Ternary scale theorems: Difference between revisions
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'''Case 3:''' {{nowrap|3 ≤ μ ≤ {{floor|''n''/2}}}}. | '''Case 3:''' {{nowrap|3 ≤ μ ≤ {{floor|''n''/2}}}}. | ||
Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where {{nowrap|''x'' {{=}} {{floor|''n''/μ}}}} {{nowrap|≥ {{floor|''n''/{{floor|''n''/2}}}}}} = 2 or {{nowrap|''x'' {{=}} {{ceil|''n''/μ}}}} {{nowrap|{{=}} {{floor|''n''/μ}} + 1}}. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that contains {{nowrap|Λ<sub>3</sub>[''n'' − : 1]}} as a subword. This chunk must be of size ''y'', where | Λ<sub>2</sub> has a chunk of β (after the first β′) of size ''x'' where {{nowrap|''x'' {{=}} {{floor|''n''/μ}}}} {{nowrap|≥ {{floor|''n''/{{floor|''n''/2}}}}}} = 2 or {{nowrap|''x'' {{=}} {{ceil|''n''/μ}}}} {{nowrap|{{=}} {{floor|''n''/μ}} + 1}}. Hence Λ<sub>3</sub> has a chunk of γ of size ''x''. Λ<sub>3</sub> also has a chunk that contains {{nowrap|Λ<sub>3</sub>[''n'' − 1 : 1]}} as a subword. This chunk must be of size ''y'', where | ||
<math>2 \lfloor\frac{n}{\mu}\rfloor - 1 {{=}} 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 {{=}} 2\lfloor\frac{n}{\mu}\rfloor + 3.</math> | <math>2 \lfloor\frac{n}{\mu}\rfloor - 1 {{=}} 2 \big(\lfloor \frac{n}{\mu} \rfloor - 1\big) + 1 \leq y \leq 2 \big(\lfloor\frac{n}{\mu}\rfloor + 1 \big) + 1 {{=}} 2\lfloor\frac{n}{\mu}\rfloor + 3.</math> | ||