Timeline for Deriving an asymptotic statement from a recursion
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2021 at 18:22 | history | edited | Testname420 |
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| Jul 5, 2021 at 12:44 | vote | accept | Testname420 | ||
| Jul 4, 2021 at 22:11 | answer | added | MathRoc | timeline score: 2 | |
| Jul 4, 2021 at 16:10 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Jul 4, 2021 at 15:06 | comment | added | Testname420 | @MathRoc Thanks for your reply! Unfortunately I only get $\theta_{n+1}=p_c\theta_n-p_c^2\theta_n^2+o(\theta_n^2)$. Its a bit hard to completely follow your proposition, since I don't speak Chinese. Does it have a name or a proof? It seems like a bit overkill, since the statement should be "not hard to see". | |
| Jul 4, 2021 at 12:58 | comment | added | MathRoc | Hint : $\theta_{n+1}=\theta_n-\frac{r-2}{2r-2}\theta_n^2+o(\theta_n^2)$ implies that $f(x)=x-Ax^2+o(x^2)$, where $A=\frac{r-2}{2r-2}>0$. | |
| Jul 4, 2021 at 12:53 | comment | added | MathRoc | Using the proposition from mp.weixin.qq.com/s/PWR8x-53_wP6qzTH-BgHTw can easily solve it. | |
| Jul 4, 2021 at 12:17 | comment | added | Testname420 | You are right, $r$ is the degree of the tree, constant and should be $r>2$. Otherwise our tree would be just a straight line and considering $p_c=1$ would be trivial. | |
| Jul 4, 2021 at 11:26 | comment | added | Carlo Beenakker | I think you need $r>2$; for $r=2$ the solution is $\theta_n=\theta_1$, which does not decay with $n$. | |
| Jul 4, 2021 at 10:46 | comment | added | MathRoc | Is $r$ a constant greater than 1 ? | |
| Jul 4, 2021 at 10:11 | review | First posts | |||
| Jul 4, 2021 at 11:01 | |||||
| Jul 4, 2021 at 10:04 | history | asked | Testname420 | CC BY-SA 4.0 |