User:Frostburn/Lens RTT: Difference between revisions

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Define lens-monzos and lens-vals.
 
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Now we can treat the exponents as a lens-linear vector L{{monzo | u⁻¹, v⁻¹, w⁻¹}}. (Again using the primes as the basis.)
Now we can treat the exponents as a lens-linear vector L{{monzo | u⁻¹, v⁻¹, w⁻¹}}. (Again using the primes as the basis.)


For now the point is that the lens-vals corresponding to these lens-monzos better capture the fact that they correspond maps of fractions of equal temperaments. E.g. L5edo <math>\oplus</math> L7edo = L12edo
For now the point is that the lens-vals corresponding to these lens-monzos better capture the fact that they correspond to maps of fractions of equal temperaments. E.g. L5edo <math>\oplus</math> L7edo = L12edo


<math>\mathrm{L}\left \langle \frac{1}{5}, \frac{1}{8}, \frac{1}{12} \right ] \oplus \mathrm{L}\left \langle \frac{1}{7}, \frac{1}{11}, \frac{1}{16} \right ] = \mathrm{L}\left \langle \frac{1}{12}, \frac{1}{19}, \frac{1}{28} \right ]</math>
<math>\mathrm{L}\left \langle \frac{1}{5}, \frac{1}{8}, \frac{1}{12} \right ] \oplus \mathrm{L}\left \langle \frac{1}{7}, \frac{1}{11}, \frac{1}{16} \right ] = \mathrm{L}\left \langle \frac{1}{12}, \frac{1}{19}, \frac{1}{28} \right ]</math>