Mathematics of MOS: Difference between revisions

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=== MOS scales are generated ===

See [[Recursive structure of MOS scales#Proofs]].

=== MOS scales aL bs with gcd(a, b) > 1 reduce to single-period MOS scales on a smaller period ===

Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period.

=== Maximally even scales are MOS ===

Let m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size floor((''i'' + ''k'')''n''/''m'') − floor(''in''/''m''), and

Let 1 < m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size floor((''i'' + ''k'')''n''/''m'') − floor(''in''/''m''), and

<math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor ≤ \lfloor(i+k)n/m \rfloor - \lfloor in/m \rfloor

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As floor((''i'' + ''k'')''n''/''m'') − floor(''in''/''m'') is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for k-steps, the two sizes must differ by 1.

~~<!--~~

Now let a and b be the number of L and s steps in ME(m, n). By the Euclidean algorithm, we have gcd(m, n) > 1 iff gcd(a, b) = d > 1. Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period. (This paragraph also reduces arbitrary aL bs MOS scales with gcd(a, b) > 1 to single-period MOS scales.)

~~We have reduced to the gcd(''m'', ''n'') = 1 case. -->~~Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.

Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.

This maximally even MOS has n % m large steps and m − (n % m) small steps.

@@ Line 237: / Line 237: @@
 === MOS scales are generated ===
 See [[Recursive structure of MOS scales#Proofs]].
+=== MOS scales aL bs with gcd(a, b) > 1 reduce to single-period MOS scales on a smaller period ===
+Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period.
 === Maximally even scales are MOS ===
-Let m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size floor((''i'' + ''k'')''n''/''m'') &minus; floor(''in''/''m''), and
+Let 1 < m < n, gcd(m, n) = d, and ME(''m'', ''n'') be as in [[Maximal evenness]]. An arbitrary k-step in ME(''m'', ''n'') has size floor((''i'' + ''k'')''n''/''m'') &minus; floor(''in''/''m''), and
 <math>\lfloor in/m \rfloor + \lfloor kn/m\rfloor - \lfloor in/m\rfloor = \lfloor kn/m\rfloor ≤ \lfloor(i+k)n/m \rfloor - \lfloor in/m \rfloor
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 As floor((''i'' + ''k'')''n''/''m'') &minus; floor(''in''/''m'') is an integer, the above proves that there are at most two possible values for it and that if there are two sizes for k-steps, the two sizes must differ by 1.
-<!--
-Now let a and b be the number of L and s steps in ME(m, n). By the Euclidean algorithm, we have gcd(m, n) > 1 iff gcd(a, b) = d > 1. Assume that gcd(m, n) > 1 and assume that (n/d)-steps came in 2 sizes, pa+qb and ra+sb. Then at least one size, say pa+qb, must differ from (a/d)L + (b/d)s. WLOG p > a/d and pa+qb occurs on degree 0. Since equave = aL + bs, on a degree k(n/d) for some integer k > 0, there must be another (n/d)-step with fewer L's than a/d. This involves more than two changes from L to s. Scooting an (n/d)-step one step at a time from degree 0 to k(n/d) changes its size one step size substitution at a time, showing that intermediate (n/d)-steps also exist. This violates the MOS property, whence (n/d)-steps have only one size, which must be the period. (This paragraph also reduces arbitrary aL bs MOS scales with gcd(a, b) > 1 to single-period MOS scales.)
-We have reduced to the gcd(''m'', ''n'') = 1 case. -->Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.
+Steps of ME(''m'', ''n'') have exactly two sizes because if it were one size, we would have m | (n/d), which is a contradiction.
 This maximally even MOS has n % m large steps and m &minus; (n % m) small steps.