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From: Samuel DEVULDER <samuel_dot_devulder@laposte_dot_net.invalid>
Newsgroups: fr.sci.maths
Subject: =?UTF-8?Q?Re=3a_Prouver_une_in=c3=a9galit=c3=a9_pour_tout_x_et_y?=
Date: Sun, 22 Aug 2021 12:15:04 +0200
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Le 22/08/2021 à 09:38, Michel Talon a écrit :
> Pour éviter de se taper le ascii-art on peut demander à maxima de le faire:
> 
> (%i1) L(x):=[1,x,x^2,x^3,x^4,x^5,x^6,x^7];
>                                      2   3   4   5   6   7
> (%o1)               L(x) := [1, x, x , x , x , x , x , x ]
> (%i2) M:matrix(L(a),L(b),L(c),diff(L(a),a),diff(L(b),b),diff(L(c),c),
> diff(L(a),a,2),diff(L(b),b,2));
>                  [        2     3     4      5      6      7   ]
>                  [ 1  a  a     a     a      a      a      a    ]
>                  [                                             ]
>                  [        2     3     4      5      6      7   ]
>                  [ 1  b  b     b     b      b      b      b    ]
>                  [                                             ]
>                  [        2     3     4      5      6      7   ]
>                  [ 1  c  c     c     c      c      c      c    ]
>                  [                                             ]
>                  [               2     3      4      5      6  ]
>                  [ 0  1  2 a  3 a   4 a    5 a    6 a    7 a   ]
> (%o2)           [                                             ]
>                  [               2     3      4      5      6  ]
>                  [ 0  1  2 b  3 b   4 b    5 b    6 b    7 b   ]
>                  [                                             ]
>                  [               2     3      4      5      6  ]
>                  [ 0  1  2 c  3 c   4 c    5 c    6 c    7 c   ]
>                  [                                             ]
>                  [                      2      3      4      5 ]
>                  [ 0  0   2   6 a   12 a   20 a   30 a   42 a  ]
>                  [                                             ]
>                  [                      2      3      4      5 ]
>                  [ 0  0   2   6 b   12 b   20 b   30 b   42 b  ]

oui

> 
> 
> Ce que tu fais est dans la ligne de la solution. Pour ma part je 
> procéderais comme ceci:
> 
> M:matrix(L(a),L(b),L(c),L(a+x),L(b+y),L(c+z),L(a+u),L(b+v));

Heu je ne pige pas le passage de L'(a) à L(a+x) c'est trop rapide pour 
moi là.

sam.