Timeline for Asymptotic form of $L^1$-norm of Hermite functions
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| May 13, 2018 at 15:55 | vote | accept | Mateus Araújo | ||
| S May 13, 2018 at 13:11 | history | suggested | user111 | CC BY-SA 4.0 |
corrected spelling, two tags added
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| May 13, 2018 at 12:40 | review | Suggested edits | |||
| S May 13, 2018 at 13:11 | |||||
| May 12, 2018 at 11:58 | answer | added | user111 | timeline score: 3 | |
| Jun 7, 2011 at 5:48 | comment | added | Pietro Majer | You could also write $\|\varphi_m\|_1$ as an alternating sum of the integral of $\varphi_m$ between consecutive zeros (for they are all simple and $\varphi_m$ changes sign). Then try using the recurrence relations and the known informations on the zeros of Hermite polynomials. | |
| Jun 6, 2011 at 19:42 | history | edited | Mateus Araújo | CC BY-SA 3.0 |
improved formatting
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| May 27, 2011 at 6:22 | comment | added | Junkie | To determine $R_m$ if you have not already, use the representation $H_m(x)=m!\sum_{l=0}^{m/2}{(-1)^l\over l!(m-2l)!}(2x)^{m-2l}$. You can ignore the first quotient in the sum as bounded by 1, and get $|H_m(x)|\le m\cdot m!\cdot (2x)^m$ for $x>1/2$. Putting this into the above formula and taking logs, it is easy to see that $x>>\sqrt{m\log m}$ will contribute negligibly. You only have to compare $\log|H_m(x)|\sim m\log m+m\log x$ with $\log e^{-x^2}\sim -x^2$. | |
| May 20, 2011 at 3:11 | comment | added | Mateus Araújo | Good idea. I'm pretty sure that $R_m$ grows as $\sqrt{m}$, although I haven't seen a proof of it, so my last equation would give $f(m,n) \le \alpha \sqrt[4]{m^2/n}$, and this is enough to prove a good upper bound. Thanks! The Hermite functions decay exponentially fast, so it is easy to bound the error by restricting the interval, but I still need some formula for $R_n$. | |
| May 20, 2011 at 2:50 | history | edited | Mateus Araújo | CC BY-SA 3.0 |
Include definition of hermite function
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| May 20, 2011 at 1:56 | comment | added | Zen Harper | Some (probably stupid) comments: maybe it would help if you explicitly gave the formula for $\varphi_n$, since there seem to be several different versions in use. Anyway, if you only need an upper bound, you could find an interval $[-R_n, R_n]$ where $\varphi_n$ is concentrated, and use Cauchy-Schwarz to bound the $L^1$ norm on this interval using the $L^2$ norm, since the $L^2$ norm is much easier to use. The only trick you'd need to make this work is an inequality telling you how rapidly the functions decay away from zero. Hopefully, $R_n$ would not grow too fast with $n$ for this to work. | |
| May 19, 2011 at 19:24 | history | edited | Mateus Araújo | CC BY-SA 3.0 |
Simplifying question and including progress
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| May 16, 2011 at 14:26 | history | asked | Mateus Araújo | CC BY-SA 3.0 |