Timeline for How to prove the sum of n squared binomial probabilities does not increase as n increases
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2018 at 17:59 | comment | added | Fedor Petrov | To check the case $p_2=q_3=0$ very limited knowledge is required:) | |
| Apr 11, 2018 at 14:15 | vote | accept | Jack | ||
| Apr 11, 2018 at 13:31 | comment | added | Jack | Thank you so much for your excellent answer to the newest updated problem. Though I haven’t gone through the whole derivation yet because my limited mathematic knowledge, I verified the numerical equality between $F(n)$ and the proposed integral form using Matlab when p1, p2, and p3 were set to some specific values. I believe your answer is correct and I really appreciate it because it helps me a lot. I will keep working on it till I fully understand it. | |
| Apr 11, 2018 at 7:20 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Mar 14, 2018 at 6:57 | comment | added | Jack | A perfect answer! Thank you very much. | |
| Mar 14, 2018 at 4:41 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Mar 11, 2018 at 10:24 | vote | accept | Jack | ||
| Apr 11, 2018 at 14:14 | |||||
| Mar 11, 2018 at 6:14 | comment | added | Fedor Petrov | simply write $\varphi(x)=\sum_k a_k x^k$ and integrate, you get $(2\pi)^{-1}\sum_k a_k\int_0^{2\pi} e^{ikt}dt=a_0$. | |
| Mar 11, 2018 at 3:34 | comment | added | Jack | Thanks for your explanation. Could you please kindly show some references or some keywords about the useful conclusion for Laurent polynomial that I can relate to? | |
| Mar 10, 2018 at 17:40 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
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| Mar 10, 2018 at 15:28 | comment | added | Jack | Thank you very much for your quick responds but please excuse me as a rookie. I didn’t quite follow you. Could you please give more details about how $[1]((p^2+q^2)+qp(x+x^{-1}))^n=(2\pi)^{-1}\int_{0}^{2\pi}((p^2+q^2)+qp(e^{it}+e^{-it}))^ndt$. Lots of thanks! | |
| Mar 10, 2018 at 14:18 | history | answered | Fedor Petrov | CC BY-SA 3.0 |