Kite's thoughts on pergens: Difference between revisions

__FORCETOC__

A '''pergen''' (pronounced "peer-gen") is a way of classifying a regular temperament solely by its period and generator(s). For any temperament, there are many possible periods and generators. For the pergen, they are chosen to use the fewest, and smallest, prime factors possible. Fractions are allowed, e.g. half-octave, but avoided if possible.

In keeping with the higher-prime-agnostic nature of pergens, untempered just intonation has a pergen of the octave, the fifth, and a list of commas, each containing only one higher prime. The higher prime's exponent in the comma's monzo must be ±1, i.e. the color depth must be 1. Furthermore, the comma should map to P1, e.g. 81/80 or 64/63. The commas are notated with microtonal accidentals: (P8, P5, ^1, /1,...).

For any comma, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents, where GCD (0,x) = x. The comma will split the octave into m parts, and if n > m, it will split some 3-limit interval into n parts.

The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is > 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.

Pergens allow for a systematic exploration of all posible rank-2 tunings, potentially identifying new musical resources.

|}

==Naming very large intervals==

Because n is a multiple of b, n/b is an integer

(a+b)·P8 = b·(M/b - P5) = b·((n/b)·G - P5)

Since the pergen is a double-split, m > 1, therefore |b| > 1, therefore c ≠ 0

P8/m = P8/|b| = sign (b) · (d·P8 - c·P5 + c·(n/b)·G)

Therefore if m = |b|, the pergen is explicitly false

Assume the pergen is a false double, and there's a comma C that splits both P8 and (a,b) appropriately. Can we prove r = 1? Let Q = the higher prime that C uses. Express P, G and C as monzos of the prime subgroup 2.3.Q, by expanding the 2x2 pergen matrix to a 3x3 matrix A:

C = (u, v, w)

Here u, v and w are integers. If GCD (u, v, w) > 1, simplify C so that it = 1. The inverse of A expresses 2, 3 and Q in terms of P, G and C. If C is tempered out, the C column can be discarded, making the usual 3x2 period-generator mapping. However, if C is not tempered out, the inverse of A is a 3x3 period-generator-comma mapping, which is simply a change of basis. For example, 5-limit JI can be generated by 2/1, 3/2 and 81/80. Here is the inverse of A:

Q = Q/1 = ((av-bu)m/wb, -vn/wb, 1/w) · (P, G, C)

Fractions are allowed in the first two rows of A but not the 3rd row. Fractions are allowed in the last column of A-inverse, but not the first two columns. Every pergen except the unsplit one requires |w| > 1, so the last column almost always has a fraction. To avoid fractions in the first two columns, A must be unimodular '''''[I think, not sure]''''', and we have wb/mn = ±1, and w = ±mn/b. Substituting for w, we have:

Q = Q/1 = (±(av-bu)/n, ±(-v)/m, ±b/mn) · (P, G, C)

to find the definition, use control-F to search for the first (bolded) occurrence of the word on this page.

liftspan

General rules for combining pergens:

However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.

Gedras can be expanded to 5-limit or higher by including another keyspan that is compatible with 7 and 12, such as 9 or 16. But a more useful approach is for the third number to be the comma 81/80. Thus 5/4 would be a M3 minus a comma, [4, 2, -1]. We can use 64/63 to expand to the 7-limit. For (a,b,c,d) we get [k,s,g,r]:

r = -d

d = -r