Neutral and interordinal intervals in MOS scales: Difference between revisions
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= #{k : 0 < k < a + b and larger k-step of basic aLbs = smaller (k + 1)-step of basic aLbs} = # of potential improprieties, | = #{k : 0 < k < a + b and larger k-step of basic aLbs = smaller (k + 1)-step of basic aLbs} = # of potential improprieties, | ||
where ''potential improprieties'' are pairs of adjacent interval classes that witness the impropriety of a hard-of-basic tuning of the MOS. | where ''potential improprieties'' are pairs of adjacent interval classes that witness the impropriety of a hard-of-basic tuning of the MOS. Part (4) immediately follows. | ||
Also recall that the following are equivalent for a MOS aLbs: | Also recall that the following are equivalent for a MOS aLbs: | ||
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* To show that these actually occur in bL(a − b)s, consider smaller and larger j-steps (1 ≤ j ≤ a − 1) in the parent MOS. These intervals also occur in the MOS aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a − 1. These j's correspond to values of k such that larger k-step < smaller (k + 1)-step. Note that we are considering “junctures” between k-steps and (k + 1)-steps in aLbs, excluding k = 0 and k = a + b − 1, so the total number of “junctures” to consider is finite, namely a + b − 2. This proves parts (1) and (2). | * To show that these actually occur in bL(a − b)s, consider smaller and larger j-steps (1 ≤ j ≤ a − 1) in the parent MOS. These intervals also occur in the MOS aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a − 1. These j's correspond to values of k such that larger k-step < smaller (k + 1)-step. Note that we are considering “junctures” between k-steps and (k + 1)-steps in aLbs, excluding k = 0 and k = a + b − 1, so the total number of “junctures” to consider is finite, namely a + b − 2. This proves parts (1) and (2). | ||
Part (3) is also immediate now: when larger k-step = smaller (k + 1)-step, larger (k + 1)-step − smaller k-step = 2(L − s) = 2s = L. The step L is 4 steps in 2n-edo | Part (3) is also immediate now: when larger k-step = smaller (k + 1)-step, larger (k + 1)-step − smaller k-step = 2(L − s) = 2s = L. The step L is 4 steps in 2n-edo. {{qed}} | ||
[[Category:MOS scale]] | [[Category:MOS scale]] | ||
[[Category:Pages with proofs]] | [[Category:Pages with proofs]] | ||