Kite's thoughts on pergens: Difference between revisions
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Because n is a multiple of b, n/b is an integer | Because n is a multiple of b, n/b is an integer | ||
M/b = (n/b)·M/n = (n/b)·G<br/> | M/b = (n/b)·M/n = (n/b)·G<br /> | ||
(a+b)·P8 = b·(M/b - P5) = b·((n/b)·G - P5) | (a+b)·P8 = b·(M/b - P5) = b·((n/b)·G - P5) | ||
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Since the pergen is a double-split, m > 1, therefore |b| > 1, therefore c ≠ 0 | Since the pergen is a double-split, m > 1, therefore |b| > 1, therefore c ≠ 0 | ||
c·(a+b)·P8 = c·b·((n/b)·G - P5)<br/> | c·(a+b)·P8 = c·b·((n/b)·G - P5)<br /> | ||
(1 - d·b)·P8 = c·b·((n/b)·G - P5)<br/> | (1 - d·b)·P8 = c·b·((n/b)·G - P5)<br /> | ||
P8 = d·b·P8 + c·b·((n/b)·G - P5) = b · (d·P8 + c·(n/b)·G - c·P5)<br/> | P8 = d·b·P8 + c·b·((n/b)·G - P5) = b · (d·P8 + c·(n/b)·G - c·P5)<br /> | ||
P8/m = P8/|b| = sign (b) · (d·P8 - c·P5 + c·(n/b)·G) | P8/m = P8/|b| = sign (b) · (d·P8 - c·P5 + c·(n/b)·G) | ||
Therefore P8 is split into m periods<br/> | Therefore P8 is split into m periods<br /> | ||
Therefore if m = |b|, the pergen is explicitly false | 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: | 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: | ||
P = (1/m, 0, 0)<br/> | P = (1/m, 0, 0)<br /> | ||
G = (a/n, b/n, 0)<br/> | G = (a/n, b/n, 0)<br /> | ||
C = (u, v, w) | 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: | 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: | ||
2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br/> | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br /> | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br/> | 3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br /> | ||
Q = Q/1 = ((av-bu)m/wb, -vn/wb, 1/w) · (P, G, C) | 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: | 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: | ||
2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br/> | 2 = 2/1 = P8 = (m, 0, 0) · (P, G, C)<br /> | ||
3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br/> | 3 = 3/1 = P12 = (-am/b, n/b, 0) · (P, G, C)<br /> | ||
Q = Q/1 = (±(av-bu)/n, ±(-v)/m, ±b/mn) · (P, G, C) | Q = Q/1 = (±(av-bu)/n, ±(-v)/m, ±b/mn) · (P, G, C) | ||
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to find the definition, use control-F to search for the first (bolded) occurrence of the word on this page. | to find the definition, use control-F to search for the first (bolded) occurrence of the word on this page. | ||
pergen<br/> | pergen<br /> | ||
split<br/> | split<br /> | ||
multigen<br/> | multigen<br /> | ||
ups and downs (the ^ and v symbols)<br/> | ups and downs (the ^ and v symbols)<br /> | ||
higher prime (any prime > 3)<br/> | higher prime (any prime > 3)<br /> | ||
color depth<br/> | color depth<br /> | ||
dependent/independent<br/> | dependent/independent<br /> | ||
square mapping<br/> | square mapping<br /> | ||
lifts and drops (the / and \ symbols)<br/> | lifts and drops (the / and \ symbols)<br /> | ||
enharmonic<br/> | enharmonic<br /> | ||
genchain<br/> | genchain<br /> | ||
perchain<br/> | perchain<br /> | ||
wide/widen (increased by an octave)<br/> | wide/widen (increased by an octave)<br /> | ||
single-split, double-split<br/> | single-split, double-split<br /> | ||
single-pair, double-pair (number of new accidentals in the notation)<br/> | single-pair, double-pair (number of new accidentals in the notation)<br /> | ||
true double, false double<br/> | true double, false double<br /> | ||
explicitly false<br/> | explicitly false<br /> | ||
unreduced<br/> | unreduced<br /> | ||
alternate vs. equivalent (generator or period)<br/> | alternate vs. equivalent (generator or period)<br /> | ||
mapping comma<br/> | mapping comma<br /> | ||
keyspan<br/> | keyspan<br /> | ||
stepspan<br/> | stepspan<br /> | ||
gedra<br/> | gedra<br /> | ||
count<br/> | count<br /> | ||
mid<br/> | mid<br /> | ||
edomapping<br/> | edomapping<br /> | ||
upspan<br/> | upspan<br /> | ||
liftspan | liftspan | ||
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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]: | 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]: | ||
k = 12a + 19b + 28c + 34d<br/> | k = 12a + 19b + 28c + 34d<br /> | ||
s = 7a + 11b + 14c + 20d<br/> | s = 7a + 11b + 14c + 20d<br /> | ||
g = -c<br/> | g = -c<br /> | ||
r = -d | r = -d | ||
a = -11k + 19s - 4g + 6r<br/> | a = -11k + 19s - 4g + 6r<br /> | ||
b = 7k - 12s + 4g - 2r<br/> | b = 7k - 12s + 4g - 2r<br /> | ||
c = -g<br/> | c = -g<br /> | ||
d = -r | d = -r | ||