Diamond tradeoff: Difference between revisions
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Next, we find the generators matrix <math>G</math> corresponding to the tuning where these two intervals are unchanged. The formula for this generators matrix is <math>G = U(MU)^{-1}</math> (see [[ | Next, we find the generators matrix <math>G</math> corresponding to the tuning where these two intervals are unchanged. The formula for this generators matrix is <math>G = U(MU)^{-1}</math> (see [[Generator embedding optimization#Generator embedding|here]] for a full explanation of this formula). Here, let's work it out for our chosen <math>U</math>. First, we multiply <math>MU</math>: | ||
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Reading the columns from <math>G</math>, the first one {{vector|1 0 0}} confirms our period of 2/1, and the second column {{vector|-1 1 0}} gives our generator 3/2. Which is unsurprising. In cents, that's {{nowrap|1200 | Reading the columns from <math>G</math>, the first one {{vector|1 0 0}} confirms our period of 2/1, and the second column {{vector|-1 1 0}} gives our generator 3/2. Which is unsurprising. In cents, that's {{nowrap|1200 × log<sub>2</sub>(3/2) ≈ 701.955{{cent}}}}. The next unchanged interval will give a more interesting result. | ||
==== Second diamond extrema ==== | ==== Second diamond extrema ==== | ||
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(MU)^{-1} \\ | (MU)^{-1} \\ | ||
\left[ \begin{array} {rrr} | \left[ \begin{array} {rrr} | ||
1 & \ | 1 & \frac{1}{2} \\ | ||
0 & \ | 0 & \frac{1}{4} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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(MU)^{-1} \\ | (MU)^{-1} \\ | ||
\left[ \begin{array} {rrr} | \left[ \begin{array} {rrr} | ||
1 & \ | 1 & \frac{1}{2} \\ | ||
0 & \ | 0 & \frac{1}{4} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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1 & 0 \\ | 1 & 0 \\ | ||
0 & 0 \\ | 0 & 0 \\ | ||
0 & \ | 0 & \frac{1}{4} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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This tells us our generator is { | This tells us our generator is <math>\monzo{0 & 0 & \frac{1}{4}}</math>, or 5<sup>{{frac|1|4}}</sup>. In cents, that's {{nowrap|1200 × log<sub>2</sub>(5<sup>{{frac|1|4}}</sup>) ≈ 696.578{{c}}}}. | ||
==== Third diamond extrema ==== | ==== Third diamond extrema ==== | ||
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(MU)^{-1} \\ | (MU)^{-1} \\ | ||
\left[ \begin{array} {rrr} | \left[ \begin{array} {rrr} | ||
1 & \ | 1 & \frac{2}{3} \\ | ||
0 & -\ | 0 & -\frac{1}{3} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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(MU)^{-1} \\ | (MU)^{-1} \\ | ||
\left[ \begin{array} {rrr} | \left[ \begin{array} {rrr} | ||
1 & \ | 1 & \frac{2}{3} \\ | ||
0 & -\ | 0 & -\frac{1}{3} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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G \\ | G \\ | ||
\left[ \begin{array} {rrr} | \left[ \begin{array} {rrr} | ||
1 & \ | 1 & \frac{1}{3} \\ | ||
0 & -\ | 0 & -\frac{1}{3} \\ | ||
0 & \ | 0 & \frac{1}{3} \\ | ||
\end{array} \right] | \end{array} \right] | ||
\end{array} | \end{array} | ||
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This tells us our generator is <math>\monzo{\frac{1}{3} & -\frac{1}{3} & \frac{1}{3}}</math>, or expressed another way, ({{frac|10|3}})<sup>{{frac|1|3}}</sup>. In cents, that's {{nowrap|1200 | This tells us our generator is <math>\monzo{\frac{1}{3} & -\frac{1}{3} & \frac{1}{3}}</math>, or expressed another way, ({{frac|10|3}})<sup>{{frac|1|3}}</sup>. In cents, that's {{nowrap|1200 × log<sub>2</sub>(({{frac|10|3}})<sup>{{frac|1|3}}</sup>) ≈ 694.786{{c}}}}. | ||
==== Determining the final range ==== | ==== Determining the final range ==== | ||