Recursive structure of MOS scales: Difference between revisions
→Preservation of the MOS property: fleshed out the proof that the reduced word is a mos |
|||
| Line 340: | Line 340: | ||
=== Preservation of the MOS property === | === Preservation of the MOS property === | ||
(Assume that the mos has length n. | (Assume that the mos has length n; the notattion w(X, Y) for a word w(L, s) in L, s means w with step sizes X substituted for L and Y substituted for s. | ||
Suppose we had three chunks L...s with r, r+1 and r+2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes. | Suppose we had three chunks L...s with r, r+1 and r+2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes. | ||
| Line 352: | Line 352: | ||
Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps: | Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps: | ||
# w₁ = the word sW₁(L<sup>r+1</sup>s, L<sup>r</sup>s) [W₁ interpreted as a subword of the original mos], with (k + 1) s's | # w₁(L, s) = the word sW₁(L<sup>r+1</sup>s, L<sup>r</sup>s) [W₁ interpreted as a subword of the original mos], with (k + 1) s's | ||
# w₂ = the first K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | # w₂(L, s) = the first K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | ||
# w₃ = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | # w₃(L, s) = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | ||
If w₂ contains fewer complete chunks than w₃, we are done. | If w₂ contains fewer complete chunks than w₃, we are done. | ||