Ternary scale theorems: Difference between revisions

Line 62:

In case 1, let {{nowrap|'''g'''1 {{=}} (2, 1) − (1, 1)|'''g'''2 {{=}} (1, 2) − (2, 1)}}, and {{nowrap|'''g'''3 {{=}} (1, 1) − ({{frac|''n''|2}}, 2)}} {{nowrap|{{=}} ((−{{frac|''n''|2}} − 1)*'''g'''1 − {{frac|''n''|2}}*'''g'''2) (mod '''e''')}}. We assume that '''g'''1, '''g'''2 and '''e''' are ℤ-linearly independent. We have the chain '''g'''1 '''g'''2 '''g'''1 '''g'''2 ... '''g'''1 '''g'''3 which visits every note in ''s''.

~~Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator {{nowrap|'''g''' {{=}} ('''g'''2 + '''g'''1)}}.~~ Since '''g'''1 and '''g'''2 subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''1 must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence {{nowrap|('''g'''3 + '''g'''1)}}, the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''2 and '''g'''3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''2 with '''g'''3 in the next part.

Since '''g'''1 and '''g'''2 subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''1 must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence {{nowrap|('''g'''3 + '''g'''1)}}, the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''2 and '''g'''3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''2 with '''g'''3 in the next part.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators:

@@ Line 62: / Line 62: @@
 In case 1, let {{nowrap|'''g'''<sub>1</sub> {{=}} (2, 1) − (1, 1)|'''g'''<sub>2</sub> {{=}} (1, 2) − (2, 1)}}, and {{nowrap|'''g'''<sub>3</sub> {{=}} (1, 1) − ({{frac|''n''|2}}, 2)}} {{nowrap|{{=}} ((−{{frac|''n''|2}} − 1)*'''g'''<sub>1</sub> − {{frac|''n''|2}}*'''g'''<sub>2</sub>) (mod '''e''')}}. We assume that '''g'''<sub>1</sub>, '''g'''<sub>2</sub> and '''e''' are ℤ-linearly independent. We have the chain '''g'''<sub>1</sub> '''g'''<sub>2</sub> '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub> '''g'''<sub>3</sub> which visits every note in ''s''.
-Since ''s'' is generator-offset it is well-formed with respect to the aggregate generator {{nowrap|'''g''' {{=}} ('''g'''<sub>2</sub> + '''g'''<sub>1</sub>)}}. Since '''g'''<sub>1</sub> and '''g'''<sub>2</sub> subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''<sub>1</sub> must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence {{nowrap|('''g'''<sub>3</sub> + '''g'''<sub>1</sub>)}}, the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''<sub>2</sub> and '''g'''<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''<sub>2</sub> with '''g'''<sub>3</sub> in the next part.
+Since '''g'''<sub>1</sub> and '''g'''<sub>2</sub> subtend the same number of steps by the AGS assumption, each is an odd-step. All multiples of the aggregate generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''<sub>1</sub> must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a MOS subset. Hence {{nowrap|('''g'''<sub>3</sub> + '''g'''<sub>1</sub>)}}, the imperfect generator of the MOS generated by '''g''', subtends the same number of steps as '''g'''. Thus '''g'''<sub>2</sub> and '''g'''<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''<sub>2</sub> with '''g'''<sub>3</sub> in the next part.
 Let ''r'' be odd and ''r'' &ge; 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators: