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From: Olivier Miakinen <om+news@miakinen.net>
Newsgroups: fr.sci.maths
Subject: =?UTF-8?Q?Re:_Calcul_de_d=c3=a9terminant?=
Date: Sat, 21 Aug 2021 17:08:54 +0200
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Le 21/08/2021 à 16:58, Olivier Miakinen a écrit :
>> 
>> | 1 a a^2 a^3 ... a^7 |
>> | 1 b b^2 b^3 ... b^7 |
>> | 1 c c^2 c^3 ... c^7 |
>> | 0 1 2a 3a^2 ... 7a^6|
>> | 0 1 2b 3b^2 ... 7b^6|
>> | 0 1 2c 3c^2 ... 7c^6|
>> | 0 0 2  6a  ... 42a^5|
>> | 0 0 2  6b  ... 42b^5|

Avant d'aller plus loin, je propose une notation. Soit m(i,j) l'élément
de la i-ème ligne et la j-ème colonne (0 ≤ i,j ≤ 7), on a pour tout j :
 m(0,j) = a^j
 m(3,j) = d(a^j)/da
 m(6,j) = d²(a^j)/da²
et :
 m(1,j) = b^j
 m(4,j) = d(b^j)/db
 m(7,j) = d²(b^j)/db²
et :
 m(2,j) = c^j
 m(5,j) = d(c^j)/dc

Pour le moment je n'ai pas d'idée sur comment continuer. Je vais y
réfléchir.

-- 
Olivier Miakinen