Journal of Mathematics and Music Vol. 01, No. 03, December 2007, 1–2
David Lewin and Maximally Even Sets Emmanuel AMIOT 1 rue du Centre, F 66570 St NAZAIRE, France (v1.0.0 released june 2007)
Supplementary II: pictures and proofs
Figure 1. These two hexachords share intervallic content
On figure 1 with the two complementary hexachords, the fifths have been signaled with arrows. Each hexachord has the same number of fifths, three in this example.
Figure 2. Etude N◦ 5 opus 10, Fr´ ed´ eric Chopin
Proof of prop. ??: Proof From Thm. ?? we have A is τ −periodic ⇐⇒ ∀t ∈ Zc
FA (t) = e−2iπτ t/c FA (t) ⇐⇒ ∀t ∈ Zc
(1 − e−2iπτ t/c )FA (t) = 0
Professor in Class Preps, Perpignan, France. Email:
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Journal of Mathematics and Music c 2007 Taylor & Francis Ltd. ISSN 1745-9737 print / ISSN 1745-9745 online http://www.tandf.co.uk/journals DOI: 10.1080/17459730xxxxxxxxx
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title on some pages
Unless e−2iπτ t/c = 1, this compels FA (t) to be 0. Now the condition e−2iπτ t/c = 1 is equivalent to c | τ t, i.e. t multiple of m = c/ gcd(c, τ ) – this makes sense for any representative of the residue classes τ and t. This is compatible with reduction modulo c, and means t ∈ m Zc ⊂ Zc . Conversely, if FA is nil except on a subgroup, say m Zc with 0 < m | c in Z (we recall all subgroups of Zc are cyclic) then, by inverse Fourier Transform ∀k ∈ Zc
1A (k) =
1X 1 X 1 0 FA (t)e2iπ k t/c = FA (t0 )e2iπ k t /c = c c 0 c t ∈m Zc
t∈Zc
00
X t00 =1...
FA (m t00 )e2iπ k t
m/c
c m
c c as a period, as each term in the sum is and this is obviously periodic with (the residue class of ) m m periodic. Proof of prop ??: Proof For transposition and inversion it is theorem ??. For complementation we see that |FZc \A (c − d)| = |FZc \A (−d)| = | − FA (d)| = |FA (d)| holds for any subset A. So the one value is maximal whenever the other is, e.g. A is a ME set iff Zc \ A is maximally even. Proof of remark ??, linking he Fourier coefficients of A and its reduction B mod c0 : FA (d) =
X k∈A
−2iπdk/c
e
=
X m−1 X k00 ∈B `=0
−2iπd(k00 +` c0 )/c
e
=
X k00 ∈B
−2iπd k00 /c
e
m−1 X `=0
e−2iπ` = m
X k00 ∈B
0
e−2iπd
k00 /c0
= m FB (d0 )
