 15:59 | comment | added | Laurent Moret-Bailly | In your first comment you wrote "we can define $f(a+b\varepsilon)$ via right derivative". What else does this mean? | |
| Jun 5, 2022 at 14:49 | comment | added | Anixx | @LaurentMoret-Bailly where did you get the first one? I never wrote this, it is wrong. | |
| Jun 5, 2022 at 13:56 | comment | added | Laurent Moret-Bailly | But according to your own definition, you get that $f(a+b\varepsilon)=f(a)+b\varepsilon f'_r(a)$, while $f(a-(-b)\varepsilon)=f(a)+b\varepsilon f'_l(a)$. These are different in general, which is absurd since $a+b\varepsilon=a-(-b)\varepsilon$. | |
| Jun 5, 2022 at 7:39 | comment | added | Anixx | @LaurentMoret-Bailly apparently, $f(a-\varepsilon)$ | |
| Jun 3, 2022 at 18:28 | comment | added | Laurent Moret-Bailly | And by the way, with your definition, what is $f(a+b\varepsilon)$ if $b=-1$? | |
| Jun 2, 2022 at 11:42 | comment | added | Laurent Moret-Bailly | Well, my point is that you can define it any way you please. What would make such a definition relevant? | |
| Jun 2, 2022 at 6:44 | comment | added | Anixx | @LaurentMoret-Bailly as stated in the post, if $f$ is not differentiable, but has right or left derivative then we can define $f(a+b\varepsilon)$ via right derivative and $f(a-b\varepsilon)$ via left derivative. | |
| Jun 2, 2022 at 6:37 | comment | added | Laurent Moret-Bailly | For a differentiable function $f$ of a real variable, you seem to take the formula for $f(a+b\varepsilon)$ as a (provable) fact. So how do you define $f(a+b\varepsilon)$? If $f$ is given by a formula (such as a polynomial) you can just plug substitute $a+b\varepsilon$ for $$x$ in the formula and see if the result makes sense (and is independent of the formula!). But in general? | |
| Jun 1, 2022 at 22:10 | answer | added | Simon Henry | timeline score: 10 | |
| Jun 1, 2022 at 21:49 | review | Close votes | |||
| Jun 6, 2022 at 3:05 | |||||
| Jun 1, 2022 at 20:54 | history | asked | Anixx | CC BY-SA 4.0 |