Timeline for Verifying a sequence that converges to pi [closed]
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2010 at 16:40 | vote | accept | CommunityBot | ||
| Dec 26, 2010 at 23:51 | vote | accept | CommunityBot | ||
| Dec 27, 2010 at 16:40 | |||||
| Dec 26, 2010 at 19:21 | comment | added | user11822 | So it seems the initial value doesn't matter just as long as its in $(0,\pi]$. | |
| Dec 26, 2010 at 15:34 | history | closed |
Gjergji Zaimi Harald Hanche-Olsen Pietro Majer Todd Trimble Gerry Myerson |
too localized | |
| Dec 26, 2010 at 15:10 | answer | added | Anixx | timeline score: 0 | |
| Dec 26, 2010 at 14:25 | comment | added | Pietro Majer | The function $f(a):=a+\sin(a)$ maps the interval $[0,\pi]$ into itself, keeping fixed the endpoints. Moreover $f(a) > a$ for $0 < a < \pi$. So any starting point $0 < a_0 \le \pi $ generates an increasing bounded sequence; hence it converges; by continuity of $f$ the limit is a fixed point; hence it's $\pi$. Since $f'(\pi)=0$ and $f''(\pi)=0$ the convergence is cubic; precisely $0 < \pi-a_ {n+1} \le (\pi- a_n)^3 /6$. | |
| Dec 26, 2010 at 13:51 | answer | added | aster | timeline score: 2 | |
| Dec 26, 2010 at 13:48 | history | asked | user11822 | CC BY-SA 2.5 |