K*N subgroups: Difference between revisions
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For any [[Harmonic_Limit|prime limit]] p, EDO N and positive integer k, the p-limit k*N subgroup is the largest [[Just_intonation_subgroups|just intonation subgroup]] of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the [[Patent_val|patent val]] for k*N-edo. This also means they temper out the same commas. | |||
A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the [[ | A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the [[Euler_genera|Euler genus]] Euler(m^i) for integers 1, 2, and so forth. Here Euler(d) for an odd integer d is the set of all divisors of d reduced to an octave, and including 2. For each such genus, select those intervals such that the k*N patent val maps the interval to a number divisible by k, and then find the corresponding [[Normal_lists|normal interval list]]. When two successive values i and i+1 lead to the same normal list, add to that a basis for the commas of the k*N patent val, and return the normal interval list for that. | ||
For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime. | For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime. | ||
[[Category:algorithm]] | |||
[[Category:math]] | |||
[[Category:theory]] | |||
Revision as of 00:00, 17 July 2018
For any prime limit p, EDO N and positive integer k, the p-limit k*N subgroup is the largest just intonation subgroup of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the patent val for k*N-edo. This also means they temper out the same commas.
A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the Euler genus Euler(m^i) for integers 1, 2, and so forth. Here Euler(d) for an odd integer d is the set of all divisors of d reduced to an octave, and including 2. For each such genus, select those intervals such that the k*N patent val maps the interval to a number divisible by k, and then find the corresponding normal interval list. When two successive values i and i+1 lead to the same normal list, add to that a basis for the commas of the k*N patent val, and return the normal interval list for that.
For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.