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| <h2>IMPORTED REVISION FROM WIKISPACES</h2>
| | {{todo|explain its xenharmonic value}} |
| This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
| | For any [[Harmonic_limit|prime limit]] ''p'', EDO ''N'', and positive integer ''k'', the ''p''-limit ''k''*''N'' subgroup is the largest [[just intonation subgroup]] of the ''p''-limit on which ''N''-edo and ''k''*''N''-edo approximate intervals to the same values using the mapping supplied by the [[patent val]] for ''k''*''N''-edo. This also means they temper out the same commas. |
| : This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-05-28 12:50:44 UTC</tt>.<br>
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| : The original revision id was <tt>232592028</tt>.<br>
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| : The revision comment was: <tt></tt><br>
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| 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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| <h4>Original Wikitext content:</h4>
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| <div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;white-space: pre-wrap ! important" class="old-revision-html">For any [[Harmonic limit|prime limit]] p, EDO N and positive integer k, the p-limit k*N subgroup is the largest [[Just intonation subgroups|just intonation subgroup]] of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the [[patent val]] for k*N-edo. This also means they temper out the same commas.
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| A procedure for finding the k*N subgroup is to take enough of the intervals of the p-limit which are mapped to a value divisible by k by the k*N patent val, and add to this set 2 and a basis for the commas of the k*N patent val, and then reduce this to a [[Normal lists|normal interval list]], giving the canonical list of generators for the subgroup. To get "enough" intervals, take the [[Diamonds|diamond]] of the primes from 2 to p, which is the tonality diamond, then the diamond of the diamond, and so forth, until the iterated diamond construction includes a basis for the commas of the k*N patent val. | | A procedure for finding the ''k''*''N'' subgroup which seems to suffice is to take the product ''m'' of the odd primes less than or equal to ''p'', and then find the [[Euler genus]] Eu(''m''<sup>''i''</sup>) for integers 1, 2, and so forth. Here Eu(''d'') for an odd integer ''d'' is the set of all divisors of ''d'' reduced to an octave, and including 2. For each such genus, select those intervals such that the ''k''*''N'' patent val maps the interval to a number divisible by ''k'', and then find the corresponding [[normal lists|normal interval list]]. When two successive values ''i'' and {{nowrap|''i'' + 1}} lead to the same normal list, add to that a basis for the commas of the ''k''*''N'' patent val, and return the normal interval list for that. |
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| For example, the 5-limit diamond is 6/5, 5/4, 4/3, 3/2, 8/5, 5/3, 2. The intervals mapped to an even integer by <12 19 28| are 5/4, 8/5 and 2, and a basis for the 5-limit commas is 81/80 and 128/125. If we add 81/80 and 128/125 to the diamond and reduce to the normal list, we get 2.81.5 for the 2*6 subgroup, which is not correct; however the diamond of the diamond (or the diamond of the diamond of the diamond, etc.) gives the correct subgroup, 2.9.5. The diamond of the diamond of the diamond contains a basis for the commas, and therefore the procedure described above would give the correct answer.</pre></div> | | For example, to find the 7-limit 2*6 subgroup we first find {{nowrap|''m'' {{=}} 3 × 5 × 7 {{=}} 105}}. The subgroup of Eu(105) consisting of those intervals mapped to an even number by {{vector|12 19 28 34}} is 2.5.7. The subgroup of Eu(105<sup>2</sup>) is 2.9.5.7, not the same. However, the subgroup of Eu(105<sup>3</sup>) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime. |
| <h4>Original HTML content:</h4>
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| <div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>k*N subgroups</title></head><body>For any <a class="wiki_link" href="/Harmonic%20limit">prime limit</a> p, EDO N and positive integer k, the p-limit k*N subgroup is the largest <a class="wiki_link" href="/Just%20intonation%20subgroups">just intonation subgroup</a> of the p-limit on which N-edo and k*N-edo approximate intervals to the same values using the mapping supplied by the <a class="wiki_link" href="/patent%20val">patent val</a> for k*N-edo. This also means they temper out the same commas. <br />
| | [[Category:Algorithms]] |
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| | [[Category:Math]] |
| A procedure for finding the k*N subgroup is to take enough of the intervals of the p-limit which are mapped to a value divisible by k by the k*N patent val, and add to this set 2 and a basis for the commas of the k*N patent val, and then reduce this to a <a class="wiki_link" href="/Normal%20lists">normal interval list</a>, giving the canonical list of generators for the subgroup. To get &quot;enough&quot; intervals, take the <a class="wiki_link" href="/Diamonds">diamond</a> of the primes from 2 to p, which is the tonality diamond, then the diamond of the diamond, and so forth, until the iterated diamond construction includes a basis for the commas of the k*N patent val.<br />
| | [[Category:Regular temperament theory]] |
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| For example, the 5-limit diamond is 6/5, 5/4, 4/3, 3/2, 8/5, 5/3, 2. The intervals mapped to an even integer by &lt;12 19 28| are 5/4, 8/5 and 2, and a basis for the 5-limit commas is 81/80 and 128/125. If we add 81/80 and 128/125 to the diamond and reduce to the normal list, we get 2.81.5 for the 2*6 subgroup, which is not correct; however the diamond of the diamond (or the diamond of the diamond of the diamond, etc.) gives the correct subgroup, 2.9.5. The diamond of the diamond of the diamond contains a basis for the commas, and therefore the procedure described above would give the correct answer.</body></html></pre></div>
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