Generator-offset property: Difference between revisions
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In case 2, let (2, 1) − (1, 1) = g<sub>1</sub>, (1, 2) − (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover generator after stacking alternating g<sub>1</sub> and g<sub>2</sub>. Then the generator circle looks like g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>2</sub> g<sub>3</sub>. Then the combinations of alternants corresponding to a step come in exactly 3 sizes: | In case 2, let (2, 1) − (1, 1) = g<sub>1</sub>, (1, 2) − (2, 1) = g<sub>2</sub> be the two alternants. Let g<sub>3</sub> be the leftover generator after stacking alternating g<sub>1</sub> and g<sub>2</sub>. Then the generator circle looks like g<sub>1</sub> g<sub>2</sub> g<sub>1</sub> g<sub>2</sub> ... g<sub>1</sub> g<sub>2</sub> g<sub>3</sub>. Then the combinations of alternants corresponding to a step come in exactly 3 sizes: | ||
# '' | # ''k''g<sub>1</sub> + (''k'' − 1)g<sub>2</sub> | ||
# (''k'' − 1)g<sub>1</sub> + '' | # (''k'' − 1)g<sub>1</sub> + ''k''g<sub>2</sub> | ||
# (''k'' − 1)g<sub>1</sub> + (''k'' − 1) g<sub>2</sub> + g<sub>3</sub>, | # (''k'' − 1)g<sub>1</sub> + (''k'' − 1) g<sub>2</sub> + g<sub>3</sub>, | ||
if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2). | if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. This proves (2). | ||