Tenney–Euclidean tuning: Difference between revisions

: This revision was by author [[User:keenanpepper|keenanpepper]] and made on <tt>2012-03-26 01:16:24 UTC</tt>.<br>

: The original revision id was <tt>314520702</tt>.<br>

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.

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 v12P equals v12; on the other hand, if we take the monzo for 81/80, which is |-4 4 -1&gt; and monzo-weight it to &lt;-4 4log2(3) - log2(5)| and multiply (either side, the matrix is symmetric) by P, we get the zero vector, corresponding to the unison.

Then the pseudoinverse V` is V`, approximately

If we use unweighted coordinates we get the Frobenius projection map instead, whose rows are [[fractional monzos]]. The unweighted pseudoinverse u12` of the 5-limit val u12 for 12 equal is the column matrix u12t/1289, where the t denotes the transpose; that is, the 1x3 matrix with column [12/1289 19/1289 28/1289]. Then u12`u12 is the 3x3 Frobenius projection map F:

Multiplying &lt;12 19 28| by F gives &lt;12 19 28| again. Multiplying the monzos for 81/80, 128/125, 648/625 and etc gives the zero monzo, corresponding to a unison. Multiplying the val for 5-limit 19 equal, &lt;19 30 44|, by F gives [24360 38570 56840]/1289, which is approximately the 19 equal val. Multiplying the 5-limit monzo for 3/2, which is |-1 1 0&gt; times F gives the fractional monzo corresponding to (2^84 3^133 5^196)^(1/1289), which equates to 698.121 cents, the tempering of 3/2 in Frobenius tuning for 5-limit 12et, the tuning with octave defined by the top row of F, which is to say by |1 0 0&gt;F, of 1196.778 cents.

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;

  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 v12P equals v12; on the other hand, if we take the monzo for 81/80, which is |-4 4 -1&amp;gt; and monzo-weight it to &amp;lt;-4 4log2(3) - log2(5)| and multiply (either side, the matrix is symmetric) by P, we get the zero vector, corresponding to the unison.&lt;br /&gt;

Then the pseudoinverse V` is V`, approximately&lt;br /&gt;

If we use unweighted coordinates we get the Frobenius projection map instead, whose rows are &lt;a class="wiki_link" href="/fractional%20monzos"&gt;fractional monzos&lt;/a&gt;. The unweighted pseudoinverse u12` of the 5-limit val u12 for 12 equal is the column matrix u12t/1289, where the t denotes the transpose; that is, the 1x3 matrix with column [12/1289 19/1289 28/1289]. Then u12`u12 is the 3x3 Frobenius projection map F:&lt;br /&gt;

Multiplying &amp;lt;12 19 28| by F gives &amp;lt;12 19 28| again. Multiplying the monzos for 81/80, 128/125, 648/625 and etc gives the zero monzo, corresponding to a unison. Multiplying the val for 5-limit 19 equal, &amp;lt;19 30 44|, by F gives [24360 38570 56840]/1289, which is approximately the 19 equal val. Multiplying the 5-limit monzo for 3/2, which is |-1 1 0&amp;gt; times F gives the fractional monzo corresponding to (2^84 3^133 5^196)^(1/1289), which equates to 698.121 cents, the tempering of 3/2 in Frobenius tuning for 5-limit 12et, the tuning with octave defined by the top row of F, which is to say by |1 0 0&amp;gt;F, of 1196.778 cents.&lt;br /&gt;