Kite's thoughts on pergens: Difference between revisions

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For example, 12edo's 3-limit edomapping is (12, 19), and 16edo's is (16, 25). The determinant of [(12 19) (16 25)] is -4, thus the pergen for 12edo and 16edo is (P8/4, P5). To make the calculations easier, octave-reduced edomappings can be used, which indicate the number of edosteps that 3/2 maps to, not 3/1. For 12 and 16, we have [(12 7) (16 9)], and d is again -4.

If |d| ≠ m or 0, P5 is not the generator, and we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree, let g/g' be the smallest ancestor of this ratio. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.

If |d| ≠ m or 0, P5 is not the generator, and we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree ([http://tallkite.com/misc_files/Scale-Tree-Complete.pdf pdf] or [http://tallkite.com/misc_files/Scale-Tree-Complete.jpg jpeg]), let g/g' be the smallest ancestor of this ratio. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.

For example, for 7-edo and 17-edo, m = 1 but d = 2, so G ≠ P5. The smallest ancestor of 7/17 is 2/5. G maps to both 2\7 and 5\17, which are both neutral 3rds. Using the other ancestor always gives the octave inverses, in this case 5/12 giving 5\7 and 12\17, which are neutral 6ths. 2\7 = 343¢ and 5\17 = 353¢, and their difference is only 1\N", where N" = LCM (N, N'). This 10¢ difference is the least difference between any 7edo note and any 17edo note, except that the two neutral 6ths differ by the same 10¢, and of course some note pairs coincide exactly, such as 0\7 and 0\17.

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 For example, 12edo's 3-limit edomapping is (12, 19), and 16edo's is (16, 25). The determinant of [(12 19) (16 25)] is -4, thus the pergen for 12edo and 16edo is (P8/4, P5). To make the calculations easier, octave-reduced edomappings can be used, which indicate the number of edosteps that 3/2 maps to, not 3/1. For 12 and 16, we have [(12 7) (16 9)], and d is again -4.
-If |d| ≠ m or 0, P5 is not the generator, and we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree, let g/g' be the smallest ancestor of this ratio. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.
+If |d| ≠ m or 0, P5 is not the generator, and we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree ([http://tallkite.com/misc_files/Scale-Tree-Complete.pdf pdf] or [http://tallkite.com/misc_files/Scale-Tree-Complete.jpg jpeg]), let g/g' be the smallest ancestor of this ratio. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.
 For example, for 7-edo and 17-edo, m = 1 but d = 2, so G ≠ P5. The smallest ancestor of 7/17 is 2/5. G maps to both 2\7 and 5\17, which are both neutral 3rds. Using the other ancestor always gives the octave inverses, in this case 5/12 giving 5\7 and 12\17, which are neutral 6ths. 2\7 = 343¢ and 5\17 = 353¢, and their difference is only 1\N", where N" = LCM (N, N'). This 10¢ difference is the least difference between any 7edo note and any 17edo note, except that the two neutral 6ths differ by the same 10¢, and of course some note pairs coincide exactly, such as 0\7 and 0\17.