Jake Huryn's scratchpad - Revision history

BudjarnLambeth: Categorised this uncategorised page

2023-04-30T07:53:25Z

Categorised this uncategorised page

← Older revision		Revision as of 07:53, 30 April 2023
Line 18:		Line 18:

	Solving for a and d (k is irrelevant here), we find that our mapping begins \|<5\|, <2\|>—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for \|0 1 0> and \|0 0 1> we can complete this mapping.		Solving for a and d (k is irrelevant here), we find that our mapping begins \|<5\|, <2\|>—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for \|0 1 0> and \|0 0 1> we can complete this mapping.
			[[Category:Todo:move to userspace]]
			[[Category:Regular temperament theory]]

Wikispaces>FREEZE at 00:00, 17 July 2018

2018-07-17T00:00:00Z

@@ Line 1: / Line 1: @@
-<h2>IMPORTED REVISION FROM WIKISPACES</h2>
+I'm trying to learn tuning theory. Please point out anything I might be missing or have wrong! My apologies for any difficult-to-understand descriptions or weird notation, this is just a personal page meant for getting my thoughts on paper. If there's anything you don't understand I'll do my best to explain better. Thanks!
 Suppose we start with the JI subgroup 2.3.5 and temper out a comma, say 81/80. In terms of abstract algebra, we can say that the result is the quotient group G{2, 3, 5}/G{81/80}, where G{a, b, c, …} represents the group generated by a, b, c, etc, under multiplication. We have thus produced a group homomorphic to 2.3.5 but in which 81/80 (and all of its powers) are the kernel, meaning that they are mapped to the identity (unison, or 1/1). The elements of this quotient group are the cosets G{81/80}p/q, or the sets produced by multiplying every element of G{81/80} by p/q. Given that a/b and c/d do not differ by an integer multiple of 81/80 (excluding 1/1), G{81/80}a/b and G{81/80}c/d will be distinct cosets. These cosets may be multiplied: G{81/80}a/b*G{81/80}c/d = G{81/80}(ac)/(bd). This works and produces a group (G{2, 3, 5}/G{81/80}, as above) because multiplication is commutative, so any subgroup of 2.3.5 is commutative and thus a normal subgroup.
@@ Line 16: / Line 9: @@
 Using this same quotient group, let's chose G{|-4 4 -1&gt;}|1 0 0&gt; and G{|-4 4 -1&gt;}|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: |&lt;1 0|, &lt;0 1|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{|-4 4 -1&gt;}|0 0 1&gt;, we can multiply |0 0 1&gt; by |-4 4 -1&gt; to produce the same coset: G{|-4 4 -1&gt;}|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{|-4 4 -1&gt;}|-4 4 0&gt; = (G{|-4 4 -1&gt;}|1 0 0&gt;)^-4 * (G{|-4 4 -1&gt;}|0 1 0&gt;)^4, so our completed mapping is |&lt;1 0 -4|, &lt;0 1 4|&gt;.
-This is not always so easy, however. Suppose we wanted to find the mapping when we use G{|-4 4 -1&gt;}|-3 2 0&gt; and G{|-4 4 -1&gt;}|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{|-4 4 -1&gt;}|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{|-4 4 -1&gt;}|1 0 0&gt; ``= (G{|-4 4 -1&gt;}|-3 2 0&gt;)^a * (G{|-4 4 -1&gt;}|4 -1 -1&gt;)^d, and expanding everything out, G{|-4 4 -1&gt;}|1 0 0&gt; =`` G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{|-4 4 -1&gt;}|1-4k 4k -k&gt; = G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):
+This is not always so easy, however. Suppose we wanted to find the mapping when we use G{|-4 4 -1&gt;}|-3 2 0&gt; and G{|-4 4 -1&gt;}|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{|-4 4 -1&gt;}|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{|-4 4 -1&gt;}|1 0 0&gt; = (G{|-4 4 -1&gt;}|-3 2 0&gt;)^a * (G{|-4 4 -1&gt;}|4 -1 -1&gt;)^d, and expanding everything out, G{|-4 4 -1&gt;}|1 0 0&gt; = G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{|-4 4 -1&gt;}|1-4k 4k -k&gt; = G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):
 -3a + 4d = 1 - 4k
 a - d = 4k
 -d = -k.
-Solving for a and d (k is irrelevant here), we find that our mapping begins |&lt;5|, &lt;2|&gt;—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for |0 1 0&gt; and |0 0 1&gt; we can complete this mapping.</pre></div>
+Solving for a and d (k is irrelevant here), we find that our mapping begins |&lt;5|, &lt;2|&gt;—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for |0 1 0&gt; and |0 0 1&gt; we can complete this mapping.

Wikispaces>jake.huryn: Imported revision 613357541 - Original comment:

2017-05-22T15:51:59Z

**Imported revision 613357541 - Original comment: **

← Older revision		Revision as of 15:51, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:46:11 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:51:59 UTC</tt>.<br>
	: The original revision id was <tt>~~613357173~~</tt>.<br>		: The original revision id was <tt>613357541</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> \= (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> \= G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> ``= (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> =`` G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; \= (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; \= G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextRawRule:00:``= (G{\|-4 4 -1&amp;gt;}\|-3 2 0&amp;gt;)^a * (G{\|-4 4 -1&amp;gt;}\|4 -1 -1&amp;gt;)^d, and expanding everything out, G{\|-4 4 -1&amp;gt;}\|1 0 0&amp;gt; =`` -->= (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; =<!-- ws:end:WikiTextRawRule:00 --> G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />
	-3a + 4d = 1 - 4k<br />		-3a + 4d = 1 - 4k<br />

Wikispaces>jake.huryn: Imported revision 613357173 - Original comment:

2017-05-22T15:46:11Z

**Imported revision 613357173 - Original comment: **

← Older revision		Revision as of 15:46, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:44:22 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:46:11 UTC</tt>.<br>
	: The original revision id was <tt>~~613357047~~</tt>.<br>		: The original revision id was <tt>613357173</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> = (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> .= G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> \= (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> \= G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; ~~<!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id~~=~~"toc0"><a name="x~~(G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; ."></a><!-- ws:end:WikiTextHeadingRule:0 --> (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; .</h1>		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; \= (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; \= G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />
	-3a + 4d = 1 - 4k<br />		-3a + 4d = 1 - 4k<br />

Wikispaces>jake.huryn: Imported revision 613357047 - Original comment:

2017-05-22T15:44:22Z

**Imported revision 613357047 - Original comment: **

← Older revision		Revision as of 15:44, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:43:33 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:44:22 UTC</tt>.<br>
	: The original revision id was <tt>~~613356999~~</tt>.<br>		: The original revision id was <tt>613357047</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> =. (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> = (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> .= G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="x. (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt;"></a><!-- ws:end:WikiTextHeadingRule:0 -->. (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; </h1>		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="x(G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; ."></a><!-- ws:end:WikiTextHeadingRule:0 --> (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; .</h1>
	G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />		G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />

Wikispaces>jake.huryn: Imported revision 613356999 - Original comment:

2017-05-22T15:43:33Z

**Imported revision 613356999 - Original comment: **

← Older revision		Revision as of 15:43, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:42:49 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:43:33 UTC</tt>.<br>
	: The original revision id was <tt>~~613356949~~</tt>.<br>		: The original revision id was <tt>613356999</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> = (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> =. (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="x(G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt;"></a><!-- ws:end:WikiTextHeadingRule:0 --> (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; </h1>		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="x. (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt;"></a><!-- ws:end:WikiTextHeadingRule:0 -->. (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; </h1>
	G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />		G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />
	-3a + 4d = 1 - 4k<br />		-3a + 4d = 1 - 4k<br />

Wikispaces>jake.huryn: Imported revision 613356949 - Original comment:

2017-05-22T15:42:49Z

**Imported revision 613356949 - Original comment: **

← Older revision		Revision as of 15:42, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:42:35 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:42:49 UTC</tt>.<br>
	: The original revision id was <tt>~~613356931~~</tt>.<br>		: The original revision id was <tt>613356949</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> \= (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> \= G{\|-4 4 -1>}\|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> = (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; \= (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; \= G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="x(G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt;"></a><!-- ws:end:WikiTextHeadingRule:0 --> (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; </h1>
			G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />
	-3a + 4d = 1 - 4k<br />		-3a + 4d = 1 - 4k<br />

Wikispaces>jake.huryn: Imported revision 613356931 - Original comment:

2017-05-22T15:42:35Z

**Imported revision 613356931 - Original comment: **

← Older revision		Revision as of 15:42, 22 May 2017
Line 1:		Line 1:
	<h2>IMPORTED REVISION FROM WIKISPACES</h2>		<h2>IMPORTED REVISION FROM WIKISPACES</h2>
	This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>		This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
	: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:41:49 UTC</tt>.<br>		: This revision was by author [[User:jake.huryn\|jake.huryn]] and made on <tt>2017-05-22 15:42:35 UTC</tt>.<br>
	: The original revision id was <tt>~~613356895~~</tt>.<br>		: The original revision id was <tt>613356931</tt>.<br>
	: The revision comment was: <tt></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>		The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
Line 16:		Line 16:
	Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.		Using this same quotient group, let's chose G{\|-4 4 -1>}\|1 0 0> and G{\|-4 4 -1>}\|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: \|<1 0\|, <0 1\|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1>}\|0 0 1>, we can multiply \|0 0 1> by \|-4 4 -1> to produce the same coset: G{\|-4 4 -1>}\|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1>}\|-4 4 0> = (G{\|-4 4 -1>}\|1 0 0>)^-4 * (G{\|-4 4 -1>}\|0 1 0>)^4, so our completed mapping is \|<1 0 -4\|, <0 1 4\|>.

	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> = (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1>}\|-3 2 0> and G{\|-4 4 -1>}\|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1>}\|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1>}\|1 0 0> \= (G{\|-4 4 -1>}\|-3 2 0>)^a * (G{\|-4 4 -1>}\|4 -1 -1>)^d, and expanding everything out, G{\|-4 4 -1>}\|1 0 0> \= G{\|-4 4 -1>}\|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1>}\|1-4k 4k -k> = G{\|-4 4 -1>}\|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

	-3a + 4d = 1 - 4k		-3a + 4d = 1 - 4k
Line 34:		Line 34:
	Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />		Using this same quotient group, let's chose G{\|-4 4 -1&gt;}\|1 0 0&gt; and G{\|-4 4 -1&gt;}\|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: \|&lt;1 0\|, &lt;0 1\|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{\|-4 4 -1&gt;}\|0 0 1&gt;, we can multiply \|0 0 1&gt; by \|-4 4 -1&gt; to produce the same coset: G{\|-4 4 -1&gt;}\|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{\|-4 4 -1&gt;}\|-4 4 0&gt; = (G{\|-4 4 -1&gt;}\|1 0 0&gt;)^-4 * (G{\|-4 4 -1&gt;}\|0 1 0&gt;)^4, so our completed mapping is \|&lt;1 0 -4\|, &lt;0 1 4\|&gt;.<br />
	<br />		<br />
	This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; ~~<!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id~~=~~"toc0"><a name="x~~(G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt;"></a><!-- ws:end:WikiTextHeadingRule:0 --> (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; </h1>		This is not always so easy, however. Suppose we wanted to find the mapping when we use G{\|-4 4 -1&gt;}\|-3 2 0&gt; and G{\|-4 4 -1&gt;}\|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{\|-4 4 -1&gt;}\|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{\|-4 4 -1&gt;}\|1 0 0&gt; \= (G{\|-4 4 -1&gt;}\|-3 2 0&gt;)^a * (G{\|-4 4 -1&gt;}\|4 -1 -1&gt;)^d, and expanding everything out, G{\|-4 4 -1&gt;}\|1 0 0&gt; \= G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k\|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{\|-4 4 -1&gt;}\|1-4k 4k -k&gt; = G{\|-4 4 -1&gt;}\|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
	<br />		<br />
	-3a + 4d = 1 - 4k<br />		-3a + 4d = 1 - 4k<br />

Wikispaces>jake.huryn: Imported revision 613356895 - Original comment:

2017-05-22T15:41:49Z

**Imported revision 613356895 - Original comment: **

New page

<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:jake.huryn|jake.huryn]] and made on <tt>2017-05-22 15:41:49 UTC</tt>.<br>
: The original revision id was <tt>613356895</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>
<h4>Original Wikitext content:</h4>
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;white-space: pre-wrap ! important" class="old-revision-html">I'm trying to learn tuning theory. Please point out anything I might be missing or have wrong! My apologies for any difficult-to-understand descriptions or weird notation, this is just a personal page meant for getting my thoughts on paper. If there's anything you don't understand I'll do my best to explain better. Thanks!

Suppose we start with the JI subgroup 2.3.5 and temper out a comma, say 81/80. In terms of abstract algebra, we can say that the result is the quotient group G{2, 3, 5}/G{81/80}, where G{a, b, c, …} represents the group generated by a, b, c, etc, under multiplication. We have thus produced a group homomorphic to 2.3.5 but in which 81/80 (and all of its powers) are the kernel, meaning that they are mapped to the identity (unison, or 1/1). The elements of this quotient group are the cosets G{81/80}p/q, or the sets produced by multiplying every element of G{81/80} by p/q. Given that a/b and c/d do not differ by an integer multiple of 81/80 (excluding 1/1), G{81/80}a/b and G{81/80}c/d will be distinct cosets. These cosets may be multiplied: G{81/80}a/b*G{81/80}c/d = G{81/80}(ac)/(bd). This works and produces a group (G{2, 3, 5}/G{81/80}, as above) because multiplication is commutative, so any subgroup of 2.3.5 is commutative and thus a normal subgroup.

(Although this makes everything appear very complicated, I will now notate intervals as monzos as they are easier to work with.)

As it turns out, this quotient group can be generated by only two elements; in tuning theory terms, tempering one comma from a three dimensional system produces a two dimensional system. Furthermore, for each two generators we chose, say p and q, we can produce a mapping of the form |<a b c|, <d e f|>, where G{|-4 4 -1>}|1 0 0> = (G{|-4 4 -1>}p)^a*(G{|-4 4 -1>}q)^d. In other words, our tempered 2/1 (its corresponding coset in the quotient group) can be written as a product of the tempered interval p to the a-th power times the tempered interval q to the d-th power. This can of course be done for |0 1 0> and |0 0 1>, corresponding to the second and third elements of each val, respectively.

Using this same quotient group, let's chose G{|-4 4 -1>}|1 0 0> and G{|-4 4 -1>}|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: |<1 0|, <0 1|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{|-4 4 -1>}|0 0 1>, we can multiply |0 0 1> by |-4 4 -1> to produce the same coset: G{|-4 4 -1>}|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{|-4 4 -1>}|-4 4 0> = (G{|-4 4 -1>}|1 0 0>)^-4 * (G{|-4 4 -1>}|0 1 0>)^4, so our completed mapping is |<1 0 -4|, <0 1 4|>.

This is not always so easy, however. Suppose we wanted to find the mapping when we use G{|-4 4 -1>}|-3 2 0> and G{|-4 4 -1>}|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{|-4 4 -1>}|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{|-4 4 -1>}|1 0 0> = (G{|-4 4 -1>}|-3 2 0>)^a * (G{|-4 4 -1>}|4 -1 -1>)^d, and expanding everything out, G{|-4 4 -1>}|1 0 0> = G{|-4 4 -1>}|-3a+4d 2a-d -d> It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{|-4 4 -1>}|1-4k 4k -k> = G{|-4 4 -1>}|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

-3a + 4d = 1 - 4k
2a - d = 4k
-d = -k.

Solving for a and d (k is irrelevant here), we find that our mapping begins |<5|, <2|>—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for |0 1 0> and |0 0 1> we can complete this mapping.</pre></div>
<h4>Original HTML content:</h4>
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>Jake Huryn's scratchpad</title></head><body>I'm trying to learn tuning theory. Please point out anything I might be missing or have wrong! My apologies for any difficult-to-understand descriptions or weird notation, this is just a personal page meant for getting my thoughts on paper. If there's anything you don't understand I'll do my best to explain better. Thanks!<br />
<br />
Suppose we start with the JI subgroup 2.3.5 and temper out a comma, say 81/80. In terms of abstract algebra, we can say that the result is the quotient group G{2, 3, 5}/G{81/80}, where G{a, b, c, …} represents the group generated by a, b, c, etc, under multiplication. We have thus produced a group homomorphic to 2.3.5 but in which 81/80 (and all of its powers) are the kernel, meaning that they are mapped to the identity (unison, or 1/1). The elements of this quotient group are the cosets G{81/80}p/q, or the sets produced by multiplying every element of G{81/80} by p/q. Given that a/b and c/d do not differ by an integer multiple of 81/80 (excluding 1/1), G{81/80}a/b and G{81/80}c/d will be distinct cosets. These cosets may be multiplied: G{81/80}a/b*G{81/80}c/d = G{81/80}(ac)/(bd). This works and produces a group (G{2, 3, 5}/G{81/80}, as above) because multiplication is commutative, so any subgroup of 2.3.5 is commutative and thus a normal subgroup.<br />
<br />
(Although this makes everything appear very complicated, I will now notate intervals as monzos as they are easier to work with.)<br />
<br />
As it turns out, this quotient group can be generated by only two elements; in tuning theory terms, tempering one comma from a three dimensional system produces a two dimensional system. Furthermore, for each two generators we chose, say p and q, we can produce a mapping of the form |&lt;a b c|, &lt;d e f|&gt;, where G{|-4 4 -1&gt;}|1 0 0&gt; = (G{|-4 4 -1&gt;}p)^a*(G{|-4 4 -1&gt;}q)^d. In other words, our tempered 2/1 (its corresponding coset in the quotient group) can be written as a product of the tempered interval p to the a-th power times the tempered interval q to the d-th power. This can of course be done for |0 1 0&gt; and |0 0 1&gt;, corresponding to the second and third elements of each val, respectively.<br />
<br />
Using this same quotient group, let's chose G{|-4 4 -1&gt;}|1 0 0&gt; and G{|-4 4 -1&gt;}|0 1 0&gt; as our generators. Clearly we can already fill out the first two elements of each val: |&lt;1 0|, &lt;0 1|&gt;. However, we still need to find a mapping for the fifth harmonic. Starting with G{|-4 4 -1&gt;}|0 0 1&gt;, we can multiply |0 0 1&gt; by |-4 4 -1&gt; to produce the same coset: G{|-4 4 -1&gt;}|-4 4 0&gt;. Now it is clear how we may write this in terms of our generators: G{|-4 4 -1&gt;}|-4 4 0&gt; = (G{|-4 4 -1&gt;}|1 0 0&gt;)^-4 * (G{|-4 4 -1&gt;}|0 1 0&gt;)^4, so our completed mapping is |&lt;1 0 -4|, &lt;0 1 4|&gt;.<br />
<br />
This is not always so easy, however. Suppose we wanted to find the mapping when we use G{|-4 4 -1&gt;}|-3 2 0&gt; and G{|-4 4 -1&gt;}|4 -1 -1&gt; (9/8 and 16/15, respectively) as our generators. Even for G{|-4 4 -1&gt;}|1 0 0&gt; it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{|-4 4 -1&gt;}|1 0 0&gt; <h1 id="toc0"><a name="x(G{|-4 4 -1&gt;}|-3 2 0&gt;)^a * (G{|-4 4 -1&gt;}|4 -1 -1&gt;)^d, and expanding everything out, G{|-4 4 -1&gt;}|1 0 0&gt;"></a> (G{|-4 4 -1&gt;}|-3 2 0&gt;)^a * (G{|-4 4 -1&gt;}|4 -1 -1&gt;)^d, and expanding everything out, G{|-4 4 -1&gt;}|1 0 0&gt; </h1>
G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt; It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k|-4 4 -1&gt;, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{|-4 4 -1&gt;}|1-4k 4k -k&gt; = G{|-4 4 -1&gt;}|-3a+4d 2a-d -d&gt;, which can now be written as a system of three linear equations (now in three variables):<br />
<br />
-3a + 4d = 1 - 4k<br />
2a - d = 4k<br />
-d = -k.<br />
<br />
Solving for a and d (k is irrelevant here), we find that our mapping begins |&lt;5|, &lt;2|&gt;—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for |0 1 0&gt; and |0 0 1&gt; we can complete this mapping.</body></html></pre></div>

Jake Huryn's scratchpad - Revision history

BudjarnLambeth: Categorised this uncategorised page

Wikispaces>FREEZE at 00:00, 17 July 2018

Wikispaces>jake.huryn: **Imported revision 613357541 - Original comment: **

Wikispaces>jake.huryn: **Imported revision 613357173 - Original comment: **

Wikispaces>jake.huryn: **Imported revision 613357047 - Original comment: **

Wikispaces>jake.huryn: **Imported revision 613356999 - Original comment: **

Wikispaces>jake.huryn: **Imported revision 613356949 - Original comment: **

Wikispaces>jake.huryn: **Imported revision 613356931 - Original comment: **