Rank-3 scale theorems: Difference between revisions

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==== Definition: LQ ====

A scale word ''S'' with ''k'' step sizes X_1, ..., X_k (with a_1 X_1's, ..., and a_k X_k's) is ''line-quantizing'' (LQ) if S, when viewed as a set of instructions tracing a path in Z^k from the origin (in which each X_i means "go 1 step in the positive x_i direction"), results in a path that is a closest approximation to the line [a_1 : a_2 : ... : a_k] intersecting the origin in R^k. (~~Closest approx in what sense?~~)

A scale word ''S'' with ''k'' step sizes X_1, ..., X_k (with a_1 X_1's, ..., and a_k X_k's) is ''line-quantizing'' (LQ) if S, when viewed as a set of instructions tracing a path in Z^k from the origin (in which each X_i means "go 1 step in the positive x_i direction"), results in a path that is a closest approximation to the line [a_1 : a_2 : ... : a_k] intersecting the origin in R^k.

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}}.

Approximation might be based on individual notes/lattice points, so it will have to be a sum of squares or max of the distances of the points R(k) to the corresponding point L(k). I don't know how a slope based formulation should work or if that would be equivalent to the pointwise formulation of closest

In a mos (MV2), 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.

In a mos (MV2), I believe that 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 finish the proof for MV2.

==== MV3 Theorem 1 ====

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 ==== Definition: LQ ====
-A scale word ''S'' with ''k'' step sizes X_1, ..., X_k (with a_1 X_1's, ..., and a_k X_k's) is ''line-quantizing'' (LQ) if S, when viewed as a set of instructions tracing a path in Z^k from the origin (in which each X_i means "go 1 step in the positive x_i direction"), results in a path that is a closest approximation to the line [a_1 : a_2 : ... : a_k] intersecting the origin in R^k. (Closest approx in what sense?)
+A scale word ''S'' with ''k'' step sizes X_1, ..., X_k (with a_1 X_1's, ..., and a_k X_k's) is ''line-quantizing'' (LQ) if S, when viewed as a set of instructions tracing a path in Z^k from the origin (in which each X_i means "go 1 step in the positive x_i direction"), results in a path that is a closest approximation to the line [a_1 : a_2 : ... : a_k] intersecting the origin in R^k.
+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}}.
+Approximation might be based on individual notes/lattice points, so it will have to be a sum of squares or max of the distances of the points R(k) to the corresponding point L(k). I don't know how a slope based formulation should work or if that would be equivalent to the pointwise formulation of closest
-In a mos (MV2), 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.
+In a mos (MV2), I believe that 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 finish the proof for MV2.
 ==== MV3 Theorem 1 ====