Ternary scale theorems: Difference between revisions

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==== Statement (2) ====

In case 2, let {{nowrap|''n'' ≥ 3}} and let {{nowrap|(2, 1) − (1, 1) {{=}} '''g'''1|(1, 2) − (2, 1) {{=}} '''g'''2}} be the two alternants. Let '''g'''3 be the closing generator after stacking alternating '''g'''1 and '''g'''2. Then the generator circle is {{nowrap|('''g'''1 '''g'''2){{floor|''n''/2}}}} '''g'''3. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:

# {{nowrap|{{ceil|''k''/2}}'''g'''1 + {{floor|''k''/2}}'''g'''2}}

# {{nowrap|{{ceil|{{frac|''k''|2}}}}'''g'''1 + {{floor|{{frac|''k''|2}}}}'''g'''2}}

# {{nowrap|{{floor|''k''/2}}'''g'''1 + {{ceil|''k''/2}}'''g'''2}}

# {{nowrap|{{floor|{{frac|''k''|2}}}}'''g'''1 + {{ceil|{{frac|''k''|2}}}}'''g'''2}}

# {{nowrap|{{floor|''k''/2}}'''g'''1 + {{floor|''k''/2}} '''g'''2 + '''g'''3}}

# {{nowrap|{{floor|{{frac|''k''|2}}}}'''g'''1 + {{floor|{{frac|''k''|2}}}} '''g'''2 + '''g'''3}}

(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 (linear) word {{nowrap|('''g'''1 '''g'''2){{floor|''n''/2}}}}, we see that the first two sizes must both occur {{sfrac|''n'' − ''k''|2}} times. This proves (2).

(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 (linear) word {{nowrap|('''g'''1 '''g'''2){{floor|{{frac|''n''|2}}}}}}, we see that the first two sizes must both occur {{sfrac|''n'' − ''k''|2}} times. This proves (2).

==== Statement (3) ====

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 ==== Statement (2) ====
 In case 2, let {{nowrap|''n'' &ge; 3}} and let {{nowrap|(2, 1) &minus; (1, 1) {{=}} '''g'''<sub>1</sub>|(1, 2) &minus; (2, 1) {{=}} '''g'''<sub>2</sub>}} be the two alternants. Let '''g'''<sub>3</sub> be the closing generator after stacking alternating '''g'''<sub>1</sub> and '''g'''<sub>2</sub>. Then the generator circle is {{nowrap|('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>{{floor|''n''/2}}</sup>}} '''g'''<sub>3</sub>. If a step is formed by stacking ''k'' generators, we may assume that ''k'' is odd, and the combinations of alternants corresponding to a step come in exactly 3 sizes:
-# {{nowrap|{{ceil|''k''/2}}'''g'''<sub>1</sub> + {{floor|''k''/2}}'''g'''<sub>2</sub>}}
+# {{nowrap|{{ceil|{{frac|''k''|2}}}}'''g'''<sub>1</sub> + {{floor|{{frac|''k''|2}}}}'''g'''<sub>2</sub>}}
-# {{nowrap|{{floor|''k''/2}}'''g'''<sub>1</sub> + {{ceil|''k''/2}}'''g'''<sub>2</sub>}}
+# {{nowrap|{{floor|{{frac|''k''|2}}}}'''g'''<sub>1</sub> + {{ceil|{{frac|''k''|2}}}}'''g'''<sub>2</sub>}}
-# {{nowrap|{{floor|''k''/2}}'''g'''<sub>1</sub> + {{floor|''k''/2}} '''g'''<sub>2</sub> + '''g'''<sub>3</sub>}}
+# {{nowrap|{{floor|{{frac|''k''|2}}}}'''g'''<sub>1</sub> + {{floor|{{frac|''k''|2}}}} '''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). By counting the length-''k'' subwords of the (linear) word {{nowrap|('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>{{floor|''n''/2}}</sup>}}, we see that the first two sizes must both occur {{sfrac|''n'' &minus; ''k''|2}} times. This proves (2).
+(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 (linear) word {{nowrap|('''g'''<sub>1</sub> '''g'''<sub>2</sub>)<sup>{{floor|{{frac|''n''|2}}}}</sup>}}, we see that the first two sizes must both occur {{sfrac|''n'' &minus; ''k''|2}} times. This proves (2).
 ==== Statement (3) ====