Timeline for Is there a way to automate the generation of these polynomials?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4 at 3:22 | comment | added | Claude Leibovici | You made my day ! Thanks again and cheers :) | |
| May 4 at 2:20 | comment | added | A. Kato | Ok, don't worry. It is my great pleasure if you are satisfied with my revised answer. | |
| May 4 at 2:15 | comment | added | Claude Leibovici | Sorry ! It does not work. I I define $g$ as your previous $f$ and $f$ as $g/g'$ | |
| May 4 at 2:00 | comment | added | A. Kato | Please tell me how, if possible. | |
| May 4 at 1:57 | comment | added | Claude Leibovici | Thanksa gain ! Meanwhile, I fixed your previous version | |
| May 4 at 1:55 | comment | added | A. Kato | I revised my answer. | |
| May 4 at 1:53 | history | edited | A. Kato | CC BY-SA 4.0 | revised answer. |
| May 3 at 15:15 | comment | added | Claude Leibovici | I think that we need to define df=D[f,x] and to use ans = Series[x - f /df /. x -> 1/2, {m, Infinity, p}]; But this does not simplify. | |
| May 3 at 14:02 | comment | added | A. Kato | @yarchik OK, I might be wrong. I'm so sleepy now. I will temporarily delete my answer and try again, maybe tomorrow. Oops, I cannot delete since it is already accepted.... | |
| May 3 at 13:59 | comment | added | yarchik | Actually I tried to understand your solution, and I found this relation was confusing. Indeed, $x=x+f(x)$ implies $f(x)=0$, so this is not wrong, but not useful either. | |
| May 3 at 13:34 | comment | added | A. Kato | @yarchik Well, I thought of using the recursion like that, too. But I found a solution without using recursion --- just a series expansion around $m=\infty$ was enough. The equation $x_{(m)}=x_{(m)} - f(x_{(m)})$ holds simply because $f(x_{(m)})=0$ by definition. | |
| May 3 at 12:16 | comment | added | yarchik | I guess the intention was $x_{m+1}=x_m+f(x_m)$ | |
| May 3 at 12:09 | comment | added | A. Kato | @yarchik Sorry, I cannot find a typo. Please edit for me. | |
| May 3 at 12:05 | comment | added | yarchik | I think there is a typo in the recursive identity. | |
| May 3 at 10:32 | comment | added | Claude Leibovici | Thank you ! This is the perfect answer to m'y question | |
| May 3 at 10:30 | vote | accept | Claude Leibovici | ||
| May 3 at 10:25 | history | edited | A. Kato | CC BY-SA 4.0 | added 5 characters in body |
| May 3 at 10:19 | history | answered | A. Kato | CC BY-SA 4.0 |