Timeline for A question about non-archimedean binomial expansion
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2021 at 1:51 | answer | added | KConrad | timeline score: 1 | |
| Oct 8, 2021 at 23:34 | comment | added | stack99 | It appears not simple to actually prove that binomial expansion converges to $1$ in stead of by above contradiction. To check the infinity of the index of $ s - 1$ at that place over $p$ may be a try but not simple. However, I also get confirmation of $ s = 1 $ from other people but without explicit argument. | |
| Oct 8, 2021 at 23:08 | comment | added | LSpice | I don't think I have anything to add to that—if you have a complete argument that $s = 1$ (with which I agree), and an incomplete argument that $s \ne 1$, then it seems that the contradiction must lie in the incomplete argument. I tried testing this with $p = 2$ to see if it was more obvious there, but had no luck. | |
| Oct 8, 2021 at 22:54 | comment | added | stack99 | Thanks for the note. The analysis about the index of each terms at the place over $p$ in the expansion of $s - 1$ is not so simple, there is term with the same index as the term $\frac{p}{q} * ( 1 - \omega_{p} )$ at the related place. So $s$ could be equal to $1$. And it looks to avoid the mentioned possible contradiction, $s$ has to be always equal to $1$. Waiting for your comment if any. | |
| Oct 8, 2021 at 7:06 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading, hopefully while preserving meaning
|
| Oct 8, 2021 at 7:02 | comment | added | LSpice | What is the simple analysis that shows $s \ne 1$? | |
| S Oct 8, 2021 at 5:50 | review | First questions | |||
| Oct 8, 2021 at 7:08 | |||||
| S Oct 8, 2021 at 5:50 | history | asked | stack99 | CC BY-SA 4.0 |