Tenney–Euclidean metrics: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

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: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-01-30 19:09:33 UTC</tt>.

: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-01-30 19:11:41 UTC</tt>.

: The original revision id was <tt>~~197259402~~</tt>.

: The original revision id was <tt>197259818</tt>.

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Using this, we find the temperamental norm of [9 13] to be sqrt([9 13]//P//[9 13]*) ~ sqrt(1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while **P** does not depend on the choice of basis vals for the temperament, //P// does; if we choose [<1 2 3|, <0 -3 -5|] for our basis instead, then 3/2 is represented by [1 -3] and //P// changes coordinates to produce the same final result of temperamental complexity.

If instead we want OE complexity, we may remove the first row of [<1 2 3|, <0 -3 -5|], leaving just [<0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [[0.1215588]]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.

If instead we want OE complexity, we may remove the first row of [<1 2 3|, <0 -3 -5|], leaving just [<0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [<0.1215588|]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.

For a more substantial example we need to consider at least a rank three temperament, so let us turn to marvel, the 7-limit temperament tempering out 225/224. The 2x4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [|1 0 0 0>, |-5 2 2 -1>]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [|1 0 0 0>, |-5 2p3 2p5 -p7>], and P = I - M`M = [|1 0 0 0>, |0 4p5^2+p7^2 -4p3p5 2p3p7>/H,

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Using this, we find the temperamental norm of [9 13] to be sqrt([9 13]P[9 13]*) ~ sqrt(1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while P does not depend on the choice of basis vals for the temperament, P does; if we choose [&lt;1 2 3|, &lt;0 -3 -5|] for our basis instead, then 3/2 is represented by [1 -3] and P changes coordinates to produce the same final result of temperamental complexity.

If instead we want OE complexity, we may remove the first row of [&lt;1 2 3|, &lt;0 -3 -5|], leaving just [&lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value &lt~~;a class="wiki_link" href="/0.1215588"&gt~~;0.1215588~~</a>~~. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.

If instead we want OE complexity, we may remove the first row of [&lt;1 2 3|, &lt;0 -3 -5|], leaving just [&lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [&lt;0.1215588|]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.

For a more substantial example we need to consider at least a rank three temperament, so let us turn to marvel, the 7-limit temperament tempering out 225/224. The 2x4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [|1 0 0 0&gt;, |-5 2 2 -1&gt;]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [|1 0 0 0&gt;, |-5 2p3 2p5 -p7&gt;], and P = I - M`M = [|1 0 0 0&gt;, |0 4p5^2+p7^2 -4p3p5 2p3p7&gt;/H,

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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:<br>
-: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-01-30 19:09:33 UTC</tt>.<br>
+: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-01-30 19:11:41 UTC</tt>.<br>
-: The original revision id was <tt>197259402</tt>.<br>
+: The original revision id was <tt>197259818</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 42: / Line 42: @@
 Using this, we find the temperamental norm of [9 13] to be sqrt([9 13]//P//[9 13]*) ~ sqrt(1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while **P** does not depend on the choice of basis vals for the temperament, //P// does; if we choose [&lt;1 2 3|, &lt;0 -3 -5|] for our basis instead, then 3/2 is represented by [1 -3] and //P// changes coordinates to produce the same final result of temperamental complexity.
-If instead we want OE complexity, we may remove the first row of [&lt;1 2 3|, &lt;0 -3 -5|], leaving just [&lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [[0.1215588]]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.
+If instead we want OE complexity, we may remove the first row of [&lt;1 2 3|, &lt;0 -3 -5|], leaving just [&lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [&lt;0.1215588|]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.
 For a more substantial example we need to consider at least a rank three temperament, so let us turn to marvel, the 7-limit temperament tempering out 225/224. The 2x4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [|1 0 0 0&gt;, |-5 2 2 -1&gt;]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [|1 0 0 0&gt;, |-5 2p3 2p5 -p7&gt;], and P = I - M`M = [|1 0 0 0&gt;, |0 4p5^2+p7^2 -4p3p5 2p3p7&gt;/H,
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 Using this, we find the temperamental norm of [9 13] to be sqrt([9 13]&lt;em&gt;P&lt;/em&gt;[9 13]*) ~ sqrt(1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while &lt;strong&gt;P&lt;/strong&gt; does not depend on the choice of basis vals for the temperament, &lt;em&gt;P&lt;/em&gt; does; if we choose [&amp;lt;1 2 3|, &amp;lt;0 -3 -5|] for our basis instead, then 3/2 is represented by [1 -3] and &lt;em&gt;P&lt;/em&gt; changes coordinates to produce the same final result of temperamental complexity.&lt;br /&gt;
 &lt;br /&gt;
-If instead we want OE complexity, we may remove the first row of [&amp;lt;1 2 3|, &amp;lt;0 -3 -5|], leaving just [&amp;lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value &lt;a class="wiki_link" href="/0.1215588"&gt;0.1215588&lt;/a&gt;. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.&lt;br /&gt;
+If instead we want OE complexity, we may remove the first row of [&amp;lt;1 2 3|, &amp;lt;0 -3 -5|], leaving just [&amp;lt;0 -3 -5|]. If we now call this 1x3 matrix A, then (AW^2A*)^(-1) is a 1x1 matrix; in effect a scalar, with value [&amp;lt;0.1215588|]. Multiplying a monzo b times A* gives a 1x1 matrix bA* whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.&lt;br /&gt;
 &lt;br /&gt;
 For a more substantial example we need to consider at least a rank three temperament, so let us turn to marvel, the 7-limit temperament tempering out 225/224. The 2x4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [|1 0 0 0&amp;gt;, |-5 2 2 -1&amp;gt;]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [|1 0 0 0&amp;gt;, |-5 2p3 2p5 -p7&amp;gt;], and P = I - M`M = [|1 0 0 0&amp;gt;, |0 4p5^2+p7^2 -4p3p5 2p3p7&amp;gt;/H, &lt;br /&gt;