Ternary scale theorems: Difference between revisions
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Since ''S'' is generator-offset it is well-formed with respect to 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, all multiples of the 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 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 (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 ''S'' is generator-offset it is well-formed with respect to 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, all multiples of the 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 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 (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'' ≥ 3. Consider the following abstract sizes for the interval class | Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators: | ||
# from g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub>, we get a<sub>1</sub> = (''r'' − 1)/2*g + g<sub>1</sub> = ceil(''r''/2) g<sub>1</sub> + floor(''r''/2) g<sub>2</sub> | # from g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub>, we get a<sub>1</sub> = (''r'' − 1)/2*g + g<sub>1</sub> = ceil(''r''/2) g<sub>1</sub> + floor(''r''/2) g<sub>2</sub> | ||
# from g<sub>2</sub> g<sub>1</sub> ... g<sub>2</sub>, we get a<sub>2</sub> = (''r'' − 1)/2*g + g<sub>2</sub> = floor(''r''/2) g<sub>1</sub> + ceil(''r''/2) g<sub>2</sub> | # from g<sub>2</sub> g<sub>1</sub> ... g<sub>2</sub>, we get a<sub>2</sub> = (''r'' − 1)/2*g + g<sub>2</sub> = floor(''r''/2) g<sub>1</sub> + ceil(''r''/2) g<sub>2</sub> | ||