Kite's thoughts on pergens: Difference between revisions

| | vvM2

| | ^^C = D

| | C - ^F=vG - C

| | Thotho, if 13/8 = M6

P8/2 = v/A4 = ^\d5

| | C - /D=\Eb - F,

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10<span style="">⋅</span>P1 = m2. It must be downed, thus E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10<span style="">⋅</span>G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x<span style="">⋅</span>m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.

<span style="display: block; text-align: center;">C -- ^<span style="vertical-align: super;">4</span>D=v<span style="vertical-align: super;">6</span>Eb -- vvF -- ^^G -- ^<span style="vertical-align: super;">6</span>A=v<span style="vertical-align: super;">4</span>Bb -- C</span>

<span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 -- P4</span>