Recursive structure of MOS scales: Difference between revisions

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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.

~~Proving that the chunk sizes form a mos is equivalent to proving that the places where the s's are in the mos form a mos in n-edo (which must be [[maximally even]] by above).~~

Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:

~~For now~~ assume r ≥ 1. Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:

# p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in W', corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's

# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking the word to the right if necessary we assume it begins in λ.

@@ Line 344: / Line 344: @@
 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.
-Proving that the chunk sizes form a mos is equivalent to proving that the places where the s's are in the mos form a mos in n-edo (which must be [[maximally even]] by above).
+Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:
-For now assume r ≥ 1. Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:
 # p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in W', corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's
 # p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking the word to the right if necessary we assume it begins in λ.