Timeline for Why polynomial $\psi^\top(t) A^{-1} \psi(t)$ attains maximum on $[-1, 1]$ at $t = \pm 1$, where $\psi_k(t) = t^k$?
Current License: CC BY-SA 3.0
7 events
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| Apr 20, 2016 at 9:22 | comment | added | Ilya Bogdanov | Then it is better to correct the mess your question... | |
| Apr 19, 2016 at 11:17 | comment | added | Sergey Dovgal | By the way, though the nature of factorization of derivatives of $P_k(t)$ is not essential for the problem, it is somehow interesting. Perhaps it could be also explained through numerous properties of legendre polynomials. | |
| Apr 19, 2016 at 11:16 | vote | accept | Sergey Dovgal | ||
| Apr 19, 2016 at 11:16 | comment | added | Sergey Dovgal | Thank you! The derivation is correct (I messed a bit with vector-columns and vector-rows, when gave the problem formulation, but finally everything collided). It is surprising that the fact holds independent on the choice of basis $\psi(t)$. As for legendre polynomials, there are proofs of the fact using some integral representation, e.g. pskgu.ru/ebooks/ts/ts_dop2_2_1.pdf page 677. | |
| Apr 19, 2016 at 10:01 | history | edited | Ilya Bogdanov | CC BY-SA 3.0 |
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| Apr 18, 2016 at 19:18 | history | edited | Ilya Bogdanov | CC BY-SA 3.0 |
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| Apr 18, 2016 at 17:56 | history | answered | Ilya Bogdanov | CC BY-SA 3.0 |