Square perfect tilings On remplace toutes ces notes par des polynômes ! Indication 1: on note (X)=1+X+X 2 Indication 2: (j k )=0 sauf si 3|k…
tois the F 2 [X], rhythmic therepattern, is noBgood the sequence map back of onsets. into Z[X]. For instance one c hmic mentioned <strong>canons</strong>today. and tiling This is cyclic related groups elsewhere ([],[]). llowing be seen that the simplification of assuming integer beats (see [], []) and the periodi factors trations are given ces) of the wholeMaths during the talk. canon are not arbitrary pour but necessary. <strong>canons</strong> anon other is (1 hand, the + X)(1 there play+ of are X the + other Xsame possible rhythmic specifications, pattern stemming different from voices, musical with consider diffe ondition. The exponentiation 2 ) = 1 + X 3 ∈ F of a 2 5X subset of N means s with augmentations, retrogradations) which where instrumental in gathering impor or the study of <strong>canons</strong> but will not be mentioned today. This is related elsewhere ([], at on every beat there is a single note playing at a time, which makes the pro minology is : pty) subset. Then we set Polynomials t. to If A (T= 0 ){0, ina and cyclotomic factors 1 Z[X]. , . . . a k−1 The} is best a subset one was of N, it is the inner voice of a rhythmic ca onentiation 1, . . . b l−1 } one andgets period an equivalentr n iff condition. The exponentiation of a subset of N me ⇐⇒ itUne is true somme in any directe Fx k p [X] A ⊕ B = Z/nZ A(x) = ∑ k∈A TION 2. Let A ⊂ N be a finite (non empty) subset. Then we set rhythmic pattern, B the sequence of onsets. eaning that the number of notes on each beat is 1 iff it is equal t Exponentiation x k ible). A(x) = ∑ k∈A n that the simplification of assuming integer beats (see [], []) and the per map the whole A × Bcanon ∋ (a, are b) not ↦→ arbitrary a + b is but onenecessary. to one) iff d that the Condition above(T statement 0 ) is different from the following assertio hand, there are other possible specifications, stemming from musical consi augmentations, ⊂ Z[X], A(x) n such × B(x) retrogradations) that= A(X) (A ⊕× B)(x) B(X) which ≡ 1 where + x + instrumental . . . x n−1 (mod in gathering X n − 1) inimZ A(x) × B(x) = (A ⊕ B)(x) p, n p such that T 0 yields in F p [X] » SITION. The sum A + B is direct (i.e. map A × B ∋ (a, b) ↦→ a + b is one to one) iff study of <strong>canons</strong> but will not be mentioned today. This is related elsewhere ( 0, 1, 3, 6 ⊕ 0, 8, 12, 4 or eomials ndition have some for any and time (T 0 rhythmic ) cyclotomic : prove canon factors condition (T 0 ) : X 6 the X 3 reciprocal, like in X 1 X 12 YONEDA X 8 philosophy SITION. forms. AThe is a pattern resultfor was a rhytmic surprising. X 4 1 canon with outer rhythm B and period n iff anon with outer rhythm (T 0 ) 15 B 14 and A(x) × 13 period B(x) X= 12 n iff e. Let fields A ⊂ N be a finite (non X 18 empty) X 15 subset. X 14 Then X1 13 + x X + we X 11 x 12 2 set + . X. . 10 Xx 11 n−1 X 9 (mod X 10 X x 8n − 1) A(x) × B(x) = 1 + x + xX 27 9 + . X. 6 . 8 x n−1 X 5 7 X (mod X 4 6 X X 3 ∑x 5n X − 2 k X1) 4 X X 3 1 X 1 ation one gets an equivalentr condition. The exponentiation of a subset of N
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