Recursive structure of MOS scales: Difference between revisions

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for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).

Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' − ''n'')''g''; any other size must leave the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators~~, which~~ does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''.

Assume ''c'' = ''b'' and ''e'' = −''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' − 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' − 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' − ''n'')''g''; any other size must leave the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' − 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' − ''k'')-step equivalent to ''−jg'', which must be distinct from an (''n'' − ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' − ''k'')-step does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''.

To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', ~~note~~ that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.

To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', observe that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.

== See also ==

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 for appropriate integers ''c, d, e, f'', where |''c''| < ''n'' and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(L, s).
-Assume ''c'' = ''b'' and ''e'' = &minus;''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' &minus; 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' &minus; 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' &minus; ''n'')''g''; any other size must leave the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' &minus; ''k'')-step equivalent to ''&minus;jg'', which must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators, which does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''.
+Assume ''c'' = ''b'' and ''e'' = &minus;''a''. [This corresponds to assuming that ''g'' is the "bright" generator.] Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness and binarity, any interval class that has at least two sizes must have sizes separated by ''ng'' (the separation corresponding to changing an L step to an s step). Since ''g'' and ''p'' are linearly independent, for each ''j'' ∈ {1, ..., ''n'' &minus; 1} there exists at most one ''k'' = ''k''(''j'') ∈ {1, ..., ''n'' &minus; 1}</sub> such that ''jg'' is ''p''-equivalent to one size of ''k''-step. Hence if an interval class that has two sizes must have the sizes ''jg'' and (''j'' &minus; ''n'')''g''; any other size must leave the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has at most two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1). Each non-''p''-equivalent class must have ''exactly'' two sizes, since the inverse of the ''k''-step that is equivalent to ''jg'' is an (''n'' &minus; ''k'')-step equivalent to ''&minus;jg'', which must be distinct from an (''n'' &minus; ''k'')-step equivalent to a positive number of ''g'' generators; note that the latter (''n'' &minus; ''k'')-step does occur in the "brightest" (most ''g'' generators stacked ''up'' rather than ''down'') mode of ''S''.
-To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.
+To show that maximum variety 2 holds for non-linearly-independent ''p'' and ''g'', observe that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with linearly independent ''p'' and ''g'', and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'' and remains ''p''-inequivalent to 0 in the limit since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.
 == See also ==