Ternary scale theorems: Difference between revisions
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==== Statement (2) ==== | ==== Statement (2) ==== | ||
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 is (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup> g<sub>3</sub>. Assuming that a step is | 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 is (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup> g<sub>3</sub>. Assuming that a step is formed by stacking ''k'' generators, where ''k'' is odd, 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''g<sub>1</sub> + (''k'' − 1)g<sub>2</sub> | ||
# (''k'' − 1)g<sub>1</sub> + ''k''g<sub>2</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> | ||
(since the scale size is odd, we can always ensure this by taking octave complements of all the generators). | (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). By counting the length-k subwords of the word (g<sub>1</sub> g<sub>2</sub>)<sup>floor(''n''/2)</sup>, we see that the first two sizes must occur the same number of times. This proves (2). | ||
(The above holds for any odd ''n'' ≥ 3.) | (The above holds for any odd ''n'' ≥ 3.) | ||