Timeline for Polynomials orthogonal w.r.t. the logarithmic weight
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Mar 22, 2021 at 8:29 | history | edited | gmvh |
Added top-level tag (post was bumped already)
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Mar 22, 2021 at 8:16 | answer | added | user111 | timeline score: 2 | |
Mar 22, 2021 at 7:35 | comment | added | Fedor Petrov | doi.org/10.3842/SIGMA.2018.056 looks related | |
Jun 2, 2019 at 17:27 | comment | added | mamiladi | i would like have the expression of $p_n$ | |
Oct 5, 2014 at 20:38 | comment | added | Twi | Sorry, I certainly assume that the degree of $p_{n}$ is equal to $n$. | |
Oct 5, 2014 at 20:06 | answer | added | Christian Remling | timeline score: 2 | |
Oct 5, 2014 at 19:37 | comment | added | Joonas Ilmavirta | @ChristianRemling, that is what I thought, but I could imagine that some other choice would come in an application. Vanishing at zero was a silly miscalculation; that would happen if the weight was not integrable near zero. | |
Oct 5, 2014 at 19:26 | comment | added | Christian Remling | @JoonasIlmavirta: There is a very standard procedure that gives unique $p_n$'s: you run Gram-Schmidt on $1,x,x^2,\ldots$ and make the leading coefficient positive. Also, there is no reason why the $p_n$ would have to vanish at zero. | |
Oct 5, 2014 at 16:44 | comment | added | Joonas Ilmavirta | The family of orthogonal polynomials with respect to a weight is not unique unless you make some restricting choices. Do you assume, for example, that $p_n$ has degree $n+1$? Note that all of your polynomials must vanish at zero. | |
Oct 5, 2014 at 16:29 | history | asked | Twi | CC BY-SA 3.0 |