Rank-3 scale theorems: Difference between revisions

Let n = a+b+c be the scale size, w = aX bY cZ be the scale word, let R be the corresponding path following the word w (R(k) = your location after taking k steps according to w), and put n+1 equally spaced points p_n on the line segment L = {(a,b,c)t : 0 <= t <= n}, i.e. the points {L(k) = (a,b,c) k : k ∈ {0, ..., n}}. Say ''S'' is ''pointwise-least-squares-LQ'' if the sum of (R(k) - L(k))^2 over k ∈ {0, ..., n} is minimized; say ''S'' is ''pointwise-minimax-LQ'' if the max distance max{|R(k)-L(k)| : k} is minimized. Say that a [[mos]] (MV2) scale ''S'' is ''slope-LQ'' if the slope between any two pair of points (representing a ''k''-mosstep) is one of the two nearest possible slopes (in the set {k/0,...,0/k}) to b/a.

Every mos is slope-LQ. This follows from the fact that every k-step is of the form pL qs or (p+1)L (q-1)s, where p/q <= a/b <= (p+1)/(q-1). This in turn follows from the fact that a mos has a generator, since every k-step is an octave reduction of either k perfect generators or k-1 perfect generators + 1 imperfect generator, and the perfect generator and the imperfect generator differ by changing one L to an s. If both the perfect generator and the imperfect generator had #L/#s > b/a (it must definitely be smaller than ceil(b/a) + 1, since the perfect generator is a collection of L...Ls chunks, and the size of such a chunk is <= ceil(b/a)), this would imply that n-1 perfect generators + 1 imperfect generator would not close at a period multiple.

There should be some Calc 1 stuff that I can do to show that mosses are pointwise-LQ.

''The following are equivalent for a non-multiperiod scale word S with steps x, y, z:''

# ''S is MV3.''

# ''S is AG, or S is of the form x'y'z'y'x' or its repetitions, or x'y'x'z'x'y'x' or its repetitions.''

TODO: account for case xyzyx.

''To be continued...''

We now prove that except in the case xyxzxyx, if the scale is pairwise MOS, then it is AG.