Je suis obligée de faire un billet parce que copié-collé de Words dans les commentaires fait perdre les illustrations.
C’est juste une représentation en deux dimensions d’ une chronophotographie en somme, comme dans le jeu de la vie de CONWAY en 2D les images successives, ici, on peut le faire en décalant dans le temps parce que ce n’est qu’en 1D. Pour m’expliquer, j’ai copié l’image sur Wikipédia.
Pour un bracelet brésilien à 6 fils, donc 5 colonnes dont chacune représenterait une des étapes de « l’itération » (pardon si ce n’est pas le mot juste), ces étapes étant jointives par ailleurs, cela donnerait :
| ligne | longueur d’un segment | nombre de points des segments noirs – essai 1 |
| colonne | longueur totale du motif : L0 | 81 points |
| 1ère | L1 = L0 / 30 | 3 * 27 = 81 |
| 2ème | L2 = L1 / 3 = L0 / 31 | 3 * 9 = 27 |
| 3ème | L3 = L2 / 3 = L 1 / 32 = L0 / 32 | 3 * 3 = 9 |
| 4ème | L4 = L3 / 3 | 3 * 1 =3 |
| 5ème | L5 = L4 / 3 | 1 |
| nème | Ln = L1 / 3n-1 = L0 / 3n-1 |
Maintenant il faut se pencher sur la longueur de la partie nouée :
7 motifs ont une longueur de 8,4 cm
1 motif est fait de 4 points mis bout à bout
7 * 4 = 28 points ont une longueur de 8,4 cm
Nous pouvons en déduire la longueur totale de la partie nouée représentative de la formation de l’ensemble de CANTOR pour le premier essai par une Règle de trois :
28 points font 8,4 cm
1 point fait 28 fois moins, soit 8,4 / 28 cm
et 81 points font 81 fois plus, soit (8,4 * 81) / 28 = 24,3 cm pour du coton perlé N°5. Il me semble que c’est un peu grand.
Pour que ce soit bien, il ne faut pas ce genre de réseau mais plutôt celui-ci, tous les calculs restant valables par ailleurs :
et voici ce que ça donnerait comme grille totale :
Monique-Mauve 
#1 par elytis le 18 septembre 2008 - 3 h 06 min
Are we talking here about a multidimentional function ,an endomorphism or is something else ?
I don\’t understand some basic parameters a) the dimension b) the space. c) If is a symmetrical or an unsymmetrical function.
The definition of determinant as a multilinear mapping on rows can be modified to provide a basis-free definition of determinant. In order to make it clear that we are not using bases. we shall speak in terms of an endomorphism of a vector space over rather than speaking of a matrix whose entries belong to . We start by recalling some preliminary facts.
Suppose is a finite-dimensional vector space of dimension over a field . Recall that a multilinear map is alternating if whenever there exist distinct indices such that . Every alternating map is skew-symmetric, that is, for each permutation , we have that , where denotes , the result of permuting the entries of .
Since the trivial map is alternating and any linear combination of alternating maps is alternating, it follows that alternating maps form a subspace of the space of multilinear maps. In the following proposition we show that this subspace is one-dimensional.
Theorem Suppose is a finite-dimensional vector space of dimension over a field . Then the space of alternating maps from to is one-dimensional.
Proof. We use a basis here, but we will throw it away later. We need the basis here because each map we will consider has exactly as many elements as a basis of . So let be a basis of .
#2 par elytis le 18 septembre 2008 - 3 h 05 min
Suppose and are nontrivial alternating maps from to . We claim that and are linearly dependent. Let . We may assume that the entries of are basis vectors, that is, that . If , then there exist distinct indices such that . Since and are alternating, it follows that , which implies that . On the other hand, if , then there is a permutation such that . Since and are skew-symmetric, it follows that
In either case we find that . Since and are fixed scalars, it follows that and are linearly dependent.
#3 par elytis le 18 septembre 2008 - 3 h 03 min
So far we have shown only that the dimension of the space of alternating maps is less than or equal to one. In order to show that the space is one-dimensional we simply need to find a nontrivial alternating form. To do this, let be the natural basis of , so that is the Kronecker delta of and for any . Define a map by
One can check that is multilinear and alternating. Moreover, , so it is nontrivial. Hence the space of alternating maps is one-dimensional.
For an alternate view of the above results, we could look instead at linear maps from the exterior product into . The proposition above can be viewed as saying that the dimension of is .
We define the determinant of an endomorphism in terms of the action of the endomorphism on alternating maps. Recall that if is an endomorphism, its pullback is the unique operator such that
Since the space of alternating maps is one-dimensional and endomorphisms of a one-dimensional space reduce to scalar multiplication, it follows that is a scalar multiple of . We call this scalar the determinant. It is well-defined because the scalar depends on but not on .
Definition Suppose is a finite-dimensional vector space of dimension over a field , and let be an endomorphism. Then the determinant of is the unique scalar such that
for all alternating maps .