Tenney–Euclidean tuning: Difference between revisions

Wikispaces>genewardsmith
**Imported revision 186492863 - Original comment: **
Wikispaces>genewardsmith
**Imported revision 196912296 - Original comment: **
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<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:genewardsmith|genewardsmith]] and made on <tt>2010-12-08 15:03:01 UTC</tt>.<br>
: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-01-28 15:03:11 UTC</tt>.<br>
: The original revision id was <tt>186492863</tt>.<br>
: The original revision id was <tt>196912296</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>
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If the vals defining V are linearly independent, then P = V*(VV*)^(-1)V. I the rows of M are independent, then we likewise have P = I - M*(MM*)^(-1)M.
If the vals defining V are linearly independent, then P = V*(VV*)^(-1)V. I the rows of M are independent, then we likewise have P = I - M*(MM*)^(-1)M.
==Examples==
The val for 5-limit 12et is &lt;12 19 28|. In weighted coordinates, that becomes v12 = &lt;12 19/log2(3) 28/log2(5)| ~ &lt;12.0 11.988 12.059|. If we take this to be a 1x3 matrix and take the pseudoinverse, we get the 3x1 matrix v12` ~ [|0.027706 0.027677 0.027842|]. Then P = v12`v12 is a projection map onto the one-dimensional subspace whose single basis vector is v12. We find that v12.P equals v12 (using a dot to denote matrix multiplication.)
</pre></div>
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<h4>Original HTML content:</h4>
<h4>Original HTML content:</h4>
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Spelling this out, if V is a matrix whose rows are vals, then P = V`V is a &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Positive-definite_matrix" rel="nofollow"&gt;positive-semidefinite&lt;/a&gt; &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Symmetric_matrix" rel="nofollow"&gt;symmetric matrix&lt;/a&gt; with rational matrix entries, which exactly specifies the regular temperament defined by the vals of V. If M is a matrix with rows of monzos which spans the subspace of interval space containing the commas, then this same matrix P is given by I - M`M.  &lt;br /&gt;
Spelling this out, if V is a matrix whose rows are vals, then P = V`V is a &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Positive-definite_matrix" rel="nofollow"&gt;positive-semidefinite&lt;/a&gt; &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Symmetric_matrix" rel="nofollow"&gt;symmetric matrix&lt;/a&gt; with rational matrix entries, which exactly specifies the regular temperament defined by the vals of V. If M is a matrix with rows of monzos which spans the subspace of interval space containing the commas, then this same matrix P is given by I - M`M.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If the vals defining V are linearly independent, then P = V*(VV*)^(-1)V. I the rows of M are independent, then we likewise have P = I - M*(MM*)^(-1)M.&lt;/body&gt;&lt;/html&gt;</pre></div>
If the vals defining V are linearly independent, then P = V*(VV*)^(-1)V. I the rows of M are independent, then we likewise have P = I - M*(MM*)^(-1)M.&lt;br /&gt;
&lt;br /&gt;
&lt;!-- ws:start:WikiTextHeadingRule:12:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc6"&gt;&lt;a name="x-Examples"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:12 --&gt;Examples&lt;/h2&gt;
The val for 5-limit 12et is &amp;lt;12 19 28|. In weighted coordinates, that becomes v12 = &amp;lt;12 19/log2(3) 28/log2(5)| ~ &amp;lt;12.0 11.988 12.059|. If we take this to be a 1x3 matrix and take the pseudoinverse, we get the 3x1 matrix v12` ~ [|0.027706 0.027677 0.027842|]. Then P = v12`v12 is a projection map onto the one-dimensional subspace whose single basis vector is v12. We find that v12.P equals v12 (using a dot to denote matrix multiplication.)&lt;/body&gt;&lt;/html&gt;</pre></div>