Pergen names: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

This is an imported revision from Wikispaces. The revision metadata is included below for reference:

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-20 01:33:36 UTC</tt>.

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-20 01:44:27 UTC</tt>.

: The original revision id was <tt>~~624089881~~</tt>.

: The original revision id was <tt>624089979</tt>.

: The revision comment was: <tt></tt>

The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.

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To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the original generator from the period and the alternate generator.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1~~. This is much too small to be half a 4th, so we'll find an alternate generator later~~. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^4M2. Next, ~~deduce~~ the ~~alternate~~ generator G~~' as we did for the period:~~ P4 ~~plus or minus some number of enharmonics must equal~~ 2*G'~~, and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y~~ is ~~either 1 or~~ -~~1. Choose~~ -1, ~~to get a smaller~~ G~~', since the other one~~ is equivalent. 2*G' ~~= P4 - E, thus~~ G' = ^5M2~~, which happens to be P~~ + ~~G. Thus P = ^4M2, G =~~ ^1~~, G'~~ = ^~~5M2, and E~~ = ~~v10m2~~.

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*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^4M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P < G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. The alternate generator is often simpler than the original generator.

P1- - ^4M2=v6m3 -- vvP4 -- ^^P5 -- ^6M6=v4m7 -- P8C -- D^4=Ebv6 -- Fvv -- G^^ -- A^6=Bbv4 -- CP1 -- ^5M2=v5m3 -- P4C -- D^5=Ebv5 -- F

P1 - ^4M2=v6m3 - vvP4 - ^^P5 - ^6M6=v4m7 - P8C - D^4=Ebv6 - Fvv - G^^ - A^6=Bbv4 - CP1 - ^5M2=v5m3 - P4C - D^5=Ebv5 - F

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).

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To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the original generator from the period and the alternate generator.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1~~. This is much too small to be half a 4th, so we'll find an alternate generator later~~. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^4M2. Next, ~~deduce~~ the ~~alternate~~ generator G~~' as we did for the period:~~ P4 ~~plus or minus some number of enharmonics must equal~~ 2*G'~~, and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y~~ is ~~either 1 or~~ -~~1. Choose~~ -1, ~~to get a smaller~~ G~~', since the other one~~ is equivalent. 2*G' ~~= P4 - E, thus~~ G' = ^5M2~~, which happens to be P~~ + ~~G. Thus P = ^4M2, G =~~ ^1~~, G'~~ = ^~~5M2, and E~~ = ~~v10m2.~~

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*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^4M2. Next, find the original half-fourth 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 ^4M2 + ^1 = ^5M2. The alternate generator is often simpler than the original generator.

P1- - ^4M2=v6m3 -- vvP4 -- ^^P5 -- ^6M6=v4m7 -- P8C -- D^4=Ebv6 -- Fvv -- G^^ -- A^6=Bbv4 -- CP1 -- ^5M2=v5m3 -- P4C -- D^5=Ebv5 -- F

P1 - ^4M2=v6m3 - vvP4 - ^^P5 - ^6M6=v4m7 - P8C - D^4=Ebv6 - Fvv - G^^ - A^6=Bbv4 - CP1 - ^5M2=v5m3 - P4C - D^5=Ebv5 - F

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).

@@ Line 1: / Line 1: @@
 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-20 01:33:36 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-20 01:44:27 UTC</tt>.<br>
-: The original revision id was <tt>624089881</tt>.<br>
+: The original revision id was <tt>624089979</tt>.<br>
 : The revision comment was: <tt></tt><br>
 The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
@@ Line 330: / Line 330: @@
 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the original generator from the period and the alternate generator.
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.
+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*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth 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 ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is often simpler than the original generator.
+&lt;span style="display: block; text-align: center;"&gt;P1- - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 -- vvP4 -- ^^P5 -- ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; -- Fvv -- G^^ -- A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; -- F&lt;/span&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 - vvP4 - ^^P5 - ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; - Fvv - G^^ - A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; - C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 - P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; - F&lt;/span&gt;
+To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.
-To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.
 A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).
@@ Line 2,071: / Line 2,070: @@
 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the original generator from the period and the alternate generator.&lt;br /&gt;
 &lt;br /&gt;
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, thus G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.&lt;br /&gt;
+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*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*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*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &amp;lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is often simpler than the original generator.&lt;br /&gt;
-&lt;br /&gt;
+&lt;span style="display: block; text-align: center;"&gt;P1- - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 -- vvP4 -- ^^P5 -- ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; -- Fvv -- G^^ -- A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; -- F&lt;/span&gt;&lt;br /&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 - vvP4 - ^^P5 - ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; - Fvv - G^^ - A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; - C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 - P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; - F&lt;/span&gt;&lt;br /&gt;
+To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.&lt;br /&gt;
-To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For (P8/2, P4/2, the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.&lt;br /&gt;
 &lt;br /&gt;
 A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).&lt;br /&gt;