Diamond tradeoff: Difference between revisions

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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 &times; log<sub>2</sub>(3/2) &#8776; 701.955{{c}}}}. The next unchanged interval will give a more interesting result.
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 & \frac12 \\
1 & \frac{1}{2} \\
0 & \frac14 \\
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 & \frac12 \\
1 & \frac{1}{2} \\
0 & \frac14 \\
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 & \frac14 \\
0 & \frac{1}{4} \\
\end{array} \right]
\end{array} \right]
\end{array}
\end{array}
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This tells us our generator is {{vector|0 0 <math>\frac14</math>}}, or 5^(1/4). In cents, that's {{nowrap|1200 &times; log<sub>2</sub>(5<sup>{{frac|1|4}}</sup>) &#8776; 696.578{{c}}}}.  
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 & \frac23 \\
1 & \frac{2}{3} \\
0 & -\frac13 \\
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 & \frac23 \\
1 & \frac{2}{3} \\
0 & -\frac13 \\
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 & \frac13 \\
1 & \frac{1}{3} \\
0 & -\frac13 \\
0 & -\frac{1}{3} \\
0 & \frac13 \\
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 &times; log<sub>2</sub>(({{frac|10|3}})<sup>{{frac|1|3}}</sup>) &#8776; 694.786{{c}}}}.  
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 ====