K*N subgroups: Difference between revisions
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Wikispaces>genewardsmith **Imported revision 232650890 - Original comment: ** |
Wikispaces>genewardsmith **Imported revision 232859204 - 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>2011-05- | : This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-05-30 07:46:56 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>232859204</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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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 genera|Euler genus]] Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 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. | 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 genera|Euler genus]] Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 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. | ||
</pre></div> | For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) 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.</pre></div> | ||
<h4>Original HTML content:</h4> | <h4>Original HTML content:</h4> | ||
<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 /> | <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 /> | ||
<br /> | <br /> | ||
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 <a class="wiki_link" href="/Euler%20genera">Euler genus</a> Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 <a class="wiki_link" href="/Normal%20lists">normal interval list</a>. When two successive values i and 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.</body></html></pre></div> | 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 <a class="wiki_link" href="/Euler%20genera">Euler genus</a> Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 <a class="wiki_link" href="/Normal%20lists">normal interval list</a>. When two successive values i and 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.<br /> | ||
<br /> | |||
For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by &lt;12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) 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.</body></html></pre></div> | |||
Revision as of 07:46, 30 May 2011
IMPORTED REVISION FROM WIKISPACES
This is an imported revision from Wikispaces. The revision metadata is included below for reference:
- This revision was by author genewardsmith and made on 2011-05-30 07:46:56 UTC.
- The original revision id was 232859204.
- The revision comment was:
The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.
Original Wikitext content:
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. 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 genera|Euler genus]] Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 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. For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) 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.
Original HTML content:
<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 /> <br /> 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 <a class="wiki_link" href="/Euler%20genera">Euler genus</a> Euler(m^i) for integers 1, 2, and so forth. Here Euler(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 <a class="wiki_link" href="/Normal%20lists">normal interval list</a>. When two successive values i and 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.<br /> <br /> For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) 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.</body></html>