Neutral and interordinal intervals in MOS scales: Difference between revisions

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 *** 1+A+B+floor(n(k+2)/(a+b)) <= |Y| <= 1+A+B+ceil(n(k+2)/(a+b))
 *** 1+C+D+floor(nk/(a+b)) <= |X| <= 1+C+D+ceil(nk/(a+b))
-*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor((n+2)k/(a+b)) - ceil(nk/(a+b))
+*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor(n(k+2)/(a+b)) - ceil(nk/(a+b))
-*** = (A+B)-(C+D)-1 + ceil((n+2)k/(a+b)) - ceil(nk/(a+b))
+*** = (A+B)-(C+D)-1 + ceil(n(k+2)/(a+b)) - ceil(nk/(a+b))
 * As s is the chroma of bL(a-b)s, it *would* be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so kX(k+1) would become neutral or semiperfect.
 * 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. 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 a+b-2. This proves part (2).